Vedanta Excel in Opt. Mathematics Book 7 Final (2078)

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7Book

Author
Piyush Raj Gosain

Hukum Pd. Dahal Editors P. L. Shah
Tara Bdr. Magar

vedanta

Vedanta Publication (P) Ltd.
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Vanasthali, Kathmandu, Nepal
+977-01-4982404, 01-4962082
[email protected]
www.vedantapublication.com.np

vEMexdcAaenlTtaiHn AEMddAiTtiIoCnSal

Author 7Book
Piyush Raj Gosain

All rights reserved. No part of this publication may
be reproduced, copied or transmitted in any way,
without the prior written permission of the publisher.

¤ Vedanta Publication (P) Ltd.

First Edition: B.S. 2077 (2020 A. D.)
Second Edition: B. S. 2078 (2021 A. D.)

Layout and Design
Pradeep Kandel

Printed in Nepal

Published by:

Vedanta Publication (P) Ltd.
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Vanasthali, Kathmandu, Nepal
+977-01-4982404, 01-4962082
[email protected]
www.vedantapublication.com.np

Preface

The series of Vedanta Excel in Opt. Mathematics for class 9 and 10 is completely based on the

contemporary pedagogical teaching learning activities and methodologies. It is an innovative

and unique series in the sense that the contents of each textbooks of the series are written and

designed to ful ill the need of integrated teaching learning approaches.

Vedanta Excel in Opt. Mathematics has incorporated applied constructivism the latest trend of

learner centered teaching padagogy. Every lesson of the series is written and designed in such

a manner that makes the classes automatically constructive and the learner actively participate

in the learning process to construct knowledge themselves, rather than just receiving ready

made information from their instructor. Even teachers will be able to get enough opportunities

to play the role of facilitators and guides shifting themselves from the traditional methods

imposing instractions. The idea of the presentation of every mathematical item is directly or

indirectly re lected from the writer's long expenena, more than two decades, of teaching optional

mathematics.

Each unit of Vedanta Excel in Opt. Mathematics series is provided with many more worked out

examples, arranged in the hierarchy of the learning objectives and they are re lective to the

corresponding exercises.

Vedanta Excel in Opt. Mathematics class 10 covers the latest syllabus of CDC, the government of

Nepal, on the subject. My honest efforts have been to provide all the essential matter and practice

materials to the users. I believe that the book serves as a staircase for the students of class 10.

The book contains practice exercises in the form of simple to complex including the varieties of

problems. I have tried to establish relationship between the examples and the problems set for

practice to the maximum extent.

In the book, every chapter starts with review concepts of the same topic that the students have

studied in previous classes. Discussion questions in each topic are given to warm up the students

for the topic. Questions in each exercise are catagorized into three groups - Very Short Questions,

Short Questions and Long Questions.

The project works are also given at the end of exercise as required. In my exprerience, the students

of class 10 require more practices on Trigonometry and Vector Geometry, the examples and the

exercise questions are given to ful ill it in the corresponding topics. The latest syllabus of the

subject speci ication grid and a model question issued by CDC are given at the end of the book.

My hearty thanks goes to Mr. Hukum Pd. Dahal, Tara Bahadur Magar and P.L. Shah, the series

editors, for their invaluable efforts in giving proper shape to the series. I am also thankful to my

colleage Mr. Gyanendra Shrestha who helped me a lot during the preparation of the book.

I am also thankful to my respected parents and my family members for their valuable support to

bring the book out in this form. I would also like to express my hearty gratitude to all my friends,

colleagues and beloved students who always encouraged me to express my knowledge, skill and

experience in the form of books. I am highly obliged to all my known and unknown teachers who

have laid the foundation of knowledge upon me to be such a person.

Last but not the least, I am hearty theankful to Mr. Pradeep Kandel, the computer and designing

senior of icer of the publication for his skill in designing the series in such an attractive form.

Efforts have been made to clear the subject matter included in the book. I do hope that teachers

and students will best utilize the series.

Valuable suggestions and comments for its further improvement from the concerned will be

highly appreciated. Piyush Raj Gosain

CONTENT

Unit 1 Functions 5
25
Unit 2 Polynomials 44
84
Unit 3 Sequence and Series 97
113
Unit 4 Linear Programming 128
154
Unit 5 Quadratic Equations and Graphs 199
269
Unit 6 Continuity 299
347
Unit 7 Matrices

Unit 8 Co-ordinate Geometry

Unit 9 Trigonometry

Unit 10 Vectors

Unit 11 Transformations

Unit 12 Statistics

Syllabus
Specification Grid
Model Questions

Relations Relations

1

1.1 Order

The arrangement of numbers or elements in a specific manner is known as an order.
For example:
(i) 12, 22, 32, 42, 52

i.e. 1, 4, 9, 16, 25
This is an order of square of the first five natural numbers.
(ii) I, II, III, IV and V
The first five Roman numerals.
(iii) 2, 4, 6, 8, 10
The first five multiples of 2.

1.2 Pair

A set of two elements is known as pair. For example, a pair of shoes, a pair of
slippers, a pair of gloves, Ram and Sita, Shiva and Parvati, Nepal and Kathmandu,
Bangladesh and Dhaka etc. In pair of elements, the order of occurrence of elements
is not important.

1.3 Ordered Pair

Ordered pair is the combination of two numbers or elements in a fixed order in such
a way that the first element represents the x-component and the second element
represents the y-component. The elements of an ordered pair are always separated
by comma and closed between round or small brackets.

For example: (4, 8), (Shiva, Parvati), (sun, moon), etc.

(2, 5) and (5, 2) are different ordered pairs we can understant it plotting there points
on a graph paper.

vedanta Excel in Additional Mathematics - Book 7 7

Relations

For example :
The following are some ordered pairs which represent the nations and their capital
cities.

(Nepal, Kathmandu)
(India, Delhi)
(USA, Washington D.C.)
The x-component is also called antecedent and y-component is called the
consequence.
We can also use the diagram to represent the ordered pairs.

is a capital of

Kathmandu Nepal
New Delhi India
Washington DC
USA

? (Kathmandu, Nepal) and (New Delhi, India, USA, Washington DC) are ordered
pair.

Note :

(a, b) and (b,a) are not equal ordered pairs as in (a, b) the first component is a and
the second is b and whereas in (b, a), the first components is b and the second
component is a.

1.4 Equality of ordered pair

The ordered pairs (a, b) and (p, q) are said to be equal if their corresponding elements
are equal. That is, the first component a should be equal to the first component p
and the second component b should be equal to the second component q.
Mathematically,

If (a, b) = (p, q) then
Ÿ a = p and b = q
For example: Are the ordered pairs (4, 5) and (4, 5) equal?
Yes, the ordered pairs are equal as their corresponding components are equal.

8 vedanta Excel in Additional Mathematics - Book 7

Relations

Worked Out Examples

Example 1. Find out the x-component and y-component of the following ordered
Solution: pairs:

(a) (4, 7) (b) (–3, 4)

(a) Here, (4, 7)

x-component = 4 y-component = 7

(b) Here, (–3, 4)

x-component = –3 y-component = 4

Example 2. Check which of the following ordered pairs are equal. Give reason
Solution: for your answer.

(a) (2, 3) and (2, 3)

(b) (4, 5) and (7, 8)

(a) Here, in (2, 3) and (2, 3)

The x-components; 2 = 2 (which is true)

y-components; 3 = 3 (which is true)

So, (2, 3) and (2, 3) are equal ordered pairs.

(b) Here, in (4, 5) and (7, 8)

The x-components; 4 = 7 (which is false)

y-components; 5 = 8 (which is false)

So, (4, 5) and (7, 8) are not equal ordered pairs.

Example 3. Find the values of x and y from the following equal ordered pairs.
Solution:
(a) (x, y) = (4, 5)

(b) (2x, 4) = 6, y
2
x y
(c) 2 , 3 = (3, 2)

(d) (3x – 1, 2) = (4, y + 2)

(a) Here, (x, y) = (4, 5)

Since, the ordered pairs are equal, the corresponding elements
are equal.

So, x = 4, y = 5

vedanta Excel in Additional Mathematics - Book 7 9

Relations

(b) Here, (2x, 4) = 6, y
2

Since the ordered pairs are equal, the corresponding elements

are equal.

So, 2x = 6

or, x = 6 = 3
2
y
and 4 = 2

or, y = 4 × 2 = 8

Hence, x = 3, y = 8

(c) Here, x , y = (3, 2)
2 3

Since the ordered pairs are equal the corresponding elements

are equal.

So, x = 3
2

or, x = 2 × 3 = 6

and y = 2
3

or, y = 3 × 2 = 6

Hence, x = 6 and y = 6.

(d) Here, (3x – 1, 2) = (4, y + 2)

Since the ordered pairs are equal the corresponding elements
are equal.

So, 3x – 1 = 4

or, 3x = 4 + 1

or, 3x = 5

or, x = 5
3

and 2 = y + 2

or, 2 – 2 = y

or, 0 = y

or, y = 0

Hence, x = 5 , y = 0
3

10 vedanta Excel in Additional Mathematics - Book 7

Relations

1.5 Cartesian Product

Let A and B be two non empty sets. Then the product A × B read as A cross B is
defined as the set of all ordered pairs in such a way that the first element is taken
from set A and the second element is taken from set B.

Symbolically,

A × B = {(x, y) : x A and y B }
For example: Let A = {x, y}, B = {a, b}

Then, A × B can be calculated as in the following table:

Ax B b
y a (x, b)
(x, a) (y, b)
(y, a)

Now, A × B in the set of all above ordered pairs and we write it as,

A × B = {(x, a), (x, b), (y, a), (y, b)}

Similarly, B × A can be written as,

Ba A y
b x (a, y)
(a, x) (b, y)
(b, x)

So, B × A = {(a, x), (a, y), (b, x), (b, y)}

From the example, it is also observed that

Since (x, a) ≠ (a, x), (y,a ) ≠ (a, y) (x, b) ≠ (b, x), (y, b) ≠ (b, y)

A×B≠B×A

The cartesian product can be also obtained by using an arrow diagram:
AB

xa

yb

It is an arrow diagram representing A × B.

From this arrow diagram also, we can see that
A × B = {(x, a), (x, b), (y, a), (y, b)}

vedanta Excel in Additional Mathematics - Book 7 11

Relations

Similarly for B × A, we place B first and A second and draw arrows to represent it.

So, B × A = {(a, x), (a, y), (b, x), (b, y)} A
B

ax

by
An arrow diagram representing B × A

Worked Out Examples

Example 4. If A = {a, b} and B = {x}, find A × B using table.

Solution: Here, A = {a, b} and B = {x}

Now, A × B is given by

B

x

Aa (a, x)

b (b, x)

? A × B = {(a, x), (b, x)}

Example 5. If P = {6, 7} and Q = {8, 9}, find P × Q using an arrow diagram.

Solution: Here, P = {6, 7} and Q = {8, 9}

Now, P × Q is given by Q
P

68

79

So, P × Q = {6, 7} × {8, 9} = {(6, 8), (6, 9), (7, 8), (7, 9)}

Example 6. If A = { 1, 2, 3} and B = {4, 5}, find:

(a) A × B (b) B × A

(c) Show that A × B z B × A

Solution: Here, A = {1, 2, 3} and B = {4, 5}

12 vedanta Excel in Additional Mathematics - Book 7

Relations

(a) Now, A × B is given by

A 1 B 5
2 4 (1, 5)
(1, 4) (2, 5)
3 (2, 4) (3, 5)
(3, 4)

Here, A × B = {(1, 4), (2, 4), (3, 4), (1, 5), (2, 5), (3, 5)}

Similarly, B × A is given by

B4 1 A 3
5 (4, 1) 2 (4, 3)
(5, 1) (5, 3)
(4, 2)
(5, 2)

So, B × A = {(4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3)}

Again,

Since, (1, 4) ≠ (4, 1), (2, 4) ≠ (4, 2), ..., (5, 3) ≠ (3, 5)

A×BzB×A

Example 7. If P = {2, 4} find P × P. P
Solution: Here, P = {2, 4} 2

Now, P × P is given by
P
2

44

? P × P = {(2, 2), (2, 4), (4, 2), (4, 4)}

Example 8. If P × Q = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}, find the sets P and Q.
Solution: Here, P × Q = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}

By the definition of Cartesian product P × Q, in ordered pair, the first
elements are from set P and the second elements are from set Q.
? P = {1, 2, 3, 4, 5} and Q = {2, 3, 4, 5, 6}

vedanta Excel in Additional Mathematics - Book 7 13

Relations

Exercise 1.1

Short Questions :

1. (a) Define an ordered pair.

(b) Define equal ordered pairs.

(c) Define cartesian product of two sets.

2. Determine x - component (antecedent) and y - component (consequence) from
the following ordered pairs:

(a) (4, 5) (b) (6, 8)

(c) (–2, 4) (d) –2, – 1
(e) (Nepal, Mt. Everest) 2

(f) (Pokhara, Phewa Lake)

3. Check which of the following ordered pairs are equal. Give reason to support
your answer:

(a) (4, 5) and (4, 5) (b) (6, 5) and (6, 5)

(c) (4, 5) and (4, – 5) (d) (3 + 4, 2 – 1) and (7, 1)

(e) 25 , 3 and 5, 9 (f) (p + q, q – p) and (q + p, p – q)
5 3

4. Find the values of x and y from the following equal ordered pairs:

(a) (x, y) = (4, 5) (b) (x, – 5) = (– 6, – y)

(c) (5x, 2y) = (10, 10) (d) x , 7y = (2, 14)
(e) (10x + 1, 2y – 2) = (11, – 6) 7

(f) (x, 3) = (4, y – 1)

(g) (x – 4, y + 6) = (3, 6) (h) x – 7, y = (4, 5)
(i) (x – 5, 9) = (4x – 5, y + 3) 2

5. Using table find A × B if

(a) A = {2, 3} and B = {4, 5} (b) A = {a} and B = {p, q}

(c) A = {6, 7} and B = {p} (d) A = {4, 6, 8} and B = {1, 2, 3}

6. Using an arrow diagram calculate B × A from the questions of Q. No 5.

7. If A = {a, b, c} and B = {p, q}. Find

(a) A × B (b) B × A

(c) Show that A × B ≠ B × A

8. If A = {d, o, g} and B = {g, o, d}, Show that A × B = B × A.

14 vedanta Excel in Additional Mathematics - Book 7

Relations

9. If A = {p, q} find A × A.

10. If A = {p, q} and B = {r, s}, find the following :

(a) A × B (b) B × A

(c) A × A (d) B × B

11. (a) If A × B = {(p, q), (r, s), (t, u), (u, v)}, find set A and set B.

(b) If P × Q = {(2, 4), (4, 16), (5, 25)}

Find (i) set P and set Q (ii) P × P

Project Work

12. Give any four examples of pair of object with relation. Write they are ordered
pair or not. Also write their uses.

1. and 2. Show to your teacher

3. In (a), (b), (d), (e) are equal ordered pairs

4. (a) x = 4, y = 5 (b) x = –6, y = 5 (c) x = 2, y = 5 (d) x = 14, y = 2

(e) x = 1, y = – 2 (f) x = 4, y = 4 (g) x = 7, y = 0 (h) x = 11, y = 10

(i) x = 0, y = 6 5.(a) {(2, 4), (2, 5), (3, 4), (3, 5)}

(b) {(a, p), (a, q)} (c) {(6, p), (7, p)}

(d) {(4, 1), (4, 2), (4, 3), (6, 1), (6, 2)}, (6, 3), (8, 1), (8, 2), (8, 3)}

6. (a) {(4, 2), (4, 3), (5, 2), (5, 3)} (b) {(p, a), (q, a)} (c) {(p, 6), (p, 7)}

(d) {(1, 4), (1, 6), (1, 8), (2, 4), (2, 6), (2, 8), (3, 4), (3, 6), (3, 8)}

7. (a) {(a, p), (a, q), (b, p), (b, q), (c, p), (c, q)}

(b) {(p, a), (p, b), (p, c), (q, a), (q, b), (q, c)} 9. {(p, p), (p, q), (q, p), (q, q)}

10. (a) {(p, r), (p, s), (q, r), (q, s)} (b) {(r, p), (r, q), (s, p), (s, q)}

(c) {(p, p), (p, q), (q, p), (q, q)} (d) {(r, r), (r, s), (s, r), (s, s)}

11. (a) A = {p, r, t, u}, B = {q, s, u, v}

(b) (i) P = {2, 4, 5}, Q = {4, 16, 25}

(ii) {(2, 2), (2, 4), (2, 5), (4, 2), (4, 4) (4, 5), (5, 2), (5, 4), (5, 5)}

vedanta Excel in Additional Mathematics - Book 7 15

Relations

1.6 Relation - Introduction

There may be various types of relations in our daily life. Let us consider some of them,
(i) Kathmandu is the capital of Nepal.
(ii) Dasharath was father of Ram.
(iii) 20 is greater than 10.
In above relations, (i) shows the relation of "the capital and country," (ii) shows the
relation of "father and son" and (iii) shows "greater than" relation.
Hence, the word relation shows an association of two objects, or people, or numbers,
and so on. Here, we focus our attention to mathematical relation.
Let A and B be any two non-empty sets. Then a relation R from A to B is a non-
empty subset of the cartesian product A × B. It is defined by

R = {(x, y) : x  A and y  B} Ž A × B
For example : Let A = {1, 2} and B = {3, 4},

A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}
Then, let us define the following relations from A × B.
(i) R1 = {(x, y) : x + y = 5}

= {(1, 4), (2, 3)}
(ii) R2 = {(x, y) : x < y}

= {1, 3), (1, 4), (2, 3), (2, 4)}
(iii) R3 = {(x, y) : x > y} = I
In all of above relations R1, R2 and R3, all of them are subsets of A × B.
Hence a relation is a subset of cartesian product A × B.

Worked Out Examples

Example 1. If A = {1, 2, 3} and B = {1, 3, 5}, find the set of ordered pairs in the
Solution: relation "is greater than" from A to B.
Here, A = {1, 2, 3} and B = {1, 3, 5}
Then,
A × B = {1, 2, 3} × {1, 3, 5}

= {(1, 1), (1, 3), (1, 5), (2, 1), (2, 3), (2, 5), (3, 1), (3, 3), (3, 5)}
Let R1 = the set of the ordered pair whose first component is greater

than the second components

16 vedanta Excel in Additional Mathematics - Book 7

Relations

i.e. R1 = {(x, y) : x > y}
= {(2, 1), (3, 1)}

1.7 Ways of Representation of a Relation

There are various ways in which a relation can be expressed. They are as follows:

(a) Roster form (b) Set builder form

(c) Arrow or mapping diagram (d) Tabulation method

(e) Graphical method

Example 2. If A = {1, 4, 9, 16}, B = {1, 2, 3, 4} and a relation R : A o B is defined
by "is square of", represent the relation by the following ways:

(a) Roster method (b) Arrow diagram method

(c) Tabulation (d) Graphical method

Solution: Here, A = {1, 4, 9, 16} and B = {1, 2, 3, 4},

A × B = {1, 4, 9, 16} × {1, 2, 3, 4}

= {(1, 1), (1, 2), (1, 3), (1, 4), (4, 1), (4, 2), (4, 3), (4, 4),

(9, 1), (9,2), (9,3), (9,4), (16, 1), (16, 2), (16, 3), (16, 4)}

Now, R : A o B i.e. R is a relation from A to B.

Let us represent above relations in the following ways

(a) Roster method

R = {(x, y) : x is square of y}

= {(x, y) : x = y2}

= {(1, 1), (4, 2), (9, 3), (16, 4)}

(b) Arrow diagram method
R

11
42
93
16 4

R{(x, y) : x = y2} 17

vedanta Excel in Additional Mathematics - Book 7

Relations

(c) Tabulation method

x149 16
y123 4

(d) Graphical method

Here, R = {(1, 1), (4, 2), (9, 3), (16, 4)}

Let us show each point of R in a graph :
Y

(9, 3) (16, 4)

(4, 2)
(1, 1)

X' O X

1.8 Y'

Domain and Range of a Relation

Let R : A o B be a relation from A to B.

Suppose R = {(1, 1), (2, 4), (3, 9), (4, 16)}

In an arrow diagram, we can show R as in A R B
diagram given below :

Then the set all the first component of R is 1 1
called domain. 2 4

? Domain of R = {1, 2, 3, 4} 3 9

Its elements are called pre-images. 4 16

The set of all the second components of R is
called range.

? Range of R = {1, 4, 9, 16}

Its elements are also called images.

1.9 Type of Relations

Let A and B be two non-empty sets. Then, a relation R from A to B may be any one
of the following types.

(a) One to one relation
A relation R from set A to set B is called one to one if different element set A are
related to different elements of set B.

18 vedanta Excel in Additional Mathematics - Book 7

Relations

Example : Let R = A o B be defined by

R = {(1, 5), (2, 6), (3, 7) A R B
Here, R is a one to one relation because different 1 5
element of set A has different image in set B. 2 6
3 7
(b) Many to one relation :

A relation R from set A set B is called many to one

relation if two or more elements A are associated with unique element of set B.

R = {(a, x), (b, x), (c, x)}

R B
A x
a
b
c

Hence, R is called many to one relation as three elements of A are associated
with unique element x of A.

(c) One to many relation

A relation R : A o B is called one to many relation if at least one element of set
A is associated with two or more elements of set B.

Here, R = {(a, x), (a, y), (b, y), (b, z)} A R B
a x
"a of A is associated with two elements x and y of B y
and b is associated with y and z of B". Hence, R is b z
called one to many relation.

1.10 Inverse Relation

Let R = {(a, x), (b, y), (c, z)}, then interchange the element of each of the ordered
pair, we get a new relation R–1 = {(x, a), (y, b), (z, c)}.

Thus new relation R–1 is called inverse relation of R.

If R = {(x, y) : x  A and y  B}

then, R–1 = {(y, x) : y  B and x  A}

R B R–1 B
A x A a
a y x b
b z y c
c z

vedanta Excel in Additional Mathematics - Book 7 19

Relations

Example 3. If R = {(1, 2), (2, 3), (3, 4), (4, 5)}, find the inverse relation of R.

Solution: Here, R = {(1, 2), (2, 3), (3, 4), (4, 5)}
Inverse relation of R is given by

R–1 = {(2, 1), (3, 2), (4, 3), (5, 4)}

Exercise 1.2

Short Questions :
1. (a) Define a relation.

(b) State any five ways of representation of a relation.
(c) Define one to one relation with an example.
2. Write the domain and range of the following relations:
(a) R1 = {(1, 4), (2, 5), (3, 6), (4, 7)}
(b) R2 = {(1, 1), (2, 8), (3, 27), (4, 64)}
(c) R3 = {(m, a), (n, b), (p, c)}
(d) R4 = {(4, 8), (5, 10), (6, 12), (7, 14)}
3. Find the inverse relation of the following relations:
(a) R1 = {(a, p), (b, q), (c, r), (d, t)}
(b) R2 = {(1, 4), (2, 5), (3, 6), (5, 8)}
(c) R3 = {(10, 5), (8, 4), (6, 3), (2, 1)}
(d) g = {(1, 1), (4, 2), (9, 3), (16, 4)}
Long Questions :
4. Find the domain, range and inverse of relation the following relations.

(a) R (b) R

ap 1 2
bq 2 4
cr 3 6
4 8

5. Let A = {1, 2, 3} and B = {4, 5, 6}, find A × B, find the relation from A to B
determined by the conditions.

(a) x < y (b) x + y = 5

(c) x > y (d) y = 2x

20 vedanta Excel in Additional Mathematics - Book 7

Relations

6. If A = {1, 2, 3} and B = {1, 8, 27} and a relation R : A o B is defined by "cube
of". Represent R by the following wags:

(a) Set of ordered pairs

(b) Arrow diagram

(c) Tabulation method R

7. If R is a relation from P to Q as shown in the 1 1
adjoining diagram, then, 2 4
3 9
(a) Find the relation R in set of ordered pair 4 16

(b) Express R in tabular form

(c) Express R in the rule method

2. (a) D = {1, 2, 3, 4}, R = {4, 5, 6, 7} (b) D = {1, 2, 3, 4}, R = {1, 8, 27, 64}

(c) D = {m, n, p}, R = {a, b, c} (d) D = {4, 5, 6, 7}, R = {8, 10, 12, 14}

3. (a) {(p, a), (q, b), (r, c), (t, d)} (b) {(4, 1), (5, 2), (6, 3), (8, 5)}

(c) {(5, 10), (4, 8), (3, 6), (2, 1)} (d) {(1, 1), (2, 4), (3, 9), (4, 16)}

4. (a) D = {a, b, c}, R = {p, q, r}, R–1 = {(p, a), (q, b), (r, c)}

(b) D = {1, 2, 3, 4}, R = {2, 4, 6, 8}, R–1 = {(2, 1), (4, 2), (6, 3), (8, 4)}

5. A × B = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}

(a) {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}

(b) {(1, 4)} (c) { } (d) {(1, 2), (2, 4), (3, 6)}

6. (a) {(1, 1), (2, 8), (3, 27)}

(b) x 1 2 3
y 1 8 27

7. (a) R = {(1, 1), (2, 4), (3, 9), (4, 16)}

(b) x 1 2 3 4 (c) R = {(x, y) : y = x2}
y 1 4 9 16

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Transformation

6Transformation

6.0 Introduction

A process in which an object changes its shape, size, or position is called a
transformation. A transformation forms an image which is congruent or similar to
given object. In a transformation each point of an object has exactly one image point
and each image point has exactly one pre-image in the same plane. In daily life, we
come across phenomenon like our image formed on mirror, rotation of wheels in
vehicles, photographs taken by smart mobile cameras, etc. These are some examples
of transformation.

Transformation sometimes leaves certain points unchanged. These points are said
to be invariant. For example, in a rotation centre of rotation is invariant point.

6.1 Types of Transformation

Transformations may or may not change the size of the given objects: on the basis of
it, the transformation is divided into following two types :

(a) Isometric Transformation

(b) Non-isometric Transformation

(a) Isometric Transformation
A transformation in which the size of an object does not change but the position
of the object changes is known as isometric transformation. Reflection, rotation,
and translation are isometric transformations. Isometric transformations are
also called congruent transformations.

(b) Non-isometric Transformation
The transformations in which the size of an object changes is known as non-
isometric transformation. In this transformation, the distance between two
points of an object and the distance between their corresponding images after
transformation are unequal Enlargement and reduction are examples of non-
isometric transformation.

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Transformation

Four Fundamental Transformations

The following are the four fundamental transformations:

(a) Reflection (b) Rotation (c) Translation (d) Enlargement

In this class, we study only about reflection and rotation.

6.2 Reflection

Let us observe the object and its image below and try to write some notes from the
figures.

P (Object)

P P'

Q Q'

Mirror (Object) (Image)
(Axis of reflection) R R'

P' (Image) Mirror
Fig. i
(Axis of reflection)

Fig. ii

Note :

(i) The image lies on the opposite side of the object with respect to the mirror
(axis of reflection)

(ii) The object distance and the image distance from the axis of reflections are
equal.

(iii) The object size and image size are equal.

(iv) This an isometric transformation.

Definition: Reflection is a rule which shifts an object to the image of the same shape
and size each being at an equal distance from a fixed line. The fixed line is called
the mirror or axis of reflection.

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Worked Out Examples

Example 1. Find the image of the objects given below after reflection in the
mirror line M.

(a) M (b) P R

P

Q

Solution: M M

(a) Here, we follow the following steps.

(i) Draw perpendiculars from PO to the

axis of reflection or mirror M.

(ii) Produce PO to opposite side of the P O P'
(Object) (Image)

mirror such that PO = OP'.

(iii) P' is called the image of P.

(b) Here, we follow the following steps. P R
C
(i) Perpendicular PA, QB and RC are drawn R'

from object points P, Q and R to the mirror A
P'
line M.

(ii) PA, QB and RC are produced to the Q
opposite side of the mirror such that PA = B
Q'
M
AP', QB = BQ' and RC = CR'.

(iii) Join P'Q', Q'R' and R'P' successively.

(iv) 'P'Q'R' is the image of 'PQR.

(v) Shade the image triangle P'Q'R'.

6.3 Use of Coordinates in Reflection

We can use coordinates to locate an object in the graph paper and its image can be
found reflecting on X-axis, Y-axis or y = x line etc.

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Transformation Y

(a) Reflection on X-axis P(4, 5)

Take a point P(4, 5) on a graph paper. Then O MX
draw a perpendicular PM to X-axis. Let it be
reflected on X-axis. Produce PM to P' such that P'(4, –5)
PM = MP'. Find the coordinates of P'. We get X'
P'(–4, 5) which is the image of P. Y'

P(4, 5) X-axis P'(4, –5)

Similarly, we get
R(–4, 5) X-axis R'(–4, –5)
N(–4, –5) X-axis N'(–4, 5)

? P(x, y) X-axis P'(x, –y)

Example 2. If P(–2, 3), Q(2, 1), and R(4, 4) are the vertices of 'PQR. Plot 'PQR on
a graph paper and find its image reflecting it on X-axis.

Solution: Here, P(–2, 3), Q(2, 1) and R(4, 4) are the vertices of triangle PQR.
Reflecting it on X-axis. we get,
P(x, y) X-axis P'(x, –y)

Now, P(–2, 3) X-axis P'(–2, –3)
Q(2, 1) X-axis Q'(2, –1)
R(4, 4) X-axis R'(4, –4)

'PQR and its image 'P'Q'R' are plotted on a graph paper and
image is shaded as shown in the graph.

Y

R(4, 4)

P(–2, 3)

Q(2, 1)

X' O X

Q'(2, –1)

P'(–2, –3)

R'(4, –4)

Y'

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Transformation

(b) Reflection on Y-axis P'(–4, 5) Y P(4, 5)

Take a point P(4, 5) on a graph paper. Then M X

draw a perpendicular PM to Y-axis. Let it be O

reflected on Y-axis. Produce PM to P' such that

PM = MP'. Find the coordinates of P. We get X'
P'(–4, 5) which is the image of P.

P(4, 5) Y-axis P'(–4, 5)

Similarly, we get, Y'
R(–4, 5) Y-axis R'(4, 5)
N(–4, –5) Y-axis N'(4, –5)

? P(x, y) Y-axis P'(–x, y)

Example 3. If P(2, –2), Q(5, 3), and R(1, 5) are the vertices of 'PQR. Reflect 'PQR
Solution: on Y-axis and find the image 'P'Q'R'. Plot both 'PQR and 'P'Q'R' on
the same graph.

Here, P(2, –2), Q(5, 3) and R(1, 5) are the vertices of ∆PQR. Reflecting
it on Y-axis, we get
P(x, y) Y-axis P'(–x, y)

Now, P(2, –2) Y-axis P'(–2, –2)
Q(5, 3) Y-axis Q'(–5, 3)
R(1, 5) Y-axis R'(–1, 5)

∆PQR and its image ∆P'Q'R' are plotted on a graph paper and the
image is shaded as shown in the graph.

Y

R'(–1,5) R(1, 5)

Q'(–5,3) Q(5, 3)

X' O X

P'(–2,–2) P(2,–2)

Y'

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Transformation

Exercise 6.1

Short Questions :

1. Find the image of P(a, b) when :

(a) reflected on X-axis (b) reflected on Y-axis

2. Complete the following by drawing the images with l as the axis of reflection.

(a) P (Object) (b) l
P

l
Q

(c) Q (d) S

P R P
l Q
Rl

3. Find the images of the following points separately under the reflection

(a) X-axes (b) Y-axis

(i) P(x, y) (ii) Q(1, 2) (iii) R(–2, 4) (iv) S(– 4, – 4)

(v) T(–5, 6) (vi) M(4, 5) (vii) U(6, – 7) (viii) V(8, –4)

Long Questions :

4. (a) P(3, 8), Q(3, 4), and R(–6, 7) are the vertices of 'PQR. Reflect it on X-axis and
find the image vertices of 'PQR. Plot 'PQR and 'P'Q'R' on the same graph.

(b) A(1, 2), B(5, 5), and C(1, 8) are the vertices of 'ABC. Find the coordinates of
the image vertices of 'ABC when it is reflected on X-axis. Plot both 'ABC and
its image 'A'B'C' on the same graph.

(c) Let M(–2, 4), N(–1, 8), and P(–8, 4) are the vertices of 'MNP. Reflect it on X-axis
and find image 'M'N'P'. Plot 'MNP and its image 'M'N'P' on the same graph.

5. (a) If A(4, 8), B(1, 1), and C(8, 2) are the vertices of 'ABC. Reflect it on Y-axis
and find the image vertices of 'ABC. Plot both 'ABC and its image 'A'B'C'
on the same graph.

(b) If P(–2, 2), Q(–4, 8), and R(–6, 1) the vertices of 'PQR, find the coordinates
of the image vertices of the triangle PQR. Plot both the triangles on the
same graph.

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Transformation

(c) A(2, –2), B(2, 2), and C(8, 6) are the vertices of 'ABC. Reflect the triangle on
Y-axis and find the image vertices of the triangle. Plot both of the triangle
on the same graph.

Project Work

6. List any five uses of reflection and discuss them in your class.

1. (a) P'(a, –b) (b) P'(–a, b) 2. Show to your teacher.

3. (a) (i) P'(x, –y) (ii) Q'(1, –2) (iii) R'(–2, –4) (iv) S'(–4, 4)

(v) T(–5, –6) (vi) M'(4, –5) (vii) U(6, 7) (viii) V'(8, 4)

(b) (i) P'(–x, y) (ii) Q'(–1, 2) (iii) R'(2, 4) (iv) S'(4, –4)

(v) T'(5, 6) (vi) M(–4, 5) (vii) U'(–6, –7) (viii) V'(–8, –4)

4. (a) P'(3, –8), Q'(3, –4), R'(–6, –7) (b) A'(1, –2), B'(5, –5), C'(1, –8)

(c) M'(–2, –4), N'(–1, –8), P'(–8, –4) 5.(a) A'(–4, 8), B'(–1, 1), C'(–8, 2)

(b) P'(2, 2), Q'(4, 8), R'(6, 1) (c) A'(–2, –2), B'(–2, 2), C'(–8, 6)

6.4 Rotation - Introduction

Let us take a point P on a plane paper and rotate it on clockwise and anti-clockwise

direction.

(a) Here, the point P is said to be rotated through 90° in anti- P'

clockwise direction or positive quarter turn about centre O. The

anti-clockwise direction is also called positive direction. P' is 90°

the image of P due to the rotation. OO P
P
(b) Here, the point P is said to be rotated through 90q in clockwise
direction or negative quarter turn about the centre O. The 90°

clockwise direction also called negative direction. P' is the image

of P due to the rotation. P'

(c) Here, the point P is said to be rotated through 180° in

anti-clockwise direction about centre O. It is also 180°
O
called positive half turn about O. P' is the image of P P' O P
–180° P
due to the rotation.

P'

(d) Here, the point P is said to be rotated through 180q in

clockwise about centre O. It is also called negative half

turn. P' is the image of P due to the rotation.

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Transformation

Definition: Rotation is a rule which shifts each point of the given object through an

equal angular displacement about a point in the given direction.

To rotate a geometrical figure, following three conditions are required:

(i) Centre of rotation (ii) Angle of rotation (iii) Direction of rotation

A figure can be rotated in two directions.

1. Anti-clockwise direction (Positive direction)

In anti-clockwise direction, an object is rotated in opposite direction of the
rotation of the hand of a clock. It is also called positive direction.

2. Clockwise direction (Negative direction)

In clockwise direction, an object is rotated in the same direction of the rotation
of the hand of a clock. It is also called negative direction.

Now, let’s learn to draw the image of a figure when it is rotated through the given
angle in the given direction about a given centre of rotation.

6.5 Rotation through 90° in anti-clockwise direction

(Positive Quarter turn about origin)

Let 'ABC be a given triangle. It is to be rotated B' Anti-clockwise
through 90° in anti-clockwise direction about O.

The following instructions are useful for this type C' C
of rotation.

(i) Join each vertex of the figure to the centre of A'

rotation O with dotted lines. B

(ii) On each dotted line, draw 90q at O with the O
help of a protractor in anti-clockwise direction. A

(iii) With the help of compasses, cut off OA' =OA,

OB' = OB and OC' = OC.

(iv) Join A'B', B'C', and C'A' successively, we get 'A'B'C'.

Here, 'A'B'C' is the image of 'ABC formed due to the rotation through 90° in anit-
clockwise direction about O.

6.6 Rotation through 90° in clockwise direction

(Negative Quarter Turn) O C

Let 'ABC be a given triangle. It is to be rotated through B
90° in clockwise direction about centre O. The following Clockwise
instructions are useful for this type of rotation. A

(i) Join each vertex of the figure to the centre of rotation A’ C’
O with dotted line.

(ii) With each dotted line draw 90° with the help of a B’
protractor in clockwise direction.

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Transformation

(iii) With the help of a compass, cut off OA' = OA, OB' = OB, OC' = OC.
(iv) Join A'B', B'C' and C'A' successively, we get triangle A'B'C.
Here, 'A'B'C' is the image of 'ABC due to the rotation through 90° in clockwise
direction.

6.7 Rotation through 180° about origin

(Half Turn) A' C
C' B
Let 'ABC be a given triangle. It is to be rotated
through 90° in anti-clockwise direction about OA
centre O. The following instructions are useful for B'
this type of rotation.

(i) Join each vertex of the figure to the centre of
rotation O with dotted line.

(ii) With each dotted line, draw 180° with the help of a protractor in clockwise
direction.

(iii) With the help to a compass, cut off OA' = OA, OB' = OB and OC' = OC.

(iv) Join A'B', B'C', and C'A' successively we get triangle A'B'C'.

Here, 'A'B'C' is the image of 'ABC due to the rotation of through 180° in clockwise
direction.

Worked out Examples

Example 1. Rotate the point P through an angle of positive 60° about the center

of rotation O. A'

Solution: Here, we join OA with centre O and make
angle 60° with OA in anti-clockwise
direction.

i.e. ‘AOA' = 60° 60°
A' is the image of A.
O A

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Transformation

Example 2. Rotate 'ABC through 90° in clockwise direction about O.
Solution:
(i) Join each vertex of the figure to the C
centre of rotation O with dotted line.

(ii) With each dotted line draw 90° with O B
Clockwise
the help of a protractor in clockwise A

direction. A’

(iii) With the help of a compass, cut off C’
OA' = OA, OB' = OB, OC' = OC.

(iv) Join A'B', B'C', and C'A' successively, B’
we get triangle A'B'C.

Here, 'A'B'C' is the image of 'ABC due to the rotation through 90° in
clockwise direction.

6.8 Use of Coordinates in Rotation

Let us take different points on a graph paper and rotate through 90° anti-clockwise,
clockwise, or 180° about centre O. Graphically, we can derive formula of rotations.

(a) Rotation through 90° in anti-clockwise direction about origin

P'(-2,4) Y Y YY

P(4,2) P(-3,2) P'(4,3)

XO X' X O X' X O X' X O X'

Y' P'(-2,-3) P(-3,-2) Y' P(3,-4)
P'(2,-3)
P(4, 2) o P' (–2, 4) Y' P (3, –4) o P' (4, 3)
Y'
P (–3, 2) o P' (–2, –3)
P (–3, –2) o P' (2, –3)

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

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

Let us take P(4, 2) in a graph and rotate it in 90° in anti-clockwise direction about
origin O. Then, we get,

P(4, 2) o P'(– 2, 4) [in figure (i)]

Similarly, P(–3, 2) o P'(–2, –3) [in figure (ii)]

P(–3, –2) o P'(2, –3) [in figure (iii)]

P(3, –4) o P'(4, 3) [in figure (iv)]

? P(x, y) o P'(–y, x)

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Transformation

(b) Rotation through 90° in clockwise direction about origin

Y YYY

P(2,3) P(-2,1) P'(1,2) P'(-2,2)

X' O X X' O X X' O X X' OP(4,-1) X

P'(3,-2) P(-2,-2)

P'(-1,-4)

Y' Y' Y' Y'

P(2, 3) o P' (3, –2) P (–2, 1) o P' (1, 2) P (–2, –2)o P'(–2, 2) P (4, –1) o P' (–1, –4)

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

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

Let us take a point P(2, 3) in a graph paper. Rotate it through 90° in clockwise
direction about origin. Then we get,

P(2, 3) o P'(3, –2) [in figure (i)]

Similarly, P(–2, 1) o P'(1, 2) [in figure (ii)]

P(–2, –2) o P'(–2, 2) [in figure (iii)]

P(4, –1) o P'(–1, –4) [in figure (iv)]

? P(x, y) o P'(y, –x)

(c) Rotation through 180° about origin

When a point is rotated through 180q in anti-clockwise or clockwise direction about
origin the image point has the same coordinates in both cases.

Y Y Y Y
P(4,3) P(-4,2) P'(3,3) P'(-3,2)

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

P'(-4,-3) P'(4,-2) P(-3,-3) P(3,-2)
Y' Y' Y' Y'
P (–4, 2) o P' (4, –2) P (3, –2) o P' (–3, 2)
P (4, 3) o P" (–4, –3) P (–3, –3) o P' (3, 3)

? 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)

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

Let us take a point P(4, 3) and rotate it through 180° in anti-clockwise or clockwise
direction about the origin. Then we get,

P(4, 3) o P'(– 4, –3) [in figure (i)]

P(–4, 2) o P'(4, – 2) [in figure (ii)]

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Transformation

P(–3, –3) o P'(3, 3) [in figure (iii)]
P(3, –2) o P'(–3, 2) [in figure (iv)]

? P(x, y) o P'(–x, –y)

Worked out Examples

Example 1. Find the image of the point A(5, 6), B(7, 8), C(–5, –6), D(2, –5) under
the rotation through the following angles about the origin:

(a) 90° (b) –90°

(c) 180°

Solution: (a) Under the rotation through +90° about origin,

P(x, y) P'(–y, x)

A(5, 6) A'(–6, 5)

B(7, 8) B'(–8, 7)

C(–5, –6) C'(6, –5)

D(2, –5) D'(5, 2)

(b) Under the rotation through –90° about origin,

P(x, y) P'(y, –x)

A(5, 6) A'(6, –5)

B(7, 8) B'(8, –7)

C(–5, –6) C'(–6, 5)

D(2, –5) D'(–5, –2)

(c) Under the rotation through 180° about origin,

P(x, y) P'(–x, –y)

A(5, 6) A'(–5, –6)

B(7, 8) B'(–7, –8)

C(–5, –6) C'(5, 6)

D(2, –5) D'(–2, 5)

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Example 2. If P(4, 1), Q(5, –1), and R(6, 4) are the vertices of 'PQR. Find the
coordinates of its image under the rotation through +90° about
origin. Show both object and image in the same graph.

Solution: Here, P(4, 1), Q(5, –1), and R(6, 4) are the vertices of 'PQR.

Under the rotation of +90° about origin.

P(x, y) P'(–y, x)

Now, P(4, 1) P'(–1, 4)
Q(5, –1) Q'(1, 5)

R(6, 4) R(–4, 6)

? 'PQR 'P'Q'R'

Y

R'(-4, 6) 7 Q'(1, 5) R(6, 4)

6

5

P'(-1, 4) 4
3

X' 2 P(4, 1) X
1

-7 -6 -5 -4 -3 -2 -1-O1 1 2 3 45678

-2 Q(5, -1)

Y'

Example 3. Let P(5, 2), Q(3, 1), and R (2, –4) be the vertices of 'PQR. 'PQR is
Solution: rotated through +90° about origin and image 'P'Q'R' is formed.
Draw 'PQR and 'P'Q'R' on the same graph.

Rotating the 'PQR through +90° about origin, we have,

P(x, y) P'(–y, x)

Now, P(5, 2) P'(–2, 5)

Q(3, 1) Q'(–1, 3)

R(2, –4) R'(4, 2)

Again rotating 'P'Q'R' through +180° about origin, we get,

P(x, y) P'(–x, –y)

Now, P'(–2, 5) P"(2, –5)

Q'(–1, 3) Q"(1, –3)

R'(4, 2) R"(–4, –2)

? 'PQR 'P'Q'R' 'P"Q"R"

'PPQR and 'P'Q'R' are plotted in the graph.

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Transformation

Y

7

P' 6
5

4

Q' 3 R' P
2 Q

1
X' -7 -6 -5 -4 -3 -2 -1-O1 1 2 3 4 5 6 7 8 X

-2

-3

-4 R

-5

Y'

Exercise 6.2

Short Questions :
1. Find the image of P(a, b) when it is rotated:

(a) through 90° in anticlockwise about origin.
(b) through 90° in clockwise about origin.
(c) through 180° about origin.
2. Draw the images of the following figures through the angles mentioned taking
O as the center of rotation:

(a) A (Through –90°) (b) A (Through +90°)

B

OO

(c) B (d) P (through –90°)

(Through +90°)

A C Q R
O O

104 vedanta Excel in Additional Mathematics - Book 7

(e) M (through –90°) (f) Transformation
R
O P P
N (through –180°) (through +180°)

(g) A Q
O

B

O
C

3. Find the images of the following points when they are rotated through 90°.

(I) Clockwise (II) Anti-clockwise about origin

(a) P(a, b) (b) A(4, 5) (c) B(–6, 6)

(d) C(– 4, – 5) (e) D(5, – 6) (f) E(4, 0)

Long Questions :

4. (a) A(3, 4), B(2, 1), and C(7, 2) are the vertices of 'ABC. Find the coordinates
of its image vertices under the rotation through +90° about the origin.
Present both 'ABC and its image 'A'B'C' on the same graph.

(b) P(3, 4), Q(6, 10), and R(8, 3) are the vertices of 'PQR. Find the coordinates
of its image vertices under the rotation through 90° in anti-clockwise
direction. Plot both triangles on the same graph.

(c) M(2, 1), N(4, 3), and R(1, 4) are the vertices of 'MNP. Find the coordinates
of image vertices of the triangle when it is rotated through 90° in
anti-clockwise about the origin.

5. (a) A(1, 2), B(6, 2), and C(5, 6) are the vertices of 'ABC. The 'ABC is rotated
through an angle of –90° about the origin to get its image 'A'B'C'. Defermine
the coordinates of A', B' and C' and plot 'ABC and 'A'B'C' in the same
graph.

(b) M(2, 5), N(3, 1), and P(5, 6) are the vertices of 'MNP. Rotate 'MNP through
90° in clockwise direction about the origin present both of the triangles on
the same graph.

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(c) U(2, 4) and V(–2, –4) are the two points of line segment UV. Rotate the line
segment UV through 90° in clockwise direction about the origin. Plot both
of the line segments on the same graph.

6. (a) P(6, 0), Q(8, 3) and R(2, 6) are the vertices of 'PQR. Find the coordinates of
its image under the rotation through 180° about origin. Present the object
and image in the same graph paper.

(b) A(2, 1), B(3, 6) and C(0, 4) are the vertices of 'ABC. Rotate the 'ABC
through 180° in anticlockwise about origin. State the coordinate of the
image vertices present both of the triangles on the same graph.

(c) A unit square with vertices O(0, 0), A(1, 0), B(1, 1) and C(0, 1) is plotted on
a graph. Rotate the square through 180° in clockwise about the origin state
the coordinates of image vertices of the square. Also present the image of
the unit square on the same graph. (Hints: take 50 small divisions - 1 unit)

Project Work

7. Prepare a report on rotation including the following points :
(a) Introduction of rotation
(b) Examples of rotations
(c) Daily use of rotations

1. (a) P'(–b, a) (b) P'(b, –a) (c) P'(–a, –b)

3. (I) (a) P'(b, –a) (b) A'(4, –5) (c) B'(6, 6) (d) O'(–5, 4)

(e) D'(–6, –5) (f) E'(0, –4) (II) (a) P'(–b, a) (b) A'(–5, 4)

(c) B'(–6, –6) (d) C'(5, –4) (e) D'(6, 5) (f) E(0, 4)

4. (a) A'(–4, 3), B'(–1, 2), C'(–2, 7) (b) P'(–4, 3), Q'(–10, 6), R'(–3, 8)

(c) M'(–1, 2), N'(–3, 4), R'(–4, 1) 5.(a) A'(2, –1), B'(2, –6), C'(6, –5)

(b) M'(5 –2), N'(1, –3), P'(6, –5) (c) U'(4, –2), V'(–4, 2)

6. (a) P'(–6, 0), Q'(–8, –3), R'(–2, –6) (b) A'(–2, –1), B'(–3, –6), C'(0, –4)

(c) O'(0, 0), A'(–1, 0), B'(–1, –1), C'(0, –1)

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11. If P = 2 1 , Q = –4 3 , find P – Q and Q – P. Are P – Q and Q – P equal?
4 5 6 4

12. If A(1, 1), B(4, 3), C(2, 1), and D(5, 3), prove that AB = CD.

13. Define cartesian product of two sets. If A = {1, 2} and B = {4, 6, 7}, find

A × B and B × A. A

14. From the given right angled triangle, find the six fundamental 2 2
trigonometric ratios taking reference angle.

15. Prove that 1 + sin + 1 cos = 2sec . C 2B
cos + sin PQR and its
PQR.
16. If P(3, 8), Q(3, 4), and R(6, 7) are the vertices of PQR. Plot
Reflect it on X-axis and find the image vertices of
image on the same graph.

17. Find arithmetic mean from the following data :

Marks 10 20 30 40 50 60 70 80

No. of students 2 4 6 15 12 6 4 3

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7

7.0 Introduction

Statistics is a branch of applied mathematics which deals with numerical data. It
generally deals with data collection, tabulation of data, analysis of data, and drawing
the required conclusion for the purpose of the studies.

Nowadays, there is hardly any field of human activities where statistics is not used.
Statistics is used in every field of studies like economics, research, administrations,
agriculture, industry, engineering, education, etc.

7.1 Frequency

The number of times an observation occurs is called the frequency of the observations.
It is denoted by f.

For example: The marks obtained by 15 students in a unit test in mathematics are
given below. The full marks was 20 and the pass marks was 10.

5, 8, 12, 14, 15, 15, 18, 18, 18, 20, 20, 5, 8, 12, 14

From the data, we can classify the data as follows:

Marks Obtained Tally Marks No. of Students
5 || 2
8 || 2
12 || 2
14 || 2
15 || 2
18 ||| 3

20 || 2
Total 15

From the data collected in a class of 15 students, we have classified the data as
shown above. The marks are called variates. The number of times the variate values
repeat is known as frequency. The total of frequency is the total number of students.

We denote the total frequency or sum of frequency by ∑f or N.

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7.2 Frequency Distribution

A tabular arrangement of data showing the frequency of each variate value is
called a frequency distribution. The table showing data with their corresponding
frequencies is called a frequency distribution table. The frequency distribution can
be divided into three types:

(a) Individual Series.

(b) Discrete Series.

(c) Continuous Series.

(a) Individual Series
Individual series are usually list of datas which are collected in a small scale.
They are also called raw data.

Example : The amount of money brought by 5 students of a school for their tiffin.

Rs. 100, Rs. 120, Rs. 150, Rs. 180, Rs. 200

These kind of data are less in number and need not be arranged in table.

Here, number of datas is denoted by n.

In the above example, n = 5

(b) Discrete Series
The series formed by discrete values are called discrete series. We usually have
variates and frequency in this discrete series.

The marks obtained by 20 students in a unit test are given below:

Marks 7 10 12 14 15 18 20
No of students 3 232523

Here, marks denote the variate values and the number of students denote the
frequencies. Total frequency is denoted by N = ¦f.

(c) Continuous Series

The series which can be represented by a continuous variable is called
continuous series. We usually use this kind of data to insert large scale of datas.
The marks obtained by 100 student in an examination is given below.

Marks 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50
15 25 30 20
No of students 10

In this series, the data lie within groups. Such interval or groups are called
class intervals. The end points of class interval are called limits.

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Example: In case of 0-10 0 is called lower limit. 10 is called upper limit.

The difference between the upper limit and the tower limit of corresponding
class is called class interval. In above example, the class interval is 10.

In continuous class, the upper limit of the class is excluded. So it is also called
exclusive method. In class (0-10), 10 is excluded. That is 10 does not lie in class
(0-10). 10 lies in (10-20).

The middle value of corresponding class is obtained by using the formula

Middle value = upper limit + lower limit
2

In above example

Mid value of class (0-10) = 0 +10 = 5
2
10 + 20
Mid value of class (10-20) = 2 = 15

7.3 Cummulative Frequency (c.f.) Table

The cummulative frequency of a class interval is the sum of frequencies all classes
upto and including that class. A frequency distribution showing the cummulative
frequencies corresponding values of the variable systematically arranged in
increasing (or decreasing) order is known as the cummulative frequency distribution.

In a cummulative frequency table, cummulative frequency corresponding to a class-
interval is the sum of all frequencies upto and including that class interval.

A table which displays the cummulative frequencies are distributive over various
classes is called cummulative frequency table.

Example: Let us consider weights of 50 students given in the table:

Weights (in kg) 20-25 25-30 30-35 35-40 40-45 45-50
8
No of students 2 8 10 12 10

Cummulative Frequency table of above data is given below:

Weights (kg) No. of students (f) Cummulative Frequency (c.f.)
20-25 2 2
25-30 8
30-35 10 2 + 8 = 10
35-40 12 10 + 10 = 20
40-45 10 20 + 12 = 32
45-50 8 32 + 10 = 42
42 + 8 = 50

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Worked out Examples

Example 1. Construct the cummulative frequency table from the following data.

Marks (x) 5 10 15 20 25 30 35
1
Frequency (f) 2 4 6 4 3 2

Solution: Cummulative frequency table of given data is given below:

Marks (x) Frequency (f) Cummulative Frequency (c.f.)
5 2 2
10 4
15 6 2+4=6
20 4 6 + 6 = 12
25 3 12 + 4 = 16
30 2 16 + 3 = 19
19 + 2 = 21
35 1
21 + 1 = 22

Example 2. The marks obtained by 30 students in mathematics are listed below
Solution: construct the cummulative frequency table from the given data.

10 20 24 48 30 32 20 40 10 24

18 35 20 40 48 30 32 20 45 10

24 48 18 30 20 40 24 18 45 30

Cummulative frequency table of given data is given below:

Marks (x) Tally marks No. of students Cummulative
(f) Frequency (c.f.)
10 ||| 3
18 ||| 3 3
20 |||| 5 3+3=6
24 |||| 4 6 + 5 = 11
30 |||| 4 11 + 4 = 15
32 || 2 15 + 4 = 19
35 19 + 2 = 21
40 | 1
45 ||| 21 + 1 = 22
48 || 3
||| 22 + 3 = 25
2
25 + 2 = 27
3
27 + 3 = 30

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Example 3. Prepare a cummulative frequency table from given data:

Marks 0-10 10-20 20-30 30-40 40-50
No. of students 5 15 25 20 5

Solution: The cummulative frequency table of above data is given below:

Marks (x) No. of students (f) Cummulative Frequency (c.f.)
0-10 5 5
10-20 15
20-30 25 5 + 15 = 20
30-40 20 20 + 25 = 45
40-50 5 45 + 20 = 65
65 + 5 = 70
Exercise 7.1

1. Construct a cummulative frequency table from the following data:

(a) Marks obtained 10 15 20 25 30 35 40
2
No. of students 2 4 6 8 5 3

(b) x 20 30 40 60 70 80 90
f 2 5 10 7 6 3 2

(c) Marks 0-10 10-20 20-30 30-40 40-50
No. of students 6 12 15 12 5

(d) Age (years) 0-10 10-20 20-30 30-40 40-50 50-60
No. of people 10 15 25 15 10 52

2. (a) Ages of 30 students of class VII are given below. Prepare a cummulative
frequency table from the given data:

12, 16, 13, 16, 13, 12, 16, 15, 13, 14,

14, 15, 15, 14, 14, 16, 14, 12, 14, 15,

11, 16, 12, 13, 11, 13, 16, 11, 14, 15

(b) The weight (in kgs) of 25 students are given below. Prepare a cummulative
frequency table:

25, 30, 35, 25, 26, 40, 25, 28, 28, 35,

28, 40, 45, 28, 40, 40, 25, 40, 45, 48,

30, 28, 48, 30, 48

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3. (a) Given data is marks of 40 students of class VII. Construct a frequency
distribution table taking the class interval of 10. Also prepare a cummulative
frequency table:
35, 40, 45, 50, 70, 80, 90,, 55, 65, 40,
40, 45, 60, 75, 80, 54, 55, 58, 60, 75,
90, 70, 40, 55, 75, 85, 90, 45, 60, 88,
45, 60, 75, 85, 55, 94, 75, 50, 45, 68

(b) Given data is weights (kg) of 30 students of class VII. Make a frequency
table taking class interval 5. Also prepare a cummulative frequency table:
45, 44, 33, 35, 30, 50, 44, 60, 52, 60,
62, 64, 44, 55, 35, 45, 44, 50, 32, 35,
42, 46, 37, 40, 44, 47, 62, 60, 58, 50

7.4 Arithmetic Mean

Average of a set of data is very important. It daily life we use 'average' to represent
arithmetic mean. Arithmetic mean of set of data can represent whole set of data. It
usually represents the central data. The percentage of any student is the average
marks of all the marks they get in the exam.

The arithmetic mean of the given data is calculated by adding all the data together
and dividing them by the total number of data.

Mathematically,

Mean = Sum of the data
X Total number of data
= ¦nX,

where X is the variate values.

¦X is the sum of all variates

n is total number of data

Indiscrete Series.

In case of discrete series or ungrouped repeated data, we find mean by using formula.

Mean (x) = ¦fx
N

Where, x is the variate

f is the frequency

fx is the product of f and x

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∑fx is the sum of fx.
N is the total number of frequency.

Worked Out Examples

Example 1. Find the mean of 10, 20, 30, 40, 50
Solution:
Here, the data are 10, 20, 30, 40, 50

number of data (n) = 5

∑x = 10 + 20 + 30 + 40 + 50

= 150 = ¦x
Now, mean (x) n

= 150
5

= 30

Example 2. The average of, 10, 20, 30, p, 50, and 60 is 35. Find the value of p.
Solution:
Here, the data are 10, 20, 30, p, 50, 60

Number of data (n) = 5

∑x = 10 + 20 + 30 + p + 50 + 60

= 170 + p

Now, we know,

mean (X) = 35

So, mean (X ) = ¦X
n
170 + p
or, 35 = 6

or, 210 = 170 + p

or, p = 40

? The value of p is 40.

Example 3. Calculate the mean from the data given below:

Marks 10 20 30 40 50

No. of students 4 6 10 6 4

Solution: Calculation of Mean,

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Marks (x) No. of students (f) fx
40
10 4 120
300
20 6 240
200
30 10 ¦fx = 900

40 6

50 4

N = 30

Now, Mean (x) = ¦fx = 900 = 30
N 30

Hence, mean marks = 30.

Example 4. Find the value of p from the data given below whose mean is 20:

x 5 10 15 20 25 30 35
f 27p 7 644

Solution: Calculation of missing frequency: fx
10
xf 70
52 15p
10 7 140
15 p 150
20 7 120
25 6 140
30 4 ¦fx = 630 + 15p
35 4

¦f = 30 + p

By using formula,

Mean (x) = ¦fx
N
630 + 15p
So, 20 = 30 + p

or, 600 + 20p = 630 + 15p

or, 30 = 5p

? p=6

Hence, the missing frequency is 6.

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Exercise 7.2

1. Find the mean from the data given below:
(a) 35, 40, 45, 50, 60
(b) 10, 11, 14, 16, 17, 19, 13, 20, 25, 21

2. (a) If the mean of 5, 12, 8, 10, and p is 8, find the value of p.

(b) If the mean of 15, 21, 25, and p is 20, find the value of p.
Long Questions :
3. (a) From the data given below calculate the mean:

Marks 20 40 60 80 100
3
No of students 6 9 4 3
500
(b) From the data given below, find the mean: 10

Wages (in Rs.) 100 200 300 400

No. of workers 10 15 10 15

(c) x 5 10 15 20 25 30
f 2 5 10 7 4 2

(d) x 58 60 62 64 66 68

f 12 14 20 13 8 5

(e) Marks 40 50 55 62 75 80

No. of students 4 6 10 8 5 2

4. (a) The mean of the data given below is 21. Find the value of p.

x 10 15 20 25 35
f 56p65

(b) The mean of the data given below is 34. Find the missing frequency.

x 10 20 30 40 50 60
f 24q432

Project Work

5. List the marks obtained by the students of your class in mathematics in the
class 6. Then find the average of the marks.

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1. (a) 46 (b) 16.6 2.(a) 5 (b) 19
3. (a) 50.4 marks (b) Rs. 300
4. (a) 3 (b) 5 (c) 17 (d) 62.17 (e) 58.31 marks

7.5 Median

The data that divides the given set of data into two equal halves is known as median.

It is denoted by Md. To calculate median, we need to arrange the data either in
ascending or descending order.

(a) For Individual Series

Median (Md) = value of n+1 th
2
item

where, n = total number of items

(b) For Discrete Series
Steps for calculation of Median

(i) First arrange the variables in ascending order.

(ii) Write commulative frequency in c.f. column.

(iii) Find N+ 1 , where N = total frequency.
2

(iv) Then observe the c.f. just greater than or equal to N+ 1 at c.f. column,
2
N+ 1
then the corresponding value of 2 in variate column is the required

median. If any c.f. is just greater than N+ 1 of c.f. column, then the
2

value corresponding to that c.f is the required median.

Worked out Examples

Example 1. Find the median from the set of data given below:
Solution: 30, 70, 90, 100, 150
Here, the given items are 30, 70, 90, 100, 150

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which are already in ascending order.

So, number of items (n) = 5

Now, median (Md) = value of (n + 1)th item
2

= value of (5 + 1)th item
2

= value of 6 th
2
item

= value of 3rd item

? Median (Md) = 90
Example 2. Find the median from the given data : 22, 18, 14, 12, 24, 26, 32, 34
Solution: Here, the given data are arranged in ascending order

12, 14, 18, 22, 24, 26, 32, 34

Now, number of items (n) = 8

So, Median (Md) = value of (8 + 1)th item
2

= value of 9 th item
2

= value of 4.5th item

Now 4. 5th item is not present

So, 4.5th item is the mean of 4th item and 5th item

So, median (Md) = value of 4th item + value of 5th item
2

= 22 + 24
2

= 46
2

= 23

? Median (Md) = 23

Example 3. 16, 18, 20, x, 24, 26, 28 are in ascending order. If the median of given
Solution: data is 22, find the value of x.

Here, the given items are 16, 18, 20, x, 24, 26, 28

the number of items (n) = 7

? Also, median (Md) = 22

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We know, = value of (n + 1)th item
Median (Md) 2

= value of (7 + 1)th item
2

= value of 4th item

=x

But, by the question,

Md = 22
? x = 22

Example 4. 5, p – 12, p + 11 and 42 are in ascending order. If the median of the
data is 19, find the value of p.

Solution: Here, the given items are 5, p – 12, p + 11 and 42

number of items (n) = 4

? Now, median (Md) = value of (n + 1)th item
2
(4 + 1)th
= value of 2 item

= value of 2.5th item

= value of (2nd item + 3rd item)
2
p – 12 + p + 11
= 2

= 2p – 1
2

But, by the question,

Md = 19

So, 2p – 1 = 19
2

or, 2p – 1 = 38

or, 2p = 39

? p = 19.5

Example 5. Find the median from given data:

x 5 10 15 20 25 30 35 40
f 4 6 10 15 10 8 6 4

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Solution: To find median, given data can be written in c.f. table.

x f Cummulative frequency (c.f.)
5 4 4
10 6
15 10 4 + 6 = 10
20 15 10 + 10 = 20
25 10 20 + 15 = 35
30 8 35 + 10 = 45
35 6 45 + 8 = 53
40 4 53 + 6 = 59
Total N = 63 59 + 4 = 63

Here, N = 63

we have, using formula,

Median (Md) = N+1 th
2
item

= 63 + 1 th
2
item

= 32nd item

c.f. just greater 32 is 35 whose corresponding value is 20.

? Median (Md) = 20

Exercise 7.3

1. Find the median from the data given below:

(a) 22, 24, 26, 28, 32 (b) 6, 8, 10, 12, 14, 16, 18

(c) 100, 200, 300, 400, 500, 600, 700

2. Find the median of the given data:

(a) 28, 21, 24, 20, 27, 25 (b) 40, 44, 48, 60, 56, 72, 65, 53

(c) 100, 400, 600, 800, 1000, 300, 200, 700

3. (a) If the median of the data 13, 14, p, 17, 20 which are arranged in ascending
order is 16, find the value of p.

(b) If the median of the data 400, 350, 300, q, 200, 150, 100 which are arranged
in descending order is 250, find the value of q.

4. (a) The median of the data p – 1, p + 1, 2p + 5, and 3p + 1 are the data in
ascending order is 18, find the value of 'p' and the given numbers.

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(b) If p, p + 4, p + 6, p + 8 are in ascending order and of the median is 20, find
the value of 'p' and the numbers.

5. Find the median from the following data:

(a) Marks 10 20 30 40 50 60 70

No. of students 2 4 6 10 4 3 1

(b) Weight (kg) 10 15 20 22 24 30 35

No. of children 8 10 16 10 4 6 3

(c) x 100 200 300 400 500 600
f 4 6 10 8 6 2

Project Work

6. Measure the heights of your friends and tabulate them as shown in the given
table.

Name of students Height (feet)
1.
2.
3.
4.
5.

Arrange the heights in ascending order. Find the median height.

1. (a) 26 (b) 12 (c) 400

2. (a) 24.5 (b) 54.5 (c) 500

3. (a) 16 (b) 250 4.(a) p = 10, 9, 11, 25, 31

(b) p = 15; 15, 19, 21, 23 5.(a) 40 (b) 20 (c) 300

7.6 Mode

The variate value which occurs maximum number of times in a distribution is
called mode. In other words, the most frequently repeated value is called the mode.
In a distribution, mode occurs with maximum frequency. It is denoted by MO.

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