Approved by the Government of Nepal, Ministry of Education, Science and Technology,
Curriculum Development Centre, Sanothimi, Bhaktapur as an Additional Learning Material
vedanta
Excel in
MATHEMATICS
Book 7
Book 7
Author
Hukum Pd. Dahal
Editor
Tara Bahadur Magar
vedanta
Vedanta Publication (P) Ltd.
]
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Vanasthali, Kathmandu, Nepal
+977-01-4982404, 01-4962082
[email protected]
www.vedantapublication.com.np
vedanta
Excel in
MATHEMATICS
Book 7
Book 7
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.
First Edition: B.S. 2077 (2020 A. D.)
Second Edition: B.S. 2078 (2021 A. D.)
Published by:
Vedanta Publication (P) Ltd.
]
jbfGt klAns;g k|f= ln=
]
Vanasthali, Kathmandu, Nepal
+977-01-4982404, 01-4962082
[email protected]
www.vedantapublication.com.np
Preface
The series of 'Excel in Mathematics' is completely based on the contemporary pedagogical teaching
learning activities and methodologies extracted from Teachers' training, workshops, seminars and
symposia. It is an innovative and unique series in the sense that the contents of each textbooks of
the series are written and designed to fulfill the need of integrated teaching learning approaches.
Excel in Mathematics is an absolutely modified and revised edition of my three previous series:
'Elementary mathematics' (B.S. 2053), 'Maths In Action (B. S. 2059)' and 'Speedy Maths' (B. S. 2066).
Excel in Mathematics has incorporated applied constructivism. Every lesson of the whole series
is written and designed in such a manner, that makes the classes automatically constructive and
the learners actively participate in the learning process to construct knowledge themselves, rather
than just receiving ready made information from their instructors. Even the teachers will be able
to get enough opportunities to play the role of facilitators and guides shifting themselves from the
traditional methods of imposing instructions.
Each unit of Excel in Mathematics series is provided with many more worked out examples.
Worked out examples are arranged in the hierarchy of the learning objectives and they are reflective
to the corresponding exercises. Therefore, each textbook of the series itself is playing a role of a
‘Text Tutor’. There is a well balance between the verities of problems and their numbers in each
exercise of the textbooks in the series.
Clear and effective visualization of diagrammatic illustrations in the contents of each and every
unit in grades 1 to 5, and most of the units in the higher grades as per need, will be able to integrate
mathematics lab and activities with the regular processes of teaching learning mathematics
connecting to real life situations.
The learner friendly instructions given in each and every learning contents and activities during
regular learning processes will promote collaborative learning and help to develop learner-
centred classroom atmosphere.
In grades 6 to 10, the provision of ‘General section’, ‘Creative section - A’ and ‘Creative
section - B’ fulfills the coverage of overall learning objectives. For example, the problems in
‘General section’ are based on the Knowledge, understanding and skill (as per the need of the
respective unit) whereas the ‘Creative sections’ include the Higher ability problems.
The provision of ‘Classwork’ from grades 1 to 5 promotes learners in constructing knowledge,
understanding and skill themselves with the help of the effective roles of teacher as a facilitator
and a guide. Besides, teacher will have enough opportunities to judge the learning progress and
learning difficulties of the learners immediately inside the classroom. These classworks prepare
learners to achieve higher abilities in problem solving. Of course, the commencement of every
unit with 'Classwork-Exercise' may play a significant role as a 'Textual-Instructor'.
The 'project works' given at the end of each unit in grades 1 to 5 and most of the units in higher
grades provide some ideas to connect the learning of mathematics to the real life situations.
The provision of ‘Section A’ and ‘Section B’ in grades 4 and 5 provides significant opportunities
to integrate mental maths and manual maths simultaneously. Moreover, the problems in ‘Section
A’ judge the level of achievement of knowledge and understanding and diagnose the learning
difficulties of the learners.
The provision of ‘Looking back’ at the beginning of each unit in grades 1 to 8 plays an important
role of ‘placement evaluation’ which is in fact used by a teacher to judge the level of prior
knowledge and understanding of every learner to make his/her teaching learning strategies.
The socially communicative approach by language and literature in every textbook especially in
primary level of the series will play a vital role as a ‘textual-parents’ to the young learners and
help them in overcoming maths anxiety.
The Excel in Mathematics series is completely based on the latest curriculum of mathematics,
designed and developed by the Curriculum Development Centre (CDC), the Government of Nepal.
I do hope the students, teachers and even the parents will be highly benefited from the ‘Excel in
Mathematics’ series.
Constructive comments and suggestions for the further improvements of the series from the
concerned will be highly appreciated.
Acknowledgments
In making effective modification and revision in the Excel in Mathematics series from my
previous series, I’m highly grateful to the Principals, HOD, Mathematics teachers and experts,
PABSON, NPABSAN, PETSAN, ISAN, EMBOCS, NISAN and independent clusters of many
other Schools of Nepal, for providing me with opportunities to participate in workshops,
Seminars, Teachers’ training, Interaction programmes and symposia as the resource person.
Such programmes helped me a lot to investigate the teaching-learning problems and to research
the possible remedies and reflect to the series.
I’m proud of my wife Rita Rai Dahal who always encourages me to write the texts in a more
effective way so that the texts stand as useful and unique in all respects. I’m equally grateful to
my son Bishwant Dahal and my daughter Sunayana Dahal for their necessary supports during
the preparation of the series.
I’m extremely grateful to Dr. Ruth Green, a retired professor from Leeds University, England
who provided me very valuable suggestions about the effective methods of teaching-learning
mathematics and many reference materials.
Grateful thanks are due to Mr. Tara Bahadur Magar for his painstakingly editing of the series.
Moreover, I gratefully acknowledge all Mathematics Teachers throughout the country who
encouraged me and provided me the necessary feedback during the workshops/interactions
and teachers’ training programmes in order to prepare the series in this shape.
I’m profoundly grateful to the Vedanta Publication (P) Ltd. to get this series published. I would
like to thank Chairperson Mr. Suresh Kumar Regmi, Managing Director Mr. Jiwan Shrestha,
Marketing Director Mr. Manoj Kumar Regmi for their invaluable suggestions and support during
the preparation of the series.
Last but not the least, I’m heartily thankful to Mr. Pradeep Kandel, the Computer and Designing
Senior Officer of the publication house for his skill in designing the series in such an attractive
form.
Hukum Pd. Dahal
Contents
S.N Unit Page
1. Set 5-22
1.1 Set – Looking back, 1.2 Membership of a set and set notation,
1.3 Methods of describing sets, 1.4 Cardinal number of a set, 1.5 Types of
sets, 1.6 Relationships between sets, 1.7 Universal set and subset, 1.8 Set
Operations, 1.9 Venn-diagrams
2. Number System in Different Bases 23-33
2.1 Whole numbers - Looking back, 2.2 Decimal or Denary
number system, 2.3 Periods and place, 2.4 Binary number system,
2.5 Conversion of binary numbers to decimal numbers, 2.6 Conversion
of decimal numbers to binary numbers, 2.7 Quinary number
system, 2.8 Conversion of quinary numbers to decimal numbers,
2.9 Conversion of decimal numbers to quinary numbers
3. Operations on Whole Numbers 34-55
3.1 Factors and multiples - Looking back, 3.2 Highest common factor
(H. C. F.), 3.3 Finding H. C. F. by Factorization method, 3.4 Finding H.C.F. by
Division method, 3.5 Lowest common multiple ( L.C.M), 3.6 Finding L.C.M.
by factorisation method, 3.7 Finding L.C.M. by division method, 3.8 Square
and square root, 3.9 Process of finding square root, 3.10 Cube and cube root
4. Real Numbers 56-77
4.1 Integers – Looking back, 4.2 Absolute value of integers,
4.3 Operations on integers, 4.4 Sign rules of addition and subtraction
of integers, 4.5 Properties of addition of integers, 4.6 Multiplication and
division of integers, 4.7 Sign rules of multiplication and division of integers,
4.8 Properties of multiplication of integers, 4.9 Order of operations,
4.10 Rational numbers – review, 4.11 Properties of Rational numbers,
4.12 Terminating and non-terminating rational numbers, 4.13 Irrational
numbers
5. Fraction and Decimal 78-98
5.1 Fraction – Looking back, 5.2 Addition and subtraction of fraction - revision,
5.3 Multiplication of fractions, 5.4 Division of fractions, 5.5 Decimal - revision,
5.6 Terminating and non-terminating recurring decimal, 5.7 Four fundamental
operations on decimals
6. Ratio, Proportion and Unitary Method 99-117
6.1 Ratio and Proportion – Looking back, 6.2 Proportion, 6.3 Types of
proportions, 6.4 Unitary method
7. Percent and Simple Interest 118-127
7.1 Percent – Looking back, 7.2 Operations on percent, 7.3 Simple interest –
Review, 7.4 Calculation of simple interest
8. Profit and Loss 128-140
8.1 Profit and Loss – Looking back, 8.2 Profit and loss per cent,
8.3 Calculation of S.P. when C.P. and profit or loss per cent are given,
8.4 Calculation of C.P. when S.P. and profit or loss per cent are given,
8.5 Discount, 8.6 Discount per cent, 8.7 Value Added Tax (VAT)
9. Algebraic Expressions 141-147
9.1 Algebraic terms and expressions – Looking back, 9.2 Types of
algebraic expressions, 9.3 Polynomial, 9.4 Degree of polynomials,
9.5 Evaluation of algebraic expressions, 9.6 Addition and subtraction of
algebraic expressions
10. Laws of Indices 148-178
10.1 Laws of indices (or exponents), 10.2 Multiplication of algebraic
expressions, 10.3 Some special products and formulae, 10.4 Division
of algebraic expressions, 10.5 Factors and factorisation – Introduction,
10.6 Simplification of rational expressions
11. Equation, Inequality and Graph 179-200
11. 1 Open statement and equation - Looking back, 11.2 Linear equations
in one variable, 11.3 Solution to equations, 11.4 Applications of equations,
11.5 Trichotomy – Review, 11.6 Inequalities, 11.7 Replacement set and
solution set, 11.8 Graphical representation of solution sets, 11.9 Function
machine and relation between the variables x and y,
12. Coordinates 201-218
12.1 Coordinates – Looking back, 12.2 Coordinate axes and quadrants,
12.3 Finding points in all four quadrants, 12.4 Plotting points in all four quadrants,
12.5 Reflection of geometrical figures, 12.7 Rotation of geometrical figures,
12.8 Rotation of geometrical figures using coordinates, 12.9 Displacement,
13. Geometry: Angles 219-231
13.1 Angels – Looking back, 13.2 Different pairs of angles – Review,
13.3 Verification of properties of angles, 13.4 Pairs of angles made by a
transversal with parallel lines
14. Triangle, Quadrilateral and Polygon 232-251
14.1 Triangles – Looking back, 14.2 Properties of triangles,
14.3 Some special types of quadrilaterals, 14.4 Verification of properties of
special types of quadrilaterals, 14.5 Interior and exterior angles of regular
polygons
15. Congruency and Similarity 252-258
15.1 Congruent figures – Introduction, 15.2 Congruent triangles,
15.3 Conditions of congruency of triangles, 15.4 Similar triangles
16. Construction 259-267
16.1 Construction of perpendicular bisector of a line segment,
16.2 Transferring angles, 16.3 Construction of different angles,
16.4 Construction of triangles, 16.5 Construction of parallelograms,
16.6 Construction of squares, 16.7 Construction of rectangles,
16.8 Construction of rhombus, 16.9 Construction of kite
17. Circle 268-270
17.1 Circle and its different parts – review
18. Perimeter, Area and Volume 271-293
18.1 Perimeter, Area and Volume – Looking back, 18.2 Perimeter of plane
figures, 18.3 Area of plane figures, 18.4 Nets and skeleton models of regular
solids, 18.5 Area of solids, 18.6 Volume of solids
19. Symmetry, Design and Tessellation 294-298
19.1 Symmetrical and asymmetrical shapes, 19.2 Line or axis of symmetry,
19.3 Rotational symmetry, 19.4 Order of rotational symmetry, 19.5 Tessellations,
19.6 Types of tessellations
20. Scale Drawing and Bearing 299-305
20.1 Scale drawing, 20.2 Scale factor, 20.3 Bearing
21. Statistics 306-318
21.1 Statistics – Review, 21.2 Types of data and frequency table, 21.3 Cumulative
frequency table of ungrouped data, 21.4 Grouped and continuous data,
21.5 Bar graph, 21.6 Average (or Mean), 21.7 Mean or average of ungrouped
repeated data
Answers 318-332
Evaluation Model 333-334
Unit Set
1
1.1 Set – Looking back
Classroom - Exercise
1. Let's write any three members of the well defined collections inside braces.
a) A collection of High mountains of Nepal ....................................................
b) A collection of mountains of Nepal which have more than 6000 m altitude.
.........................................................................................................................
c) A collection of fruits. .....................................................................................
d) A collection of tasty fruits. ............................................................................
2. If W = {0, 1, 2, 3, 4, 5} and A = {2, 4, 6}, let's tell and write 'true' or 'false'.
a) A W = ................. b) 2 W = .................
c) 6 W = ................. d) {2, 4} W = .................
3. Let's tell and list the elements of these sets.
a) {letters of the word ‘SUCCESS’} ...................................................................
b) {x : x <10, x odd number} .......................................................................
4. The set-builder form of A = {2, 4, 6, 8} .............................................................
It is a collection of whole numbers less than 5. Then, it
0 definitely includes the numbers 0, 1, 2, 3 and, 4. Because the
1 2
statement ‘whole numbers less than 5’ defines the distinct and
3 4 distinguishable objects which are to be included in the collection.
Therefore, it is a well-defined collection and it is called a set.
On the other hand, is it possible to write the members of the collection
of tall students of your class? Is it possible to write the members of
the collection of tasty fruits? Discuss why these collection are not the
well - defined. Therefore, the collections which are not well-defined are not sets.
1.2 Membership of a set and set notation
A member or an element of a set is any one of the distinct objects that make up that
set.
For example, in a set N = {1, 2, 3, 4, 5} , the members or elements of the set N are
1, 2, 3, 4, and 5.
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Set
The membership of a member of a set is denoted by the symbol ‘'. For example,
1N or 1{ 1, 2, 3, 4, 5}. We read it as '1 belongs to set N' or '1 is a member of set N'
or 1 is an element of set N. Similarly, 2N, 3N, 4N and 5N.
However, when any element is not a member of a given set, it is denoted by the
symbol . For example: In N = {1, 2, 3, 4, 5}, 6 N, 7 N, ... and so on.
Set notation
We denote sets by capital letters, such as, A, B, C, W, N, etc. The members of a set
are enclosed in braces { } and they are separated by commas. For example,
A = {a, e, i, o, u}, W = {0, 1, 2, 3, 4, 5}, and so on.
1.3 Methods of describing sets
We usually write the members of a set by the following four methods:
(i) Diagramatic method
2 3
In this method, we write the members of a set inside a circular
or oval diagram. A set of prime numbers less than 10 is shown 5 7
in the diagram.
(ii) Description method
In this method, we describe the common property (or properties) of the
members of a set inside the braces. For example:
N = {natural numbers less than 10}
P = {prime numbers between 10 and 20}
V = {vowels of English alphabets}, and so on.
(iii) Listing/Roster/Tabular Method
In this method, we list the members of a set inside the braces and members are
separated by commas. For example:
N = {1, 2, 3, 4, 5, 6, 7, 8, 9}
P = {11, 13, 17, 19}
V = {a, e, i, o, u}, and so on.
(iv) Set-builder/Rule Method
In this method, we use a variable such as x, y, z, p, q, etc. to represent the
members of a set and the common property (or properties) of the members is
described by the variable. For example:
We describe, N ={1, 2, 3, 4, 5, 6, 7, 8, 9}= {x : x is a natural number less than 10}
and read as: “N is the set of all values of x, such that x is a whole number less
than 10.”
Similarly, P = {11, 13, 17, 19} = {y : 10 < y < 20, y prime number}
V = {a, e, i, o, u} = {z : z is a vowel of English alphabets}
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Set
1.4 Cardinal number of a set
Let's take a set of even numbers less than 12, i.e., E = {2, 4, 6, 8, 10}.
There are five members or elements in this set E. Therefore, the cardinal number of
the set E, denoted by n (E) = 5. Similarly, in A = {m, a, t, h}, its cardinal number
n (A) = 4. Thus, the number of members or elements contained by a set is known
as its cardinal number.
1.5 Types of sets
According to the number of elements contained by sets, there are four types of sets.
(i) Empty or null set (ii) Unit or singleton set (iii) Finite set (iv) Infinite set
(i) Empty or null set
There can be certain sets that do not contain any element at all. For example,
the sets of months with 35 days, the set of triangles with 4 sides, etc. do not
contain any element. So, the listing for these sets contains no element at all
and we call it an empty or null set.
Thus, a set containing no elements is called an empty or a null set. It is
represented by the symbol { } or I (Phi). Similarly,
if A = {natural numbers between 9 and 10}, A = { } or I and n(A) = 0.
if B = {whole numbers less than 0}, B = {} or I and n(B) = 0, and so on.
(ii) Unit or singleton set
Let's take a set A = {odd numbers between 8 and 10}.
Here, A = {9}. Thus, the set A contains exactly one element. Therefore, the
set A is called a unit or singleton set. A set containing exactly one element is
called a unit or singleton set.
Similarly,
if P = {prime and even numbers}, P = {2} and n(P) = 1
if Q = {x : 6 < x < 8, x ∈ N}, Q = {7} and n(Q) = 1, and so on.
(iii) Finite sets
Let's consider a set C = {composite numbers less than 10}.
Here, C = {4, 6, 8, 9} and we can count the number of elements of this set,
i.e. n(C) = 5. Hence, the set C is said to be a finite set.
Thus, a set containing finite number of elements is called a finite set.
Similarly,
if A = {1, 2,3, …, 20}, n(A) = 20
if B = {x : 2 < x < 31, x ∈ prime number}, n(B) = 9. It is a finite set.
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Set
(iv) Infinite sets
Let’s take a set W = {0, 1, 2, 3, 4, 5, ...}. It is the set of whole numbers. This
set is so large that we can never finish counting its elements. It has infinite
number of elements. So, it is an infinite set.
Thus, a set containing infinite number of elements is called an infinite set.
EXERCISE 1.1
General Section -Classwork
1. Let's tick () the well-defined collections. Also, list any three members of
well-defined collections.
a) A collection of greater natural numbers. ............................
b) A collection of natural numbers less than 4. ............................
c) A collection of name of English months starting ............................
with 'J' letter.
d) A collection of favourite English months. ............................
2. If A = { 0, 1, 2, 3, 4, 5 } and P = {2, 3, 5, 7}, let's insert the correct symbol ''
or '' in the blanks..
a) 4 .................. A b) 7 .................. A c) 1 .................. P
d) 5 .................. P e) {2, 3, 5, 7}.............. A f) {2, 3, 5} .................. P
3. Let's tell and write the cardinal numbers of these sets as quickly as possible.
a) A = { 2, 3, 5, 7}, n (A) = .................
b) B = { even numbers less than 7 }, n (B) = .................
c) P = { letters of the word 'PUPIL' }, n (P) = .................
d) O = { odd numbers between 3 and 5 }, n (O) = .................
4. Let's tell and write whether these sets are empty, unit, finite, or infinite as
quickly as possible.
a) A = { odd numbers less than 100 } ...................................................
b) B = { odd numbers more than 100 } ...................................................
c) C = { composite number between 1 and 5 } ...................................................
d) D = { prime number between 7 and 10 } ..................................................
Creative section - A
5. a) Define set. Is the collection of 'tall'' students' in a class a set? Why?
b) Write four methods of describing sets. Give one examples of each method.
Vedanta Excel in Mathematics - Book 7 8 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Set
c) Define cardinal number of a set with an example.
d) Write four types of sets on the basis of cardinal numbers. Write one example
of each.
6. Let's rewrite these sets in description method.
a) A = {1, 2, 3, 4} b) B = {5, 10, 15, 20, 25}
c) C = {7, 11, 13, 17, 19} d) D = {1, 2, 3, 6}
7. Let's rewrite these sets in listing method.
a) P = { prime numbers between 10 and 20 }
b) A = { letters of the word ‘FOOTBALL’ }
c) F = {x : x is a A factor of 18 }.
d) M = {y : y is a multiple of 3, 5 < y < 10}.
8. Let's rewrite these sets in set builder method.
a) A= {1, 2, 3, 4, 5} b) B = { 2, 3, 5, 7}
c) C = {1, 4, 9, 16, 25} d) D = {1, 2, 4, 8}
9. Let's list the elements and write the cardinal numbers of these sets.
a) A = { composite numbers between 10 and 20}.
b) B = { all possible factors of 12 }.
c) Z = { x : x is an integer, –2 ≤ x ≤ 2}
d) W = { x : x is a whole number, x < 1 }
It's your time - Project work!
10. a) Let's write the whole numbers from 90 to 100. Select the appropriate
numbers to form the following sets. Then, write the types of sets.
(i) A={even numbers} (ii) B={odd numbers}
(iii) C={x : x is a prime number} (iv) D={y : y is a composite number}
(v) E={z : z is a square number} (vi) F={cube numbers}
(vii) G={multiples of 7} (viii) H={x : x is divisible by 11}
b) Let's observe around the kitchen of your house and select any five objects
as the members of a set. Then, express the set in description, roster, and
rule methods.
1.6 Relationships between sets
According to the types and number of elements contained by two or more sets, there
are various types of relationships between the sets, such as equal sets, equivalent
sets, disjoint sets, and overlapping sets.
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(i) Equal sets
Let’s take any two sets, A = {s, v, u, 3, ª} and B = {ª, u, s, 3, v}.
Here, n (A) = 5 and n (B) = 5. Thus, they have the equal cardinal number and
both the sets have exactly the same elements. Therefore, sets A and B are said
to be the equal sets and written as A = B.
Thus, two or more than two sets are said to be equal if they have exactly the
same elements and equal cardinal number.
(ii) Equivalent sets
Let’s take any two sets, A = { c, o, w } and B = { g, o, d}. Here,
n (A) = 3 and n (B) = 3. They have the equal cardinal number.
However, the elements c, w of set A are not contained by the set B and the
elements g, d of the set B are not contained by the set A. So, they are not equal
and they are said to be equivalent sets.
We write it as A ~ B.
Thus, two or more than two sets are said to be equivalent if they have the equal
cardinal number but they do not have exactly the same elements.
(iii) Overlapping sets
Let’s take any two sets: A = {1, 2, 3, 6} and B = { 1, 2, 4, 8}.
In these two sets, the elements 1 and 2 are common to both the sets. Therefore,
sets A and B are overlapping sets.
A B
Thus, two or more than two sets are said to be overlapping 3 1 4
if they contain at least one element common. The common 2
elements of overlapping sets are shown in the shaded 6 8
region of the two intersecting diagrams.
(iv) Disjoint sets
Let’s take any two sets : A = { 3, 6, 9, 12} and B = { 5, 10, 15, 20 }
In these two sets, there is no any common element. Therefore, sets A and B are
disjoint sets. Of course, non overlapping sets are the disjoint sets.
Thus, two or more than two sets are said to be A B
disjoint if they do not have any element common. 3 6 5 10
The elements of disjoint sets are shown in 9 12
non-intersecting diagrams. 15 20
1.7 Universal set and subset
Let’s take a set of natural numbers less than 15.
N = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14}
Now, let's select certain elements from this set and make a few other sets.
Vedanta Excel in Mathematics - Book 7 10 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Set
A = {even numbers less than 15 = {2, 4, 6, 8, 10, 12, 14}
B = {prime numbers less than 15} = {2, 3, 5, 7, 11, 13}
C = {square number less than 15} = {1, 4, 9}
D = {possible factors of 12} = {1, 2, 3, 4, 6, 12}
N A N B 14 N C 14 N D
1
1 2 4 14 13 1 2 3 12 2 1 13 12 5 2 3 4 14
3 6 12 5 13 7 10 3 4 11 7 6 13
5 8 10 11 4 11 9 5 9 8 10 8 12 11
7 9 6 8 6 7 910
Here, the set of natural numbers less than 15 is known as the universal set.
Furthermore, every element of the sets A, B, C, and D is also an element of the set
N. In such a case, sets A, B, C and D are called the subsets of the set N. We use the
symbol ‘’ to represent a set as a subset of another set. For example:
‘A is a subset of N’ is written as A N .
‘B is a subset of N’ is written as B N, and so on.
Remember that, every set is a subset of itself and an empty set (I) is a subset of very
set.
On the other hand, if a set is not the subset of a given set, we denote it by the symbol ''.
Thus, a set under the consideration from which many other subsets can be formed is
called a universal set. The set of teachers of a school is a universal set from which the
subsets like set of Maths teachers, set of Science teachers, set of English, etc. can be
formed. We usually denote a universal set by the capital letter 'U'.
Super set
If the set A is a subset of N, N is called the super set of A. It is denoted as N A and
read as ‘N is a super set of A’.
Proper subset
Let's take any two sets, A = {p, e, n, c, i, l} and B = {p, e, n}.
Here, B is a subset of A and B is not equal to A. In such a case, set B is said to be a
proper subset of A.
Thus, the set B is said to be a proper subset of the set A if it contains at least one
element less than A.
We use a symbol ‘’ to represent a set as a proper subset of another set. For example:
B A and we read it as ‘B is a proper subset of A’.
Improper subset
]
]
Let's take any two set, A = {g, kf, n} and B = {kf, n, g}
Here, B is a subset of A and B is equal to A. In such a case, the set B is said to be an
improper subset of A.
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Set
Thus, the set B is said to be an improper subset of A, if B is equal to A, i.e., B = A. We
use the symbol to represent a set as an improper subset of another set. For example,
B A and we read it as 'B is an improper subset of A'.
Number of subsets of a given set
Let's study the following table and draw the conclusion about the number of subsets
of a given set.
Set Subsets No. of elements No. of subsets
A = { } { } n(A) = 0 1 m 2°
B = {a} {a}, I n(B) = 1 2 m 2 1
C = {a, b} {a}, {b}, {a, b}, I n(C) = 2 4 m 2 2
D = {a, b, c} {a}, {b}, {c}, {a, b}, 3
{b, c}, {a, c} {a, b, c} I n(D) = 3 8 m 2
From the above table, we conclude that the number of subsets of a set is given by the
formula 2 , where n is the cardinal number of the given set.
n
EXERCISE 1.2
General section - Classwork
1. Let's say whether these pairs of sets are equal or equivalent sets and write
A = B or A ~ B in the blank spaces.
(a) A = {1, 2, 4, 8 } and B = { 1, 4, 9, 16}, .......................................
(b) A = { 4, 6, 8, 9, 10} and B = { 6, 9, 4, 8, 10}, ......................................
2. Let's say and write whether these pairs of sets are overlapping or disjoint sets.
(a) A = { 1, 3, 5, 15 } and B = {5, 10, 15, 20}, A and B are .........................
(b) P = { 1, 3, 5, 7, 9 } and Q = { 4, 8, 12 }, P and Q are ........................
3. Let's say and fill in the blanks with appropriate words.
(a) If each element of set A is also an element of B, A is said to be
a ............................... of B and B is said to be a ............................... of A.
(b) Every set is a ............................... of itself.
(c) The empty set is a ............................... of every set.
(d) The number of possible subsets of a set containing 'n' number of elements
is given by the formula ...............................
4. Let's say and write the possible number of subsets of these sets.
(a) In {1, 2}, number of subsets = .................................................................
(b) In {g, o, d }, number of subsets = .................................................................
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Set
5. Let's say and write the possible number of proper subsets of these sets.
(a) In {k} number of proper subsets = .....................................................
(b) In { 1, 4, 9 } number of proper subsets = .....................................................
6. Let's say and write the possible subsets of these sets.
(a) {P} ...............................................................................................................
(b) { 2, 7} ..............................................................................................................
7. Let's say and write which one is the universal set and its subset in the given
pairs of sets below.
a) A = { students of class 7 } and B = { boys of class 7 }
Universal set is .................. and subset is ..................
b) X = { even numbers less than 20 } and W = {whole numbers less than 20}.
Universal set is ....................... and subset is.......................
Creative section
8. a) Define the following types of sets with examples.
(i) Equal and equivalent sets (ii) Overlapping and disjoints sets
b) Define universal set and subset with examples.
c) Define proper and improper subsets with examples.
9. Let's list the elements of the following sets and write with reasons whether
they are equal or equivalent sets:
a) A = {days of the week starting with the letter S} and
B = {days of the week starting with the letter T}
b) M = {letters of the word 'FOLLOW'} and
N = {letters of the word 'WOLF'}
10. Let's list the elements of the following sets and write with reasons whether
they are overlapping or disjoint sets:
a) O = {odd numbers less than 10} and
S = {perfect square numbers less than 10}
b) X = {x : x is a prime number which exactly divides 30}
Y = {y : y is a prime number which exactly divides 77}
11. Let's write the possible subsets of the following sets and list the proper subsets
separately:
(a) F = {apple} (b) A = {1, 2} (c) E = {a, b, c} (d) T = {1, 2, 3, 4}
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Set
12. At first, let's list the elements of each of these sets from the diagrams. Then,
select the common elements and list them in separate sets.
a) A B b) P Q
3 1 4 m t i n
6 2 8 a h k
13. Let's list the elements and common elements of these pairs of sets and show
them in diagrams.
(a) A = { x : x is a whole number less than 10 }and B = { factors of 24 }
(b) P = { x : x is a multiple of 4, x ≤ 20 } and
Q = { x : x is a multiple of 5, x ≤ 20 }
It's your time - Project work!
14. a) Let's take a universal set under your consideration. Then, write as many
subsets as possible from your universal set.
b) Let's conduct a survey inside your classroom among your friends. Then,
list the name of your friends and make separate sets in the following cases:
(i) Sets of friends who like tea, coffee or milk.
(ii) Sets of friends who like Mo:Mo, chowmein or Thukpa.
(iii) How many overlapping sets are formed? Show them in diagrams.
(iv) How many disjoint sets are formed? Show them in diagrams.
(v) How many equal sets are formed?
(vi) How many equivalent sets are formed?
c) Is there any possibility to form overlapping sets of the sets of teachers
who are teaching different subjects in your school? If so, make these sets
and show in diagrams.
15. Let's make groups of 3 students and play game!
Each student of each group should make two disjoint sets with maximum
5 members taking the natural numbers from 1 to 20. Next day, from the sets
of numbers made by the students of each group, form as many number of
overlapping sets as possible. The group which has the maximum number of
overlapping sets is the winner!
1.8 Set Operations
Sets can be combined in a number of different ways to make another set. It is known
as set operations. There are four basis set operations. They are:
(i) Union of sets (ii ) Intersection of sets
(iii) Difference of sets (iv) Complement of a set
(i) Union of sets
Let's make a sport committee P with 3 members Ram, Sita, Laxmi and a
cultural committee Q with 4 members Sita, Laxman, Ajay, and Sudip. When
the committees P and Q have a joint meeting and a new committee R is
Vedanta Excel in Mathematics - Book 7 14 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Set
formed; then, clearly the new committee R (the union of committees P and Q)
has 6 members: Ram, Sita, Laxmi, Laxman, Ajay, and Sudip. It does not have
7 members.
When the elements of two or more sets are combined and listed together in a
single set, it is called the union of these sets. Now, let's take another example.
If A = {1, 3, 5, 7, 9} and B = {2, 3, 5, 7},
the union of sets A and B = {1, 2, 3, 5, 7, 9},
it is denoted as A B = {1, 2, 3, 5, 7, 9}.
Thus, the union of two sets A and B is the set consisting of all elements that
belong to A or B (or both), and it is denoted by A B (read as ‘A union B’).
The symbol ‘’ (cup) denotes the union of sets.
It is noted that while making the union of sets, the common elements should
be listed only once.
(ii) Intersection of sets
In the above example, set P = {Ram, Sita, Laxmi} and set Q = {Sita, Laxman,
Ajay, Sudip}. Here, Sita is the member common to the sets P and Q. So, the
intersection of these two sets is {Sita}.
When the common elements of two or more sets are listed in a separate set, it
is called intersection of sets. Now, let's take another example.
If A = {1, 2, 3, 4, 5, 6} and B = {1, 3, 5, 7, 9, 11},
the intersection of sets A and B = {1, 3, 5}
It is denoted as A B = {1, 3, 5}
Thus, the intersection of sets A and B is the set consisting of all elements that
belong to A and belong to B. It is denoted by A B (read as ‘A inter section B’).
The symbol ‘’ (cap) denotes the intersection of sets.
(iii) Difference of sets
The difference of two sets A and B denoted by A – B is the set of all elements
contained only by A but not by B. For example:
If A = {2, 4, 6, 8, 10} and B = {1, 2, 4, 8},
the difference of A and B = {6, 10}, which is only A.
It is denoted as A – B = {6, 10}
Similarly, the difference of B and A denoted by B – A is the set of all elements
contained only by B but not by A.
So, B – A = {1}, which is only B.
(iv) Complement of a set
When your teacher asks students of odd roll numbers to raise their hands, the
students of even roll numbers who do not raise hands are the complement
of odd roll numbers. Thus, if a set A is the subset of a universal set U, the
c
complement of A denoted by A or A' or A is the set which is formed due to
the difference of U and A, i.e. U – A.
For example, if U = {1, 2, 3, … 8} and A = {1, 2, 4, 8}, the complement of
A = U – A = {3, 5, 6, 7} and it is denoted as A = {3, 5, 6, 7}.
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1.9 Venn-diagrams
We can show the sets and set operations by using diagrams. The concept was at
first introduced by Euler, the Swiss Mathematician and it was further developed by
the British Mathematician John Venn. So, the diagrams are famous as Venn Euler
diagrams or Venn diagrams.
In Venn diagrams, a universal set is represented by a rectangle and its subsets are
represented by circles or ovals.
Let’s learn to represent sets and set operations by using Venn diagrams.
U U U
A A B A B
U U U
A B A B A B
U U U
A B A B A B
The shaded region The shaded region The shaded region
represents A – B represents A – B represents B – A
U U U
A B A B
A
A A B
U U U
A B A B A B
The shaded region The shaded region The shaded region
represent A B. represent A – B. represent B – A .
Vedanta Excel in Mathematics - Book 7 16 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Set
Worked-out examples
Example 1: If M = {m, a, t, h, s} and N = {m, a, g, i, c}, find
(a) M N (b) M N (c) M – N, (d) N – M. Also, represent them in
Venn diagrams.
Solution:
Here, M = {m, a, t, h, s} and N = {m, a, g, i, c} M N
Now, t m g
h i
a) M N = {m, a, t, h, s} {m, a, g, i, c} = {m, a, t, h, s, g, i, c} a c
s
The shaded region represents M N. M N
t m g
b) M N = {m, a, t, h, s} {m, a, g, i, c} = {m, a} h i
a
s c
The shaded region represents M N.
M N
t m g
c) M – N = {m, a, t, h, s} – {m, a, g, i, c} = {t, h, s}
h i
a
s c
The shaded region represents M – N.
M N
d) N – M = {m, a, g, i, c} – {m, a, t, h, s} = {g, i, c} t m g
h i
a c
The shaded region represents N – M. s
Example 2: If U = {0, 1, 2, …, 10}, A = {1, 2, 3, 4, 5}, and B = {1, 3, 5, 7, 9},
complete the following set operations and represent them in
Venn-diagrams.
a) A B b) A B c) A – B d) B – A
e) A B f) A B g) A – B h) B – A
Solution:
Here, U = {0, 1, 2, …, 10}, A = {1, 2, 3, 4, 5} and B = {1, 3, 5, 7, 9}
U
Now, A B
2 1 7
a) A B = {1, 2, 3, 4, 5} {1, 3, 5, 7, 9} and 0 3
4 5 9
B = {1, 2, 3, 5, 7, 9} 6 8 10
U
The shaded region represents A B.
A B
2 1
0 3 7
b) A B = {1, 2, 3, 4, 5} {1, 3, 5, 7, 9} = {1, 3, 5} 4 5
The shaded region represents A B. 6 8 9 10
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Set
U
c) A – B = {1, 2, 3, 4, 5} – {1, 3, 5, 7, 9} = {2, 4} A B
The shaded region represents A – B. 0 2 1 7
3
4 5 9
6
8 10
U
d) B – A = {1, 3, 5, 7, 9} – {1, 2, 3, 4, 5} = {7, 9} A B
The shaded region represents the elements of B – A. 0 2 1 7
3
4 5 9
6
8 10
U
e) A B = U – (A B) A B
= {0, 1, 2, …, 10} – {1, 2, 3, 4, 5, 7, 9} 0 2 1 7
3
= {0, 6, 8, 10} 4 5 9
The shaded region represents A B . 6 8 10
U
f) A B = U – (A B) A B
= {0, 1, 2, …, 10} – {1, 3, 5} 2 1 7
= {0, 2, 4, 6, 7, 8, 9, 10} 0 4 3
5
The shaded region represents A B . 6 9
8 10
U
g) A – B = U – (A – B) A B
= {0, 1, 2, …, 10} – {2, 4} 2 1 7
= {0, 1, 3, 5, 6, 7, 8, 9, 10} 0 4 3
5
The shaded region represents A – B . 6 9
8 10
U
h) B – A = U – (B – A)
A B
= {0, 1, 2, …, 10} – {7, 9} 2 1
0 3 7
= {0, 1, 2, 3, 4, 5, 6, 8, 10} 4 5 9
6
The shaded region represent B – A 8 10
Example 3: From the Venn-diagram given alongside, list the P Q U
elements of the following sets.
a) P Q b) P Q c) P – Q d) Q – P 6 2 4 1
e) P Q f) P Q g) P – Q h) Q – P 10 8
Solution:
3 5 7 9
Here, U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
a) P Q = {1, 2, 4, 6, 8, 10} b) P Q = {2, 4 ,8}
c) P – Q = {6, 10} d) Q – P = {1}
e) P Q = {3, 5, 7, 9} f) P Q = {1, 3, 5, 6, 7, 9, 10}
g) P – Q = {1, 2, 3, 4, 5, 7, 8, 9} h) Q – P = {2, 3, 4, 5, 6, 7, 8, 9, 10}
Vedanta Excel in Mathematics - Book 7 18 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
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Example 4: If U = {1, 2, 3, … 15}, A = {2, 4, 6, 8, 10, 12, 14}, B = {3, 6, 9, 12, 15}, and
C = {2, 3, 5, 7, 11, 13} , list the elements of the following sets.
a) A B C b) A B C c) (A B) C d) (A B) C
e) (A B) – C f) A – (B C) g) A B C h) A B C
Solution:
Here, U = {1, 2, 3, …15}, A = {2, 4, 6, 8, 10, 12, 14}, B = {3, 6, 9, 12, 15}
C = {2, 3, 5, 7, 11, 13}
a) A B = {2, 4, 6, 8, 10, 12, 14} {3, 6, 9, 12, 15}
= {2, 3, 4, 6, 8, 9, 10, 12, 14, 15}
? A B C = (A B) C
= {2, 3, 4, 6, 8, 9, 10, 12, 14, 15} {2, 3, 5, 7, 11, 13}
= {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}
b) A B = {2, 4, 6, 8, 10, 12, 14} {3, 6, 9, 12, 15}
= {2, 4, 6, 8, 10, 12, 14} {3, 6, 9, 12, 15} = {6, 12}
? A B C = (A B) C
= {6, 12} {2, 3, 5, 7, 11, 13} = I
c) A B = {2, 3, 4, 6, 8, 9, 10, 12, 14, 15}
? (A B) C = {2, 3, 4, 6, 8, 9, 10, 12, 14, 15} {2, 3, 5, 7, 11, 13}
= {2, 3}
d) A B = {6, 12}
? (A B) C = {6, 12} {2, 3, 5, 7, 11, 13}
= {2, 3, 5, 6, 7, 11, 12, 13}
e) (A B) = {2, 3, 4, 6, 8, 9, 10, 12, 14, 15}
? (A B) – C = {2, 3, 4, 6, 8, 9, 10, 12, 14, 15} – {2, 3, 5, 7, 11, 13}
= {4, 6, 8, 9, 10, 12, 14, 15}
f) B C = {3, 6, 9, 12, 15} {2, 3, 5, 7, 11, 13} = {3}
? A – (B C) = {2, 4, 6, 8, 10, 12, 14} – {3}
= {2, 4, 6, 8, 10, 12, 14}
g) A B C = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 ,15}
? A B C = U – (A B C)
= {1, 2, 3, …, 15} – {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}
= {1}
h) A B C = I
? A B C = U – {A B C} = {1, 2, 3, …15} – I
= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}
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Set
EXERCISE 1.3
General Section - Classwork
1. Let's say and write the members of the following set operations.
U = {a, b, c, d, e} A = {a, b, c}, and B = { b, c, d}
(i) A B = ............................. (ii) A B = .............................
(iii) A – B = .............................. (iv) B – A = ..............................
(v) A = .................................. (vi) B = ....................................
(vii) A B = ............................. (viii) A B = .............................
2. Let's say and write the required set operations represented by the shaded
regions of Venn-diagrams.
a) b) c) d)
U U U U
A B P Q X Y A B
.......................... .......................... .......................... ..........................
3. Let’s identify and shade the region to show the given operations.
(i) (ii) (iii) (iv)
U U U U
A B P Q M N X Y
A B P – Q M N Y – X
Creative Section - A
4. S is the set of students in sport-club and C is the set of students in
cultural club of a school. If S = {Ram, Sita, Krishna, Shiva, Abdul} and
C = {Sita, Laxmi, Shova, Hari, Joseph}, answer the following questions:
a) Make a separate set of students who are in both the clubs.
b) Make a separate set of students at the time of their joint meeting and
write the number of members in the joint meeting.
c) Make a set of students who are only the member of Sport-club.
d) Make a set of students who are only the member of Cultural-club.
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Set
5. a) If A = {a, e, i, o, u} and B = {a, b, c, d, e}, find: (i) A B (ii) A B
(iii) A – B, (iv) B – A and show these operations in Venn-diagrams.
b) If P ={1, 3, 5, 7, 9} and Q = {1, 3, 6, 9, 18}, find: (i) P Q (ii) P Q
(iii) P – Q (iv) Q – P and show these operations in Venn-diagrams.
c) If M = {1, 2, 3, 6} and N = {factor of 8}, find: (i) M N (ii) M N
(iii) M – N (iv) N – M and show these operations in Venn-diagrams.
d) If A = {x : x ≤ 5, x N} and B = {2, 4}, find: (i) A B (ii) A B
(iii) A – B (iv) B – A and show these operations in Venn-diagrams.
6. a) If U = {1, 2, 3, 4, … 15}, A = {1, 2, 3, 4, 5, 6} and B = {2, 4, 6, 8, 10, 12},
find: (i) A (ii) B and show them in Venn - diagrams.
b) If U = {1, 2, 3, ... 10}, P = {1, 4, 9} and Q = {2, 3, 5, 7}, find:
(i) P (ii) Q and show these operations in Venn-diagrams.
7. a) If U = {1, 2, 3, ... 10}, A = {2, 4, 6, 8}, and B = {2, 3, 5}, find:
(i) A B (ii) A B and show these operations in Venn-diagrams.
b) If U = {1, 2, 3, ... 12}, A = {3, 6, 9, 12}, and B = {2, 4, 6, 8}, find:
(i) A B (ii) A B and show these operations in Venn-diagrams.
c) If U = {1, 2, 3, ... 10}, A = {1, 2, 4, 8}, and B = {2, 4, 8, 10}, find:
(i) A – B (ii) A –B and show these operations in Venn-diagrams.
d) If U = {1, 2, 3, ... 10}, A = {1, 2, 4, 8}, and B = {2, 4, 8, 10}, find:
(i) B – A (ii) B – A and show these operations in Venn-diagrams.
U
8. From the given Venn – diagram, list the elements of X Y
the following sets. 1
2
3 9
a) U b) X Y and X Y c) X Y and X Y 4 5
6 7 11
8
d) X – Y and X Y e) Y – X and Y X 10 12 13 14 15
Creative Section - B
9 . If U = { 1, 2, 3, ... 12}, A = { 3, 4, 5, 6, 7}, B = { 2, 4, 6, 8, 10 } and
C = { 2, 3, 4, 6, 12}, list the elements of the following set operations.
a) A B C and A B C b) A B C and A B C
c) (A B) C d) (A B) C
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Set
10. If U = {x : x ≤ 15, x N}, A = {even numbers less than 15}, and
B = {factors of 12}, list the elements of the following set operation:
a) A (A B) b) A (A B) c) A – (A B) d) B – (A B)
e) A B f) A B g) A B h) A B
11. If A = { 1, 2, 3, 4, 5}, B = { 1, 3, 5, 7}, and C = { 2, 3, 5, 7 }, show that:
a) A (A B) = A B b) B (A B) = B A
c) A (B C) = (A B ) C d) (A B ) C = A ( B C)
12. a) If F and F denote the sets of all possible factors of 6 and 8 respectively,
6 8
list the elements of F and F , then find F F
6 8 6 8 .
b) F and F are the sets of all possible factors of 12 and 18 respectively.
18
12
List the elements of F F
12
18
13. a) If M and M denote the sets of the first six multiples of 6 and 8
6 8
respectively, list the elements of M and M and find M M .
6 8 6 8
b) M and M are the sets of the first five multiples of 5 and 10 respectively.
5 10
List the elements of M M
5 10
It's your time - Project work!
14. Let's make groups of 10 students and conduct a survey in your class to find how
many students like apple or orange in each group. Then make two separate sets
of students who like apple or orange and answer the following questions.
a) Make a set of students of each group who like apple or orange.
b) Make a set of students of each group who like apple as well as orange.
c) Make a set of students of each group who like only apple.
d) Make a set of students of each group who like only orange.
e) Are there any students in your group who do not like apple or orange? Make
a set to show these students.
15. Write a universal set 'U' of the natural numbers from 1 to 15. Write any two
overlapping subsets 'A' and 'B' of this universal set U. Then, draw Venn-diagrams
to show the members of the following set operations by shading to show the
members of the following set operations by shading.
a) A B b) A B c) A – B d) B – A e) A B
f) A B g) A – B h) B – A i) A j) B
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Unit Number Systems in Different Bases
2
2.1 Whole numbers - Looking back
Classroom - Exercise
1. Let's tick () the correct numeral of the number name.
a) Seven lakh nine thousand: (i) 79000 (ii) 700900 (iii) 709000
b) Three crore six hundred: (i) 30006000 (ii) 30000600 (iii) 30060000
c) Eight million ninety-nine thousand: (i) 8099000 (ii) 899000 (iii) 8990000
d) If x is the digit at tens and y is at ones place, the number is:
(i) xy (ii) x + y (iii) 10x + y
e) The place name of five in 950038120 is:
(i) Ten-lakhs (ii) Crores (iii) Ten-crores
f) The place value of 9 in 20900470850 is:
(i) 90000000 (ii) 9000000 (iii) 900000000
2. Let's say and write the answers as quickly as possible.
a) The place name of 6 in 2460178953 in Nepali system is ................................
and in International system is ...........................................
b) Rewrite 5639240000 using commas and express in words in Nepali system.
............................................................................................................................
c) Rewrite 8719480000 using commas and express in words in International
system. ..............................................................................................................
Now, let’s study about the following sets of whole number and natural numbers.
W = {0, 1, 2, 3, 4, 5, ...} is the set of whole numbers. The set of natural numbers,
i.e. N = {1, 2, 3, 4, 5, ...} is a subset of the set of whole numbers.
Zero (0) is the least whole number, whereas 1 is the least natural number. The
greatest whole number or natural number is infinite.
2.2 Decimal or Denary number system
Hindu-Arabic number system is based on decimal or denary number system. In this
system, we use ten digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. Any whole number can be
written using a combination of these ten digits. The system is based on grouping of
tens. So, it is also known as the Base Ten or Decimal system.
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Number Systems in Diff erent Bases
Let’s take 17 blocks of cubes and regroup them into the group of 10 blocks.
17 = 10 + 7 = 1 × 10 + 7 × 10°
1
Let’s take 39 pencils and regroup them into the group of 10 blocks.
1
39 = 30 + 9 = 3 × 10 + 9 × 10°
0
Similarly, 594 = 500 + 90 + 4 = 5 × 10 + 9 × 10 + 4 × 10 .
2
1
In this way, whole numbers can be regrouped into the base of 10 with some power
of 10. It is called the decimal numeration system or denary system.
Each digit of a numeral has its own place and its place value is obtained multiplying
the digit by its place. For example, let's take a numeral 7425.
7425
It is at ones place and place value is 5 × 1 = 5
It is at tens place and place value is 2 × 10 = 20
It is at hundreds place and place value is 4 × 100 = 400
It is at thousands place and place value is 7 × 1000 = 7000
Now, we can write the numeral 7425 in the expanded form in the following way.
7425 = 7 × 1000 + 4 × 100 + 2 × 10 + 5 × 1
= 7 × 10 + 4 × 10 + 2 × 10 + 5 × 10°
3
2
1
In this way, if x, y and z are the digits at hundreds, tens and ones place respectively
in a number, then the number can be expressed as 100x + 10y + z.
2.3 Periods and place
The tables given below show the periods and places in Nepali system and
International system of numerations.
9 1 6 7 2 4 3 8 5 6 3 0
9 1 6 7 2 4 3 8 5 6 3 0
Vedanta Excel in Mathematics - Book 7 24 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Number Systems in Diff erent Bases
From the table, the number name of 916724385630 in Nepali system is:
Nine kharab sixteen arab seventy-two crore forty-three lakh eighty-five thousand
six hundred thirty.
In International system the number name is:
Nine hundred sixteen billion seven hundred twenty-four million three hundred
eighty-five thousand six hundred thirty.
Comparison between Nepali and International numeration system.
100 thousand = 1 lakh , 1 million = 10 lakhs, 10 million = 1 crore
1 billion = 1 arab, 100 billions = 1 kharab
Placement of commas
We can read and write the larger numbers more easily and comfortably when the
periods of the digits are separated by using commas. Let's take a numeral 507490680
and rewrite it using commas in both Nepali and International system.
Nepali system International system
50,74,90,680 507,490,680
Separating unit period Separating unit period
Separating thousands period
Separating lakhs period Separating thousands period
The number name is fifty crore The number name is five hundred
seventy-four lakh ninety thousand seven million four hundred ninety
six hundred eighty thousand six hundred eighty.
EXERCISE 2.1
General Section - Classwork
1. Let's say and write the number names in Nepali system and in International
system.
Number name in Nepali Number name in
Numerals
system International system
a) 100000
b) 4000000
c) 70000000
d) 200000000
e) 5000000000
f) 90000000000
g) 800000000000
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 25 Vedanta Excel in Mathematics - Book 7
Number Systems in Diff erent Bases
2. Let's write the numerals which have the following expanded forms.
a) 2 × 10 + 4 × 10 + 7 × 10 + 5 × 10 = ..........................................
0
2
3
1
b) 7 × 10 + 5 × 10 + 9 × 10 + 8 × 10 = ..........................................
1
4
0
2
c) 4 × 10 + 6 × 10 + 1 × 10 + 7 × 10 = ..........................................
0
5
4
1
6
2
d) 9 × 10 +3 × 10 + 5 × 10 0 = ..........................................
3. a) If x is the digit at tens place and y is the digit at ones place, the two-digit
number formed by these digits is ..............................................
b) If y is the digit at tens place and x is the digit at ones place, the two-digit
number formed by these digits is ..............................................
c) If a is digit at hundreds place, b is at tens place and c is at ones place, the
three-digit number formed by these digit is ..............................................
4. Rewrite these numerals using commas in Nepali system and in International
system.
Numerals Nepali system International system
a) 18576390
b) 420198675
c) 99999999999
Creative Section A
5. a) The estimated cost of construction of a hydro-power project is
Rs 2562880000. Rewrite this cost in words in Nepali as well as International
numeration system.
b) The world population estimated by US Census Bureau in June 2019 was
7577130400. Rewrite this population in words according to Nepali numeration
system.
c) The annual budget of Government of Nepal was Rs 1315161700000 in the
fiscal year 2075/076. Rewrite it in words according to Nepali and International
system.
6. Rewrite these number names in numerals using commas both in Nepali system
and International system.
a) Six kharab eighty-four arab ninety-one crore forty-seven lakh five thousand
three hundred forty.
b) Thirty-seven billion eight hundred fifteen million six hundred sixty-eight
thousand five hundred twenty-one.
7. Let's write the following numbers in the expanded forms:
a) 52063709 b) 400801530 c) 7502600048 d) 23900068407
Vedanta Excel in Mathematics - Book 7 26 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Number Systems in Diff erent Bases
8. 2.5 arab = 2500000000 = Two arab fifty crore = Two billion five hundred million.
7.38 billion = 7380000000 = Seven billion three hundred eighty million
= Seven arab thirty-eight crore
Now, let's write the values of the following numbers and rewrite the number
names in Nepali or in International systems.
a) 1.6 crore b) 4.75 crore c) 3.4 arab d) 5.13 arab
e) 2.7 million f) 6.99 million g) 7.2 billion h) 8.36 billion
Creative Section B
9. a) Find the difference between one million and two hundred fifty thousand.
Express the difference in words in International and Nepali Systems.
b) Find the difference between two billion and eighty-five million. Express the
difference in words in International and Nepali System.
c) Find the difference between seven crore and seventy-five lakh. Express the
difference in words in Nepali and International Systems.
d) Find the difference between three arab and forty-eight crore. Express the
difference in words in Nepali and International Systems.
10. a) By how much is Rs 3.6 million more than Rs 3.6 lakh? Express in Nepali
System of numeration.
b) By how much is Rs 7.5 crore more than Rs 7.5 million? Express in
International System of numeration.
It's your time - Project work!
11. Using the digits from 0 to 9, let's write the greatest and the least ten-digit
numerals.
a) Express these numerals in words in Nepali and International Systems.
b) Express the difference between these numerals in Nepali and International
Systems.
12. Let's visit the available and reliable website. Then, search and find the
following facts and figures:
a) Today's live population of the three most populated countries in the world.
Express the population in Nepali and International Systems of numeration.
b) Find the total population of these three countries and express in words in
Nepali and International System of numeration.
c) The distance between the Sun and its four nearest planets. Express the
distance in words in Nepali and International System of numeration.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 27 Vedanta Excel in Mathematics - Book 7
Number Systems in Diff erent Bases
2.4 Binary number system
In denary number system, we use ten digits 0 to 9 to write any number. However, in
binary number system, we use only two digits 0 to 1 to express any number. Binary
number system has its broad applications in digital electronics.
Computers and hand held calculators actually use the binary system for their
internal calculations since the system consists of only two symbols, 0 and 1. All
numbers can then be represented by electronic “switches”, of one kind or another,
where “on” indicates 1 and “off” indicates 0.
2.5 Conversion of binary numbers to decimal numbers
To convert a binary number into decimal, it is expanded in the power of 2. Then, by
simplifying the expanded form of the binary number, we obtain a decimal number.
For example:
1
3
2
(i) 11011 = 1 u 2 + 1 u 2 + 0 u 2 + 1 u 2 + 1 u 1 0
4
2
= 1 u 16 + 1 u 8 + 0 u 4 + 1 u 2 + 1 u 1
= 16 + 8 + 0 + 2 + 1 = 27
? 11011 = 27
2
(ii) 1011001 = 1 u 2 + 0 u 2 + 1 u 2 + 1 u 2 + 0 u 2 + 0 u 2 + 1 u 2 0
1
2
4
3
6
5
2
= 1 u 64 + 0 u 32 + 1 u 16 + 1 u 8 + 0 u 4 + 0 u 2 + 1 u 1
= 64 + 0 + 16 + 8 + 0 + 0 + 1 = 89
? 1011001 = 89
2
2.6 Conversion of decimal numbers to binary numbers
Again, let’s take 15 blocks of cubes and regroup them into the group of 2 blocks.
7 pairs blocks of cube and 1 cube
Now, let’s arrange the groups of 2 blocks into the base of 2 with the maximum
possible powers.
8 4 2 1
2 2 2 2 1 2 0
3
So, 15 = 1 × 2 + 1 × 2 + 1 × 2 + 1 × 2° = 1111
2
3
1
2
In this way, the denary number 15 can be expressed in binary number as 1111 .
2
We can convert a decimal number into a binary number by using the place value
table of the binary system. For example:
Vedanta Excel in Mathematics - Book 7 28 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Number Systems in Diff erent Bases
Convert 19 into binary system.
Power of base 2 2 6 2 4 2 3 2 2 2 1 2 °
Decimal Equivalent 32 16 8 4 2 1
Binary number 1 0 0 1 1
There is one 16 in 19. So There is no 4 in 3. So,
insert 1. insert 0.
Remainder = 19– 16 = 3 Remainder is again 3.
There is one 2 in 3. So
There is no 8 in 3. So, insert 1.
insert 0. Remainder =3 – 2 = 1.
Remainder is still 3.
There is one in 1. So
insert 1.
From the table, 19 = 1 × 16 + 0 × 8 + 0 × 4 + 1 × 2 + 1 × 1
= 1 × 2 + 0 × 2 + 0 × 2 + 1 × 2 + 1 × 2 °
3
4
1
2
= 10011
2
Convert 25 into binary system
Power of base 2 2 5 2 4 2 3 2 2 2 1 2 0
Decimal equivalent 32 16 8 4 2 1
Binary number 1 1 0 0 1
From the table, 25 = 1 u 16 + 1 u 8 + 0 u 4 + 0 u 2 + 1 u 1
= 1 u 2 + 1 u 2 + 0 u 2 + 0 u 2 + 1 u 2q = 11001
2
3
4
1
2
? 25 = 11001
2
Alternative method
We can convert a decimal number into binary system also by another method. In
this method, we should divide the given number successively by 2 until the quotient
is zero. The remainder obtained in each successive division is listed in a separate
column. For example:
2 19 Remainder
2 9 1
2 4 1
2 2 0 Arranging the remainders in reverse order:
2 1 0 10011 2
0 1
? 19 = 10011 2
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Number Systems in Diff erent Bases
2.7 Quinary number system
We have already discussed on denary (or decimal) number system as the base
ten system and binary number system as the base two system. Likewise, quinary
(or pental) number system is known as the base five system. In this system, we
use only five digits: 0, 1, 2, 3, and 4. The numbers in the quinary system can be
expressed in the power of 5. For example:
1
1
2
14 = 1 u 5 + 4 u 5q, 321 = 3 u 5 + 2 u 5 + 1 u 5q, and so on.
For the proper identification of quinary numbers, they are written with their base 5
in the suffix. Some examples are 34 , 102 , 31240 , etc.
5 5 5
2.8 Conversion of quinary numbers to decimal numbers
To covert a quinary number into decimal number, it is expanded in the power of 5.
Then, by simplifying the expanded form of the quinary number, we get a decimal
number. For example:
(i) 32 = 3 u 5 + 2 u 5q (ii) 1234 = 1 u 5 + 2 u 5 + 3 u 5 + 4 u 5q
1
1
2
3
5 5
= 3 u 5 + 2 u 1 = 1 u 125 + 2 u 25 + 3 u 5 + 4 u 1
= 15 + 2 = 17 = 125 + 50 + 15 + 4 = 194
2.9 Conversion of decimal numbers to quinary numbers
We can convert a decimal number into quinary number by using the place value
table of the quinary system. For example:
Convert 134 into quinary system.
Power of base : 5 5 4 5 3 5 2 5 1 5 °
Decimal Equivalent 625 125 25 5 1
178 1×125 2 × 25 0 × 5 3 × 1
Quinary number 1 2 0 3
There is one 125 in 178. There is no 5 in 3.
So insert 1. So, insert 0.
Remainder = 178 – 125 = 53 Remainder = 3 – 0 = 3
There is two 25 in 53. There are three 1 in 3.
So insert 2. So, insert 3.
Remainder = 53 – 50 = 3.
From table, 178 = 1 u 125 + 2 u 25 + 0 u 5 + 3 u 1
1
2
0
3
= 1 × 5 + 2 × 5 + 0 × 5 + 3 × 5 = 1203 5
Alternative method
In this method, to convert a decimal number into quinary number, we should divide
the given number successively by 5 until the quotient is zero. The remainders of
each successive division are then arranged in the reverse order to get the required
quinary number. For example:
Vedanta Excel in Mathematics - Book 7 30 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Number Systems in Diff erent Bases
5 178 Remainders
5 35 3
5 7 0
5 1 2 Arranging the remainders in reverse order:
0 1 1203 5
? 178 = 1203 5
EXERCISE 2.2
General Section - Classwork
1. The glowing lamp indicates 1 and the lamp which is glowing out indicates
0 of the binary system. Let's say and write the binary numbers indicated by
these lamps.
a) b) c)
........................... ........................... ...........................
d) e) f)
........................... ........................... ...........................
2. Let’s study the given examples carefully and write the correct numbers in
the blank space as quickly as possible after shading the appropriate blank
circles.
Decimal Number 1×2 3 1×2 2 1×2 1 1×2° Binary numbers
4 2 1
8
1 = 1 2
2 = 10 2
3 = 11 2
4 = 100 2
5 = 101 2
6 = ?
7 = ?
8 = ?
9 = 1001 2
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Number Systems in Diff erent Bases
10 = ?
11 = ?
12 = ?
13 = 1101 2
14 = ?
15 = 1111 2
3. a) Let's say and write the binary values of these powers of twos.
2
3
2° = 1, 2 = 10, 2 = ................, 2 = 1000, 2 = ..................
1
4
b) Let's say and write the quinary values of these powers of fives.
5° = 1, 5 = 10, 5 = ................., 5 = .................. , 5 = .................
1
2
4
3
Creative section - A
4. Let's convert each of the following binary numbers into decimal number.
a) 11 2 b) 101 2 c) 110 2 d) 1010 2 e) 1100 2
f) 10010 g) 11111 h) 111010 i) 111001 j) 1100101
2 2 2 2 2
5. Let's convert each of the following decimal numbers into binary numbers .
a) 7 b) 18 c) 25 d) 72 e) 99
f) 124 g) 145 h) 216 i) 308 j) 417
6. Let's convert each of the following quinary numbers into decimal numbers.
a) 13 5 b) 24 5 c) 32 5 d) 120 e) 203
5
5
f) 340 g) 1203 h) 2113 i) 1234 j) 4444
5 5 5 5 5
7. Let's convert each of the following decimal numbers into quinary numbers.
a) 9 b) 13 c) 27 d) 55 e) 126
f) 212 g) 512 h) 670 i) 1126 j) 4025
Creative section - B
8. Let's convert each of the binary numbers into quinary numbers and quinary
number into binary number.
a) 1110 2 b) 11011 2 c) 34 5 d) 42 5
9. Lets change each pair of numbers in denary numbers and compare them.
a) 111 and 12 5 b) 10101 and 32 5
2
2
It's your time - Project work!
10. Let's learn with fun to convert a decimal number in to binary system.
Process 1: If the given decimal number is even, write zero (0) just below it. If it is
odd, write 1 just below it.
Process 2: Now divide the given number into half and write to the left of the
number.
Vedanta Excel in Mathematics - Book 7 32 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Number Systems in Diff erent Bases
Process 3: For each half write 0, if it is even and write 1 if it is odd.
For example 1: Let’s convert 14 and 67 into binary number.
? 14 = 1110 ? 67 = 1000011
2 2
Now, let's try for the numbers: a) 12 b) 43 c) 216
11. a) Let's take a few number of matchsticks and represent the following binary
numbers by pairing the required number of matchsticks.
(i) 10 2 (ii) 11 2 (iii) 101 2 (iv) 111
2
(v) 110 (vi) 1110 (vii) 1111
2 2 2
b) Let's draw the required number of pairs of dots in the table to convert the
given denary numbers into binary numbers.
Denary numbers 2 4 2 3 2 2 2 1 2 0
3
11
17
c) Let's draw the required number of groups of 5 dots in the table to convert
given denary numbers into quinary numbers.
Denary numbers 5 2 5 1 5 0
7
8
10
14
28
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 33 Vedanta Excel in Mathematics - Book 7
Unit Operations on Whole Number
3
3.1 Factors and Multiples - Looking back
Classroom - Exercise
1. Let's say and write the factors of the following numbers. Then circle the prime
factors.
a) Factors of 4 are ................................ b) Factors of 6 are ................................
c) Factors of 12 are ................................ d) Factors of 18 are ................................
2. Let's say and write the first five multiples of the following numbers.
a) 2 o .............................................. b) 3 o ..............................................
c) 4 o .............................................. d) 5 o ..............................................
3. Let's say and write the square and cube numbers of the following numbers.
a) square of 2 = ....... Cube of 2 = ....... b) square of 3 = ....... Cube of 3 = .......
c) square of 4 = ....... Cube of 4 = ....... d) square of 5 = ....... Cube of 5 = .......
e) square of 10= ....... Cube of 10= ....... f) square of 20= ....... Cube of 20= .......
4. Let's say and write the square roots and cube roots of the following numbers.
3
3
a) 4 = .................... 8 = .................... b) 9 = .................... 27 = ....................
3
3
c) 16 = .................. 64 = ................. d) 25 = .................. 125
= ................
3
e) 100 3 = ................ 8000 = ................
= ................ 1000 = ................ f) 400
Let’s take a whole number 24.
The possible factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24. Among these factors, 2 and
3 are the prime numbers and they are called the prime factors. Thus, the factors of
whole number divide it without leaving a remainder.
Again, let’s take a whole number 9.
A few multiples of 9 are: 1 × 9 = 9, 2 × 9 = 18, 3 × 9 = 27, 4 × 9 = 36, 5 × 9 = 45, ...
Thus, multiple of a whole number is the number obtained multiplying the whole
number by any natural number.
Vedanta Excel in Mathematics - Book 7 34 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Operations on Whole Numbers
3.2 Highest Common Factor (H. C. F.)
Let’s take F and F are the sets of all possible factors of 12 and 18 respectively.
12
18
Here, F = { 1, 2, 3, 4, 6, 12 } and F = { 1, 2, 3, 6, 9, 18}
12 18
Now, let’s make another set of the common factors of F and F .
12 18
F F = { 1, 2, 3, 6}
18
12
Among these common factors, 6 is the highest one. So, 6 is called the Highest
Common Factor (H.C.F) of 12 and 18.
Let's remember the following steps to find H. C. F. from all possible factors of the
given numbers.
Step 1: Find all possible factors of given numbers.
Step 2: List the factors common to the given number.
Step 3: Write the highest/greatest one as H.C.F.
3.3 Finding H. C. F. by Factorization Method
In this method, we should find the prime factors of the given numbers. Then,
the product of the common prime factors is the H.C.F. of the given numbers. For
example:
Find the H.C.F. of 24 and 36.
Finding the prime factors of 24 and 36.
2 24 2 36
2 12 2 18 Here, 24 = 2 × 2 × 2 × 3
2 6 3 9 36 = 2 × 2 × 3 × 3 H.C.F. = Product of common
3 3 ? HCF = 2 × 2 × 3 = 12 prime factors
2
3.4 Finding H.C.F. by Division Method
In this method, we divide the larger number by the smaller one. Again, the first
remainder so obtained divides the first divisor. The process is continued till the
remainder becomes zero. The last divisor for which the remainder becomes zero is
the H.C.F. of the given numbers. For example:
Find the H.C.F. of 28 and 48.
Dividing the greater number
28) 48 (1 by the smaller one.
–28
20) 28 (1 Since the remainder is not zero, dividing
st
–20 the 1 divisor 28 by the remainder 20.
8) 20 (2 Since, the remainder is not zero, dividing
–16 divisor 20 by the remainder 8.
4) 8 (2 Since, the remainder is not zero, dividing
–8 the divisor 8 by the remainder 4.
? H.C.F. = 4 0 Since, the remainder is zero for the divisor 4,
the H.C.F. is 4.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 35 Vedanta Excel in Mathematics - Book 7
Operations on Whole Numbers
Worked-out examples
Example 1: Write the sets of all possible factors of 18, 30, and 36. Then make a
set of their common factors and find their H.C.F.
Solution:
Here, F = {1, 2, 3, 6, 9, 18} I’ve remembered!
(18)
F = {1, 2, 3, 5, 6, 10, 15, 30} To find the common elements
(30) (here, factors) of the sets, we
F = {1, 2, 3, 4, 6, 9, 12, 18, 36} should find their intersection!!
(36)
Now, F F F = {1, 2, 3, 6}
(18) (30) (36)
? H.C.F. of 18, 30 and 36 is 6.
Example 2: Find the H.C.F. of 56, 84, and 140 by prime factorisation method.
Solution:
2 56 2 84 2 140 Now, it’s clear to me!
2 28 2 42 2 70 H.C.F. is the product of
2 14 3 21 5 35 common factors of the
7 7 7 given numbers!!
Now, 56 = 2 u 2 u 2 u 7
84 = 2 u 2 u 3 u 7
140 = 2 u 2 u 5 u 7
? H.C.F. = 2 u 2 u 7 = 28
Example 3: Find the greatest number that divides 40 and 56 without leaving a
remainder.
Solution:
Here, the required greatest number is the H.C.F. of 40 and 56.
40 56 1
-40 Possible factors of 40 are 1, 2, 4, 5, 8, 10, 20, 40
16 40 2 Possible factors of 56 are 1, 2, 4, 7, 8, 14, 28, 56
Common factors of 40 and 56 are 1, 2, 4, and 8
- 32 40 ÷ 8 = 5 and 56 ÷ 8 = 7
8 16 2 So, 8 is required greatest number.
-16
0
? H.C.F. = 8
Hence, the required greatest number is 8.
Example 4: Three jars of milk contain 45 l, 60 l and 120 l of milk respectively.
Find the greatest capacity of a vessel that can empty out each jar
with the exact number of fillings.
Solution:
Vedanta Excel in Mathematics - Book 7 36 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Operations on Whole Numbers
Here, the greatest capacity of the vessel is the H.C.F. of 45 l, 60 l and 120 l.
45 60 1 15 120 8
-45 - 120
15 45 3 0
-45
0
? H.C.F. = 15
Hence, the required vessel with the greatest capacity is 15 l.
Example 5: What is the greatest number of students to whom 84 books, 140
exercise books, and 252 pens can be distributed equally? Also, find
the shares of each item among them.
Solution:
Here, the required greatest number of students is the H.C.F. of 84, 140 and 252.
2 84 2 140 2 252
2 42 2 70 2 126
3 21 5 35 3 63
3 21
7 7 7
Now, 184 = 2 u 2 u 3 u 7
140 = 2 u 2 u 5 u 7
252 = 2 u 2 u 3 u 3 u 7
?H.C.F. = 2 u 2 u 7 = 28
Again, to find the shares of each item,
84 ÷ 28 = 3, 140 ÷ 28 = 5, and 252 ÷ 28 = 9
Hence, the required greatest number of student is 28 and each student shares
3 books, 5 exercise books, and 9 pens.
EXERCISE 3.1
General Section - Classwork
1. Between 2 and 4, 2 is a factor of 4. So, 2 is the H.C.F. of 2 and 4.
Between 6 and 18, 6 is a factor of 18. So, 6 is the H.C.F of 6 and 18.
Let's investigate the idea from the above facts. Then say and write the H.C. F.
of the following pairs of numbers.
(a) 3 and 6 is ............. (b) 7 and 28 is ............. (c) 6 and 30 is .............
(d) 9 and 63 is ............. (e) 12 and 48 is ............. (f) 15 and 90 is .............
2. Let's take any two numbers 12 and 18.
The smaller number 12 = 2 × 6
6 can also divide 18 exactly. So, the H.C.F of 12 and 18 is 6.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 37 Vedanta Excel in Mathematics - Book 7
Operations on Whole Numbers
Let's take other two numbers 36 and 45.
The smaller number 36 = 2 × 18 18 cannot divide 45 exactly.
= 3 × 12 12 cannot divide 45 exactly.
= 4 × 9 9 can also divide 45 exactly.
So, the H.C.F. of 36 and 45 is 9.
Let's investigate the idea from the above facts. Say and write the H.C.F. of the
following pairs of numbers.
a) 6 and 9 is .............. b) 10 and 15 is .............. c) 12 and 30 is ..............
d) 18 and 24 is .............. e) 24 and 32 is .............. f) 21 and 28 is ..............
Creative Section - A
3. Let's write the sets of all possible factors and the sets of common factors of
these pairs of numbers, then find their H.C.F.
a) 9, 12 b) 15, 18 c) 12, 16 d) 18, 24 e) 36, 48
4. Let's find the H.C.F. of the following numbers by factorisation method.
a) 18, 24 b) 40, 50 c) 54, 72 d) 24, 36, 42
e) 35, 56, 70 f) 30, 75, 90 g) 32, 64, 80 h) 96, 144, 216
5. Let's find the H.C.F. of the following numbers by division method.
a) 12, 18 b) 28, 32 c) 36, 42 d) 52, 65
e) 12, 15, 18 f) 30, 45, 60 g) 156, 221, 390 h) 360, 456, 696
Creative Section - B
6. a) Find the greatest number that divides 45, 60 and 75 without leaving a
remainder.
b) Find the greatest number by which 60, 90 and 120 are exactly divisible.
7. a) Three jars of milk contain 40 l, 50 l and 60 l of milk. Find the greatest
capacity of a vessel which can empty out each jar with the exact number of
fillings.
b) What is the maximum number of students to whom 48 apples, 60 bananas
and 96 guavas can be distributed equally? Also find the shares of each
fruit.
8. a) A rectangular floor is 20 feet long and 16 feet broad. If it is to be paved
with squared marbles of same size, find the greatest length of each squared
marble.
b) The length, breadth and height of a room are 9 m 80 cm, 8 m 40 cm and
4 m 20 cm respectively. Find the longest tape which can measure the
all three dimensions of the room exactly.
9. a) Shaswat distributed 54 snickers, 72 kit kats and 90 cadburies equally to his
classmates on the occasion of his birthday. What was the greatest number
of his classmates? How many chocolates of each type did every one get?
Vedanta Excel in Mathematics - Book 7 38 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Operations on Whole Numbers
b) Sunayna is making identical balloon arrangements for a party. She has 56
orange balloons, 40 maroon balloons, and 24 white balloon. She wants each
arrangement to have the same number of each colour. What is the greatest
number of arrangements that she can make if no balloon is left over? Also
find the shares of each coloured balloon.
c) Find the greatest number of old people of a geriatric care centre to whom
50 sweaters, 75 kambals, and 100 warm jackets can be equally distributed.
Also, find the share of each item among them.
d) There are 21 apples, 28 pears, and 49 oranges. These are to be arranged in
heaps containing the same number of fruits. Find the greatest number of
fruits possible to keep in each heap. How many heaps are formed by this
arrangement?
3.5 Lowest common multiple ( L.C.M)
Let’s take, M and M as the sets of a few multiples of 6 and 9 respectively.
6
9
Here, M = {6, 12, 18, 24, 30, 36, 42, 48, 54, 60, ...}
6
M = { 9, 18, 27, 36, 45, 54, 63, 72, 81, 90,...}
9
Now, let’s make another set A of the common multiples of M and M ,
6 9
A = M M = { 18, 36, 54, ...}
6 9
Among these common multiples, 18 is the lowest one. So, 18 is called the Lowest
Common Multiple (L.C.M.) of 6 and 9.
Thus, the lowest common Multiples (LCM) of two or more natural numbers is the
least natural number which is exactly divided by the given numbers.
3.6 Finding L.C.M. by Factorisation Method
In this method, we should find the prime factors of the given numbers. Then the
product of the common prime factors and the remaining prime factors (which are
not common) is the L.C.M. of the given numbers. For example:
Find the L.C.M. of 36 and 60.
2 36 2 60 Here, 36 = 2 u 2 u 3 u 3
2 18 2 30 60 = 2 u 2 u 3 u 5
3 9 3 15
3 5 ? L.C.M. = 2 u 2 u 3 u 3 u 5 = 180
3.7 Finding L.C.M. by Division Method
In this method, we arrange the given numbers in a row and they are successively
divided by the least common factors till all the quotients are 1 or prime factors.
Then, the product of these prime factors is the L.C.M. of the given numbers.
For example:
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 39 Vedanta Excel in Mathematics - Book 7
Operations on Whole Numbers
Find the L.C.M of 28, 30, and 35
2 28, 30, 35 2 is the least common prime factor of 28 and 30.
5 14, 15, 35 5 is the least common prime factor of 15 and 35.
7 14, 3, 7 7 is the least prime factor of 14 and 7.
2, 3, 1
? L.C.M. = 2 × 5 × 7 × 2 × 3 = 420
Workedout examples
Example 1: Write the sets of a few multiples of 6, 9, and 12. Make a set of their
common multiples and find their L.C.M.
Solution:
Here, M = {6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, …}
(6)
M = {9, 18, 27, 36, 45, 54, 63, 72, 81, 90, …}
(9)
M (12) = {12, 24, 36, 48, 60, 72, 84, 96, 108, 120, …}
Now, M M M = (36, 72, …)
(6) (9) (12)
? L.C.M. of 6, 9, and 12 is 36.
Example 2: Three measuring rods are 30 cm, 45 cm, and 60 cm long. Find the
shortest length of cloth which can be measured exactly by any one
of these rods.
Solution:
Here, the shortest length of the cloth is the L.C.M. of 30 cm, 45 cm, and 60 cm.
2 30, 45, 60
3 15, 45, 30
5 5, 15, 10
1, 3, 2
? L.C.M. = 2 × 3 × 5 × 3 × 2 = 180
Hence, the required shortest length of the cloth is 180 cm.
Example 3: Find the least number which leaves a remainder 7 when it is divided
by any one of the numbers 20, 30, and 40.
Solution:
Here, the least number divisible by 20, 30, and 40 is their L.C.M. Then, the required
least number = (L.C.M. of 20, 30, 40) + 7.
2 20, 30, 40
2 10, 15, 20
5 5, 15, 10
1, 3, 2
? L.C.M. = 2 × 2 × 5 × 3 × 2 = 120
Hence, the required least number = 120 + 7 = 127.
Vedanta Excel in Mathematics - Book 7 40 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Operations on Whole Numbers
EXERCISE 3.2
General Section -Classwork
1. i) 6 is a multiple of 3. So, L.C.M. of 3 and 6 is 6.
ii) 2 and 3 are prime numbers. So, L.C.M. of 2 and 3 = 2 × 3 = 6.
iii) 4 is a composite number and 5 is a prime number. They are not exactly
divisible to each other. So, L.C.M. of 4 and 5 = 4 × 5 = 20.
iv) 4 and 9 are composite numbers and they do not have any common prime
factors So, L.C.M. of 4 and 9 = 4 × 9 = 36.
Let's investigate the ideas from the above facts. Tell and write the L.C.M. of
a) 2 and 6 is ............. b) 4 and 8 is ............. c) 6 and 18 is .............
d) 2 and 5 is ............. e) 5 and 7 is ............. f) 7 and 11 is .............
g) 5 and 6 is ............. h) 3 and 8 is ............. i) 9 and 10 is .............
j) 4 and 9 is ............. k) 9 and 8 is ............. l) 10 and 21 is .............
2. Let's take any two composite numbers 10 and 15. They have a common prime
factor. Take the bigger number 15 and think of its multiples.
15 × 1 = 15 is not the multiple of 10.
15 × 2 = 30 is the multiple of 10.
So, L.C.M. of 10 and 15 is 30.
Let's follow the similar process and tell and find the L.C.M. of
a) 6 and 9 is ....................... b) 8 and 12 is .....................
c) 10 and 15 is ..................... d) 12 and 18 is .....................
e) 8 and 20 is ................... f) 15 and 25 is .....................
Creative Section - A
3. Let's write the sets of the first ten multiples and the sets of common multiples
of these pairs of numbers. Then, find their L.C.M.
a) 2, 3 b) 3, 4 c) 4, 6 d) 6, 9 e) 8, 12
4. Let's find the L.C.M. of the following numbers by factorization method.
a) 6, 9 b) 9, 12 c) 12, 18 d) 14, 21
e) 4, 6, 8 f) 8, 12, 18 g) 10, 15, 25 h) 14, 21, 35
5. Let's fnd the L.C.M. of the following numbers by division method.
a) 8, 12 b) 14, 21 c) 30, 45 d) 12, 15, 20
e) 12, 16, 24 f) 20, 30, 40 g) 45, 60, 90 h) 72, 90, 120
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 41 Vedanta Excel in Mathematics - Book 7
Operations on Whole Numbers
Creative Section - B
6. a) Find the least number which is exactly divisible by 12, 16, and 24.
b) Find the smallest number which is when divided by 25, 30, and 75 leaves
no remainder.
7. a) If three buckets of capacities 10l, 12l, and 15l can fill a drum in exact number
of fillings, find the least capacity of the drum.
b) Three measuring rods are 40 cm, 50 cm and 80 cm long. Find the shortest
length of cloth which can be measured exactly by any one of these rods.
c) After travelling every 400 km, a motorcycle needs to fill petrol. After every
1600 km, it needs to change mobil. After every 2400 km, it needs servicing.
If these works are done together on a day. After how many kilometres of
travelling are all theses works repeated again ?
8. a) Three bells ring at an interval of 10, 15, and 20 minutes respectively. If they
all ring together at 6:00 a.m., at what time do they ring again together?
b) From Kathmandu bus park, buses moved to Pokhara in an interval of every
30 minutes, Birgunj in every 45 minutes, and Mahendranagar in every
1 hour. If the buses move to these places at the same time at 3:00 p.m. at
what time do the buses again move from the bus park at a time?
9. a) Find the least number which leaves a remainder 5 when it is divided by
any one of the numbers 10, 12 , and 15.
b) Find the least number which leaves a remainder 4 when it is divided by any
one of the number 12, 16, and 24.
10. a) Find the least number with which when 5 is added, the sum is exactly
divisible by 12, 18, and 20.
b) Find the least number with which when 20 is added, the sum is exactly
divisible by 30, 40, and 60.
It’s your time : Project work!
11. Let's do the following activities and answer the given questions.
a) Let's take 4 green cards and 6 yellow cards.
(i) Divide these cards equally between your four friends. Is it possible?
(ii) Divide these cards equally between your three friends. Is it possible?
(iii) Divide these cards equally between your two friends. Is it possible?
(iv) What is the H.C.F. of 4 and 6?
b) Let's take the number of the following pairs of coloured card and repeat the
similar activities to find the H.C.F. of the each pair of numbers.
(i) 2 green and 4 yellow cards (ii) 6 green and 9 yellow cards
(iii) 8 green and 12 yellow cards (iv) 10 yellow and 15 green cards.
Vedanta Excel in Mathematics - Book 7 42 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Operations on Whole Numbers
12. a) Let's take two call bells and mark them as A and B. Now, let's do the following
activities and answer the given questions.
(i) Let's ring both the bells at a time and see at what time you rang the bell.
(ii) Then, ring bell A after each interval of 2 minutes and bell B after each
interval of 3 minutes. In this way, after how many minutes do you ring
both the bells at a time?
(iii) Again, ring bell A in every 4 minutes intervals and bell B in every 6
minutes intervals.
In this way, after how many minutes do you ring both the bells at a
time? Now, did you understand one of the uses of L.C.M. in our real life
situations?
b) Let's take any two pairs of composite numbers between 5 and 15.
(i) Let's find the H.C.F. and L.C.M. of each pair of your composite numbers.
(ii) Find the product of H.C.F. and L.C.M. of each pair of numbers. Also,
find the product of each pair of numbers.
Could you investigate a fact about the H.C.F. and L.C.M. of any pair of
numbers? Write a report about it and discuss in the class.
3.8 Square and square root
Prefect square numbers
Let's study the following examples and investigate the idea of perfect square
numbers.
m 1 squared = 1 = 1 × 1 = 1 (1 is a perfect square number)
2
m 2 squared = 2 = 2 × 2 = 4 (4 is a perfect square number)
2
m 3 squared = 3 = 3 × 3 = 9
2
(9 is a perfect square number)
m 4 squared = 4 = 4 × 4 = 16 (16 is a perfect square numbers),
2
and so on.
Thus, a perfect square number is the product of two identical numbers. 1, 4, 9, 16,
25, 36, 49, …, etc. are the examples of perfect square numbers.
Square root
Let's study the following examples and investigate the idea of square root.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 43 Vedanta Excel in Mathematics - Book 7
Operations on Whole Numbers
1 = 1 = 1 × 1 o So, 1 is the square root of 1. o 1 = 1
2
4 = 2 = 2 × 2 o So, 2 is the square root of 4. o 4 = 2
2
9 = 3 = 3 × 3 o So, 3 is the square root of 9. o 9 = 3
2
16 = 4 = 4 × 4 o So, 4 is the square root of 16. o 16 = 4
2
Thus, a perfect square number is the product of two identical numbers and one of
the identical numbers is the square root of the square number.
The radical symbol ( ) is used to denote the square root of a number.
3.9 Process of finding square root
We usually use fractorisation method and division method to find the square root of
the given perfect square number.
Factorisation method to find square root
We usually use this method to find the square root of smaller perfect square numbers.
Let’s learn the process from the example given below.
Find the square root of 225 by factorisation method.
3 225 To find the square root, we should make the pairs of
3 75 two identical factors and the product of one of
the factors taken from each pair is the square
5 25 root of the number.
5
Now, 225 = 3 × 3 × 5 × 5
? 225 = 3 × 5 = 15
Division method to find square root
We use division method to find the square roots of the larger numbers. It is also
useful to find the square root of decimals. Let’s learn the process from the example
given below. (i) Starting from the unit place make the number pairs by using
bar mark just over each pair.
Find the square root of (ii) Take the first pair 68 and think of the largest perfect square
(i) 6889 (ii) 64516 which is equal to 68 or less than 68. It is 64.
(iii) Think of the square root of 64, it is 8.
8 68 89 83 (iv) Write 8 as the divisor and quotient. Multiply 8 u 8 = 64 and
+ 8 – 64 write just below 68, then subtract. The remainder is 4.
163 489 (v) Add 8 + 8 = 16 in the divisor side. It is the trial divisor.
+ 3 – 489 (vi) Bring down the other pair 89. Now, 489 is the new dividend.
166 0
(vii) 48 is 3 times divisible by 16. So, write 3 in the quotient and
6889 = 83 also in the divisor.
(viii) Find the product of 163 u 3 which is 489, and write it just
below 489 and subtract. The remainder is 0. Thus, 83 is the
square root of 6889.
Vedanta Excel in Mathematics - Book 7 44 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Operations on Whole Numbers
2 645 16 254 The greatest perfect square less than 6 is 4
and its square root is 2.
+ 2 – 4
24 ÷ 4 = 6, 46 × 6 = 276 > 245
45 245 so, 45 × 5 = 225
+ 5 –225
504 2016 20 ÷ 5 = 4 , 504 × 4 = 2016.
+ 4 –2016
508 0
Worked out examples
Example 1: Find the square of a) 24 b) 85 c) 120
Solution:
a) Square of 24 = 24 = 24 × 24 = 576
2
2
b) Square of 85 = 85 = 85 × 85 = 7225
2
c) Square of 120 = 120 = 120 × 120 = 144000 12×12 = 144, then 120×120=14400
Example 2: Find the square root of 1764 by prime factorisation method.
f
Solution: I got it!
2 1764 Now, 1764 = 2 u 2 u 3 u 3 u 7 u 7 The product of one of the
2 882 identical factors taken from
2
2
3 441 = 2 u 3 u 7 2 each pair is the square root.
3 147 ? 1764 = 2 u 3 u 7 = 42
7 49
7
Example 3: Find the square root of 11449 by division method.
Solution:
1 114 49 107
+ 1 –1
20 14 While making the number pairs in
+ 0 –0 11449 from the unit place, 1 is left
unpaired. So, it is divided at first.
207 1449
+ 7 –1449
214 0
? 11449 = 107
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 45 Vedanta Excel in Mathematics - Book 7
Operations on Whole Numbers
Example 4: Simplify a) 8 × 2 18 × 3 75 (b) 144
441
Solution:
a) 8 × 2 18 × 3 75
Think of the factors of 8, 18 and 75 such
= 4 × 2 × 2 9 × 2 × 3 25 × 3 that one of the factors is a perfect square.
= 2 2 × 2 × 3 2 × 3 × 5 3 Find the square root of each perfect square factor.
= 180 12 2 × 6 × 15 = 180 and 2 × 2 × 3 = 12
= 180 × 4 × 3 Think of the factors of 12 such that one of
the factors is a perfect square.
= 180 u 2 3
= 360 3
144 = 2 × 2 × 2 × 2 × 3 × 3
2
2
2
2
b) 144 = 2 × 2 × 3 2 = 2 × 2 × 3 2
2
2
441 3 × 7 441 = 3 × 3 × 7 × 7
2
= 2 × 2 = 4 = 3 × 7 2
7 7
Example 5: Find the square root of 3.24
Solution:
2 324 2 100
Here, 3.24 = 324 2 162 2 50
100
3 81 5 25
? 3.24 = 324 3 27 5
100
3 9
2 × 3 × 3 2 3
2
2
= 2 × 5 2 324 = 2 × 2 × 3 × 3 × 3 × 3
2
2
2
= 2 × 3 × 3 = 2 × 3 × 3 2
2 × 5 100 = 2 × 2 × 5 × 5
2
= 18 = 1.8 = 2 × 5 2
10
Example 6 : There are 40 pupils in class 7. If every pupil collects as much as
money as their numbers for a charity show, how much money do
they collect altogether?
Solution:
Here, Number of pupils = 40
sum of money collected from each pupil = Rs 40
Thus, the required sum of money = Rs (40 )
2
= Rs ( 40 × 40) = Rs 1600.
Hence, they collect Rs 1,600 altogether.
Vedanta Excel in Mathematics - Book 7 46 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Operations on Whole Numbers
Example 7. The commander arranged his 784 soldiers in the square form for
march pass. How many soldiers were there in each row.
Solution: 2 784
Let the number of soldier in each row be x. 2 392
Then, total number of soldiers = x × x 2 196
2 98
or, x = 784
2
7 49
or, x = 2 × 2 × 7 2
2
2
2
7
or, x = (2 × 2 × 7) 2 ? 784 = 2 × 2 × 2 × 2 × 7 × 7
2
2
2
? x = 2 × 2 × 7 = 28 = 2 × 2 × 7 2
784 = 2 × 2 × 7 = 28
Hence, the number of soldiers in each row was 28.
Example 8. Find the smallest number by which when 180 is multiplied it becomes
a perfect square.
Solution:
2 180
2 90 In 180 = 2 × 3 × 5, two
2
3
3 45 numbers 2 and 3 are perfect
2
2
3 15 square but 5 is not. So, to
5 make 5 a perfect square it is
multiplied by 5.
? 180 = 2 × 2 × 3 × 3 × 5 = 2 × 3 × 5
2
2
Hence, the required smallest number is 5.
Example 9. Find the least number which is to be subtracted from 6090 to make
it a perfect square.
Solution:
7 60 90 78
+ 7 – 49 I understood!
148 1190 6090 is not a perfect square because of the
remainder 6. So, the remainder 9 should be
+ 8 –1184 subtracted to make it a perfect square.
156 6
So, the required least number to be subtracted is 6.
Example 10. Find the least number which is to be added to 3993 to make it a
perfect square.
Solution: 2
Here, 6 39 93 63 3993 is greater than 63 . So, by
adding the least number it should be
+ 6 – 36 equal to 64 = 4096.
2
123 393 Now, what should be added to 3993
+ 3 –369 to make it 4096?
126 24 It is 4096 – 3993.
Here, 64 = 4096
2
? The required least number to be added = 4096 – 3993 = 103
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 47 Vedanta Excel in Mathematics - Book 7
Operations on Whole Numbers
EXERCISE 3.3
General Section - Classwork
1. Let's investigate the idea from the given examples and apply it to find the
squares of the given numbers.
2
2
2
2
Example: 3 = 9 , 30 = 900, 300 = 90000, 3000 = 9000000
30 has one zero. So its square has two zeros. 300 has two zeros . So, its square
has four zeros !!
2
2
a) 4 = ........ , 40 = ........ , 400 = ........, 4000 = ........
2
2
2
2
2
2
b) 5 = ........ , 50 = ........ , 500 = ........, 5000 = ........
c) 7 = ........ , 70 = ........ , 700 = ........, 7000 = ........
2
2
2
2
d) 8 = ........ , 80 = ........ , 800 = ........, 8000 = ........
2
2
2
2
2. Let's investigate the idea from examples and apply it to find the square
roots of the given numbers.
Example : = 9 = 3, √900 = 30, √90000 = 300 , √9000000 = 3000
900 has two zeros and its square root has one zero. 90000 has four zeros and
its square root has two zeros.
a) 4 = ........ , 400 = ........ , 40000 = ........, 4000000 = ........
b) 16 = ........ , 1600 = ........ , 160000 = ........, 16000000 = ........
c) 36 = ........ , 3600 = ........ , 360000 = ........, 36000000 = ........
d) 81 = ........ , 8100 = ........ , 810000 = ........, 81000000 = ........
Creative Section - A
3. Lets find the squares of a) 45 b) 88 c) 124
4. Let's observe the example given below and learn to identify whether the given
numbers are perfect square or not.
196 By factorization of 196 = 2 × 2 × 7 × 7 = 2 × 7 2
2
As pairing of identical factor is possible, 196 is a perfect square.
216 By factorization on 216 = 2 × 2 × 2 × 3 × 3 × 3 = 2 × 2 × 3 × 3
2
2
As 2 and 3 are left unpaired, 216 is not a perfect square number.
Let's apply the above process to test whether the following numbers are perfect
squares.
a) 225 b) 392 c) 324 d) 432 e) 576
5. Let’s find the square roots of these numbers by factorization method.
a) 64 b) 100 c) 144 d) 324 e) 784 f) 1225
Vedanta Excel in Mathematics - Book 7 48 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
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