Prime

M a t h e m a t i c s Series

9

Approved by

Government of Nepal, Ministry of Education, Science and

Technology Curriculum Development Centre Sanothimi,

Bhaktapur as and additional learning materials.

Raj Kumar Mathema Authors

Dirgha Raj Mishra Bhakta Bahadur Bholan

Uma Raj Acharya Yam Bahadur Poudel

Naryan Prasad Shrestha Bindu Kumar Shrestha

Editors

Anil Kumar Jha

Dhurba Narayan Chaudhary

Hari Krishna Shrestha

Language Editor

Mrs. Tara Pradhan

Pragya Books Pragya Books and Distributor Pvt. Ltd.

and

Distributors Pvt. Ltd.

Printing history

First Edition 2074 B.S.

Revised Edition 2077 B.S.

Layout and design

Ramesh Maharjan

© Publisher

All rights reserved. No part of this book, or designs and illustrations here within, may be

reproduced or transmitted in any form by any means without prior written permission.

ISBN : 978-9937-9170-3-2

Printed in Nepal

Pragya Books Published by

and

Pragya Books and Distributors Pvt. Ltd.

Distributors Pvt. Ltd.

Kathmandu, Nepal

E-mail : [email protected]

Preface

Prime Mathematics Series is a distinctly outstanding mathematics series designed in compliance

with Curriculum Development Centre (CDC) to meet international standards. The innovative, lucid

and logical arrangement of the content makes each book in the series coherent. The presentation

of ideas in each volume makes the series not only unique, but also a pioneer in the evolution of

mathematics teaching.

The subject matter is set in an easy and child-friendly structure so that students will discover

learning mathematics a fun thing to do. A lot of research, experimentation and careful gradation have

gone into the making of the series to ensure that the selection and presentation is systematic, innovative

and both horizontally and vertically integrated.

Prime Mathematics Series is based on child-centered teaching and learning methodologies,

so the teachers will find teaching this series equally enjoyable.We are optimistic that this series shall

bridge the existing inconsistencies between the cognitive capacity of children and the course matter.

We owe an immense debt of gratitude to the publishers for their creative,thoughtful and inspirational

support in bringing about the series. Similarly, we would like to acknowledge the tremendous support

of teachers, educationists and well-wishers for their contribution, assistance and encouragement in

making this series a success.

We hope the series will be another milestone in the advancement of teaching and learning

mathematics in Nepal.We solicit feedback and suggestions from teachers, students and guardians alike

so that we can refine and improvise the series in the future editions.

Our team would like to express our special thanks to Mr. Nara Bahadur Gurung, Mr. Ram

Narayan Shah, Mr. Tulsi Kharel, Mr. Mani Ram Khabas, Mr. Umesh Acharya, Mr. J. Phuldel, Mr.

Kamal Raj Tripathee, Mr. Rudra Prasad Pokharel, Mr. Uttam Prasad Panta, Mr. L.N. Upadhyaya,

Mr. Shakti Prasad Acharya, Mr. Upendra Subedi, Mr. Kul Narayan Chaudhary, Mr. Bishonath

Lamichhane, Mr. Harilal Lamichhane, Mr. Govinda Paudel, Mr. Krishna Aryal, Mr. Nim Bhujel,

Mr. Santosh Simkhada, Mr. Pashupati Upadhyaya, Mr. Dipak Adhikari, Mr. Mukti Adhikari,

Mr. Dipendra Upreti, Mr. Dipak Khatiwada, Mr. Narayan Nepal, Mr. Raj Kumar Dahal, Mr. Bhim Raj

Kandel, Mr. Prem Giri,Mr. Iswor Khanal, Mr. Balram Ghimire, Mr. Om Kumar Chhetri, Mr. Ram Hari

Bhandari, Mr. Krishna Kandel, Mr. Madhav Atreya, Mr. Harihar Adhikari, Mr. Chura Gurung, Mr. Shiva

Devkota, Mr. Chandra Dev Tiwari, Jivan K.C., Raghu Kandel and Baikuntha Marhattha and Subash Bidari

for their Painstaking effort in peer reviewing of this book.

Contents

1. Sets

1.1 Sets................................................................................................................. 2

1.2 Venn Diagrams and Set operations ......................................................... 5

- Exercise 1.1.............................................................................................13

1.3 Application of Sets ....................................................................................15

- Exercise 1.2 ............................................................................................18

2. Arithmetic

2.1 Profit and Loss............................................................................................ 26

- Exercise 2.1 ........................................................................................... 32

2.2 Commission .................................................................................................. 33

- Exercise 2.2........................................................................................... 35

2.3 Discount ....................................................................................................... 36

- Exercise 2.3........................................................................................... 39

2.4 Taxation ....................................................................................................... 40

- Exercise 2.4........................................................................................... 42

2.5 Bonus............................................................................................................. 43

- Exercise 2.5........................................................................................... 45

2.6 Share and Dividend ................................................................................... 46

- Exercise 2.6........................................................................................... 49

2.7 Home Arithmetic ....................................................................................... 50

2.7.1 Electricity Billing............................................................................ 50

- Exercise 2.7.1 ............................................................ 53

2.7.2 Water Billing ................................................................................... 55

- Exercise 2.7.2............................................................ 56

2.7.3 Telephone Billing ............................................................................ 58

- Exercise 2.7.3............................................................ 60

2.7.4 Taxi Fare...........................................................................................61

- Exercise 2.7.4............................................................ 62

3. Mensuration

3.1 Area of Pathways........................................................................................71

- Exercise 3.1 ........................................................................................... 75

3.2 Surface Area and Volume of Prisms ..................................................... 77

- Exercise 3.2........................................................................................... 80

3.3 Area of Four Walls, Floor and Ceiling .................................................. 83

- Exercise 3.3........................................................................................... 89

3.4 Volume of Walls ......................................................................................... 90

- Exercise 3.4........................................................................................... 95

4. Alebra

4.1 Factorisation............................................................................................. 100

- Exercise 4.1.1 ....................................................................................... 107

- Exercise 4.1.2 ...................................................................................... 109

4.2 Indices ........................................................................................................110

- Exercise 4.2 ..........................................................................................116

4.3 Exponential Equation................................................................................118

- Exercise 4.3 ..........................................................................................121

4.4 Ratio and Proportion ............................................................................... 122

4.4.1 Ratio ................................................................................................ 122

- Exercise 4.4.1 ........................................................... 126

4.4.2 Proportion ...................................................................................... 128

- Exercise 4.4.2 ........................................................... 137

4.5 Simultaneous Equations.......................................................................... 140

4.5.1 Linear Equation ............................................................................. 140

- Exercise 4.5.1 ........................................................... 145

4.5.2 Graphical Method......................................................................... 146

- Exercise 4.5.2 ........................................................... 148

4.6 Quadratic Equation ................................................................................. 148

4.6.1 Solving a Quadratic Equation by Factorization Method .... 149

- Exercise 4.6.1 ........................................................... 151

4.6.2 Solving Quadratic Equation by Completing the Square ...... 153

- Exercise 4.6.2 ........................................................... 156

4.6.3 Solving Quadratic Equation by using Formula....................... 157

- Exercise 4.6.3 ........................................................... 160

5. Geometry

5.1 Triangle ...................................................................................................... 168

- Exercise 5.1 ...........................................................................................171

- Exercise 5.2 ......................................................................................... 180

5.2 Parallelogram ............................................................................................ 183

- Exercise 5.3 ......................................................................................... 190

5.3 Mid-Point Theorem.................................................................................. 193

- Exercise 5.4 ......................................................................................... 196

5.4 Similarity ...................................................................................................200

- Exercise 5.5 .........................................................................................205

5.5 Pythagoras Theorem ...............................................................................209

- Exercise 5.6 ......................................................................................... 213

5.6 Construction of Quadrilaterals............................................................ 215

- Exercise 5.7 ......................................................................................... 219

5.7 Circle...........................................................................................................220

- Exercise 5.8 .........................................................................................227

6. Trigonometry

6.1 Trigonometric Ratios ..............................................................................234

- Exercise 6.1 ..........................................................................................239

6.2 Trigonometric Ratios of Some Standard Angles.............................242

- Exercise 6.2 .........................................................................................246

7. Statistics

7.1 Graphs and Pie Charts ............................................................................252

- Exercise 7.1 ..........................................................................................260

7.2 Measure of Central Tendency ..............................................................262

7.2.1 Mean ................................................................................................262

- Exercise 7.2.1 ...........................................................................265

7.2.2 Median.............................................................................................266

- Exercise 7.2.2 ..........................................................................268

7.2.3 Mode................................................................................................269

- Exercise 7.2.3 ..........................................................................269

7.2.4 Quartiles ........................................................................................270

- Exercise 7.2.4 ..........................................................................272

8. Probability

8.1 Probability .................................................................................................276

- Exercise 8.1 ..........................................................................................282

8.2 Empirical probability...............................................................................285

- Exercise 8.2 .........................................................................................286

n Model Question Set ................................................. 291

n Prime questions for more practice ................................. 303

1Chapter

Sets

Objectives:

At the end of this chapter, the

students will be able to:

find the set relation using set

operation between the sets.

represent the set operations in

the Venn diagram.

solve the word problems related

to cardinality of sets using Venn

diagram.

Teaching Materials:

Chart paper, flass cards, different

colour board markers, charts of related

formula, geometrical instruments

Historical facts

1. The set theory was developed by a German mathematician Georg Cantor

(1845-1918). The “New Mathematics” which helps in solving problem in

some case, more easily and with more pleasure, makes much use of sets.

Sets provide a useful way of representing groups of things and correlating them.

2. Augustus De Morgan (1806-1871; England) was born blind in one eye in

1806 in Madras, India. His famous laws on set theory states that if A and B are subsets of a

universal set then the complement of union of A and B is the intersection of the complements

of A and B i.e. (AB) = AB and the complement of the intersection of A and B is the

union of complements of A and B i.e. (AB) = AB.

1.1 Sets

A collection or aggregate of well defined objects is called a set. e.g. the set of natural numbers, the set

of vowels of English alphabets. The objects of a set are called its elements or members. If x is an

element of the set A, then we write xA (x belongs to A).

Set Notation

The elements of a set are enclosed with curly brackets { } and a set is denoted by capital letters. A, B,

C, N, V, ……… etc.

For example:

N = {1, 2, 3, …….. }

V = { a, e, i, o, u}

A set can be described mainly by three methods

1. Description method: In this method, the set is described in words with the common properties

of the elements.

For example:

(i) A is a set of digits of Hindu-Arabic number system,

(ii) V is a set of vowels of English alphabets.

2. Listing or Roster Form (tabular form): In this method, we list all the elements.

For example:

(i) A set of digits in Hindu-Arabic number system, A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

(ii) A set of vowels of English alphabets, V = {a, e, i, o, u}

(iii) A set of integers, Z = {…….. -2, -1, 0, 1, 2, ……….}

(iv) A set of whole number, W = {0, 1, 2, 3, …….}

2 | Mathematics - 9 Sets

3. The Rule method or Set builder form: In this form, we specify the set with ‘defining

property’ giving a variable, like A = {x : x has property p}. Which is read as: A is the set of x

such that x has property p.

For example:

(i) N = {x : x is a natural number}

(ii) W = {x : x is a whole number}

(iii) A = {x : x 20, x N}

(iv) S = {x : (x – 1)( x + 1) = 0}

Note: Specially, we denote the numbers as:

(i) Natural numbers, {1, 2, 3, ………… } = N

(ii) Whole numbers, {0, 1, 2, 3, ……….. } = W

(iii) Integers, {……………. -3, -2, -1, 0, 1, 2, 3, ……….. } = Z

(iv) Rational number = Q

(v) Real number = R

Various Sets

The Empty Set or Null Set or Void Set: A set which contains no elements is called the empty

set and is denoted by or { }.

For example:

(i) A = {x : x + 1= 0, x N} =

(ii) B = {Boys of class IX of Saint Mary’s School} =

Note:

(i) Since there is one and only one empty set with different descriptions, hence we say ‘the empty set’

or ‘the null set’ or ‘the void set’ instead of ‘an empty set’, ‘a null set’.

(ii) The empty set is denoted by which is not Greek letter ‘phi’ but Danish letter ‘oe’.

(iii) {0}, since {0} is a set whose element is 0.

(iv) {}, since {} is a set whose element is .

Finite Set: A set with finite number of elements is called a finite set.

For example:

(i) A = {1, 2, 3, …………., 100}

(ii) B = {x : x is a district of Nepal}

Infinite Set: A set which is neither a null nor a finite set is called an infinite set.

For example:

(i) N = {1, 2, 3, ……………….}

(ii) A = {Rational number between 1 and 2}

Sets Mathematics - 9 |3

Singleton Set: A set containing only one element is called a singleton set.

For example: {}, {0}, {1}, {x : x is an even number between 1 and 3}.

Universal Set: A set which contains all the sets under consideration as sub-sets is called a

Universal set. In other way, a set which contains all the possible elements under consideration is

called a universal Set. It is denoted by U. In some texts, we also find a universal set denoted by

(pxi) or X or [ ].

For example: Let's consider two sets A = {1, 2, 3, 4, 5} and B ={1, 3, 5, 7}, for these two sets,

universal set may be {1, 2, 3, 4, 5, 6, 7, 8, 9} or {x : x N} or {x : x W} or {x : x Z} or

{x : x Q}.

Note: The choice of a universal set is not unique.

Relations between the sets:

Disjoint Sets: Two sets are disjoint if they have no element in common.

For example:

(i) If A = {1, 3, 5, 7, 9} and B = {2, 4, 6, 8}, then A and B are disjoint sets.

(ii) If M = {x : x N} and N = {x : x Z, x < 0}, then M and N are disjoint sets.

Overlapping / Joint Sets: If two sets have some common elements, they are called overlapping sets.

For example: If A = {1, 2, 3, 4, 5} and B = {2, 4, 6, 8}, two sets A and B have common elements 2

and 4. So the sets A and B are overlapping sets.

Equal Sets: Two sets are said to be equal, if they contain the same elements.

For example:

(i) If A = { x : x is letter of the word LATENT} and B = { x : x is letter of the word TALENT}

then A =B or AB.

(ii) If P = { x : x is a digit of number 123} and Q = { x : x is a digit of the number 321} then P = Q

or PQ.

Equivalent Sets: Two sets are equivalent, if and only if a one-to-one correspondence exists

between them. In other words A & B are equivalent sets if they have equal number of elements.

For example: If A = {1, 2, 3} and B = {p, q, r}, then A and B are equivalent sets.

Note: If two sets are equal they must be equivalent, however two equivalent sets need not be equal.

Sub Sets: Set A is a subset of B, if every elements of set A is also an element of Set B. It is denoted

as A B, [read as A is subset of B, or A is contained in B] or B A [read as B contains A]. And B is

called the super set of A.

For example:

(i) If A = {a, b} and B = {a, b, c} then AB.

(ii) If P = {1, 2, 3} and Q = {1, 2, 3}, then PQ.

4 | Mathematics - 9 Sets

Power Set: The set of all subsets of a set A is called the power set of the set A.

For example: If A = {a, b}, then the power set of A = {{ }, {a}, {b}, {a, b}}. If A = {a, b, c}, then the

power set of A = {, {a}, {b}, {c}, {a, b}, {b, c}, {a, c}, {a, b, c}}.

Note:

(i) Every set is a subset of itself i.e. AA.

(ii) is subset of every set i.e. A.

(iii) Number of subsets of a set A containing n elements is 2n.

For A = {a}, sub-sets of A are:

and {a}, where number of subsets = 2 = 21

For A = {a, b}, subsets of A are:

, {a}, {b}, {a, b}, where number of subsets = 4 = 22.

For A = {a, b, c}, subsets of A are:

, {a}, {b}, {c}, {a, b}, {b, c}, {a, c}, {a, b, c}.where number of subsets = 8 = 23.

If number of elements of A is n then number of subsets = 2n.

Proper sub-set: Set A is a proper subset of set B, if every element of set A is an element of set B

and at least an element of B is not an element of A, it is denoted as AB.

If N is a set of natural numbers, W whole numbers, Z integers, Q rational numbers and R real numbers

then N W Z Q R.

Note:

(i) A set is not a proper subset of itself.

(ii) is a proper subset of every set.

(iii) is not a proper subset of itself.

(iv) Number of sub-sets of a set A containing n elements is 2n – 1.

1.2 Venn Diagrams and Set operations

Venn-diagrams: John Venn (1834-1923, England) first illustrated sets graphically in 1870 in a paper

on “Boole’s logical system.”

A Venn-diagram consists of a rectangle representing universal set U and subsets of U by circles or

ovals inside it. The sets of elements representing different operations like Union, intersection,

difference and complements are shown by shading the required regions.

Leonard Euler (1707 – 1783, Switzerland) also used diagrams to represent sets. So, Venn-diagrams

are sometimes called Euler-Venn-diagrams. UU

BAB

U

AA

AU A and B are C

disjoint sets

A, B, C are

Sets overlapping sets

Mathematics - 9 |5

U U U

AB AB BA

A and B are AB BA or AB

overlapping sets

Set Operations

There are four set operations:

a. Union

b. Intersection

c. Difference

d. Complement

a. Union of Two or More Sets

The union of two sets A and B is the set of the elements that belong either to A or to B or to

both A and B.

Mathematically, AB = {x : x A or x B}.

For two disjoint sets For three disjoint sets

U A U

AB BC

AB ABC

For two overlapping sets For three overlapping sets

U U

AB AB

AB C

6 | Mathematics - 9 ABC

Sets

For AB For A and B overlapping and B and C overlapping

U A U

AB BC

AB = B ABC

b. Intersection of Sets

Intersection of two sets A and B is the set of the elements which belong to both A and B.

Mathematically, AB = {x : x A and x B}.

When A and B are disjoint When A and B are overlapping When AB

U U U

AB AB AB

AB = AB AB = A

When A, B, C are overlapping U U

AB AB

U

AB

C CC

ABC (AB)C (AB)(AC)(BC)

c. Difference between of Two Sets

Difference between two sets A and B denoted A-B is the set of elements which belong to A but

don’t belong to B.

Thus, A – B = {x : x A, x B}.

When A and B are disjoint When A and B are overlapping

U U

AB AB

A–B=A A–B

Sets Mathematics - 9 |7

U U

AB AB

CC

(AB) – C A – (BC)

d. Complement of a Set

Complement of a Set A is denoted as A¯ or A or Ac which is the set of the elements which

belong to the universal set U but do not belong to A.

Thus, A¯ or Aor Ac = U – A A U U

BA B

U

A

_ ____ ____

A AB AB

A UA U A U

B B B

A-B

ABC

C C ABC

____ _______

A–B ABC

Special Properties of Complements of Sets

1. (A) = A or A= = A

Proof: _

A= = {x : x A }

= {x : x A }

A=

=A

2. De Morgan’s law

______ __

(a) (AB) = A B

______ __

(b) (AB) = A B

8 | Mathematics - 9 Sets

Proof:

______

(a) (AB) = {x : x (AB)}

= {x : x A and x B}

__

= {x : x A and x B}

______ __

(AB) = A B

______

(b) (AB) = {x : x (AB)}

= {x : x A or x B}

__

= {x : x A or x B}

__

= {x : x ( A B)}

______ __

(AB) = ( A B)

Worked Out Examples

Example 1: If U = {0, 1, 2, ………, 9}, A = {1, 2, 3, 4} and B = {1, 3, 5, 7}, find AB and

Solution: illustrate in a Venn-diagram.

Here,

U = {0, 1, 2, ………, 9}

A = {1, 2, 3, 4}

B = {1, 3, 5, 7}

AB = {1, 2, 3, 4} {1, 3, 5, 7}= {1, 2, 3, 4, 5, 7}

Represnting the above information in a Venn-diagram,

U

A2 B

15

64 37

89 0

The shaded region denotes AB.

Example 2: If U = {x : x 12, x N}, A = {x : x is a multiple of 4, 1 < x 12} and

B = {x : x is an even number, 2 x < 11}, find AB and illustrate in a

Venn-diagram.

Sets Mathematics - 9 |9

Solution: Here,

U = {x : x 12, x N} = {1, 2, 3, ……,12}

A = {x : x is a multiple of 4, 1 < x 12} = {4, 8, 12}

B = {x : x is an even number, 2 x < 11} = {2, 4, 6, 8, 10}

Now, AB = {4, 8, 12} {2, 4, 6, 8, 10} = {4, 8}

Represnting the above information in a Venn-diagram,

The shaded region denotes AB.

Example 3: If U = {1, 2, 3, 4, 5, 6, 7, 8}, A = {1, 2, 3, 4} and B = {3, 4, 5, 6}, find A – B

Solution: and B – A.

Here,

U = {1, 2, 3, 4, 5, 6, 7, 8}, A = {1, 2, 3, 4}, B = {3, 4, 5, 6}

Now, A – B = {1, 2, 3, 4} - {3, 4, 5, 6}

A - B = {1, 2}

And B – A = {3, 4, 5, 6} - {1, 2, 3, 4}

B - A = {5, 6}.

Example 4: If U = {0, 1, 2, 3, ………., 12}, A = {1, 2, 3, 5, 6, 8, 9}, B = {3, 4, 5, 6, 7, 10}

Solution: and C = {5, 6, 8, 9, 10, 11}, find (AB) – C and illustrate it in a Venn-

diagram.

Here,

U = {0, 1, 2, 3, ………., 12}

A = {1, 2, 3, 5, 6, 8, 9}

B = {3, 4, 5, 6, 7, 10}

and C = {5, 6, 8, 9, 10, 11},

then AB = {1, 2, 3, 5, 6, 8, 9} {3, 4, 5, 6, 7, 10}

= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Now, (AB) – C = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {5, 6, 8, 9, 10, 11}

(AB) – C = {1, 2, 3, 4, 7}

10 | Mathematics - 9 Sets

Represnting the above information in a Venn-diagram,

A BU

1 34

2 8 5 7

9 6

10

11 12

C0

(AB) – C

The shaded region denotes (AB) – C.

Example 5: If U = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 5, 7} and B = {2, 4, 5, 6}, find

the following sets and illustrate them in a Venn-diagram.

(a) A (b) (AB)' (c) (AB) (d) (A – B)

Here,

Solution:

U = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 5, 7}, B = {2, 4, 5, 6}

a. A = U – A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 5, 7} = {0, 4, 6, 8, 9}

It is illustrated in a Venn-diagram as

A'

b. AB = {1, 2, 3, 5, 7} {2, 4, 5, 6}

AB = {1, 2, 3, 4, 5, 6, 7}

Now,

(AB) = U - (AB)

= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 4, 5, 6, 7}

= {0, 8, 9}

It is illustrated in a Venn-diagram as

Sets Mathematics - 9 |11

c. (AB) = {1, 2, 3, 5, 7}{2, 4, 5, 6}

AB = {2, 5}

Now, (AB) = U - (AB) = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 5}

(AB) = {0, 1, 3, 4, 6, 7, 8, 9}

Which is illustrated in a Venn-diagram as

d. A – B = {1, 2, 3, 5, 7} - {2, 4, 5, 6}

A - B = {1, 3, 7}

Now, (A - B) = U – (A – B) = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 3, 7}

(A - B) = {0, 2, 4, 5, 6, 8, 9}

Which is illustrated in a Venn-diagram as

Example 6: If A = {1, 2, 3, 4}, B = {1, 3, 5, 7, 9} and U = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} show

Solution: that (AB) = A B.

Here,

A = {1, 2, 3, 4}, B = {1, 3, 5, 7, 9} and U = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

(AB) = {1, 2, 3, 4} {1, 3, 5, 7, 9}

= {1, 2, 3, 4, 5, 7, 9}

Now, (AB) = U - (AB)

= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} - {1, 2, 3, 4, 5, 7, 9}

(AB)= {0, 6, 8} ... (i)

And A = U – A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} - {1, 2, 3, 4}

A = {0, 5, 6, 7, 8, 9}

B = U – B = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} - {1, 3, 5, 7, 9}

12 | Mathematics - 9 Sets

B = {0, 2, 4, 6, 8}

Now, A B = {0, 5, 6, 7, 8, 9} {0, 2, 4, 6, 8}

A B = {0, 6, 8} ... (ii)

Hence, from (i) and (ii), we get

(AB) = A B Proved.

Exercise 1.1

1. If U = {1, 2, 3, ……….., 12}, A = {2, 3, 5, 7, 11} and B = {1, 2, 3, 4, 5}, list the following sets

and illustrate in a Venn-diagram.

(a) AB (b) AB

(c) A – B (d) A

2. If U = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 3, 5}, B = {1, 4, 7, 8}, and C = {1, 3, 5, 7, 9}, find

(a) A B (b) BC

(c) (C – B)A (d) (AB)C

3. If U = {a, b, c, d, e, f, g, h, i, j, k, l}, A = {a, b, c, d, e}, B = {b, c, d, e, f, g, h, i} and

C = {b, d, e, g, h, i, j}, find and show the following sets in a Venn-diagram.

(a) (ABC) (b) (AB)C

(c) ABC (d) (AB) (AC) (BC)

4. Using set symbols, write down expressions for the shaded portion in the following Venn-diagrams.

(a) U (b) U

A A B

B

(c) U (d) U

A B A B

Sets Mathematics - 9 |13

5. Using the symbols, write down expressions for the shaded portion in the following Venn-diagrams.

(a) U (b) U

A B A B

(c) C (d) C

A U A U

B B

CC

6. Study the given Venn-diagram and list the elements and shade in the Venn-diagram for each of

the following sets.

____ ____ (c) B'A

(a) A – B (b) B – A

_____ _____ (f) U – B'

(d) A B (e) A B

7. Study the given Venn-diagram and list the elements and shade the region of each of the

following sets.

_______ (b) (AB)C

(a) (ABC )

(c) C – (AB) _____

(d) (AB) – C

8. If A = {a, b, c}, B = {b, c, d, e} and C = {c, d, e, f} are subsets of the universal set

U = {a, b, c, d, e, f, g, h}, verify the following.

(a) A(B – C) = (AB) – (AC) (b) A(BC) = (AB) (AC)

_ _ _ ____ ____ _

(c) A ( B – C ) = AB – AC (d) A B = B – (AB)

9. A and B are two subsets of a universal set U such that AB .

____

(a) Draw a Venn-diagram to represent the relation between A and B and shade AB .

__

(b) Re-draw the Venn-diagram and shade A B .

____ __

(c) Write down the relation between (AB ) and ( A B ).

(d) What is this law specially known as?

14 | Mathematics - 9 Sets

10. If Q = {quadrilateral}, T = {Trapezium}, P = {Parallelogram}, U

R = {Rectangle}, S = {Square} and X = { Rhombus}, label the

given sets with appropriate symbols.

11. Given U = {Triangles}, I = {Isosceles triangles},

E = {Equilateral triangles}, R = {Right angled triangles}

(a) Draw a Venn-diagram to show the relationship between these sets.

(b) What are the measures of the angles of a triangle belonging to IR?

1.3 Application of Sets

Cardinality of a set

The number of elements in a set is known as the cardinal number or cardinality of the set. The cardinal

number of a set A is denoted as n(A). Cardinal numbers of the sets AB, AB, A – B, A are denoted

as n(AB), n(AB), n(A – B), n(A) etc. If A = {a, e, i, o, u}, then n(A) = 5.

Cardinality relation of union of two sets

If A and B are two overlapping sets in a universal set U, then n(AB) = n(A) + n(B) - n(AB)

Proof:

If n(A) = a, n(B) = b and n(AB) = c, A U

then number of elements belonging to A only, no(A) = a – c. B

Number of elements belonging to the set B only, no(B) = b – c.

From the Venn-diagram, we get

n(AB) = no(A) + no(B) + n(AB)

=a–c+b–c+c=a+b–c

n(AB) = n(A) + n(B) - n(AB) ... (i) no(A) = a - c no(B) = b - c

If the two sets A and B are disjoint, then n(AB) = 0 n(AB) = c

n(AB) = n(A) + n(B) - n(AB)

= n(A) + n(B) – 0

n(AB) = n(A) + n(B) ... (ii)

... (iii)

In this case n(AB) is maximum.

If AB, then n(AB) = n(A), then

n(AB) = n(A) + n(B) – n(AB)

= n(A) + n(B) – n(A) = n(B)

n(AB) = n(B)

In this case, n(AB) is minimum.

Sets Mathematics - 9 |15

Some other important relations:

1. n(AB) = n(U) - n(AB)

2. no(A) = n(A – B) = n(AB)

also no(A) = n(A) – n(AB)

no(B) = n(B – A) = n(AB)

also no(B) = n(B) – n(AB)

Worked Out Examples

Example 1: If A = {a, b, c, d, e, f} and B = {b, d, f, h, j}, find n(A), n(B), n(AB) and n(AB).

Solution: Here,

A = {a, b, c, d, e, f}

n(A) = 6

B = {b, d, f, h, j}

n(B) = 5

AB = {b, d, f}

n(AB) = 3

And n(AB) = n(A) + n(B) - n(AB)

=6+5–3=8

Example 2: If n(A) = 16, n(AB) = 7 and n(AB) = 22, find n(B).

Solution: Here,

n(A) = 16, n(AB) = 7 and n(AB) = 22, n(B) = ?

We have,

n(AB) = n(A) + n(B) - n(AB)

or, 22 = 16 + n(B) – 7

n(B) = 13.

Example 3: If n(U) = 50, n(A) = 20, n(B) = 26 and n(_A__B_) = 8, find

Solution: (i) n(A – B) (ii) n(AB) (iii) n(AB)

Here,

n(U) = 50, n(A) = 20, n(B) = 26, n(AB) = 8

(i) n(A – B) = n(A) - n(AB) = 20 – 8

n(A – B) = 12

(ii) n(AB) = n(A) + n(B) – n(AB)

= 20 + 26 – 8

n(AB) = 38

____

(iii) n(AB) = n(U) – n(AB)

= 50 – 38 = 12.

16 | Mathematics - 9 Sets

Example 4: In a class of 25 students, 17 like volleyball, 15 like basketball and 10 like

Solution: both the games. Illustrate the above information in a Venn-diagram and

find the number of students who do not like any of the games.

Let V and B be the sets of students who like volleyball and basketball respectively.

Here, total number of students n(U) = 25, n(V) = 17, n(B) = 15 and n(VB) = 10

n(VB) = n(v) + n(B) - n(VB) U

VB

= 17 + 15 – 10 = 22.

____ n() = 15

n(V)

Now, n(VB) = n(U) – n(VB) n(V) = 17 n(V) = 10

= 25 – 22 = 3

Therefore, the number of students who

don’t like both the games is 3.

Example 5: In a class of 50 students, all the students participate in at least one of

Solution: the activities music or dance. If 35 students are taking part in music and

45 in dance, find how many students are taking part in both the

activities. Show the information in a Venn-diagram as well.

Let A and B be the sets of students who participate in music and dance respectively.

Here, total number of students n(U) = n(AB) = 50, n(A) = 35 and n(B) = 45.

We have, U

n(AB) = n(A) + n(B) - n(AB) AB

n(AB) = n(A) + n(B) - n(AB)

n() = 45

= 35 + 45 – 50 n(A) = 0

= 80 – 50 n(A) = 35 n(A) = 30

= 30.

Therefore, 30 students are taking part in both the activities.

Example 6: In a group of 60 people, 18 drink tea but not coffee, 35 drink tea. If 4 of

Solution: them do not drink any of the drinks, find

(i) how many people drink both tea and coffee?

(ii) how many people drink coffee but not tea?

(iii) illustrate the above information in a Venn-diagram.

Let A and B be the sets of people who drink tea and coffee respectively, then set of people

who drink tea but not coffee is (A – B) and those who drink coffee but not tea is (B – A).

____

Here, n(U) = 60, n(A) = 35, n(A – B) = 18, n(AB) = 4.

Now,

(i) n(AB) = n(A) - n(A - B)

= 35 – 18 = 17

17 people drink both tea and coffee.

(ii) n(U) _____

= n(A) + n(B) - n(AB) + n(AB)

60 = 35 + n(B) - 17 + 4

Sets Mathematics - 9 |17

or, n(B) = 60 - 22 = 38

n(B - A) = n(B) - n(AB)

= 38 - 17 = 21

21 people drink coffee but not tea.

(iii) Venn-diagram:

U

AB

n() = 18

n() = 21

n(B) = 4

n(A) = 35 n(A) = 17

Example 7: In a survey of people 70% like football, 50% like volleyball, 35% like both

Solution: the games and 30 people like neither of such games. Find the total

number of people who took part in the survey by drawing venn diagram.

Also find who like only one game.

Let, F and V denotes the number of people who like football and volleyball respectively.

n() = 100% = x (say) n() = x

n(F) = 70 x of x = 70x 70x 50x

100 100 100

n(F) = n(V) =

50x

n(V) = 50% of x = 100

n(FV) = 35% of x = 35x 35x 35x 15x

100 100 100 100

_____ ____

n(FV) = 20 n(FV) = 30

Here, we have, 35x + 35x + 15x + 30 = x

100 100 100

or, 15x = 30

100

x = 200

Then, Total no. of people = 200

Number of people who like only one game = 35x + 15x = 50×200 = 100

100 100 100

Exercise 1.2

1. (a) If A = {2, 4, 6, 8} and B = {7, 8, 9, 10} are two subsets of a universal set

____

U = { x : x 10, x w}, find n(A), n(B), n(AB) and n(AB) .

(b) If A = {a, b, c, d} and B = {c, d, e, f, g}, verify that n(AB) = n(A) + n(B) - n(AB).

(c) If A = { x : x is a letter in the word “difference”} and B = {x : x is a letter in the word

“Complement”}, find n(AB).

18 | Mathematics - 9 Sets

2. (a) If n(A) = 10, n(B) = 8 and n(AB) = 3, find n(AB).

(b) For two sets A and B, if n(A) = 35, n(B) = 25 and n(AB) = 48, find n(AB).

(c) A and B are subsets of a universal set U. If n(U) = 100, n(A) = 55, n(B) = 65 and

____

n(AB) = 30, find (i) n(AB) (ii) n(only A) (iii) n(AB) .

(d) M and N are subsets of a universal set U. If n(U) = 95, no(M) = 35, no(N) = 25 and

____

n(MN) = 15, find n(MN).

U

3. (a) The cardinal numbers of the respective sets are shown in the P Q

given Venn-diagram. Find the cardinal number of the 28

universal set U.

10

25 13

(b) Cardinal numbers of the respective sets are shown in the U

B

Venn-diagram given alongside. Find n(AB). A

no(B) = 5x

n(AB) = 5

no(A) = 3x n(AB) = 14

(c) Study the Venn-diagram and find no (A) and no (B),

where n(U) = 200 is provided.

(d) The cardinalities of the respective sets are given in

____

the Venn-diagram. If n(AB) = 18, find the

cardinality of the universal set U.

4. (a) If AB, n(A) = 20 and n(B) = 45, find the value of n(AB) and n(AB).

____

(b) If AB, n(B) = 40 and n(AB) = 15, find n(U).

____

(c) If UBA, n(U) = 60 and n(B) = 15, find the value of n(AB) .

(d) If n(A) = 10 and n(B) = 15, what may be the minimum and maximum value of

(i) n(AB) and (ii) n(AB)?

(e) If n(A) = 9 and n(B) = 6, find the maximum and minimum value of (i) n(AB)

(ii) n((AB).

Sets Mathematics - 9 |19

5. (a) In a survey of 100 persons of a locality, it was found that 63 could speak Tamang, 59

could speak Newari and 32 could speak both the languages.

(i) How many could speak none of the two languages?

(ii) How many could speak only Tamang language?

(iii) Illustrate the above information in a Venn-diagram.

(b) Out of 90 civil servants, 65 were working in the office, 50 were working in the field,

and 35 were working in both the premises.

(i) How many civil servants were absent?

(ii) How many civil servants were working in the field only?

(iii) Represent the above information in a Venn-diagram.

(c) A survey of 200 students revealed that 100 didn't like to play football, 120 did not like

to play basketball and 40 did not like to play any of the games.

(i) How many students liked to play both the games?

(ii) How many students were interested in these games?

(d) In a survey of a group of people, it was found that 62 people didn't like to drink tea, 85

didn't like coffee, 33 liked both and 36 didn’t like both the beverages.

(i) Illustrate the above information in a Venn-diagram.

(ii) How many people were surveyed?

(iii) How many people liked to drink tea?

6. (a) In a survey of 500 tourists who visited Nepal, it was found that 300 visited Pokhara and

340 visited Lukla and every tourist visited at least one place.

(i) How many tourists visited both places?

(ii) How many of them visited only one place?

(iii) Represent the above information in a Venn-diagram.

(b) Students having computers at their homes own at least one of the popular computer

games Alien Invaders and Race Car Derby, 40 have Alien Invaders, 45 have Race Car

Derby and 25 have both the games.

(i) Illustrate the above information in a Venn-diagram.

(ii) How many students have only one game?

(iii) How many students have computers?

(c) A class of 42 students take part in at least one of the activities sports and music. If 30

take part in sports and 24 in music, find how many students take part in both activities.

Also find how many students take part in sports only and music only. Represent the

result in a Venn-diagram.

(d) A group of students play either cricket or football or both. 28 play cricket, 30 play

football, and 18 play both the games,

(i) Illustrate the above information in a Venn-diagram.

(ii) How many students are there in the group?

20 | Mathematics - 9 Sets

7. (a) In a class of 60 students, 15 students liked Maths only, 20 liked English only and 5 did

not like both of the subjects.

(i) Find the number of students who like both.

(ii) Find the number of students who liked at least one subjects.

(iii) Illustrate above information in a Venn-diagram.

(b) Out of 100 students, 80 passed in science, 71 in Mathematics, 10 failed in both subjects

and 7 did not appear in the examination. Find the number of students who passed in

both subjects by representing the above information in a Venn-diagram.

(c) In a survey of 53 people, 20 like Nepali TV Serial, 25 like the Hindi TV serial and 10

did not like both TV Serials. If 3 people did not use to watch the TV, find the number of

people who like both serials by using a Venn-diagram.

(d) In an examination, 35% of students passed in social studies only, 25% passed in health

only. If 10% failed in both subjects,

(i) what percent of students passed in both subjects?

(ii) what percent of students passed in social studies?

(iii) represent the result in a Venn-diagram.

8. (a) In a survey of some customers, it was found that 65% preferred laptop and 85% preferred

desktop computers. If there were not any customer who did not like both computer,

(i) what percent were there who preferred both the computers?

(ii) illustrate the above information in a Venn-diagram.

(b) In a survey among some people of a group. It was found that 25% of them liked

Literature only, 50% of them liked Music only and 10% of them liked none.

(i) What percent of people liked both?

(ii) What percent of people liked music?

(iii) Illustrate these information in a venn-diagram.

(c) In an examination, 70% of the students passed science, 60% passed mathematics and 12

students passed both subjects. If 10% failed in both the subjcts.

(i) find the percent of students who passed both the subjects.

(ii) find the total number of students.

(iii) represent the above information in a Venn-diagram.

(d) In a survey of community who watch television, 40% like Jire Khurshani, 50% like

'Tito Satya' and one fifth people like none of them.

(i) Draw a Venn-diagram of the above information.

(ii) What percentage of them like both serial?

9. (a) In a survey of bus travelers, 60% liked to take Day bus while 75% liked Night bus for

long tour. If 72 travelers liked to take Night bus only,

(i) how many travelers were surveyed?

(ii) what percent liked both busses?

(iii) illustrate the above information in a Venn-diagram.

(b) In a group of students, 50% liked tea, 70% liked coffee, 10% did not like both and 120

like both. By using a Venn-diagram, find the total number of students.

Sets Mathematics - 9 |21

(c) In a survey of a group of tourists it was found that 70% liked to visit Chitwan, 60%

liked to visit Pokhara, 40% liked to visit both the places and 40 tourists did not like to

visit both places. By using a Venn-diagram, find the total number of tourists, who took

part in the survey.

(d) In a survey of internal tourists, 20% of tourists liked Sauraha only, 50% liked Lumbini

only and 10% tourist did not like both the place. If 200 tourists liked Sauraha, find the

total number of tourists in the survey by drawing a Venn-diagram.

10. (a) In a survey of students of New Horizon School the number of students who liked maths

is two thirds of the number of students who liked Nepali, 10% liked maths only, 30%

liked Nepali only. Find the percent of the students who did not like both the subjects by

using a Venn-diagram.

(b) In a survey of a school, the number of students who liked maths is one third of students

who liked science, 2% liked Maths only, 30% liked science only. Find the percent of

student who did not like both by using a Venn-diagram.

(c) In a survey of 20,000 students of class ten in Chitwan district, 18% of the students take

tuition before S.L.C. examination. Out of them, 40% take tuition in English only, 10%

in maths only and 20% in other subjects, find the number of students who take tuition in

English and maths both subjects.

11. (a) In a survey, it was found that the ratio of the people who like modern songs and folk

songs is 8:9 out of which, 50 people liked both songs, 40 liked folk songs only and 80

liked none of the songs.

(i) Show the above information in a Venn-diagram.

(ii) Find the number of people who participated in the survey.

(b) In a survey of 120 people, the number of people who like to rear. Only cows is three

times the number of people who like to rear only buffalo. If 10 people liked to rear both

and 30 people did not like to rear both then,

(i) find the number of people who liked to rear cows.

(ii) illustrate these information in a Venn-diagram.

(c) Out of 50 people, the number of people who like only Dashain Festival is 10 less than

two third of people who like only Tihar Festival. If 10 people like both Festivals and 20

did not like both festival then,

(i) represent the above information in a Venn-diagram.

(ii) find the number of people who like Dashain.

(d) In a survey, one-fifth students like maths only, 30 students didn’t like maths, If 35

students like science and 25 like none of them,

(i) show the above information in a Venn-diagram.

(ii) how many students like maths?

12. (a) In a survey two third people like to eat 'Dhido' only, 12 people did not like 'Dhido', 20%

people like to eat rice and 8 people did not like to eat them.

(i) Show the above information in a Venn-diagram.

(ii) How many people like to eat both the things?

22 | Mathematics - 9 Sets

(b) From the adjoining Venn-diagram, if n() = 60, find UA B

4x y x

(i) n(AB) ii) n(A – B)

____ 39 24

(iii) n(AB)

(c) In a survey of some people, the ratio of people who like milk but not curd and who like

curd but not milk is 6:5. The ratio of people who like both and dislike both is 7:9. If

50% did not like milk and 44 people like only one of them, find the number of people

that participated in the survey by using a Venn-diagram.

Unit Test

Time: 40 minutes F.M.- 24

1. In a class of 120 students, 95 like account, 80 like biology. If there are none who don’t like

both subjects, find

(a) The number of students who like both subjects.

(b) The number of students who like account only.

(c) Show the above information in a venn- diagram.

2. In a survey of 100 people, it was found that 65 like folk song, 55 like modern song and 35

like folk as well as modern song. Then

(a) Draw a Venn-diagram to show the above information.

(b) How many people did not like both the songs?

3. Out of 100 students in an examination of class 9, 70 passed in Mathematics, 60 passed in

Science and 20 failed in both subjects. Find the number of students who passed in both

subjects by using a Venn-diagram.

4. In a survey of 200 students, 30 liked neither to sing nor to dance, 60 liked to song only and

50 liked to dance only. Then

(a) Show the above information in a Venn-diagram.

(b) Find the number of students who can do both the sing as well as dance.

_____

5. If n(A) = 65, n(B) = 50, n(AB) = 30 and n(AB) = 17, find n(U) and and show the result

in a Venn-diagram.

_____

6. If n(U) = 120, n0(A) = 50, n0(B) = 40, n(AB) = 25, find n(AB) and n(AB).

Sets Mathematics - 9 |23

Answers ____________________________________________________________

Exercise 1.1

1. (a) {1, 2, 3, 4, 5, 7, 11} (b) {2, 3, 5} (c) {7, 11} (d) {1, 4, 6, 8, 9, 10, 12}

(c) {3, 5} (d) {1, 3, 5, 7}

2. (a) {0, 1, 4, 6, 7, 8, 9} (b) {4, 8} (c) {b, e} (d) {b, c, d, e, g, h, i}

3. (a) {k, l} (b) {b, d, e, g, h, i}

4. (a) B – A (b) (AB) (c) (A) (d) (AB) – (AB)

5. (a) A(BC) (b) (¯A¯¯¯B¯¯¯C¯) (c) (AB)(AC)(BC) (d) (AB)(BC)

6. (a) {1, 2, 3, 4, 5, 9, 10} (b) {1, 4, 5, 6, 7, 8, 9, 10} (c) {6, 7, 8} (d) {1, 9, 10}

(e) {1, 2, 3, 6, 7, 8, 9, 10} (f) {2, 3, 4, 5}

7. (a) {k, l} (b) {c, e, g, j} (c) {g, h, i, j} (d) {a, f, k, l}

_____ __ __

9. (c) (AB) = A B (d) De-Morgan’s law

11. (b) 45, 45°, 90°

Exercise 1.2

1. (a) 4, 4, 1, 4 (c) 12 (c) (i) 90 (ii) 25 (iii) 10 (d) 20

2. (a) 15 (b) 12 (c) 58, 78 (d) 72

3. (a) 50 (b) 94 (c) 45 (d) (i) 15, 25 (ii) 0, 10

4. (a) 45, 20 (b) 55

(c) 20, 160 (d) 144, 82

(e) (i) 15, 9 ( ii) 6.0 (b) 10, 15 (c) 12, 18, 12 (d) 40

5. (a) 10, 31 (b) 35, 60 (b) 68 (c) 5

6. (a) 140, 360 (ii) 55

7. (a) (i) 20 (c) (i) 40% (ii) 30 (d) 10%

(b) (i) 15% (ii) 65% (c) 400 (d) 500

(d) (i) 30% (ii) 65% (b) 400 (c) 1080

8. (a) 50% (b) 56% (c) 12 (d) 45

9 (a) (i) 180 (ii) 40% (b) 70 (c) 76

10. (a) 30% (b) (i) 19 (ii) 20 (iii) 16

11. (a) 200

12. (a) 8

n

24 | Mathematics - 9 Sets

2Chapter

Arithmetic

Objectives:

At the end of this chapter, the

students will be able to:

collect and solve the problems

related to discount, profit and

loss, commission, taxation,

dividend and bonus.

calculate the electricity bill,

water bill, telephone bill and taxi

fare.

Teaching Materials:

Bills of grocery with discount, charts

of rule of tax published by government,

electricity bill, water bill, telephone

bill of a household, rate list of taxi

fare published by government.

2.1 Profit and Loss

The terms used in the transaction of goods while buying and selling are:

Cost price (C.P) : The price for which an article is purchased.

Selling price (S.P) : The price for which an article is sold.

If the selling price is more than the cost price, there is profit.

Profit (P) = S.P – C.P

or, S.P = C.P + P

or, C.P = S.P – P

If the selling price is less than the cost price, there is loss.

Loss (L) = C.P – S.P

or, S.P = C.P – L

or, C.P = S.P + L

To distinguish between a profit and profit percentage, let’s consider an example:

Raj bought an article for Rs. 50 and sold for Rs. 75. Then he made a profit of Rs. 25. Similarly, Jay

purchased an article for Rs. 200 and sold for Rs. 225. Then, he also made a profit of Rs. 25. Though,

both of them made same amount of profit but their profit percentage is different. Let's see how,

When C.P is Rs. 50, Raj made a profit of Rs. 25.

When C.P is Re. 1, Raj made a profit of Rs. 25 .

50

When C.P is Rs. 100, Raj made a profit of Rs. 25 × 50

50

2 100

Profit percentage of Raj is 50%.

Similarly,

When C.P is Rs. 200, Jay made a profit of Rs. 25.

When C.P is Re. 1, Jay made a profit of Rs. 25

200

When C.P is Rs. 100, Jay made a profit of Rs. 25 × 100

200

Profit percentage of Jay is 12.5%

Therefore, profit and profit percentage are two different concepts. Similarly loss and loss

percentage are two different things.

Profit Percentage = Profit × 100%

C.P

26 | Mathematics - 9 Arithmetic

If we simplify it, we get

S.P = 100 + P %

C.P 100

Similarly,

Loss percentage = Loss × 100%

C.P

and S.P = 100 - L %

C.P 100

From the illustration above it is understood that profit percentage or loss percentage is calculated on

the cost price.

Things to remember

Overhead expenses like repair, maintenance, transportation and other expenditure are added to the cost price to

get the total cost price. The profit percent or loss percent is calculated on the total cost price.

To calculate the profit or loss when CP and SP of different number of articles are given, we have to calculate the

CP and SP of equal number of articles.

While calculating profit or loss in case of break, leakage, lost on the way, damage etc, SP should be made with

number of articles in good condition.

Worked Out Examples

Example 1: If the cost price of 8 articles is equal to selling price of 6 articles, find the

Solution: profit percent.

Here,

Cost price of 8 articles = Selling price of 6 articles = Rs. x (say)

C.P of 1 article = Rs. x ,

8

And

S.P of 6 articles = Rs. x

S.P of 1 article = Rs. x

6

As, S.P > C.P, there is profit.

Profit (P) = S.P – C.P

= x – x = 4x - 3x = Rs. x

6 8 24 24

x

Then, Profit percent = P × 100 = 24 × 100

CP x

8

Arithmetic Mathematics - 9 |27

= x × 8 ×100 = 3313 %

24 x

3

Hence, the profit percentage is 3331 %.

Example 2: A stationer bought 3000 books at Rs. 125 each. Later, he found that 500

Solution: books were damaged and he sold the remaining books at Rs. 175 each.

Find his gain or loss percent.

Here,

Cost price of 1 book = Rs. 125

C.P of 3000 books = Rs. 125 × 3000 = Rs. 3,75,000

No. of books damaged = 500.

No. of good books = 3000 – 500 = 2500

Selling price of 1 book = Rs. 175

S.P of 2500 books = Rs. 175 × 2500 = Rs. 4,37,500

As S.P > C.P, there is profit.

Profit (P) = S.P – C.P

= 4,37,500 – 3,75,000= Rs. 62,500

Then,

Profit percent = P × 100 = 62500 × 100 = 1632

C.P 375000

Hence, the profit percentage is 16. 2 %

3

Example 3: Avay sold a mobile set for Rs. 32,900 and made 6% loss. At what price

Solution:

should he sell the mobile set to make 6% profit?

Here,

Case I Case II

Selling price (S.P) = Rs. 32,900 Cost price (CP) = Rs. 35,000

Loss percent = 6 Profit percent (P%) = 6

We have, Selling price (SP) =?

S.P = 100 - L % We have,

C.P 100

S.P = 100 + P %

C.P 100

or, 32900 = 100 - 6

CP 100

or, SP = 100 + 6

35000 100

or, 32900 = C.P

0.94 S.P 106

or, 35000 = 100

or, C.P = Rs. 35,000

or, S.P = Rs. 37,100

Hence, Avay should sell the mobile set for Rs. 37,100 to have 6% profit.

28 | Mathematics - 9 Arithmetic

Example 4: Yash bought two music systems for Rs. 52,000. He sold them making 6%

Solution: profit on one and 6% loss on the other. If their selling price is same, find

his total gain or loss percent.

Here,

For the 1st music system For the 2nd music system

Let, Cost price (CP) = Rs. x Cost price (CP) = Rs. (52,000 - x)

Profit percent (P%) = 6. Loss percent (L%) = 6

We know, We know,

SP = 100 + P% S.P = 100 - L%

CP 100 C.P 100

or, SP = 100 + 6 or, SP = 100 - 6

x 100 52000 - x 100

or, SP = 106x or, S.P = 94(52,000 - x)

100 100

But, their selling price is same.

106x = 94(52,000 - x)

100 100

or, 106x = 4888000 - 94x

or, 200x = 48,88,000

or, x = Rs. 24440

S.P of the music system = 106x = 106 × 24440 = Rs. 25906.40

100 100

Total selling price (SP) = Rs. 51,812.80

Total cost price (CP) = Rs. 52,000

As C.P > S.P, there is loss.

Loss = C.P - S.P

= Rs. 52,000 – Rs. 51,812.80 = Rs. 187.20

Loss percent (L%) = L × 100 = 187.20 × 100 = 0.36%

C.P 52000

Hence, the loss percentage is 0.36%.

Example 5: A sold an article to B at a profit of 10%. B sold the same article to C at a

Solution: profit of 15%. If C paid Rs. 3162.50 to B, how much did A pay for it?

Here,

For B (According to question)

Profit percent (P%) = 15

Selling price of B is the cost price of C.

S.P = Rs. 3162.50.

Arithmetic Mathematics - 9 |29

We know

S.P = 100 + P %

C.P 100

or, 3162.50 = 115

C.P 100

or, C.P = 3162.50

1.15

or, C.P = Rs. 2750.

Hence, A paid Rs. 2500 for the article.

For A (According to question)

Selling price of A is the cost Price of B

S.P = Rs. 2750

Profit percent (P%) = 10

Now,

S.P = 100 + P %

C.P 100

or, 2750 = 100 + 10

C.P 100

or, 2750 = C.P

1.1

C.P = Rs. 2500.

Example 6: An article when sold at a profit of 15% yields Rs. 380 more than when

Solution: sold at a loss of 1623 %. What was the cost price of the article?

Here,

Case I Case II

Let, Cost price (CP) = Rs. x Loss percent (L%) = 1632 = 50

3

Profit percent (P%) = 15

C.P = Rs. x

We know,

We know,

S.P = 100 + P %

C.P 100 S.P = 100 - L %

C.P 100

S.P 100 + 15

or, x = 100 100 - 530x

or, S.P = 115x or, S.P = 100

100

or, S.P = 250x

300

But, the difference in the S.P is Rs.380.

115x – 250x = 380

100 300

30 | Mathematics - 9 Arithmetic

or, 345x - 250x = 380

300

or, 95x = 380 × 300

or, x = Rs. 1200

Hence, the cost price of the article is Rs. 1200.

Example 7: Sunny bought 100 pens at Rs. 75 each. He sold first 30 pens at Rs. 80

Solution: each, and next 50 pens at Rs. 90 each. At what price should he sell the

remaining pens so as to make 20% profit on the total outlay?

Here, Cost price (CP) of 1 pen = Rs. 75

C.P of 100 pens = Rs. 75 × 100 = Rs. 7500

Selling price (SP) of 1 pen = Rs. 80

S.P of first 30 pens = Rs. 80 × 30 = Rs. 2400

S.P of 1 pen = Rs. 90

S.P of next 50 pens = Rs. 90 × 50 = Rs. 4500

Number of remaining pens = 100 – (50 + 30) = 20

Let, S.P of 1 pen = Rs. x

Then S.P of remaining 20 pens = Rs. 20x

Total S.P = Rs. (2400 + 4500 + 20x) = Rs. (6900 + 20x)

P % = 20

We have,

S.P = 100 + P %

C.P 100

or, 6900 + 20x = 100 + 20

7500 100

or, 6900 + 20x = 120 × 75

or, 20x = 2100

or, x = 105

Hence, the remaining 20 pens should be sold at the rate of Rs. 105 per pen so as to

have a profit of 20% in total outlay.

Example 8: A crooked shopkeeper uses false balance. He cheats 20% while

Solution: purchasing goods and 20% while selling the goods. What is his profit

percentage by cheating?

Here, Let, the shopkeeper wants to buy 100 kg goods using a false balance. At the

time of purchasing he cheats 20% means he receives 120 kg goods.

Again, he uses false balance and cheats 20% while selling the goods. So his profit

= 20% of 120kg = 20 × 120 kg= 24 kg

100

His total profit by cheating = (20 + 24) = 44 kg

i.e. Profit percentage = 14040×100 % = 44%

Arithmetic Mathematics - 9 |31

Exercise 2.1

1. (a) Jay bought an item for Rs. 50 and sold making a profit of 10%. Find out the selling

price of the item?

(b) A retailer sold a shirt for Rs. 1000 and he made a profit of 25%. Find the purchased

price of the shirt.

(c) Raj bought a guitar for Rs. 1500. After using for a few days, he sold it and had a loss of

20%. Find the price for which the guitar was sold.

(d) A sold an article for Rs. 810 so that he could have a loss of 10%. Find his cost price.

2. (a) Chunu bought a sound system for Rs. 25,000. He sold it for Rs. 30,000. Find his profit

percentage.

(b) Amar bought a motor-cycle for Rs. 1,50,000. After using for a few months, he sold it

for Rs. 1,25,000. Find his loss percentage.

(c) If the C.P of 10 articles is equal to S.P of 15 articles, find the loss percent.

(d) If the C.P of 12 pens is equal to S.P of 8 pens, find the gain percent.

3. (a) A shopkeeper bought 2000 eggs at Rs. 5 each. On the way, 400 eggs were broken and

he sold the remaining eggs at Rs. 8 each. Find his gain or loss percent.

(b) A fruit seller bought 1500 apples at Rs. 4 per apple. Later on he found that 300 apples

were bad. He sold the remaining apples and gained 20%. Find the price at which each

apple was sold.

(c) Jay sold a T-shirt for Rs. 660 and made 10% profit. At what price should he sell to have

15% profit?

(d) Bobby sold a fan for Rs. 950 and had 5% loss. At what price should he sell the fan to

have 6 % profit?

4 (a) By selling a watch in Rs 150 and 200 loss and profit are happened respectively. If the

loss and the profit are equal, find it's cost price.

(b) By selling an article in Rs 16000 and Rs 19000 some loss and some profit are happened

respectively. If loss is one – fourth of profit, find its cost.

(c) An article is sold for Rs 260 at a gain. Had it been sold for Rs 200 there would have a

loss equal to 50% of the original gain, find the cost price of the article.

5. (a) By selling 75 apples a seller gains the selling price of 15 apples , find his gain

percentage.

(b) Rabin bought a second hand bike for Rs 50,000. He spent Rs 5000 on its repair. If he

sold it for Rs 60000, find his gain or loss percentage.

(c) An article is bought and sold at a profit one – third of selling price. Find the profit

percentage.

(d) A man sells an article at Rs 600 and makes a profit of 20% of selling price. Calculate

his actual profit percentage.

6. (a) A man bought two shirts for Rs. 1200 each. On selling the first shirt, he had 10% profit

and 5% loss on the second shirt. Find his gain or loss percent on whole.

32 | Mathematics - 9 Arithmetic

(b) A man bought a pair of shoes for Rs. 1500. He sold it to Yash at a profit of 15%. Yash

sold it to Karan at a loss of 10%. How much did Karan pay for it?

(c) Jay bought two watches for Rs. 10,000. He sold them to have 5% profit on one and 5%

loss on the other. If their selling price was the same, find his total gain or loss percent.

(d) A sold an article to B at 20% profit. B sold the same article to C at 25% profit. Again, C

sold the same article to D at 10% loss. If D paid Rs. 4500 to C, at what price did A

purchase the article?

7. (a) A shopkeeper sells a pant at 10% profit. If he would have sold it with Rs. 150 more, the

profit would have been 15%. Find the purchased price of the pant.

(b) An article when sold at 10% profit yields Rs. 200 more than when sold at 10% loss.

What was the cost price of the article?

(c) A man bought some oranges at the rate of 20 for Rs. 100. If he sold all of them at a

profit of 20%, then how many oranges did he sell for Rs. 30?

(d) Yash bought a certain number of apples at Rs. 100 per 20 apples and equal number at

Rs. 180 per 30 apples. He mixed them and sold them at Rs. 175 per 25 apples. Find his

gain or loss percent.

8. (a) A shopkeeper bought 150 books and sold 100 of them at a profit of 30% and the rest at

10% loss. If he had sold all the books at once at 25% profit, he would have received

Rs. 300 more. Find the cost price of each book.

(b) A man bought 120 pens for Rs. 7200. He sold 40 pens at Rs. 70 each and 60 pens at

Rs. 80 each. At what price should he sell the remaining pens so as to make a profit of

20% in total investment?

(c) A dealer bought 80 purses at Rs. 70 each. He sold 20 purses at Rs. 80 each and next 40

purses at Rs. 75 each. At what price should he sell the remaining purses so as to have

30% profit on the total outlay?

(d) Manju bought a bale of cloth at Rs. 120 per meter. She sold 40m cloth at Rs.105/m and

the remaining at Rs. 150/m and made 20% profit. How many meters of cloth had she

bought in total?

2.2 Commission

In many business, a third party plays a role between a buyer and seller. The third party is known as

agent. The agent is paid certain sum for his service. The sum paid to the agent is known as

commission. An insurance company pays 4% commission to the agent. The T.V manufacture pays

15% commission to the dealer. It means commission is calculated as a certain percentage of selling

price.

Commission amount = Commission % of selling price

Net S.P. = S.P. – Commission % of S.P.

Arithmetic Mathematics - 9 |33

Worked Out Examples

Example 1: The monthly salary of a salesman is Rs 5000. The departmental store

Solution: gives 2% commission on the total sales. If the total sales of a certain

month is Rs 1,50,000, what is the monthly income of that month of the

salesman?

Commission rate = 2%

Total sales = Rs 1,50,000

Commission = 2% of Rs. 1,50,0000

= 2 × 1,50,000 = Rs 3000

100

Monthly salary = Rs 5000

Total income of the month = Salary + Commission

= 5000 + 3000 = Rs 8000

Example 2: A land broker sold a plot of land for Rs 95,00,000. What is the

Solution: commission received by him from the land owner at the rate of 6%. Also,

find the net sum received by the land owner.

Selling price of the land = Rs 95,00,000

Commission rate = 6%

Commission = 6% of 95,00,000

= Rs 5,70,000

Net sum received by the land owner = 95,00,000 – 5,70,000

= Rs 89,30,000.

Example 3: A motor cycle dealer sold a motor cycle for Rs 1,50,000. The dealer gave

Solution: Rs 7500 as commission to the salesman. Find the rate of commission.

Selling price = Rs 1,50,000

Commission = Rs 7500

Commission rate = x % (say)

We have,

Commission = x % of 1,50,000

or, 7500 = x × 1,50,000

100

x = 5.

Hence, the commission rate is 5%.

Example 4: A landowner sold a plot of land for Rs 90,00,000. He had an agreement

with land broker to pay the commission as 1% for first Rs 30,00,000.

0.5% for the next Rs 45,00,000 and 0.25% for the remaining sum. What is

the commission paid by the land owner to the broker and also the sum

received by the land-owner?

34 | Mathematics - 9 Arithmetic

Solution: Selling price = Rs 90,00,000

= Rs 30,00,000 + Rs 45,00,000 + Rs 15,00,000

Commission = 1% of 30,00,000 + 0.5% of 45,00,000 + 0.25% of 15,00,000

= 30,000 + 22,500 + 3750

= Rs 56,250

Net sum received by the land owner = Rs. 90,00,000 – Rs. 56,250

= Rs 89,43,750

Hence, the commission received by the land broker is Rs 56250 and the net sum

received by the land owner is Rs 89,43,750.

Exercise 2.2

1. (a) A real estate agent sold a piece of land for Rs 32,00,000. What was the commission

received by him at the rate of 10%?

(b) An insurance agent paid a yearly premium of Rs 14,000 of his customer. What

commission did he receive at the rate of 5%?

(c) A company salesman’s monthly salary is Rs 12,000. He gets 5% commission on the

total sales. If the total sales of a certain month is Rs 4,50,000, find his income of that

month.

(d) A girl working in a departmental store has monthly salary Rs 6000. She gets 2%

commission on total sales. Find her income of that month if the total sales is Rs

2,50,000.

2. (a) An agent receives Rs. 4,50,000 as commission on selling a Japanese car. If the

commission rate is 10%, at what price was the car sold?

(b) A land broker received Rs. 5,12,000 as a commission. At what price did he sell the land

if the commission rate was 8%?

(c) A mechanics was able to sell a customer’s motor-cycle to a motor cycle exchange

company for Rs. 45,000. If the mechanics received Rs. 2700 as commission from

exchange company, find the rate of commission.

(d) An agent sold a Toyota car for Rs 70,00,000. He got Rs. 3,50,000 as commission. Find

the commission rate.

3. (a) Yash’s house was sold for Rs. 65,00,000 by an agent. The agent was paid 5%

commission by the seller Yash. What is the net sum received by Yash?

(b) A land owner’s land was sold by a broker for Rs. 80,00,000. The broker received 9%

commission from the land owner. What is the net sum received by land-owner?

(c) A sales man’s monthly salary is Rs. 8000. He gets 8% commission on the total sales. If

his income of a certain month is Rs. 17,000, what is the total sales of that month?

(d) The monthly salary of a boy working in a company is Rs 6000. He gets extra 6%

commission on the total sales. If he gets Rs 12,000 on a certain month, what is the sales

of that month?

Arithmetic Mathematics - 9 |35

4. (a) A man sold his plot of land for Rs. 50,00,000 with the help of a broker. They had an

agreement of paying commission of 2% on first Rs. 15,00,000, 1% on next

Rs. 15,00,000 and 0.5% on rest of the money. What is the commission received by the

broker?

(b) The company’s rule for commission is as follows.

Monthly sales Commission rate

Re 1 – Rs 10,00,000 1%

Rs 10,00,000 – Rs 30,00,000 1.5%

Above Rs 30,00,000 2.5%

Calculate the commission received by the sales agent from the sales of Rs 50,00,000.

2.3 Discount

Discount

The price on which a seller wishes to sell an article is known as marked price or tag price or quoted

price or list price or labelled price.

The deduction or concession given in the marked price is known as discount.

Discount (D) = Discount percentage of marked price and discount = marked price – selling price

Therefore, discount is calculated on the marked price.

Rate of discount = Discount amount × 100%

Marked price

But, if there is no discount, then selling price = marked price. So, if there is discount,

S.P = MP – D.

Worked Out Examples

Example 1: An article is marked to sell at Rs. 1300; if 10% discount is allowed, find

Solution: the discount amount and the selling price.

Here,

Marked price (M.P) = Rs. 1300

Discount rate = 10%

Discount amount (D) =?

Selling price (S.P) =?

We have,

Discount amount = 10% of M.P = 10 ×1300 = Rs. 130

100

Selling price, (S.P) = M.P – D

= 1300 - 130

= Rs. 1170.

36 | Mathematics - 9 Arithmetic

Example 2: From the given condition, find the discount amount and selling Price.

Discount 10% Rs. 25,000/-

Solution: Here,

Marked price of the mobile set (M.P.) = Rs. 25,000

Discount rate = 10%, Discount amount =? and Selling Price =?

Now,

Discount amount = 10% of M.P.

= 10 × 25000 = Rs. 2500

100

Again,

Selling Price (S.P.) = M.P. – Discount

= 25000 – 2500 = Rs. 22,500

Example 3: A motor-bike is marked to sell for Rs. 1,80,000. The buyer got 10%

Solution: discount on the occasion of Tihar and later on 5% further discount on

cash payment. Find the price of the motor-bike purchased.

Here,

Marked price (M.P) = Rs. 1,80,000

S.P after 10% discount = MP – 10% of MP

= 1,80,000 – 10 ×1,80,000

100

= 1,80,000 – 18000 = Rs. 1,62,000

The customer got 5% further discount on cash payment. This discount is given on S.P

after 10% discount.

Final S.P = 1,62,000 - 5 ×1,62,000

100

= 1,62,000 – 8100

= Rs. 1,53,900

Hence, to purchase the motor-bike, the buyer has to pay Rs. 1,53,900.

Example 4: If a shopkeeper allows a discount of 20% in an article, he loses Rs. 600.

Solution: But if he allows a discount of 5%, he gains Rs. 1575. Find the marked

price and the cost price of the article.

Here,

Case I

Let, Marked price = Rs. x

Discount = 20%, Loss = Rs. 600

Selling price = M.P – D

= x – 20% of x = Rs. 0.80x

Now, L = C.P – S.P

Arithmetic Mathematics - 9 |37

or, 600 = C.P – 0.80x

or, C.P = Rs. (600 + 0.8x).

C.P = (600 + 0.8x)

= 600 + 0.8 × 14500

= Rs. 12,200

Case - II

Marked price = Rs. x

Cost price = Rs. (600 + 0.8x)

Discount = 5 %

Profit = Rs. 1575

Now, S.P = MP - D

= x – 5% of x = Rs. 0.95x

Again, P = S.P - C.P

or, 1575 = 0.95x – 600 – 0.8x

or, 2175 = 0.15x

or, x = 14,500

MP = Rs. 14,500

Example 5: Anish sold a computer at the marked price, making a profit of 40%. If

Solution: 10% discount was allowed the profit would have been Rs. 6500. Find the

cost price of the computer.

Here,

Case – I Case– II

As he sold the computer on marked D = 10%

price, there is no discount.

P = Rs. 6500

Selling price = marked price MP = Rs. 7x

5

Profit percent = 40

Let, cost price = Rs. x CP = Rs. x

We have 7x 2

5 10 7x

S.P 100 + P % SP = MP – D = - 100 × 5

C.P 100

= 50

or, MP = 100 + 40 = 70x -7x

x 100 50

MP 140 7 = Rs. 63x

x 100 5 50

or, =

or, MP = Rs. 7x

5

Now,

Profit = S.P – C.P

38 | Mathematics - 9 Arithmetic

or, 6500 = 63x -x

50

or, 500 13x

50

6500 =

or, C.P = Rs. 25000.

Hence, the cost price of the computer is Rs. 25,000.

Exercise 2.3

1. (a) An article is marked to sell at Rs. 1000. If 20% discount is allowed, find the discount

amount and the selling price.

(b) A toy bought for Rs. 600 is marked to sell at Rs. 900. If 15% discount is allowed, what

is the profit?

(c) A pant bought for Rs. 1000 is marked to sell at Rs. 1500. Because of small damage,

40% discount was given. Find the loss percentage.

2. (a) In an article, 20% discount is given. If the customer paid Rs. 640 for the article, find the

marked price.

(b) An electric iron labelled Rs. 1200 was sold for Rs. 900 giving a certain discount, find

the discount rate.

3. (a) The marked price of a pressure cooker is Rs. 1800. Two successive discounts 10% and

5% are given. Find its selling price.

(b) A sound system is marked to sell at Rs. 28000. Two successive discounts 12% and 5%

are allowed. Find its selling price.

4. (a) If a shopkeeper allows a discount of 25%, he loses Rs.500. But if he allows a discount

of 5%, he gains Rs. 1500. Find the marked price and the cost price.

(b) A man sells a D.V.D player at a discount of 8% and makes a profit of Rs. 780. If he

doesn't allow a discount he would gain a profit of Rs. 1500. Find the marked price and

cost price of the D.V.D player.

(c) Raj marked the price of a mobile set to have a gain of Rs. 500. If he allows a discount of

8%, he will have Rs. 140 loss. Find the marked price of the mobile set.

5. (a) If an i-pod is sold at the marked price, there is a profit of 16%. If 15% discount is

allowed, there is a loss of Rs. 350. Find the cost price of the i-pod.

(b) Yash bought an item for Rs. 8000 after a discount of 20% on marked price. If he sells

the item at the marked price, find his profit percent.

(c) Mana paid Rs. 6800 to purchase an article after a discount of 15% on the marked price.

If he sells the article taking Rs. 500 more than the marked price, what is the profit

percent?

6. (a) A shopkeeper bought a shirt for Rs. 2500. He fixed the price of the shirt to make a profit

of 20% after allowing a discount of 20%. Calculate the marked price.

Arithmetic Mathematics - 9 |39

(b) A suit piece after allowing a discount of 10% was sold at 20% profit. Had it been sold

after allowing 15% discount, there would have been Rs. 800 profit. Find the cost price

of the suit piece.

(c) An article is sold at Rs. 2080 with 20% discount. Find the cost price where marked

price is 30% above the cost price.

7. (a) A trader fixes his goods at 20% above the cost price. He allows 15% discount. If the

marked price is Rs. 14400, find his profit percentage or loss percentage.

(b) A T.V set is marked to sell at a profit of 10%. If the shopkeeper sells it at a discount of

Rs. 500, then there is a profit of 632 %, find the cost price.

(c) Price of an article is marked 40% above the cost price. If it is sold giving 20% discount,

what will be the profit percentage?

2.4 Taxation

The sum of money paid by people or by organization, company etc. to the government for public

purposes is known as tax. There are many types of taxes, eg. income tax, excise tax, property tax etc.

We are going to deal with income tax.

Income tax

If the income of a person exceeds a specified amount, a tax is imposed on the income above the

specified amount. The specified amount on which tax is not imposed is known as tax allowance. Also,

provident fund, citizen investment fund and insurance premium are tax free amount. The income

above the tax allowance is known as taxable income.

Taxable income = Total income – Tax allowance.

Also, Income tax = Tax rate × Taxable income.

And, Net income = Annual income – Income tax.

Worked Out Examples

Example 1: The annual income of a person’s is Rs. 2,00,000. If his tax allowance is

Solution: Rs. 1,40,000, what is the tax that he should pay per year at the rate of 15%?

Example 2: Here,

Solution: Annual income = Rs. 2,00,000

Tax allowance = Rs. 1,40,000

Taxable income = Rs. 2,00,000 – Rs. 1,40,000 = Rs. 60,000

Income tax = 15% of Rs. 60,000 = Rs. 9000

Hence, his income tax is Rs. 9000 per year.

If Rs. 17,000 is left after paying an income tax at the rate of 15%, what is

the income?

Here,

Tax rate = 15%

Income left after paying the tax = Rs. 17,000

40 | Mathematics - 9 Arithmetic

Let the income be Rs. x

If the tax allowance is not given then the income is considered as the taxable income.

x - 15% of x = Rs. 17000

or, 85x = Rs. 17,00,000

x = Rs. 20,000

Hence, the income is Rs. 20,000.

Example 3: The monthly income of a serviceman is Rs. 25,000. His tax allowance is

Solution: Rs. 75,000. How much does he pay as income tax per month, if the tax

rate of the 1st two lakh taxable income is 10% and 15% for the remaining

taxable income?

Here,

Monthly salary = Rs. 25,000

Annual income = Rs. 12 × 25,000 = Rs. 3,00,000.

Tax allowance = Rs. 75,000

Taxable income = Rs. 300,000 - Rs. 75,000

= Rs. 2,25,000 = Rs. 2,00,000 + Rs. 25,000

Tax per annum = 10% of 2,00,000 + 15% of 25,000

= 20000 + 3750 = Rs. 23750.

Tax per month = Rs. 23750 = Rs. 1979.17

12

Example 4: The monthly income of Rajani is Rs. 31,000. Her tax allowance is

Solution: Rs. 82,000. After paying income tax, if Rs. 28,100 is her net income per

month, what is the tax rate?

Here,

Monthly income = Rs. 31,000

Annual income = Rs. 12 × 31,000 = Rs. 3,72,000

Tax allowance = Rs. 82,000

Taxable income = Annual income – Tax allowance

= Rs. 3,72,000 – 82,000 = Rs. 2,90,000.

Net income per month = Rs. 28,100

Net income per year = Rs. 12 × 28,100 = Rs. 3,37,200.

Income tax = Annual income – Annual net income

= 3,72,000 – 3,37,200 = Rs.34,800

Let the tax rate be x %

x % of taxable income = Rs. 34,800

or, x × 2,90,000 = Rs. 34,800

100

x = 12

Hence, the tax rate is 12%

Arithmetic Mathematics - 9 |41

Exercise 2.4

1. (a) Raj’s annual income is Rs. 2,25,000. The tax allowance for him is Rs. 1,60,000. If the

tax rate is 14%, how much does he have to pay as income tax per year?

(b) Ajay’s annual income is Rs. 2,40,000. The tax allowance is Rs. 1,60,000. If the tax rate

is 15%, what is his income tax per month? Also, find his net-income.

(c) Yash’s annual income is Rs. 3,20,000.. His provident fund is Rs. 36,000, citizen

investment fund is Rs. 60,000 and insurance premium is Rs. 14,000, which are tax free

amounts. If the tax rate for the first taxable two lakh is 1% and for the taxable income

above Rs. 2,00,000 is 15%, find the income tax per year.

(d) Kabita’s monthly salary is Rs. 25000, her tax allowance is Rs. 50,000. The tax rate for

the first taxable income Rs. 2,00,000 is 1% and 15% for the taxable income above

Rs. 2,00,000. Find income tax she should pay per month and also her net income.

2. (a) If Rs. 24,000 is left after paying an income tax at the rate of 20%, what is the income?

(b) After paying an income tax at the rate of 25%, income left is Rs. 30,000. Find the

income.

(c) Raj’s annual income is Rs. 4,00,000 in which Rs. 1,00,000 is tax allowance. If he pays,

Rs. 45,000 as income tax per year, what is the tax rate?

(d) Yash’s monthly income is Rs. 30,000. His tax allowance is Rs. 60,000. If he pays

Rs. 3,000 as income tax per month, what is the tax rate?

3. (a) Kabita pays income tax Rs. 32,500 per year at the rate of 13% p.a. If her tax allowance

is Rs. 86,000, find her annual income.

(b) Amar pays income tax Rs. 3125 per month. His tax allowance is Rs. 62,000. If the

income tax rate is 15%, find his monthly income.

(c) The yearly income of Arjun is Rs. 3,24,000 and his tax allowance is Rs. 74,000. After

paying income tax, if his net income per year is Rs. 2,86,500, find the income tax rate.

(d) The monthly income of Prem is Rs. 32,000. His tax allowance is Rs. 84,000. After

paying the income tax if Rs. 29,500 is his net income per month, what is the tax rate?

4. (a) Anmol’s monthly salary is Rs. 15,000. He deposits 10% to provident fund and 15% to C.I.F.

which are tax free amounts. What is the yearly tax on the remaining sum at the rate of 1%?

(b) Arya’s monthly income is Rs. 20,000. She pays a premium of Rs. 20,000 to an

insurance company and has medical allowance Rs. 5000 which are tax free. What

income tax, should she pay p.a. if the tax rate for the first two lakhs taxable income is

1% and 15% for the remaining sum?

42 | Mathematics - 9 Arithmetic

2.5 Bonus

When a business company goes in profit, it distributes certain percentage of the annual profit to its

employees as an incentive. This is other than their salary. This incentive amount is known as Bonus.

Bonus amount = bonus % of yearly net profit

If the number of employees is N, bonus is equally distributed then bonus received by an employee

= Bonus amount

N

Worked Out Examples

Example 1: A company makes a profit of Rs 2,00,00,000 in a certain fiscal year. The

Solution: management decided to distribute 6% of profit as bonus equally among

20 employees. Find the bonus amount that each employee will receive.

Here,

Number of employees = 20

Profit = Rs 2,00,00,000

Bonus = 6% of profit

= 6 × 2,00,00,000 = Rs 12,00,000

100

Bonus amount that each employee receive = (12,00,000) = Rs 60,000

20

Hence, each employee will receive Rs 60,000 as bonus.

Example 2: A company made a profit of Rs. 85,00,00,000. The bonus is distributed

Solution: equally among 170 employees and each received Rs. 4,00,000. Find the

bonus rate.

Here,

Profit = Rs. 85,00,00,000

Number of employees = 170

Bonus received by each employee = Rs. 4,00,000

Thus,

Total bonus distributed = 170 × Rs. 4,00,000 = Rs. 6,80,00,000

Now,

Bonus = x % of 85,00,00,000

or, 6,80,00,000 = x × 85,00,00,000

100

x = 8%

Hence, the bonus rate is 8%.

Arithmetic Mathematics - 9 |43