9 Green C Math Class

Approved by Government of Nepal, Ministry of Education, Curriculum
Development Centre, Sanothimi, Bhaktapur as an additional material

Green

Mathematics

9

Editors/Authors
Omkar Nath Pant
Medani Prasad Kadel
Raj Kumar Mathema
Udgam Lama
Arjun Pd. Ghimire
Roshan Ngakhusi
Sabindra Dangol
Binod Kumar Shah
Gobinda Raj Khaniya

Lalitpur, Nepal, Tel: 977-1-5529899
E-mail: [email protected]
www.greenbooks.com.np

Green

Mathematics 9

Publisher
Green Books

Copyright
Authors (2074 BS)

Edition
First : B.S. 2074 (2017 AD)
Reprint : B.S. 2075 (2018 AD)

Layout
The Focus Computer
[email protected]

Printed in Nepal

Preface

Green Mathematics is a series of ten books for teaching mathematics
from the primary to the secondary level (Grade one to ten). The
books of this series have been designed according to the mathematics
curriculum prescribed by the Curriculum Development Centre (CDC)
of Nepal and strictly follow its syllabus, guidelines and specification
grid patterns.

Each book contains a wide range of activities, additional information,
explanations, solved problems, well-graded questions, teaching tips
and assessment sheets; and adopts a constructive approach to make
this series easy to understand and fun to follow for both students and
teachers.

While materializing and grading the series, a rigorous effort has been
taken to maintain a good balance between pedagogical concerns and
those of a theoretical and abstract nature for teaching mathematics.

Here are some salient features of the series:

• The series has a correct balance between concepts and practice. It
assists students in understanding mathematics concepts and helps
them make connections and understand ideas and theories.

• The series follows a correct balance of teaching methods and
approaches - from step-by-step approach to trail-and-error
method.

• The series aims to develop problem solving skills in children along
with logical and lateral thinking rather than mere memorization
of facts, rules and procedures.

• The series aims to foster a positive attitude in students and
encourage them to appreciate the applicability of mathematics in
day-to-day life.

We are thankful to the whole team of Green Books for helping us in
materializing and making this series classroom friendly in all respects
for students and teachers.

We are looking forward to your response to the series; and any
constructive suggestions for the betterment of this series will be
gratefully acknowledged.

Authors

Content

Unit 1 : Sets 1-17

1.1 Revision/Introduction 18-35
36-50
1.2 Set relations 51-69

1.3 Set operations 70-99

1.4 Cardinality of sets 100-111
Unit 2 : Profit and Loss 112-122
123-143
2.1 Review 144-152
153-162
2.2 Problems related to profit and loss 166-186

2.3 Discount
Unit 3 : Commission and Taxation

3.1 Commission

3.2 Bonus (dividend)
Unit 4 : Household Arithmatics

4.1 Domestic and Day to Day Life Activities

4.2 Telephone bill

4.3 Water Bill

4.4 Taxi meter
Unit 5 : Mensuration

5.1 Review

5.2 Area of pathways

5.3 Area, quantities and cost

5.4 Area of four walls, floor and ceiling

5.5 Area and volume of solids

5.6 Estimation of number of bricks and cost required for building wall
Unit 6 : Algebraic Expressions
6.1 Factorization
6.2 Factorization of a4 + a2b2 + b4
6.3 Introduction / Revision
Unit 7 : Laws of Indices

7.1 Indices

7.2 Exponential equation
Unit 8 : Ratio and Proportion

8.1 Ratio and proportion

8.2 Proportional (in proportion)

8.3 Properties of proportion
Unit 9 : Simultaneous, Linear Equations

9.1 Equation

9.2 Solving equation by graphical method

Unit 10 : Quadric Equation

10.1 Quadratic equation

10.2 Completing square method

10.3 Solving quadratic equation by using formula

Unit 11 : Triangle

11.1 Triangle

11.2 Theorems

Unit 12 : Parallelogram 186-210

12.1 Quadrilateral 211-221
222-232
12.2 Trapezium 233-244
245-268
12.3 Mid-point Theorems
269-301
12.4 Pythagoras Theorems
Unit 13 : Constructions 302-310

13.1 Construction of triangle 311-326
327-343
13.2 Construction of quadrilateral 344-346
Unit 14 : Similarity

14.1 Introduction

14.2 Similar Triangles

14.3 Similar Polygons
Unit 15 : Circle

15.1 Introduction

Unit 16 : Trigonometry

16.1 Introduction

16.2 Relation between trigonometric rations

16.3 Trigonometrical Ratios of 0°

16.4 Problems relating to 0°, 30°, 45°, 60° and 90° of sine, cosine and tangent.
Unit 17 : Statistics

17.1 Revision

17.2 Frequency and frequency distribution table

17.3 Grouped frequency distribution table

17.4 Frequency table (with inclusive method)

17.5 Cumulative frequency distribution

17.6 Measure of central tendancy

17.7 Median

17.8 Mode

17.9 Range

17.10 Statistical data in graph
Unit 18 : Probability

18.1 Introduction

18.2 Some basic terms used in probability

18.3 Empirical probability
No. of favourable events n(E)
a. Probability (P)= No. of total events = n(S)

Revision questions for examination
Answers
Model questions

Secondary Education Examination

Specification Grid - 2069 | Sub : Compulsory Mathematics Class - 9

S.N. Topics Units K C A HA Total Remarks
1. Sets -
1. Sets - 1×4 - 4
2. Airthmetics 2. Profit and Loss 1×1
3. Commission and Taxation At least 2 marks
4. Household Airthmetics
2×2 1×4 1×5 14 questions from

each units.

At least 3 marks

3. Mensuration 5. Mensurations 1×1 3×2 1×4 1×5 16 questions from

each units.

6. Algebraic expression

7. Surds At least 3 marks

4. Algebra 8. Indices 1×1 5×2 2×4 1×5 24 questions from

9. Quadratic Equations each units.

10. Equations

11. Triangle / parallelogram At least 4 marks

5. Geometry 12. Construction 2×1 3×2 3×4 1×5 25 questions from

13. Circle each units.

6. Trigonometry 14. Trigonometry - 1×2 1×4 - 6

7. Statistics 15 Statistics 1×1 1×2 1×4 - 7

8. Probability 16. Probability - 2×2 - - 4

Total question numbers 6 17 10 4 37

Total weightage of marks 6 34 40 20 100

1

Sets

Estimated Teaching Periods : 8

John Venn (1834-1923) was an English logician and philosopher noted for
introducing the Venn-diagram, used in the fields of set theory, probability,
logic, statistics and computer Science.

Contents

1.1 Revision/Introduction
1.2 Set relations
1.3 Set operations
1.4 Cardinality of sets

Objectives

At the end of this unit, students will be able to:
know the definition of set, types of sets, representations of set, venn-diagram.
find the relation between sets.
perform the set operation using a venn-diagram.
represent cardinality of a set in a venn-diagram.
solve the verbal problems related to two sets using venn-diagram.

Materials

Cello tape, scissors, scale, chart paper or news print for showing the relation of sets,
venn-diagram, flash cards, etc.

GREEN Mathematics Book-9 1

1.1 Review/Introduction

In the previous classes, we have studied about set and Example:
it general types.
A set of even numbers less
In 1865 AD, George Cantor introduced sets by new than 10.
vistas in modern mathematical thought and process.
i.e. E = {2, 4, 6, 8}
"A well defined collection of distinct items or objects is
known as sets." n(E) = 5

Generally, capital letters of English alphabets are used to denote the sets and
small letters are used to denote its elements.

Points to remember
The symbol ∈ stands for "is the member of"
The symbol ∉ stands for " is not the member of "

Representation of sets

There are three methods to represent a set :
a. Descriptive method: A = {the set of prime numbers less than 5}
b. Listing method: A = {2, 3}
c. Set builder method: {x : x is the prime numbers less than 5}
as A = {x:x ∉ 6 as prime number}

Types of sets

Types of sets Identification Example
M = {2, 4, 6}
Finite set It is a set containing finite number of N = {1, 3, 5, 7,......}
elements. P={1}
Q = { } or ∅
Infinite set It is a set containing infinite number of
elements.

Unit set or It is a set containing only one element.
Singleton set

Null set or Empty It is a set containing no element.
set or void set

2 GREEN Mathematics Book-9

1.2 Set Relation If A = {2, 3} and B = {2, 3, 4, 5} then
A ⊂ B (A is the sub-set of B)
a. Sub sets: a. If A ⊂ B, B is called super set of A.
b. If A ≠ B, A is called proper subset of B.
If A and B are two non empty sets,
such that every element of A is in
B, then set A is the sub set of B. It is
denoted by A⊂B.

b. Disjoint sets: Example: A U
If A = {2, 4, 6} and 2 B
If A and B are two non - B = {1, 3, 5},
empty sets such that they Hence, A and B are disjoint sets. 4 1
have no common elements, 6 35
then A and B are called
disjoint sets.

c. Overlapping sets: Example: U
AB
If A and B are two non - If A = {1, 2, 3} and B = {2, 4, 5},
empty sets such that they then A ∩ B = { 2 }. So, A and B 1 4
have at least one element in are overlapping sets. 2
common, then A and B are
overlapping sets. 35

d. Universal sets: Example: A U
B
The set which includes all If U = {1, 2,....., 9}, A = {1, 3, 5, 7, 13
elements of the sets taken 9} and B = {2, 4, 6, 8} then, 2
consideration is known as 57
universal set. The universal set is A ⊆ U and B ⊆ U. 9 46
denoted by the symbol U. 8

e. Equal sets: Example: A B
If A = {2, 4, 6} and 24
Two sets A and B are said to be equal = 24
if they have the same elements. 6
Equal sets A and B are denoted by 6
A = B.
B = {2, 4, 6}, then A = B = {2, 4, 6}

GREEN Mathematics Book-9 3

1.3 Set Operations

a. Union of sets:

If A and B are two non empty sets then Example: U
the union of set A and set B is the set of If A = {1, 2, 3} and AB
all elements which contains either in B = {2, 4, 6} then,
A or B or in both. Symbolically union A∪B = {1, 2, 3, 4, 6} 124
of sets A and B are denoted by A ∪ B. 36

(A∪B)

b. Intersection of sets:

If A and B are two non empty sets Example: U
then the intersection of A and A = {1, 2, 3} and AB
B is the set of elements which
are common to both A and B. 124
Symbolically they are denoted by
A ∩ B. B = {2, 4, 6} then, A ∩ B = { 2 } 3 6

(A∩B)

c. Difference of sets: Example: A U
B
If A and B are two non empty sets If A = {2, 4, 6} and 42
then the only elements which 6 1
belong to A but not belong to B, is
called the difference of A and B. i.e. B = {1, 2} then, A – B = {4, 6} A–B
A–B
Example: A U
Similarly, if A and B are two non B
empty sets then set of all those If A = {a, b, c} and
elements that belong to B but do not ab
belong to A is called difference of B B = {c, b, z} then B – A = { z } cz
and A. i.e. B – A
B–A

d. Complement of set: Example: A 2 U
1
The complement of the set A U = {1, 2.......9} and 76 35
with respect to universal set U is A = {1, 2, 7, 4, 6} 4 89
the set of all those elements of U A' = Ac = {3, 5, 8, 9}
which are not in A. Symbolically A=U–A A
it is denoted by A' or Ac or A or
U – A.

4 GREEN Mathematics Book-9

Worked Out EXAMPLES

EXAMPLE 1 If A = {a, b, d, e}, B = {b, c, e, f} and C = {d, e, f, g}
i. Verify A∩(B∪C) = (A∩B)∪(A∩C)
ii. Verify A∪(B∩C) = (A∪B)∩(A∪C)

Solution : i. A∩(B∪C) = (A∩B)∪(A∩C)

L.H.S. = A∩(B∪C)

B∪C = {b, c, e, f} ∪{d, e, f, g}

= {b, c, d, e, f, g}

∴ A∩(B∪C) = {a, b, d, e} ∩{b, c, d, e, f, g}

= {b, d, e} ...................... (i)

R.H.S. = (A∩B)∪(A∩C)

A∩B = {a, b, d, e,} ∩{b, c, e, f}

= {b, e}

A∩C = {a, b, d, e,} ∩{d, e, f, g} = {d, e}

∴ (A∩B)∪(A∩C) = {b, e} ∪{d, e}

= {b, d, e} ...................... (ii)

From (i) and (ii), we conclude that

∴ A∩(B∪C) = (A∩B)∪(A∩C)

ii. A∪(B∩C) = (A∪B)∩(A∪C)

L.H.S. = A∪(B∩C)
B∩C = {b, c, e, f} ∩ {d, e, f, g}

= {e, f}

∴ A∪(B∩C) = {a, b, d, e} ∪{e, f}

= {a, b, d, e, f} ............... (i)

R.H.S. = (A∪B)∩(A∪C)

A∪B = {a, b, d, e} ∪ {b, c, e, f}

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

A∪C = {a, b, d, e} ∪ {d, e, f, g}

= {a, b, d, e, f, g}

∴(A∪B)∩(A∩C) = {a, b, c, d, e, f} ∩{a, b, d, e, f, g}

= {a, b, d, e, f} .................. (ii)

From (i) and (ii), we conclude that

A∪(B∩C) = (A∪B)∩(A∪C)

GREEN Mathematics Book-9 5

EXAMPLE 2 Use venn-diagram in different situations to find the followng sets.
Solution :
a. A∪B b. A∩B c. A' A U
d. B – A e. (A∩B)' f. (A∪B)' a
d e B
b h

U = {a, b, c, d, e, f, g, h, i, j} cfg i
A = {a, b, c, d, f} j

B = {d, e, f, g}

a. A∪B = {Elements which are in A or in B or in both}

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

b. A∩B = {Elements which are common in both}

= {d, f}

c. A' = {Elements of U which are not in A}

= {e, g, h, i, j}

d. B – A = {Elements which are in B but not in A}

= {e, g}

e. (A∩B)' = {Elements of U, which are not in A∩B}

= {a, b, c, e, g, h, i, j}

f. (A∪B)' = {Elements of U, which are not in A∪B}

= {h, i, j}

EXERCISE: 1.1

A. Very Short Questions
1. What do you mean by a set ?
2. Fill in the gaps:

a. If A = {2, 3, 4, 5, 6} and B = {2, 4, 5}, then, A – B = {...................}
b. If A = {0, 2} and C = { 0 } then A ∩ C = {...................}.
c. If P = {1, 2, 3,........} is called ..........
d. If Q = {x, y, z}, so it is called ..........
e. If U = {1, 2, .........9} and A = {6, 8, 9, 1, 2}, A' or Ac = {.......................}.

6 GREEN Mathematics Book-9

f. U = {1, 2, 3, .........9}

A U The set A = {...............}.

2, 4, 6, 8

g. If A= {x, y, z} and B = {w, z}, find (A∪B) = {......................... } and A∩B = {....................}

h. If P = {The mathematics teachers of a school} is ........... method.
i. If A = { 5, 7, 9} and B = {6, 7, ......10}, then A – B {..................................} and
B – A = {......................}.
j If B = {6, 7................10} and O = {5, 8, 9}, then O – B = {......................}.

B. Short Questions
3. Identify whether the given sets are empty or single.

a. The set of students of height 15 inches.

b. The set of natural numbers between 9 and 11.

4. State whether set A = {3, 6, 9, 12} and set B {1, 5, 7, 11} are equal or equivalent. Write
a reason.

5. Write the possible subsets of:

a. A = {2, 4, 6} b. O = {a, b}

6. Express the following sets in description or set builder form:

a. D = {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}
b. B = {a, e, i, o, u}

7. If U = {1, 2,............9}, A = {1, 2, 3, 4, 5} and B = {2, 3, 4, 8}, find (A ∪ B).

8. If M = {2, 6, 9} and N = {1, 6, 15}, find (M ∩ N).

9. If U = {1, 2, 3, 4, ................. 9}, A = {1, 3, 5, 7, 9} and B = {2, 4, 6, 8} Find Ac and Bc.

10. Study the given diagram and answer the questions below. A 2 U
1 5
a. Find A' b. Find (A ∩ B) 4 B
6 7

c. Find (A – B) and (B – A) 3 89

d. Find (A ∪ B) e. Find (A ∪ B)

11. If A = {1, 2, 4, 5}, B = {2, 3, 5, 6} and C = {4, 5, 6, 7}, then verify the given relations.

a. A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) b. A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

c. (A ∩ B) ∩ C = A ∩ (B ∩ C)

GREEN Mathematics Book-9 7

12. If A = {a, b} and B = {x, y, z}, find (A ∪ B).

A. Long Questions

13. If U = {1, 2, .........., 10}, A = {1, 3, ,5, 7, 9}, B = {3, 4, 6, 8} and C = {3, 6, 9}. List the
elements of the following:

a. A ∪ (B – C) b. A ∪ (B – C) c. (A ∩ B ∩ C)
d. (A – B) ∩ C e. (A∪ B )

14. If U = {1, 2, ..............., 12}, A = {1, 2, ............, 5} and B = {2, 4, 6, 8}, List the element of
the following:

a. (A ∪B) and (A ∪ B) b. (A ∩ B) c. (B – A) d. (A – B) e. (A∩B)

15. i. From the given diagram, list the element of the following. A U
B
a. (A∪ B) = {.............} b. (B – A) = {............} 1 5
28
47

c. (B – A) = {...........} d. (A – B) = {.............} 36 9

e. (A∪C) ∪ (B∩C) = {.............}

ii. From the given venn diagram list the elements of the following:

U

A 2 B a. (A ∩ B ∩ C) b. (A ∪ B ∪ C)
1 5 c. (A ∩ B ∩ C) d. A ∪ B ∩ C .
8

3
6 47

9 e. A ∪ (B ∩ C)

16. If U = {1 ≤ x ≤ 10, x ∈ N}, A = {x : x is a prime numbers} and B = {x :x is an odd
numbers} then find,

a. (A ∪ B) b. (A ∩B)' c. (A ∩ B)
d. (A∪ B) e. (A ∩ B)' –(A ∪ B)'

17. If U = {1, 2 ..... 7} O = {odd natural number less than 7} and P = {prime factors of 12},
then find,

a. (P – O) b. (O ∩ P) c. (O – P)
d. (O – P)' e. (P – O)'

8 GREEN Mathematics Book-9

1.4 Cardinality of sets

John Venn Cardinality of a set

In 1880AD, John Venn introduced the Let A be finite and non-empty set, the
diagram in his journal of an article about total number of elements in A is known
the diagrammatic representation of sets as cardinal number of the set A and it is
and reasoning. denoted by n(A). Example: If A = {1, 2, 3,
4, 5}, n(A) = 5.

The number of elements of given sets is called its cardinality. For example: A = {1, 2, 3, 4},
B = {5, 6, 7}. Set A contains 4 elements and sets B contain 3 elements. So, cardinal number
of sets are n(A) = 4 and n(B) = 3

Method for finding cardinality sets

Look at the following example: B P U
Q
U

A

a e 16
bd f 2 47

c 3 5 89

10 11

Fig. I Fig. II

In fig I, both sets A and B doesn't have any common elements. So, they are called
disjoint sets. Then n(A) = 4, n(B) = 2; n(A∪B) = 4 + 2 = 6 and n(A∩B) = 0.

In the figure II, P and Q are two intersecting sets. The elements 10 and 11 doesn't
belong to P and Q. So universal set (U) contains all element with P and Q.

So, U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}, then n(∪) = 11

Here, {10, 11} doesn't belong to set P and Q.

So, (P∪Q) = {10, 11}, then n(P∪Q) = 2

We know,

n(U) = n(A∪B) + n(A∪B)

GREEN Mathematics Book-9 9

Proof: A n(A ∩B) U
n0 (A) B
Let U be the universal set. Let A and B be two non empty sub
sets of U then, n0 (B)

n(A) = n0 (A) + n(A ∩ B)
i.e, n0 (A) = n(A) – n(A ∩B)
n(B) = n0 (B) + n(A ∩ B)
i.e, n0 (B) = n(B) – n(A ∩ B)
From above Venn-diagram

n(A ∪ B) = n0 (A) + n0 (B) + n(A ∩ B)
= n(A) – n(A ∩ B) + n(B) – n(A ∩ B) + n(A ∩ B)

= n(A) + n(B) – n(A ∩ B)

Some relations:
a. n0 (A) = n(A) – n(A ∩ B) = n(A – B)
b. n0 (B) = n(B) – n(A ∩ B) = n(B –A)
c. Exactly one = n0 (A) + n0 (B)
d. At least one = n(A ∪ B) = n(A) + n(B) = either A and B are disjoint.
e. n(U) = n(A) + n(B) – n(A ∩ B) + n(A∪B ) or n(A∪B) + n(A∪B ).

Note :
i. If two sets A and B are non empty and disjoint sets then n(A ∪ B) = n(A) + n(B)
ii. If two sets A and B are non empty and overlapping sets then,
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

Worked Out EXAMPLES

EXAMPLE 1 In the given Venn-diagram A and B are two A U
subsets of an universal set where n(U) = 50, 15 B
n(A) = 23, n(B) = 20 and n(A ∩ B) = 8. Find the
followings. 8 12

a. n(A ∪ B) b. (A ∪ B) c. n(B – A)
d. n0 (A) e. Exactly one = ?

10 GREEN Mathematics Book-9

Solution : a. n(A ∪ B) = n(A) + n(B) – n(A∩B) n(A∩B) = 8
= 23 + 20 – 8
= 43 – 8 A U
= 35 15 B
b. n(A ∪ B)
8 12

c. n(B – A) = n(U) – n (A∪B) 15 n0(B) = 12
= 50 – 35 n(B) = 20
= 15 n(A∪B) n(A) = 23
= n(B) – n(A∩B)
= 20 – 8 n0(A) = 15

= 12

d. n0 (A) = n(A) – n(A∩B)
= 23 – 8

= 15

e. Exactly one = n0 (A) + n0 (B)

= 15 + 12

= 27

EXAMPLE 2 In an examination 80% of students passed in mathematics, 85% science
Solution :
and 75% in both. Find the doesn't read percentage of both subjects

with Venn-diagram. Venn-diagram
Let, total number of student n(U) = 100% U

MS
= 58%0% – 75%
75% 75%

Number of students who passed in maths, n(M) = 80% –

85%
10%

=

Number of students who passed in science, n(S) = 85% n (M∩S)= x

Number of students who passed in both subjects, n(M ∩ S) = 75%

Number of students who don't read in both subjects, n(M ∪ S) = x

We know,

n(U) = n(M) + n(S) – n(M ∩ S) + n(M ∪ S)

or, 100% = 80% + 85% – 75% + x

or, 100% – 80% – 85% + 75% = x

or, x = 175% – 165%

∴ x = 10%

So, the percent of student who don't read in both subjects = 10%.

GREEN Mathematics Book-9 11

EXAMPLE 3 In a group of 100 people, 72 can speak English and 43 can speak Nepali.
How many can speak English only? How many can speak Nepali only

and how many can speak both Nepali and English?

Solution : Let 'A' be the set of people who speak English and 'B' be the set of people
who speak Nepali.

Then,

A – B is the set of people who speak English only but not Nepali,

B – A is the set of people who speak Nepali only but not English

A∩B is the set of people who speak both Nepali and English.

We have,

n(A) = 72, n(B) = 43, n(A∪B) = 100

Now,

n(A∩B) = n(A) + n(B) – n(A∪B)

= 72 + 43 – 100

= 115 – 100

= 15

∴ 15 people speak both Nepali and English

n(A) = n(A – B) + n(A∩B)

or, n(A – B) = n(A) – n(A∩B)

= 72 – 15

∴ n0(A) = 57
And n(B – A) = n(B) – n(A∩B)

= 43 – 15

∴ n0(B) = 28

∴ 57 people speak English only, 28 people speak Nepali only and 15
people speak both Nepali and English.

EXAMPLE 4 In a group of total students 35 enrolled in art class and 57 enrolled in
dance class. Find the number of students who are either in art class or
in dance class, if the situation is :

i. when two classes meet at the same hour

ii. when two classes meet at different hours and 12 students are
enrolled in both activities.

12 GREEN Mathematics Book-9

Solution : Let 'A' be the set of students in art class and 'B' be the set of students in
dance class. n(A) = 35, n(B) = 57

i. When two classes meet at the same hours then n(A∩B) = O and by
using formula :

n(A∪B) = n(A) + n(B) – n(A∩B)

= n(A) + n(B) – 0

= 35 + 57 = 92

∴92 students are either in dance or in art class.

ii. When two classes meet at different hours then n(A∩B) = 12, using

formula: n(A∪B) = n(A) + n(B) – n(A∩B)

= 35 + 57 – 12

= 92 – 12 = 80

∴80 students are either in dance or in art class.

EXAMPLE 5 In a survey of 100 farmers, the ratio of the farmers who grew paddy and
Solution : wheat is 2 : 3. If 10 farmers grew both and 20 grew none of them, find
the numbers of farmers who grew paddy only by using Venn-diagram.

Let U be the total numbers of farmers in a survey P and W be the sets of
farmers who grew paddy and wheat respectivelly.

Here,

n(U) = 100,

n(P) 2 x be the multiple of ratio
= 3 = 2:3
n(W)

∴ n(P) = 2x

n(W) = 3x

n(P∪W ) = 20

and n(P∩W) = 10

Using formula, n(U) = n(P) + n(W) – n(P ∩ W) + n(P∪W )

or, 100 = 2x + 3x – 10 + 20

or, 5x = 100 – 10 n(P) = 36

∴ x = 18 n0(P) = 44
U
Now, p
W
n(P) = 2x = 2 × 18 = 36
= 3x = 3 × 18 = 54 26 n(P∩W)
10 44
n(W)
Finally,
20 n(P∪W)
n0 (P) = n(P) – n(P∩W)
= 36 –10 n0(P) = 26 n(W) = 54
= 26

GREEN Mathematics Book-9 13

EXAMPLE 6 In a survey , it was found that 80% people liked oranges, 85% liked
Solution :
mangoes and 75% liked both but 45 people liked none of them. By

drawing a Venn-diagram, find the number of people who were in the

survey. n(O∩M) = 75%

Let U be the total number of people in surveys n(M) = 85%

O and M denotes the set of people who like U
M
oranges and mangoes respectively. O

Then, 75%5% 10%

n(O) = 80%, n(M) = 85% 45
n(O ∩ M) = 75%
n(O ∪ M) = 45% n(O∪M) = 45

n(O) = 80%

n(O∪M) = n(O) + n(M) – n(O∩M)

= 80% + 85% – 75%

= 90%

∴ n(O∪M) = (100 – 90)%

= 10%

Number of the people who did not like both orange and mangoes = 45

Here, let the total number of the people be 'x' who were in the survey

10% of x = 45

10
or, 100 × x = 45
or, 10x = 45 × 100

or, x = 450

∴ Number of people who were in the survey = 450.

EXERCISE: 1.2

A. Very Short Questions
1. Define cardinal number of set.
2. Fill in the gaps.

i. If n(A) = 50, n(B) = 40 and n(A ∩ B) = 20, then n(A ∪ B) = ...................
ii. n0 (A) = n(A) – ..............
iii. n(∪) = n(A ∪ B) + (.....................).

14 GREEN Mathematics Book-9

iv. From the adjoining Venn-diagram, find A U
40 B
a. n0(A) = .......... b. n(A ∩ B) = ...............
20 30

c. (...........) = n(B) – n(A ∩ B)

d. n(A ∩ B) = ...........

v. If n(U) = 10, n(A) = 4, n(B) = 5 n(U) = 40, n(A∪ B ) = 11 and n(A ∪ B) = 8, then
n(A ∩ B) = ................

vi. If n(P) = 60, n(Q) = 50, n(P ∪ Q) = 90, then no (Q) = ................
vii. If n(U) = 10, n(P) = 3, n(Q) = 4 and n(P∩Q) = 2, then n(P∪ Q ) = ...........

viii. Mathematician Jhon-Eular Venn developed a diagram to represent the sets is ..

ix. If n(P) = 80, n(Q) = 90, n(P∪Q) = 160 then n0(Q) = ...........
x. If n(U) = 100%, n(E) = 80%, n(M) = 85%, n(E ∩ M) = 75% and n(E ∪ M) = 30 then

total number of students = ..................

B. Short Questions

3. a. In a survey of 60 students, 30 drink milk, 25 drink curd and 10 students drink
milk as well as curd then.

i. Draw a Venn diagram to illustrate the above information.

ii Find the number of students who drink neither of them.

b. In a class of 50 students, 25 students like to play football, 35 like to play cricket
and 15 like to play both the games. How many students are there who do not
like to play any games? Illustrate the above information by a Venn-diagram.

c. In class IX of 130 students, 70 students liked tea, 40 students liked coffee and
30 students did not like both. Find the number of students who liked both and
illustrate this information by means of a Venn-diagram.

d. In class IX of 55 students, 15 students liked Maths but not English and 18
students liked English but not Maths. If 5 students did not like both subject,
find the number of students who liked both subjects. Also represent the above
information in a Venn-diagram.

4. a. In group of 200 students, 120 like cricket and 105 like football, each student like
at least any one game. By drawing a Venn-diagram, find how many students
like both game.

b. In a group of 70 people, 37 like tea, 52 like milk and each person like at least one
of the two drinks. Using a Venn-diagram, find how many people like both tea
and milk.

GREEN Mathematics Book-9 15

5. a. In an examination 80 students secured the first class marks in English or
Mathematics, out of them 50 students the obtained first class marks in
Mathematics and 10 in English and Mathematics both. How many students
have secured first class marks in English only?

b. In group of 120 people, 90 play volley ball, 72 play football and 10 play neither
of games. Then, using a Venn diagram find

i. how many people play both games?
ii. how many people play only one game?

C. Long Questions

6. a. In a class of 60 students, 15 students liked Maths only, 20 liked English only
and 5 did not like any subjects then,

i. represent the above information in a Venn-diagram.
ii. find the number of students who liked both the subjects.
iii. find the number of students who liked at least one subject.
b. In a school all students play either volley ball or football 300 play football, 250

play volley ball and 110 play both the games. Draw a Venn-diagram to find:
i. the number of students who play football only.
ii. the number of students who play volley ball only
iii. the total number of students of the school.
c. In a survey of a community, 55% of the people like to listen to the radio 65%

like to watch the television and 35% like to listen to the radio as well as to watch
television.
i. Show the above information in a Venn-diagram.
ii. Find the percentage of the people who do not like to listen the radio as

well as to watch television.
d. In a survey of 350 students of a school, 200 liked Chitwan, 220 liked Lumbini

and 120 liked both places.
i. Show the above information in a Venn-diagram.
ii. Find the number of students who liked neither of two places.
e. Out of 100 students, 80 passed in science, 71 in mathematics, 10 failed in both

subjects and 7 did not appear the examination. Find the number of students
who passed in both subjects by representing the given information in a Venn-
diagram.
f. In a survey of a community, it was found that 85% of the people like winter
season, 65% like summer season and there were no people who do not like
both seasons.

16 GREEN Mathematics Book-9

i. Present the above information in a Venn-diagram
ii. What percentage of people like both the season?
iii. What percent of people like winter season only?
g. In an examination 35% of students passed in Nepali only and 25% passed in

Mathematics only and 20% of students were failed in both subjects.
i. What percent of students who passed in both subject?
ii. What percent of students who passed in Mathematics?
iii. Represent the above information in a Venn-diagram.
h. In a class of 120 students, 95 like Account, 80 like Biology and there are some

students who like both subjects.
i. Find the number of students like Account only.
ii. Find the number of students like Biology only
iii. Show the above information in a Venn-diagram.
7. If U={x:x is positive number less than or equal to 20}
P = {y:y is a prime number less than 20}
Q = {z:z is factors of 18}, R = {p:p is multiple of 3 less than 20}.
Draw the above information in venn diagram and calculate:

i. n(P∪Q) ii. n(P∪Q∪R) iii. n(P∩Q∩R)

iv. n(Q∪R) v. n(P∪R) vi. n0(P)
vii. n0(Q) viii. n0(R) ix. n(P–Q)

x. n(Q–R) xi. n(P∪Q∪R) 

8. In a survey of 2000 people 60% of people visit country A and 30% in country B. If
15% of total people didn't visit in A andB, calculate.

a. How many people visit both countries?

b. How many people visit only one contry?

c. Show the above information in Venn-diagram.

GREEN Mathematics Book-9 17

2

Profit and Loss

Estimated Teaching Periods : 8

Johann Carl Friedrich Gauss (1777-1855) was a German mathematician
who contributed significantly to many fields, including number theory,
algebra, statistics, analysis, differential geometry, geodesy, geophysics.

Contents

2.1 Review
2.2 Problems related to profit and loss
2.3 Discount

Objectives

At the end of this unit, students will be able to:
define profit and loss.
calculate profit, loss, C.P., S.P. profit percent and loss percent.
find whether business man are in profit or in loss in their business.
compare between S.P. and C.P.
use related formula to the problem.

Materials

Formula chart, relation of S.P., C.P., profit and loss.
Picture of market with seller and buyer.

18 GREEN Mathematics Book-9

2.1 Review

We have already learnt the profit and loss in previous classes. Let us revise some of the
relations and terms used in profit and loss.

Cost Price (C.P.) Selling Price (S.P.)

It is the price of an article or a thing It is the price for which an article or the
which is purchased. thing is sold.

Actual Profit Actual Loss

It is an additional money gained by It is the money which is lost by selling
selling the things or articles at a price the things or articles at a price lower
higher than the cost price. than the cost price.

2.2 Problems related to profit and loss

Remember
While solving problem we must use the following steps:
i. Understand problem and making planing.
ii. Solving problems according to planing.
iii. Review solving problem and recheck it.
iv. At last write the conclusion according to questions.

Concept
A man buys a mobile for Rs. 10,000 and sells it for Rs. 12,000. Then he makes Rs. 2000
profit. If he sells the mobile for Rs. 8000, then he makes a loss of Rs. 2000.
In this way a shopkeeper or a merchant first purchases goods and then sells them. The
price which he pays for an article is known as the cost price (C.P.).
In given example,
The cost price of mobile set C.P. = Rs. 10,000 and SP = Rs. 12000, when profit = Rs. 2000.
Similarly,
The price for which a shopkeeper sells an article is called the selling price (S.P.). In the
above example: the selling price of mobile set (S.P.) = Rs. 8000, when loss = Rs. 2000.

GREEN Mathematics Book-9 19

Now, in short,

a. A profit (P) is made if S.P > C.P.

i.e. P. = S.P. – C.P.

b. A loss (L) is made, if S.P. < C.P.

i.e. L = C.P. – S.P.

c. Profit percent and loss percent are calculated by the following relations.

i. P% =q Actual profit × 100r
C.P.

ii. L% =qActuCa.Pl.loss × 100r

d. If S.P. and P% or L% are given then C.P. can be calculated from the following
formulae:

i. C.P. = S.P. × 100 ii. C.P. = S.P. × 100
100 + P% 100 – L%

e. If C.P. and P%, or L% are given, then S.P. is calculated by using the following
formulae:

i. S.P. = 100 + P% × C.P.
100

ii. S.P. = 100 – L% × C.P.
100

The following worked out examples will make you more clear about this chapter.

Worked Out EXAMPLES L% = ............................

EXAMPLE 1 Fill in the blanks
a. P = ............................. b.
c. S.P. = ............................

Solution : a. P = S.P. – C.P.

b. L% =qActuCa.Pl.loss × 100r

i.e. L% =qC.PC. –.PS. .P. × 100r

20 GREEN Mathematics Book-9

c. S.P. = 100 + P% × C.P., in the case of P%.
100

S.P. = 100 – L% × C.P., in the case of L%.
100

EXAMPLE 2 Sadan sold 65 kg of Basmati rice at Rs 70 per kg and earned a profit of
Rs 1,000. What was his cost price of the rice?

Solution : S.P. of 1 kg of rice = Rs. 70

S.P. of 65 kg of rice = Rs. 70 × 65 = Rs. 4,550

Profit = Rs. 1,000

C.P. of rice = ?

We have, P = S.P. – C.P.

or, Rs. 1000 = Rs. 4550 – C. P.

or, C.P. = Rs. 4550 – Rs. 1000

= Rs. 3550

∴ C.P. = Rs. 3550

EXAMPLE 3 Ramita bought an article for Rs. 40 and sold it for Rs. 50. Find the gain
Solution : percent ?

C.P. = Rs. 40, S.P. = Rs 50
We have,

Actual profit (P) = S.P. – C.P. = Rs 50 – Rs 40 = Rs 10

Profit percent (P%) = q Actual profit × 100r
C.P.

= q Rs. 10 25
Rs 40 × 100r = 25

∴ Her profit percent = 25%

EXAMPLE 4 By selling 10 sweets for a rupee, 20% is lost. At what rate the sweets
were purchased ?

Solution : S.P. of 10 sweets (S.P.) = Re. 1

Loss percent (L%) = 20%

We have, (C.P.) = S.P. × 100
100 – L%

Rs. 1 × 100
= (100 – 20)

GREEN Mathematics Book-9 21

= Rs. 100
80

= Rs. 45

Now,

∴ Rs. 5 is the C.P. of 10 sweets
Re. is the C.P. of 10 4
4 × 5 sweets = 8 sweets
1

∴ 8 Sweets were purchased for Re. 1.

EXAMPLE 5 If an article which costs Rs. 90 and is sold at a profit of 6%, what is the
S.P.?

Solution : C.P. of the article = Rs. 90

Profit percent (P%) = 6%

S.P. = ?
We have,

S.P. = 100 + P% × C.P.
100

= 100 + 6 × Rs. 90
100

= Rs. 106 × 90 = Rs 9540
100 100

S.P. = Rs. 95.40

∴ The selling price of an article is Rs. 95.40

EXAMPLE 6 When I sold 20 tables for the same price as I had paid for 25, what was
my gain %?

Solution : Let, S.P. of 20 tables = Rs. x

Then, C.P. of 25 tables = Rs. x
x
C.P. 1 table = Rs. 25

∴ C.P. 20 tables = Rs. x × 20 = Rs. 4x ∴ SP of 20 tables >
25 5 CP of 20 tables.
Here, S.P. > C.P.
4x x
So, P = S.P. – C.P. = Rs. x – 5 = 5

Again, P% = q P. × 100r
C.P.

x

q r 5
= 4x × 100 ∴ P% = 25 or, P = 25%

5

22 GREEN Mathematics Book-9

EXAMPLE 7 If an article is sold for Rs. 240 at a profit of 10%, at what price should
it be sold so to gain of 14%.

Solution : In first case:

S.P. of an article = Rs. 240

Profit percent (P%) = 10

Cost price (C.P.) = ?

We have, S.P. × 100
C.P. = 100 + P%

= Rs. 240 × 100
100 + 10

= Rs. 24000
In second case: 110 = Rs. 218.18

Again, C.P. = Rs. 218.18, P% = 14%

S.P. = ?

We have, S.P. = q 100 + P × C.P.r
100

C.P. = 100 + 14 × Rs. 218.18
100

= Rs. 114 × Rs. 218.18
100

= Rs. 248.73
∴ S.P. = Rs. 248.73

EXAMPLE 8 A shopkeeper sold an article at a loss of 10%. Had he sold it for Rs. 150
more he would have made a profit of 20%. What was the C.P. of the
article?

Solution : Let, C.P. of the article be Rs x.

Now,
L oss percentage (L) = 10%

∴ Selling price (S.P.) = 100 – L% × C.P.
100

= Rs. 100 – 10 × x
100

S.P. = Rs. 9x
10

GREEN Mathematics Book-9 23

Again,

Profit percent (P%) = 20

S.P.2 = 100 + P% × C.P.
100

100 + 20
= 100 × x

6x
S.P.2 = Rs. 5

Finally, According to questions,

S.P.2 – S. P.1 = Rs. 150

or, 6x – 9x = Rs. 150
5 10

or, 12x – 9x = Rs. 150
10

or, 3x = Rs. 1500

or, x = Rs. 500

∴ Cost price of articles is (C.P.) = Rs. 500

EXERCISE : 2.1

A. Very Short Questions b. Loss
1. Define:

a. Profit

2. Fill in the blanks.

a. L = ............................. b. P% = .............................

b. If S.P. < C.P. then it will be ............................

c. Sabin bought a watch for Rs. 420 and sold it for Rs. 520, then his profit will be
............................

d. Meaning of x% profit, S.P. = ............................

e. Calculate unknown value of the following:

S.N. Name of goods C.P. (Rs.) S.P. (Rs.) Actual profit/loss Profit/Loss

a. Pen 80 90 - -

b. Copy 50 45 - -

c. Cycle 1300 - Rs. 200 -

d. Cap 500 - - 10% (profit)

e. Bag - 850 - 15% (less)

24 GREEN Mathematics Book-9

B. Short Questions

3. a. A watch which was bought for Rs. 1000 was sold at Rs. 1200. What will be the
gain in percent?

b. Shyam bought a pendrive for Rs. 800 and sold it for Rs. 760. Calculate his profit
or loss percent.

4. a. Lemons are bought at 4 for Re. 1 and sold 3 for Re. 1. Find the profit or loss
percent.

b. After buying some lemons at the rate of 5 for Re. 1 was sold them at the rate of
4 for Re. 1. Find the gain or loss percent.

c. If the cost price of 5 pens is equal to the selling price of 4 pens of the same kind,
find gain percent.

5. a. An article is bought for Rs. 300 and sold it at a profit of 25%. What is the gain?
b. Urmila bought a sari for Rs. 3500 and she sold it at loss of 10%. Find her net loss.
c. A shopkeeper sold a radio for Rs. 7700 at a profit of 10%. At what price was it

purchased?
d. If Sadan sold a pen for Rs. 252 at a profit of 5% , find his C.P.

e. A LED bulb was sold for Rs. 261 at a loss of 13%. At what price was the bulb
purchased?

f. At a loss of 10% the selling price of an article is Rs. 810. Calculate the C.P. of the
article.

6. a. Anu bought 250kg rice at rate of Rs. 60 per kg. She spent Rs. 20 per kg for
transportation. If she sold all rice for Rs. 16000, what percent of profit or loss did
she get?

b. Bibek bought 6 ropani land at the rate of Rs. 3,25,000 per ropani. He spent Rs.
7,25,000 for boundary wall. If he wants to take 20% of profit, find selling price
of each ropani.

c. Hari bought 100 kg oranges for Rs. 225 per kg from Biratnagar and 200 kg for
Rs. 160 per kg from Pokhara. He spent Rs. 3000 for transportation. At what rate
did he sell to take a profit of 30%.

C. Long Questions

7. a. Avipsa sells a radio to Sabin at a gain of 10% and Sabin sells it to Ramita at a
gain 5%. If Ramita pays Rs. 462 for the radio what did it cost to Avipsa?

b. A radio is sold for Rs. 880 at a loss of 12%. At what price should it be sold to gain
profit of 10%?

c. A wholesaler sold wind proof jacket for Rs. 8330 and made 2% loss. For what
price should he sell it to gain 7%?

8. a. An item was sold at a loss of 25%. Had it been sold for Rs. 2100 more then the
profit would have been 10%. What is the C.P. of the item?

b. Ansu sells an article at a gain of 15%. If she had bought it at 10% less and sold it
for Rs. 4 more, then he would have gained 30%. Find her cost price (C.P.)

GREEN Mathematics Book-9 25

c. A Naptaul Bazzar sold a induction cooker at a loss of 10%. Had it been sold for
Rs. 1500 more he would have made a profit of 20%, what was its cost price?

9. a. A shopkeeper bought two watches for Rs. 400. He sold them to gain 5% on one
and at a loss of 5% on the other. Calculate his final gain or loss percent if the S.P.
of both watches are the same.

b. Two pens are sold for Rs. 80 each. If there is a gain of 20% on one and a loss of
20% on the other find the net loss percent.

c. Ujjal sold two televisions for Rs. 9919 each. On one he gained 9% and on the
other he lost 9%. Find his gain or loss percent?

10. a. An item was sold at a loss of 20%. Had it been sold for Rs. 50 more then the
profit would have been 5%. What is the cost price of the item?

b. Ramita sold a facewash cream for Rs. 550 and made a profit of 10%. If Ramita
had sold it for Rs. 495, what would be her loss percent?

c. Sapoz sold a dozen dolls for Rs. 1368 at 5% loss. Find his profit or loss percent
if she had sold it for Rs. 1512.

d. By selling an article for Rs. 450, a man gets a loss of 10%. Find his gain or loss
percent if he sells it for Rs. 540.

11. a. Sabin buys a pencil for Rs. 20 and sells it at profit of 20% . For what price should
Sabin buy it so that he can make 25% profit by selling for the same SP.

b. Suson sold a desktop at a loss of 5%. If he had sold it at Rs. 5200 more he would
have a profit of 8%, find the C.P. of the desktop computer.

12. a. Dipa bought 1200 copies for Rs. 60 each. If 300 copies were lost and she sold rest
copies at a loss of 5%, find the selling price of each copy.

b. Rajan bought 120 glasses and 20 glasses were broken. If he sold remaining
glasses at Rs. 300 each he would have a profit of 30%. Find the cost price of each
glass.

Project Work:

Collect the cost price and selling price of your books, copies, pencil and geometry box
from bookshop and fill the following table and find profit or loss by selling them.

S.N. Articles Marked price Discount price Selling price
1. Books
2. Copies
3. Peincils
4. Geomtry box

26 GREEN Mathematics Book-9

2.3 Discount

Marked Price Discount

The price which is listed on the article or At the time of selling, a shopkeeper
objects or items is called marked price reduces the original price of articles
or face value or labelled price or printed or goods to the customer. The reduced
price or original price. price on the marked price of goods is
called a discount.

Note:

Discount is usually expressed as a percentage of original price.

1. Discount (D) = discount rate (d%) × MP (marked price)

2. Selling price (S.P.) = M.P. – Discount.

3. Discount Amount (D) = M.P. – S.P.

4. Discount percentage (D) = Real discount amount × 100%
marked price

5. Marked price (M.P.) = S.P. + discount

Worked Out EXAMPLES

EXAMPLE 1 How much do you have to pay to buy the item shown
in the figure whose marked price is Rs. 5000?

Solution : Here, LED TV Rs. 5000
Discount rate = 5%,
Discount = 5%

Marked price (M.P.) = Rs 5000

We know,

Selling price (to pay) = M.P. – D% of M.P.

= Rs. 5000 – 5% of Rs. 5000

= Rs. 5000 – 5 × Rs. 5000
100

= Rs. (5000 – 250)

= Rs. 4,750

GREEN Mathematics Book-9 27

EXAMPLE 2 What percent of discount should be given in a doll costing Rs 180 such
that a customer has to buy it for Rs. 160?

Solution : Here,
Marked price (M.P.) = Rs. 180
Selling price (S.P.) = Rs. 160
We know,
Discount Amount = M. P. – S.P.
= Rs. 180 – Rs. 160 = Rs. 20

Now,
Discount percent (D) = oM.P. – S.P. × 100p%

M.P.

= o 20 × 100p%
180

= 100 %
9

= 11 1 %
9

EXAMPLE 3 A watch was sold for Rs. 595. If the watch was sold at 15% discount on
the marked price, find the marked price of the watch.

Solution : Here,
Discount rate (D) = 15%
Marked price (M.P.) = Rs. x (suppose)
We know,

S.P. = M.P. – D% of M.P
or, Rs. 595 = x – 15 × x

100
or, Rs. 595 = 100x – 15x

100
or, Rs. 595 × 100 = 85x

or, x = Rs. 700

∴ The marked price of the watch = Rs. 700.

28 GREEN Mathematics Book-9

EXAMPLE 4 The cost of an electric bulb is Rs. 150. If a shopkeeper sells the bulb by
allowing 10% discount, find the real selling price of the bulb.

Solution : Here,

Marked price (M.P.) = Rs. 150

Discount rate (D) = 10%

We have, S.P. = M.P. – D% of M.P.
= Rs. 150 – 10 × Rs. 150
100


= Rs. 150 – Rs. 150 = Rs. 135
10

∴ The real (actual) selling price of an electric bulb is Rs. 135.

EXAMPLE 5 Dipendra sold 6 books at the cost price of 8 books. What percent does
he get profi or loss?

Solution : Let, S.P. of 6 books = Rs. x

∴ S.P. of a book (S.P.) = Rs. x
6

∴ Cost price of a book = Rs. x

∴ Cost price of a books (C.P.) = Rs. x
8

Now, S.P. > C.P.

Profit percent (P) = o S.P.– C.P. × 100p%
C.P.

= o x – x × 100p%
6 x 8

8

4x–3x
24
= x × 100

8

= 33 1 %
3

∴ He sold a book at profit of 33 1 %
3

GREEN Mathematics Book-9 29

EXAMPLE 6 The marked price of a calculator is 20% more than its purchased price.
If the calculator was sold after allowing 15% discount, there will be a
gain of Rs. 40, find its marked price and cost price.

Solution : Let, C.P. of a calculator (C.P.) = Rs. x

∴ Marked price (M.P.) = Rs. x + 20% of Rs. x
= Rs. x + 20x

100
= Rs. 120x

100
= Rs. 6x

5
Discount percent (D%) = 15
Again,
S elling price of the calculator (S.P) = M.P. – D% of M.P.
= 6x – 15 × 6x

5 100 5
= 6x × 85

5 100
= 102x

100
Gain amount (G) = Rs. 40

We have, G = S.P. – C.P.


or, Rs. 40 = 102x – x
100

or, Rs. 40 = 102x – 100x
100

or, Rs. 4000 = 2x

∴ The cost price of the calculator (C.P.) = Rs. 2000

∴ The marked price of the calculator (M.P.) = Rs. 6x
5

= Rs. 6 × 2000
5

= Rs. 2400

30 GREEN Mathematics Book-9

EXAMPLE 7 After allowing 20% discount on the marked price of an article, there
is a profit of 28%. If the cost price of the article is Rs. 500, calculate the
discount amount of the article.

Solution : Here,

Let M.P. be Rs. x then (D) = 20%, P = 28%

C.P. = Rs. 500

Now, S.P. = M.P. – D% of M.P.

= Rs. x – 20 × Rs. x
100
= 4x
5

Profit amount = 28 × Rs. 500
100

= Rs. 140

By using formula,

P = S.P. – C.P.
or, Rs. 140 = 4x – Rs. 500

5

or, Rs. 640 = 4x
5

or, Rs. 640 × 5 = x
4

∴ x = Rs. 800

Discount amount = 20% of M.P.
= 20 × Rs. 800 = Rs. 160

100

EXAMPLE 8 When Binesh sold a mobile set at discount of 15%, he would have
gained Rs. 500. If he allowed discount of 30%, he would have lost Rs.
1000, calculate the marked price of the mobile set.

Solution : Let, marked price of a mobile set be Rs. x
First use:

Discount percent (D) = 15%

Gain amount (G) = Rs. 500

GREEN Mathematics Book-9 31

Now,
S elling price of the mobile set (S.P.) = M.P. – D% of M.P.
= Rs. x – 15 × Rs. x

100
= 17x

20
∴ The cost price (C.P.1) = S.P. – Gain

= Rs. o 17x – 500p
20

In second case:
∴ M.P. = Rs. x
D = 30%
Loss = Rs. 100
S.P. of the mobile set (S.P.) = M.P. – D% of M.P.

= x – 30 × x
100

= 7x
10

C.P. of the mobile set (C.P.2) = S.P. + Loss

= Rs. o 7x – 1000p
10

Now, C.P.1 = C.P.2
or, 17x – 500 = 7x + 1000

20 10
or, 17x – 7x = 1000 + 500

20 10
or, 17x – 14x = 1500

20
or, 3x = 1500

20
or, x = Rs. 10,000

∴ The marked price of the mobile set is Rs. 10,000

32 GREEN Mathematics Book-9

EXERCISE : 2.2

A. Very Short Questions
1. What do you mean by marked price?
2. Define discount.
3. The discount on worth Rs. 10 at 8% is ......
4. Fill in the gaps

a. Discount amount = ........................
b. M.P. – D% of M.P. = .......................
c. M.P. = S.P. .....................................

d. o Real discount amount × 100p% = ...........................
M.P.

e. Discount amount + S.P. = .............................

f. Fill in the given table

Marked price S.P. Discount Discount amount
a. Rs. 400 ... ..... ..... 20% ... ..... ..... ..
b. ... ..... ..... Rs. 6000 Rs. 40
Rs. 4750 ... ..... ..... ... ..... .....
c. Rs. 5000 Rs. 600 5% Rs. 200
d. Rs. 800
... ..... .....

B. Short Questions
5. a. What is the cash payment of a ball whose marked price is Rs. 1800 if a discount

of 5% is given.
b. The marked price of a calculator is Rs. 260. What is the selling price of it, if 5%

discount is allowed?
c. The cost of an object is Rs. 550. If a shopkeeper sells the object by allowing 10%

discount, find the actual selling price of the object.
6. a. Sabin bought a Samsung mobile set for Rs. 15750 after getting the discount of

10%. Find the price of the mobile set before discount.
b. Find the total selling price of a shirt whose marked price is Rs. 2700 when 13%

VAT is levied.
c. Marked price of an article is Rs. 1500. What is selling price with 10% VAT?

7. a. The marked price of a radio is Rs. 5000. What will be price of the radio if 10%
VAT is levied after allowing 15% discount on it?

b. The marked price of a motor bike was Rs. 1,50,000. What was the price of the
motorbike after allowing 10% discount with 13% VAT on it.

c. Allowing 20% discount on the marked price of a watch the value of the watch

GREEN Mathematics Book-9 33

will be Rs. 2376. When the sales tax of 10% is added find its marked price.

d. After allowing 25% discount on the marked price of a computer, an 15% tax was
levied on it. If its price became Rs. 17250 how much amount was given in the
discount?

8. a. After allowing 20% discount on the marked price of an article, there is a profit of
28%. If the cost price of the article is Rs. 500, calculate the discount amount.

b. Bhola allows 10% discount on the marked price of an item, if Ramita pays Rs.
4,068 with 13% value added tax then find marked price of item.

c. The market price of a LED tube is Rs. 1500 and 10% discount is allowed then
how much should a customer pay for it with 13% VAT.

d. After allowing 5% discount on the face value of a desktop table including 10%
VAT its price becomes Rs. 1672. How much amount was given in the discount?

C. Long Questions

9. a. After allowing 30% discount on the marked price of an article, there is a profit of
10%. If the cost price of the article is Rs. 1500, find its marked price of it.

b. Hari sold a radio after allowing discount of 20% on the marked price, there is a

loss 5 15 %. If the cost price of the radio is Rs. 3400, find marked price.
17

10. a. The marked price of T.V. set is 30% above the cost price. If the T.V. was sold after
allowing discount of 20%, there would be a gain Rs. 800, calculate the marked
price and its cost price.

b. The marked price of a book is 25% above the cost price of it. If the book was sold
after allowing 22% discount there would be a loss of Rs. 12.50. Calculate the
marked price and its cost price.

11. a. If Rekha sold a sari at a discount of 20%, she had gained Rs. 400. If she had
allowed a discount of 35%, she would have lost Rs. 50. Find the marked price of
the sari.

b. If Supreme sold a watch at a discount of 30%, he would loss Rs. 200. If he had
allowed a discount of 15%, he would have gained of Rs. 400. Find the marked
price of the watch.

12. a. The marked price of an article was fixed to Rs. 1380 by increasing 15% on its
actual price. Find its actual price.

b. The price of an item with 13% sales tax is Rs. 1356. Find the price of the item
excluding sales tax.

13. a. The price of an article with 15% sales tax is Rs. 690. What will be the price
excluding sales tax.

34 GREEN Mathematics Book-9

b. The price of an item with 3.3% sales tax is Rs. 5000.75. What will be its price
excluding sales tax.

14. a. A bicycle is sold at Rs. 9040 after allowing 20% discount and imposing 13% sales
tax. Find its marked price.

b. Sadan bought an article at a discount of 15% and paid Rs. 17289. If 13% sales tax
was levied on it, what was the marked price of the article?

15. a. A shopkeeper sold his goods for Rs. 16950 allowing 25% discount and then
levied 13% sales tax. Find the amount of discount?

b. The face value of a bag is Rs. 1400 and 10% discount is allowed on it, then find
the total cost price with 15% sale tax.

c. After allowing 20% discount on the face value (marked price) of a laptop with
15% VAT levied the and the price became 22,080. What amount of VAT was
levied.

Project Work:
Visit a shop in near your home and ask about listed price, cost price and discount
amount of different articles. Show the data to your teacher.

GREEN Mathematics Book-9 35

3

Commission and Taxation

Estimated Teaching Periods : 6

Johann Carl Friedrich Gauss (1777-1855) was a German mathematician
who contributed significantly to many fields, including number theory,
algebra, statistics, analysis, differential geometry, geodesy, geophysics.

Contents

3.1 Commission
3.2 Bonus (dividend)

Objectives

At the end of this unit, students will be able to:
solve the various problems related to commission.
define discount and its types
solve the various problems related to discount
solve the various problems related to taxation for example : income tax, value added
tax, etc.
tackle the problems on bonus (dividend)

Materials

Rate list of taxation, formula of bonus, dividend, discount, VAT, relation chart, etc.

36 GREEN Mathematics Book-9

3.1 Commission

In general, a sum of money paid to an agent or a seller by the company or a
manufacture for its service is known as commission. It is calculated in percent.

Note:
1. Wages or income = Fixed salary (basic salary) + commission.

2. Commission amount = commission rate × total sale

3. Commission rate = q Real commission amount × 100r%
Total sale

Worked Out EXAMPLES

EXAMPLE 1 A plot of land is sold for Rs. 30000000 through an estate agent who
1 1
charges a commission 3 % on the first 15 lakh and 4 % for the rest.

What sum does the land owner get?

Solution : Here,

Total selling price of land = Rs 30000000

= Rs. (1500000 + Rs. 2,8500000)

Now,

( ) 1 1
Total commission = Rs. 3 % at 1500000 + 4 % of 2,8500000

( ) = Rs. 1 + 1
3 × 100 × 15,00000 4 × 100 × 2,85,00000

= Rs. 5000 + Rs. 71,250

= Rs. 76,250

Finally, the sum of money received by the land owner is

= Rs. 30000000 – total commission amount

= Rs. 3000000 – Rs. 76250

= Rs. 2,99,23,750



GREEN Mathematics Book-9 37

EXAMPLE 2 Green Books pays to an agent the following commission based on the
daily sales.

Sales Commission rate
Up to Rs. 10,00,000
Up to Rs. 15,00,000 0.5%
Above Rs. 15,00,00 1%
2%

Calculate the commissions received by the agents in the sales of

a. Rs. 8,50,000 b. 12,50,000 c. Rs. 19,50,000

Solution : a. Here,

Sales amount = Rs. 8,50,000

Provided commission = 0.5%

Commission amount = 0.5 × Rs. 8,50,000
100

= 10500 × Rs. 8,50,000

= Rs. 4250

∴ The required commission is Rs. 4250

b. Here
Sales amount = Rs. 12,50,000
P rovided commission upto Rs. 10,00,000= 0.5%
Commission = 0.5% of Rs. 10,00,000
= 10500 × Rs. 10,00,000

= Rs. 5000
Remaining sales = Rs. (12,50,000–10,00,000)
= Rs. 2,50,000

Commission provided on sales above Rs. 10,00,000 = 1%
Commissions above Rs. 10,00,000 = 1% of Rs. 2,50,000
= 1010 × Rs. 2,50,000

= Rs. 2500
T otal commission on sales of Rs. 12,50,000 = Rs. (5000 + 2500)
= Rs. 7500

∴ The required commission is Rs. 7500

38 GREEN Mathematics Book-9

c. Here, Sales amount = Rs. 19,50,000

P rovided commission upto Rs. 10,00,000= 0.5%
Commission = 0.5% of Rs. 10,00,000
= 10500 × Rs. 10,00,000

= Rs. 5000
Commission of next Rs. 5,00,000 = Rs. (15,00,000–10,00,000)
= Rs. 5,00,000

Commission provided on sales above Rs. 5,00,000 = 1%
Commissions above Rs. 10,00,000 = 1% of Rs. 5,00,000
= 1010 × Rs. 5,00,000

= Rs. 5000
Remaining sales = Rs. (19,50,000–15,00,000)
= Rs. 4,50,000
Rate of commissions for Rs. 4,50,000 = 2%
Commission amount of Rs. 4,50,000 = 2% of Rs. 4,50,000
= 1020 × Rs. 4,50,000

= Rs. 9000
Total commission on sales of Rs. 19,50,000 = Rs. (5000 + 5000 + 9000)
= Rs. 19,000

EXAMPLE 3 ABC Company pays salary for a staff Rs. 12,000. If a staff got Rs. 21,000
in a month. If he gets commissions on Rs. 10,00,000, find the rate of
commissions.

Solution : Total income of a staff = Rs. 21,000

Salary of a staff = Rs. 12,000

total sales = Rs. 10,00,000

Let x be the rate of commission.

So, total income = Salary + Commission amount
x
or, Rs. 21000 = Rs. 12,00 + 100 × Rs. 10,00,000

or, Rs. 21000 – Rs. 12000 = Rs. 10000x

or, 9000
or, 10000 = x

0.9 = x

∴ The rate of commission is 0.9%

GREEN Mathematics Book-9 39

EXERCISE: 3.1

A. Very Short Questions

1. Fill in the gaps.
a. Real commission amount × 100 % ...........................
c. Wage or iTnoctoaml sea=lebasic salary + .............................
e. The commission on Rs. 10000 at rate 0.0008% is .............................

B. Short Questions is paid Rs. 40000 per month as salary and a commission of 1 % on
2 A branch manager 6

the total transaction is added to his salary. Find his income if the total transaction is

Rs. 2000000.

3. A commission agent is able to sell goods for Rs. 3000000. What commission does he
get at the commission rate of 13%.

4. An agent (broker) sold a piece of land for Rs. 6000000. If he got 2.5% commission, find
how much did the owner get?

5. A boarding school distributes 5% of its actual profit Rs. 97,448 among 20 teachers as
a bonus. Find the bonus amount that each teacher got.

6. A broker receives Rs. 4060 as a commission at rate of 4% by selling an article. What is
the price of the article?

C. Long Questions

7. Sadan sold his house at a discount of 5% on his marked price due to the payment
of cash. He further gave 5% commission of remainder to the broker and he gets Rs.
4000000. Find the marked price of his house.

8. Sabin allowed 10% discount to the buyer of his land and offered 5% commission to
Raman. If he received Rs. 800000, what was the fixed price of the land?

9. A bus owner has sold a bus for Rs. 19,50,000 through a broker. If he paid a commission
of 0.5% for the first Rs. 10,00,000 and 1.5% for the rest. What sum of money does the
bus owner receive?

10. Based on the monthly sales, a super market pays the following commission to an
agent

Sales Commission rate

Sales up to Rs. 50,000 0.1%
Up to next Rs. 80000 0.5%
0.99%
Above Rs. 130,000

40 GREEN Mathematics Book-9

Find the commission received by an agent for the sales of

a. Rs. 30,000 b. Rs. 75000 c. Rs. 2,50,000

11. The airline company distributes the yearly bonus to staff on the basis of their salary
scale.

Salary scale Bonus %
Upto Rs. 18,000 0.8%
Up to next Rs. 30,000 0.7%
Above Rs. 30,000 0.05%

If the total transaction is 20,00,000 in the fiscal year 072/073, find the bonus received
by the staff with the following basic salaries.

a. Rs. 15,000 b. Rs. 25,000 c. Rs. 38000

12. A wholseller get 0.5% commission upto 10 lakhs sales, 1% commission upto 10 lakhs
to 15 lakhs on the sales and above 15 lakhs he gets 5% commission from the company.
If wholeseller sales Rs. 27,50,000 in this month, how much commission does he get?

13. Tata bricks company allowed the following rate of commissions to the agents.

Sales in Rs. Commissions
Upto 15 lakhs 0.5%
15 lakhs to 25 lakhs 1%
25 lakhs to 40 lakhs 1.5%
40 lakhs and above 2%

Using the above rate of commission, find the total commission amount recieved by
agent on the following sales.

a. Rs. 18,00,000 b. Rs. 32,00,000 c. Rs. 48,00,000 d. Rs. 67,00,000

14. Solve the following problems:

a. Find the total sales of agent Mr. A, if he gets Rs. 20,500 at the end of the month
when his montly salary was Rs. 10,000 and rate of commission is 4%.

b. Mr. Binesh monthly salary is Rs. 15,000, if he gets altogehter Rs. 35,500 as salary
and commission. Find his total sales where rate commission is 10%.

c. Ramita works in a press company, her salary is Rs. 12,500. If she gets Rs. 1,55,000
as a salary and commission in a year, what was the total sales if the rate of
commission is 5%.

GREEN Mathematics Book-9 41

3.2 Bonus

In general, if a factory or a company or any organization increases its yearly profit on
the whole transaction it may give a certain percentage of the profit to their employees
as an incentive besides the fixed salaries. This incentive amount is known as bonus.

Note:
1. Bonus amount = Bonus rate × net profit
2. Bonus rate = q Real bonus amount × 100r%

Net profit
3. Bonus of each employee = Net bonus

No. of employee

3.3 Dividend

A share of the after-tax profit of a company, distributes to its shareholders according
to the number and class of shares held by them.
Smaller companies typically distribute dividends at the end of an accounting year,
whereas larger, publicly held companies usually distribute it every quarter. The
amount and timing of the dividend is decided by the board of directors, who also
determine whether it is paid out of current earnings or the past earnings kept as
reserve. Holders of preferred stock receive dividend at a fixed rate and are paid first.
Holders of ordinary shares are entitled to receive any amount of dividend, based on
the level of profit and the company's need for cash for expansion or other purposes.

Note:
1. Dividend amount = Number of share × rate of dividend
2. Rate of dividend = q Dividend amount × 100r%

Net dividend
3. Dividend of each share holder = Net dividend

Total no. of share

42 GREEN Mathematics Book-9

Worked Out EXAMPLES

EXAMPLE 1 A company made yearly profit of Rs. 150000. If the firm decided to dis-
Solution : tribute 10% of the yearly profit as the bonus to all 20 staffs equally, how
much bonus does each staff receive?

Here,

Yearly profit of the company = Rs. 150000

Now, Total bonus = 10% of 150000


= 11000 × 150000

= Rs. 15,000

Total number of staffs = 20

Finally,
Bonus received by each staff = Net bonus

No. of employee

= Rs. 15000
20

= Rs. 750

EXAMPLE 2 ABC Company distributes the yearly bonus to its employee on their
salary in the given scale.

Salary per month (Rs.) Bonus Rate
Rs. 12,000 – Rs. 15000
Rs. 15000 – Rs. 27000 0.30%
Rs. 27000 – Rs. 40,000 0.20%
0.16%

If the profit is Rs. 75,00,000 in a fiscal year, find the bonus receivd by
each of the employee drawing a salary of:

a. Rs. 12400 b. Rs. 15500 c. Rs. 28500

Solution : a. Here,
Total profit = Rs. 75,00,000
Bonus received by the employee drawing
Salary Rs. 12400 = 0.30% of Rs. 75,00,000
= 01.0300 × Rs. 75,00,000
= Rs. 22,500

GREEN Mathematics Book-9 43

b. Here,

Total profit = Rs. 75,00,000
B onus received by the employee drawing
Salary Rs. 15500 = 0.20% of Rs. 75,00,000

0.20
= 100 × Rs. 75,00,000
= Rs. 15000
c. Here,

Total profit = 75,00,000
Bonus received by the employee drawing
Salary Rs. 15500 = 0.16% of Rs. 75,00,000
= 01.0106 × Rs. 75,00,000
= Rs. 12000
∴ The required bonus are Rs. 22,000, Rs. 15000 and Rs. 12000

EXAMPLE 3 A finance company distributed 2,000 shares for Rs. 100 per share. The
net profit of this finance company is Rs. 2,00,000. The management of
company decided to distribute 25% dividend. How much amount of
dividend did per share receive?

Solution : Here,
The total number of share = 2,000

Price of each share = Rs. 100

Total net profit = Rs. 2,00,000

Profit amount for dividend = 25% of Rs. 2,00,000

= 12050 × Rs. 2,00,000

= Rs. 50,000

∴ The required dividend for each share = Net dividend
No. of share

= Rs. 50000
2000

= Rs. 25

44 GREEN Mathematics Book-9