Green C Math Class 10

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

Green

Mathematics

10

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 10

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 1-9

1.1 Revision

1.2 Number of elements in the union of two sets

1.3 Number of elements in the union of three sets 10-22
Unit 2 : Profit and Loss 23-41
2.1 Problems involving VAT, Discount and Marked Price
2.2 Money exchange
Unit 3 : Compound Interest

3.1 Compound Interest

3.2 Population growth and compound depreciation 42-76
Unit 4 : Mensuration

4.1 Introduction

4.2 Prism

4.3 Area and volume of cylinder

4.4 Pyramid : Introduction

4.5 Cone : Introduction 77-89
Unit 5 : H.C.F. and L.C.M.

5.1 Highest Common Factor ( H.C.F )

5.2 Lowest Common Multiple ( L.C.M. ) 90-126
Unit 6 : Indices and Surds 127-136
6.1 Introduction
6.2 Laws of indices
6.3 Exponential equations
6.4 Surd
6.5 Rationalization of surds
6.6 Equation involving surds
Unit 7 : Simplification of Algebraic Expressions

7.1 Introduction 137-152
Unit 8 : Equations

8.1 Word problems on two variables equations

8.2 Word problems on quadratic equations 153-169
Unit 9 : Area of Triangle and Quadrilateral

9.1 Review

9.2 Theorems

i. Area of parallelograms on the same base between same parallel lines are equal in area.

ii. Area of triangle is one half area of parallelograms on the same base between same parallel line.

iii. Area of triangles on the same base between same parallel line

Unit 10 : Construction 170-177
178-211
10.1 Construction
Unit 11 : Circle 112-234

11.1 Review 235-266

11.2 Important axiom 267-286
287-304
11.3 Theorems related to arcs and the angles subtended by them 305-322

11.4 Tangents to a circle
Unit 12 : Trigonometry

12.1 Area of triangles and quadrilateral

12.2 Hight and distance

Test questions
Unit 13 : Statistics

13.1 Introduction

13.2 Quartiles-Introduction

13.3 Cumulative frequency curves or ogives
Unit 14 : Probability

14.1 Review/Introduction

14.2 Tree diagram

Revision problems for examination

Answers

Basic Level Final Examination

Specification Grid - 2069 | Sub : Compulsory Mathematics Class - 10

S.N. Topics Units K C A HA Total Remarks

1. Sets 1. Sets - - 1×4 - 4

2. Taxation and currency At least 2 marks

2. Airthmetics 3. Compound Interest 1×1 2×2 1×4 1×5 14 questions from

4. Populaton growth/Depreciation each units.

5. Plane surface At least 3 marks

3. Mensuration 6. Cylinder and sphere 1×1 3×2 1×4 1×5 16 questions from

7. Prism and Pyramid each units.

8. L.C.M. and H.C.F.

9. Surds At least 3 marks

4. Algebra 10 Indices 1×1 5×2 2×4 1×5 24 questions from

11. Algebraic expressions each units.

12. Equations

13. Area of triangle and quadrilateral At least 4 marks

5. Geometry 14. Construction 2×1 3×2 3×4 1×5 25 questions from

15. Circle each units.

6. Trigonometry 16. Trigonometry - 1×2 1×4 - 6

7. Statistics 17 Statistics 1×1 1×2 1×4 - 4

8. Probability 18. 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
1.2 Number of elements in the union of two sets
1.3 Number of elements in the union of three sets

Objectives

At the end of this unit, students will be able to:
define set and its types, represent sets in Venn-diagram
find the relation between sets
perform the set operation using a Venn-diagram
represent cardinality of a set in a Venn-diagram
research and 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,
flash cards, etc.

GREEN Mathematics Book-10 1

1.1 Revision

Problem involving sets using Venn-diagram
In 1880 AD, John Venn introduced this diagram in his journal article which is very useful
to find the cardinality of different subsets.

1.2 Number of elements in the union of two sets

Let U be the universal set and A and B be two non-empty subsets of U, then

n(A) = n0 (A) + n(A∩B) A U
or, n0 (A) = n(A) – n (A∩B) n0(A) B
Similarly, n(B) = n0 (B) + n (A∩B)
or, n0(B) = n(B) – n (A∩B) n0(B)
From the given Venn-diagram.
n(A∩B)

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 are,

(a) n(A – B) = n0 (A) = n(A) – n (A∩B)
(b) n (B – A) = n0 (B) = n(B) – n (A∩B)
(c) n(U) = n (A) + n(B) – n (A∩B) + n (A∪B)

∴ n(U) = n(A∪ B) + n (A∪B)

(d) Exactly one = n0 (A) + n0(B)
(e) At least one = n(A∪B)

Worked Out EXAMPLE

EXAMPLE 1 Out of 50 students of class IX, 25 students liked to play football, 35
liked to play basketball and 15 liked to play both the games. How
many students do not like to play any game? Show the above infor-
mation in a Venn-diagram.

Solution : Let U, B and F represent the universal set, B U
students who like to play basketball and stu- F
dents like to play football respectively.
20 15 10
n(U) = 50, n(B) = 35
n(B∪F) = 5
n(F) = 25, n(B∩F) = 15

2 GREEN Mathematics Book-10

Let, n (B∪F) = x
We have,
n(U) = n(B) + n(F) – n (B∩F) + n(B∪F)
or, 50 = 35 + 25 – 15 + x
or, x = 50 – 35 – 25 + 15
∴ x = 5
∴ Total number of students who do not like to play any game is '5'.

1.3 Number of elements in the union of three sets

Let U be universal set where A, B, and C are non-empty subsets of U.
Representing the three sets A, B and C in a Venn-diagram, we have

A orn, (0(BB–C) ∪A)U n0(A∩B)
n0(A) B or, n(A∩B) – C

or, (A–B∪C)

n0(A∩C) n0(C) C n0(B∩C)
or, n(A∩C) – B or, n(B∩C) – A
or, (C–A∪B)

Now, n(A∩B∩C)

n(A∪B∪C) = n [A∪(B∪C)]

= n(A) + n(B∪C) – n [A∩(B∪C)]

= n(A) + [n(B) + n(C) – n(B∩C)] – n [(A∩B) ∪ n(A∩C)]

= n(A) + n(B) + n(C) – n(B∩C) – [n(A∩B) + n (A∩C) – n[(A∩B)∩n(A∩C)]

= n(A) + n(B) + n(C) – n(A∩B) – n (B∩C) – n (C∩A) + n(A∩B∩C)

∴ n(A∪B∪C) = n(A) + n(B) + n(C) – n(A∩B) – n(B∩C) – n(A∩C) + n(A∩B∩C)

Some relations are,

a. n(U) = n (A∪B∪C) + n(A∪B∪C)
= n(A) – n(B∩C)
b. n0 (A) = n(B) – n(A∩C)
n0 (B) = n(C) – n(A∩B)
n0 (C) = n(A) – (A∩B) – n(A∩C) + n(A∩B∩C)
c. n0(A) = n(B) – (A∩B) – n(B∩C) + n(A∩B∩C)
n0(B) = n(C) – n(A∩C) – n(B∩C) + n(A∩B∩C)
n0(C)

GREEN Mathematics Book-10 3

d. At least one = n(A∪B∪C)

e. Exactly two = n0 (A∩B) + n0 (B∩C) + n0 (C∩A)
f. n0 (A∩B) = n(A∩B) – n(A∩B∩C)
n0 (B∩C) = n(B∩C) – n (A∩B∩C)
n0(C∩A) = n(C∩A) – n (A∩B∩C).

EXAMPLE 2 In a survey of group of people, 60 liked tea, 45 liked coffee, 30 liked milk,
25 liked coffee as well as tea, 20 liked tea as well as milk, 15 liked coffee
as well as milk and 10 liked all three. How many people were asked this
question? Solve by using Venn diagram.

Solution : Let C, T and M represent the set of people who liked coffee, tea and

milk respectively, then,

n(T) = 60 U
CT

n(C) = 45 n(C) = 45 15 2515 n(T) = 60
n(M) = 30 5 10 10

n(C∩T) = 25, n(T∩M) = 20 5M

n(C∩M) = 15, n(C∩T∩M) = 10 n(M) = 30
Now,

n(T∪C∪M) = n(T) + n(C) + n(M) – n(C∩T) – n(T∩M)

–n(C∩M) + n(T∩C∩M)

= 60 + 45 + 30 – 25 – 20 – 15 + 10

= 85

∴ n(C∪T∪M) = 85
Hence, the required number of people = 85.

EXAMPLE 3 In a group of 100 students the ratio of students who like geometry and
trigonometry is 3:5. If 30 of them like both subjects and 10 of them like
none of the subjects then by drawing a Venn-diagram, find how many
students of them like geometry only and trigonometry only.

Solution : Let G and T denote the set of students who like geometry and trigo-
nometry respectively. Let n(G) = 3x and n(T) = 5x.

n (U) = 100 (say), n(G∩T) = 30 U
GT

n (G∪T) = 10 n(G) = 3x 3x–30 30 5x–30 n(T) = 5x

n(U) = n(G) + n(T) – n (G∩T) + n(G∪T) n(G∪T) = 10

4 GREEN Mathematics Book-10

or, 100 = 3x + 5x – 30 + 10
∴ x = 15
Now, n(G) = 3x = 3 × 15 = 45
and n(T) = 5x = 5 × 15 = 75
Finally, n0 (G) = n(G) – n(G∩T) = 45 – 30 = 15
n0(T) = n(T) – n(G∩T) = 75 – 30 = 45

EXAMPLE 4 In Kumaripati, out of 140 houses 60 believe in Buddhism, 70 in Hindu-
Solution :
ism and 45 in other religions. Among them, 17 houses believe in Hindu-

ism and Buddhism, 18 houses believe in Hinduism and other religions,

and 16 houses believe in Buddhism and other religions. 6 houses don't

believe in any religion. Find the number of houses believing all three

religions. Also show them in a Venn-diagram . U

B H n(B) = 60
7 45 n(H) = 70
Let H, B and O represent the set of houses 37
who believe in Hinduism, Buddhism and 10
68
other religion respectively.

Now, n(U) = 140 21 n(O) = 45
6O

n(B) = 60, n(H) = 70, n(O) = 45, n(B∩H) = 17

n(H∩O) = 18, n(B∩O) = 16 and n(B∪H∪O) = 6

We have,

n(U) = n(B) + n(H) + n(O) – n(B∩H) – n (H∩O) – n(B∩O) +
n(B∩H∩O) + n(B∪H∪O)

or, 140 = 60 + 70 + 45 – 17 – 18 – 16 + n(B∩H∩O) + 6

or, 140 = 130 + n(B∩H∩O)

or, n(B∩H∩O) = 10

∴ 10 houses believe in all three religion.

EXAMPLE 5 In a survey of group of students, it was found that 39% of students like
Solution : English only, 27% of students like Maths only and 15% of students do
not like both subjects. Find :

i. What percent of the students like English subject?

ii. What percent of the students like Maths subject?

iii. Represent the above information in Venn-diagram. U

Let E and M denote the number of students who E M
like English and Maths respectively,

Then, n(U) = 100 39 x 27

n0(E) = 39 15

GREEN Mathematics Book-10 5

n0(M) = 27

n0(M) = 27

n0(E∪T) = 15

Let n(E∩M) = x

By using Venn-diagram

n(U) = n0(E) + n0(M) + n(E∩M) + n(E∪M)
or, 100 = 39 + 27 + x + 15

or, 100 = 81 + x

∴ x = 19 i.e. n(E∩M) = 19

Now,

i. n(E) = n0(E) + n(E∩M)
= 39 + 19

= 58

ii. n(M) = n0(M) + n(E∩M)
= 27 + 19

= 46

EXAMPLE 6 In a group of students, 15 read English, 12 read Nepali, 10 read Account,
6 Nepali and English only, 4 read Nepali and Account only, 3 read Eng-
lish and Account only and there is no one who reads Nepali only. Draw
a Venn-diagram for this information and find :

i. Total number of students in the group.

ii. Number of students reading Account only.

Solution : Let E, N and A denote the number of students who read English,
Nepali and account respectively.

Now, n(E) = 15 E U n(N) = 12
n(E) = 15 N n(A) = 10
n(N) = 12 46
n(A) = 10 x 0
n0(E∩N) = 6
n0(N∩A) = 4 3 4
n0(E∩A) = 3
n0(N) = 0 1
Let n(E∩N∩A) be x A

6 GREEN Mathematics Book-10

We know,

n(N) = n0(E∩N) + n(E∩N∩A) + n0(N∩A) + n0(N)
or, 12 =6+x+4+0

or, x = 12 – 10

or, x =2

∴ n(E∩N∩A) = 2

n(E) = n0(E∩N) + n(E∩N∩A) + n0(E∩A) + n0(E)

or, 15 = 6 + 2 + 3 + n0(E)

or, 15 = 11 + n0(E)

or, n0(E) = 15 – 11

∴ n0(E) = 4

n(A) = n0(E∩A) + n(E∩N∩A) + n0(N∩A) + n0(A)

or, 10 = 3 + 2 + 4 + n0(A)

or, 10 = 9 + n0(A)

or, n0(A) = 10 – 9

or n0(A) = 1

∴ 1 student read Account only.

Now,

Total number of students n(U) = 4 + 6 + 2 + 0 + 4 + 3 + 1

= 20

EXERCISE 1.1

A. Very Short Questions

1. What do you mean by sets?
2. Fill in the gaps:

a. n(only A) is denoted by ............................
b. Exactly one or only one in 3 sets is ............................
c. n0(B) = n (B) – [............................]
d. ............................ = n(A) – n(A∩B) – n (A∩C) + n (A∩B∩C)

GREEN Mathematics Book-10 7

e. If n(P) = 40, n(Q) = 60 and n(P∩Q) = 20 then the value of n(P∪Q) is ............................

f. If n(U) = 100, n(A) = 65, n(B) = 50, n(C) = 35 n(A∩B) = 25, n(B∩C) = 20, n(C∩A) =
15 and m(A∩B∩C) = 5 then the value of n (A∪B∪C) is ............................

g. n0 (B∩C) = ............................

h. The value of A – (B∪C) = ............................

i. The value of (B∩C) = ............................

j. n(U) – n (A∪B∪C) = ............................

B. Short Questions
3. In a survey of some students, it was found that 60% of the students studied Science

and 70% studied Arts. If 40% students studied both the subject and 10% did not study
any subject, by drawing a Venn-diagram calculate percent of students who studied-
Science only.

4. Out of 250 tourist arriving to visit Nepal 40% of them have already visited Pokhara
and 30% visited Lumbini and 10% of them have already visited both the places. Cal-
culate the number of tourists who have not visited Pokhara and Lumbini and tourists
who have visited Lumbini only by using a Venn-diagram.

5. If n(U) = 120, n(A) = 48, n(B) = 51, n(C) = 40, n(A∩B) = 11, n(B∩C) = 10, n(C∩A) = 9 and
n(A∩B∩C) = 4 find value of n(A∪B∪C).

6. If n(U) = 50, n0(P) = 20, n0(Q) = 6, n0(R) = 5, n0(P∩Q) = 4, n0(Q∩R) = 3, n0(P∩R) = 2 and
n(P∪Q∪R) = 6, find the value of n (P∩Q∩R) .

C. Long Questions

7. In a survey, it was found that 80% people liked orange, 85% liked mangoes and 75%
liked both. If 45 people liked none of them, by drawing a Venn-diagram, find the
number of people who were in the survey.

8. In a group of 95 students the ratio of students who like mathematics and science
is 4:5. If 10 of them like both the subjects and 15 of them like none of the subjects
then by drawing a Venn-diagram, find how many of them like maths only and sci-
ence only.

9. In a survey, it was found that the ratio of the people who liked 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. Represent the above information in a Venn-diagram.
ii. Find the number of people who participated in the survey.

10. In a survey of 195 people it was found that 25 liked tea only and 150 liked coffee. The
number of people who liked coffee is 3 times the number of people who liked tea.
By using a Venn-diagram, find the number of people who liked both and who liked
none of them.

8 GREEN Mathematics Book-10

11. Of the total candidates in an examination 40% students passed in Maths, 45% in
Science and 55% in Health. 10% students passed in both subjects Maths and Science,
20% in Science and Health and 15% in Health and Maths then:

i. draw a Venn-diagram to show the given information
ii. Calculate the percent of candidates who passed in all three subjects.
12. In a survey of 100 people, 65 read Kantipur, 45 read Gorkhapatra, 40 read Himalayan

times, 25 read Kantipur and Gorkhapatra, 20 read Kantipur and Himalayan times, 15
read Gorkhapatra and Himalayan times and 5 read all three subjects now:
i. draw the Venn-diagram to represent the above information.
ii. find how many people do not read all the subjects.
13. Out of 1350 candidates of PSC, 600 passed in administration, 700 in revenue, 350 in
account, and 50 failed in all three levels. If 200 passed in administration and revenue,
150 in administration and account, 100 in revenue and account, find how many can-
didates passed in all groups using a Venn-diagram.
14. In a group of students of a school, 21 are in the football team, 26 in the basketball
team and 29 of them in volleyball team. If 14 played basketball and volleyball, 12
played volleyball and football, 15 played basketball and football and 8 played all
three games, using Venn-diagram find the total number of players and no. of players
who play volleyball only.
15. A survey of village showed that 50% drink milk, 42% drink tea, 48% drink coffee, 20%
drink only milk and coffee, 15% drink only milk and tea, 15% drink only coffee and
tea and 10% drink all three drinks.
i. Find the percent of people who do not drink any one of the above.
ii. Illustrate the above information in Venn-diagram.
16. In a group of students, 34 study Maths, 40 study Biology, 24 study Physics, 8 study
Maths only, 14 study Biology only, 6 study Biology and Physics only and 12 study
Maths and Biology only. Find :
i. How many study all three subjects?
ii. How many students were there in the group?

GREEN Mathematics Book-10 9

2

Taxation and Discount

Estimated Teaching Periods : 10

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 Problems involving VAT, Discount and Marked Price
2.2 Money exchange

Objectives

At the end of this unit, students will be able to:
find the selling price/discount cost price.
exhcnage currency of different countries.

Materials

Chart paper, chart of formulae, list of rate of currency.

10 GREEN Mathematics Book-10

2.1 Problems involving VAT, Discount and Marked Price

Marked Price
The price which is listed on the article or an objects or items is called the marked price
or face value or levelled price or printed price or original price.

Discount

At the time of selling good or items, shopkeeper reduces the original price and sells
the items or goods to costumer. The reduced price in the marked price of the goods
is called a discount.

Note :
i. Discount is usually expressed in percentage of the original price

ii. Discount = discount rate × marked price

In generall,

– S.P. = Marked price – discount

– S.P. = M.P. – d% of M.P. = M.P. o1 – d p
100

– Discount Amount = M.P. – S.P.

– Discount percentage = q Real discount amount × 100r%
Marked price

Tax

Generally tax means, the yearly payment of a certain sum of money made by an
individual or an organization to the local or central government or related institutions
for the services or speciall work is known as tax. There are various kinds of taxes.
Income tax, Road tax, Value Added Tax (VAT), etc. are types of taxes.

VAT (Value Added Tax)

The full form of VAT is Value Added Tax which is generally levied on the actual
selling price (sales tax). The rate of VAT varies from country to country. In Thailand
7% VAT is levied, but in our country, it is 13%.

Note:

i. VAT amount = Rate of VAT% of selling price

ii. S.P. with VAT = S.P. + VAT% of S.P. = S.P. o1 + V1A00Tp

Generally,

VAT = Real VAT amount × 100%
Selling price

GREEN Mathematics Book-10 11

Worked Out EXAMPLES

EXAMPLE 1 Sadan bought a mobile set for Rs. 15750 after allowing the discount of
Solution : 10%. Find the price of the mobile set before the discount.

Discount rate = 10%

Price after discount = Rs. 15750

Let marked price be x

We have, S.P. = M.P. – D% of M.P.


or, 15750 = x – 10% of x

or, 15750 = x – 10x
100
9x
or, 10 = 15750

or, x = 15750 × 10
9

= Rs. 17500

∴ The marked price = Rs. 17500.

EXAMPLE 2 How much should you have to pay to buy the volleyball shown in
Solution : the figure.

Discount rate = 10%

Marked price = Rs. 1000 MRP = Rs.1000
We have,

S.P. = M.P. – D% of M.P. Discount = 10%

or, S.P. = Rs. 1000 – 10 × 1000
100

or, S.P. = Rs. 900.

∴ S.P. = Rs. 900

EXAMPLE 3 Find the total selling price of shirt whose marked price was Rs. 2700
Solution : when 13% VAT is levied?

Marked price (M.P.) = Rs. 2700

VAT = 13%

S.P. = Rs. x (Suppose)

We have, S.P. = M.P. + V% of M.P.
or,
= Rs. 2700 + 13 × 2700
100

S.P. = 2700 + 351

∴ S.P. = Rs. 3051

12 GREEN Mathematics Book-10

EXAMPLE 4 What percent of discount should be given in a doll costing Rs. 180
Solution : such that a customer has to buy it for Rs. 160?

Marked price (M.P.) = Rs.180

Selling price (S.P.) = Rs.160

We have,

Discount Percentage = Real discount amount × 100%
Marked price

= (M.P. – S.P.) × 100%
M.P.

= (Rs. 180 – Rs. 160) × 100%
Rs. 180

= 100 %
9

= 11 1 %
9

∴ Discount percent = 11 1 %
9

EXAMPLE 5 The marked price of a camera is Rs. 3200 and a shopkeeper announces
a discount of 8%. How much will a customer have to pay for buying
the camera if 10% VAT was levied on it?

Solution : Marked price (M.P.) = Rs. 3200

Discount = 8%

VAT = 10%

Now,

Selling price after discount

S.P. = M.P. – D% of M.P.

= Rs. 3200 – 8 × Rs. 3200
100

= Rs. 3200 – Rs. 256

= Rs. 2944

Again, S.P. with VAT.

S.P. = Rs. 2944 + 10 × Rs. 2944
100

= Rs. 2944 + Rs. 294.4

∴ S.P. = Rs. 3238.40

GREEN Mathematics Book-10 13

EXAMPLE 6 After allowing 15% discount on the marked price of a radio 13% VAT
Solution : is levied, then the cost of radio becomes Rs. 1997.84. Find the marked
price.

Discount = 15%

Selling Price = ?

Marked price (M.P.) = x (say)

We have,

S.P. = M.P. – D% of M.P.

= x – 15% of x

Again, S.P. with VAT = x – 15x = 17x
100 20

S.P. = 17x + 13 × 17x
20 100 20

= 17x + 221x
20 2000
Finally,
= 1921x
2000

S.P. = 1921x
2000

Selling price with VAT = Rs. 1997.84

or, Rs. 1997.84 = 1921x
2000

or, 2000 × Rs. 1997.84 = 1921x

x = M.P. = Rs. 2080.

EXAMPLE 7 A machine is sold at Rs. 99000 after 10% discount and 10% VAT on the
marked price. Find the discount and VAT amount.

Solution : VAT rate = 10%

Selling price with VAT = Rs. 99000

Discount rate = 10%

Marked price (M.P.) = x (say)

S.P. = M.P. – D% of M.P.

= x – 10% of x

= x – 10x = 9x
100 10

14 GREEN Mathematics Book-10

Again, = 9x + 10 × 9x
S.P. with VAT 10 100 10

= 9x + 9x
10 100

= 99x
100

S.P. with VAT = 99x
100

or, Rs. 99000 = 99x
100

or, 1000 × 100 = x

∴ x = Rs. 100000
Finally,
Discount = 10% of 100000

= 10 × 100000
100

= Rs. 10000

Now, S.P. = M.P. – discount

= Rs. (100000 – 10000)

= Rs. 90000
VAT amount
= 10 × Rs. 90000
100

= Rs. 9000

EXAMPLE 8 After allowing 20% discount on the marked price and then levying
Solution : 10% VAT, a radio was sold. If buyer had paid Rs. 320 for VAT, how
much was the discount?

Discount rate (D) = 20%

Marked price (M.P.) = x (say)

S.P. after discount (S.P.) = M.P. – D%of x

S.P. = x – 20 x
100

= 4x
5

∴ VAT = VAT% of S.P.

= 10 × 4x = 2x
Now, 100 5 25

VAT Amount = Rs. 320

320 = 2x
25

or, x = 320 × 25 = Rs. 4000
Finally, 2

Discount amount = x = 4000 = Rs. 800.
5 5

GREEN Mathematics Book-10 15

EXAMPLE 9 The marked price of an article is Rs. 4,500. After allowing some percent
of discount and adding 10% VAT, the article is sold for Rs. 4400. Find
the discount percent.

Solution : Marked price (M.P.) = Rs. 4,500

Discount amount (D) = x(say),

SP = MP – D

SP = Rs. (4500 – x)
S.P. with VAT = S.P. + VAT% of S.P.
Now,


or, 4400 = (4500 – x) + 10 (4500 – x)
100

or, 4400 = 4500 – x + 450 – x
10

or, 4400 – 4950 = – o 10x + x p
10

or, Rs. 5500 = x
11

∴ x = Rs. 500

Discount amount = Rs. 500

Finally,

Discount percent = Real discount amount × 100%
Marked price

= Rs. 500 × 100%
Rs. 4500

= 11.11%

EXAMPLE 10 The marked price of an article is 25% above the selling price and the
cost price is 20% below the selling price. Find the rate of discount and

the profit percentage.

Solution : Let, selling price of article (S.P.) = Rs. x

Marked price (M.P.) = x + 25% of x = x + 25 × x = Rs. 5x
100 4

Again, Cost price (C.P.) = x – x × 20 = 5x – x = Rs. 4x
100 5 5

∴ Discount percent = M.P. – S.P. ×100% = 5x –x
M.P. 4 ×100% = 20%
5x

4

∴ Profit percent = S.P. – C.P. x– 4x ×100% = 25%
C.P. ×100% = 4x 5

5

16 GREEN Mathematics Book-10

EXERCISE 2.1

A. Very Short Questions

1. Define Value Added Tax.
2. Fill in the blanks.

a. .................... amount = marked price – selling price.

b. Selling price with VAT ....................

c. VAT percentage = q ........................ × 100r%
Selling price

d. Actual selling price after discount = M.P. – [..........................]

e. Actual selling price after VAT = [........................] + [VAT% of S.P.]

f. .............................. amount = VAT% of S.P.

g. S.P. = M.P. – [..............................]

h. S.P. with VAT = S.P. + [..............................]

i. .............................. amount = D% of marked price.

j. If A = B, B = C, and C = x then the value of x = [..............................]

k. If the marked price of bag is Rs 1000, after 10% discount its price will be = Rs ...
............................

l. If the selling price of a jacket is Rs. 5000, after including 13% VAT its price will
be = Rs. ...............................

B. Short Questions

3. a. What will be the cash payment of the bill of amount Rs. 1800 allowing a discount
of 5%.

b. The cost of an article is Rs. 550. If a shopkeeper sells the article by allowing 10%
discount, find the actual selling price of the article.

c. The selling price of a watch is Rs. 595. If the watch was sold at 15% discount on
the marked price, find the marked price of the watch.

d. A cycle costs Rs. 12650 with 15% VAT, what will be the price excluding VAT?

e. A student bought a calculator for Rs. 565 including 13% VAT, what is the price
of the calculator before VAT?

f. The price of an article with 13% VAT is Rs. 1356. Find the price of the article
excluding VAT.

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

h. Selling price of an article is Rs. 1500. What is the actual selling price with 10%
VAT?

GREEN Mathematics Book-10 17

i. Find the total selling price of a pair of pants whose selling price is Rs. 2700 with
13% VAT.

j. The lebelled price of an electric bulb is Rs. 150. If a shopkeeper sells the bulb by
allowing a 10% discount, find the selling price of bulb.

k. The cost price of an article is Rs. 550. If a shopkeeper sells the article by allowing
10% discount, find the actual selling price of the article.

l. The marked price of a calculator is Rs. 260. What is the selling price of it, if 5%
discount is allowed?

C. Long Questions
4. The marked price of a bicycle was Rs. 5550. What will the price of the bicycle be if

15% VAT was levied after allowing 10% discount on the marked price?

5. The marked price of a bicycle is Rs. 1,50,000. What will be the price if after 10%
discount 13% VAT is included in the price?

6. Sabin bought an article at a discount of 15% and paid Rs. 17289. If 13% value added
tax was levied on it, what was the marked price of the article? Find it.

7. A bicycle is sold at Rs. 9040, after allowing 20% discount and imposing 13% VAT.
Find the marked price of the bicycle.

8. After allowing 10% discount on the marked price of an article 15% VAT is levied.
Find the marked price if its price becomes Rs. 16720.

9. After allowing 25% discount on the marked price of a computer 15% VAT was levied
on it. If its price becomes Rs. 10350, how much was given in the discount?

10. After allowing 10% discount on the marked price of an article and levying 13% value
added Tax (VAT), an article was sold for Rs. 9429. Find the VAT amount.

11. A tourist buys an article at 15% discount on the marked price and he pays 5% VAT. If
the VAT amounts to Rs. 850, find the marked price and selling price including VAT
amount .

12. The marked price of a bicycle is Rs. 2000. If the shopkeeper allows some discount and
a customer bought it for Rs. 1921 including 13% VAT, how much was given in the
discount.

13. After allowing 25% discount on the marked price and then levying 10% VAT, a cycle
was sold. If the discounted amount was Rs. 750, how much VAT was levied on the
price of the cycle?

14. After allowing 20% discount on the marked price and levying 13% VAT, a cycle was
sold. If the discounted amount was Rs. 850, how much VAT was levied on the price
of the cycle?

14. The marked price of an article is Rs. 2000 and shopkeeper allows 10% discount.
After levying VAT on it a customer pays Rs. 2,034 for it. What percentage of VAT is
charged?

15. The marked price of an article is Rs. 5000 and shopkeeper allows some % of discount.

18 GREEN Mathematics Book-10

After levying 13% VAT on it, a customer pays Rs. 5100. What percentage of discount
is charged?

16. A tourist buys a Nepalese flag at a discount of 15% and pays 10% VAT. If he pays Rs.
170 for VAT, then find:

a. the marked price of the flag

b. the amount he paid for the flag while buying.

17. The marked price of an article is Rs. 2200. After allowing a certain percent of discount
13% value added tax (VAT) was levied. If the article is sold for Rs. 2063.38, find the
discount percentage.

18. After allowing 10% discount on the marked price of a car, 15% VAT is charged on it.
If its price becomes Rs. 450300, what amount was given in the discount?

19. a. When an article is sold at a discount of 10% on its marked price, a profit of Rs.
8 is earned by the seller. If the same article is sold without allowing a discount
there will be a profit of Rs. 20. What should be the cost price of the article?

b. A shopkeeper selling an article at discount of 20%, loses Rs. 200. If he allows
10% discount, he gains Rs. 150. Find the marked price and cost price of the
article.

c. Ram bought two calculators for Rs. 800. He sold them to gain 20% on one and
loses 20% on the other. Calculate his final gain or loss percent if the selling price
of both the calculator is the same.

d. The marked price of a radio is 25% above the selling price and the cost price is
30% less than its marked price, find the discount percent and gain percent.

e. A watch was sold on its marked price at a gain of 20% but after allowing 5%
discount, there would have been Rs. 140 gained. At what price the shopkeeper
purchased the watch?

20. a. The list price of an article is 25% above the selling price and the cost price is 40%
below the list price. Find the rate of discount and profit percent.

b. The marked price of an article is 25% above its selling price and the cost price is
30% less than its marked price. Find the discount percent and gain percent.

21. a. The price of an article is marked at 25% above the cost price. If it is sold at a
profit of Rs. 125 after allowing 15% discount, what will be its selling price?

b. A computer after allowing a discount of 20% on its marked price was sold at
a gain of 12%. Had it been sold after allowing 30% discount there would have
been loss of Rs. 500. Find the marked price and the cost price of the computer.

c. A shopkeeper marked the price of an article at 30% above the cost price and
allows 15% discount at the time of selling. Find his profit percentage.

d. A tourist buys an article at 15% discount on the mark price and he pays 5%
VAT. If the VAT amounts to Rs. 850, find the marked price and the selling price
including the VAT.

GREEN Mathematics Book-10 19

2.2 Money exchange

In the world each country has its own currency. For example, Currencies of Thailand,
UK and Nepal are Bhat, Pound Sterling and rupees respectively.

In this chapter, we study about exchange of Nepali currency with the currency of
other countries.

Note: The bank or money exchanger will make some profit from the differences in
i. selling and buying rate

ii. Calculating or converting currency using chain rule are as follows :

If A = B, B = C and C = x (say)

A × B × C = B × C × x

∴ x = o A× B × C p
B × C

Exchange Rates Fixed by Nepal Rastra Bank

Currency Unit Buying/Rs. Selling/Rs.
Indian Rupee 100 160.00 160.15

Open Market Exchange Rates

(For the purpose of Nepal Rastra Bank)

Currency Unit Buying/Rs. Selling/Rs.

U.S. Dollar 1 106.09 106.69

European Euro 1 115.80 116.45

UK Pound Sterling 1 131.68 132.43

Swiss Franc 1 107.51 108.12

Australian Dollar 1 81.79 82.25

Canadian Dollar 1 78.99 79.44

Singapore Dollar 1 75.60 76.03

Japanese Yen 10 10.02 10.07

Chinese Yuan 1 15.62 15.71

Saudi Arabian Riyal 1 28.29 28.45

Qatari Riyal 1 29.14 29.30

Thai Baht 1 3.02 3.04

UAE Dirham 1 28.89 29.05

Malaysian Ringgit 1 24.88 25.02
South Korean Won 100 9.20 9.25
Swedish Kroner 1 11.74 11.81
Danish Kroner 1 15.56 15.65
Hong Kong Dollar 1 13.68 13.76
Kuwaity Dinar 1 349.91 351.89
Bahrain Dinar 1 281.37 282.97

Note : Under the present system the open market exchange rates quoted by different banks may differ.

20 GREEN Mathematics Book-10

Worked Out EXAMPLES

EXAMPLE 1 If 100 Indian rupees = NRs 160, convert NRs 3200 to Indian rupees by
using chain rule.

Solution : IC 100 rupees = Rs. 160 NC

NC 3200 rupee = x (say) IC

By using chain rule

i.e. (100 × 3200) = (160 × x)

or, o 100 × 3200 p = x
160

∴ Rs. 2000 IC = x

We can exchange NRs. 3200 for Rs. 2000 Indian rupees.

EXAMPLE 2 If the money exchanger charges a 0.5% commission, find Nepali
rupees required to exchange 3200 Euro if 1 Euro = Rs. 140.44

Solution : 1 Euro = Rs. 140.44
or, 3200 Euro = Rs. 3200 × 140.44

= Rs. 449408

Commission = 0.5% of Rs. 449408

= 0.5 × Rs. 449408
100

Finally, = Rs. 2247.04

Total currency = Rs. 449408 + Rs. 2247.04
= Rs. 451655.04

EXAMPLE 3 If the cost of 5 beds is equal to 8 tables, 3 tables cost as much as 11
chairs then how many chairs cost as much as 30 beds?

Solution : Let, x number of chairs cost as much as 30 beds

Now, x chairs = 30 beds

5 beds = 8 tables

3 tables = 11 chairs

B y using chain rule, x × 5 × 3 = 30 × 8 × 11
x = 176

Required number of chair is = 176

GREEN Mathematics Book-10 21

EXERCISE 2.2

A. Very Short Questions

1. Solve the following questions with the help of above exchange rates.

a. How much Nepali money can you exchange with $10?
b. How much Nepali currency reduced to exchange with £5?
c. How much IC can be exchanged with NRs. 500?
2. Fill in the blanks.

a. Rs. 500 = £........ b. $ ........... = NRs. 100 c. IC 525 = NRs. ............

B. Short Questions

3. Solve the following.

a. If $30 = Rs. 2160 and £15 = Rs. 1980, how many pounds can be exchanged for $44?

b. If £1 = Rs. 110.64 and $ 1 = Rs. 76.30, find the dollars ($) for 1 pound (£).

c. If $6 = £ 3.5 (Pound) and Rs. 220 = $3, how many pounds will be equal to Rs. 4400?

d. If Rs. 1800 = $25 and Rs. 1584 = £12, how many pounds can be exchanged from $33?

e. If £52 = $91 and Rs. 2112 = £16, how many dollars will be equal to Rs. 5280?

f. If £15 = $20 and $5 = Rs. 390, how many pounds can be exchanged for
Rs. 31200?

g. Using the rate of Rs. 70 per $1, calculate Nepali Rs. 50540 in dollars.

h. If Thai Bhat 1 = Rs. 3.08 and $ 1 = Rs. 107.70, find the Bhat for one doller.
4. a. How many dollars are equal to Rs. 63000 if the amount of Rs. 1000 is equal to £6.06

and £7 is equal to 11 dollars.
b. If Rs. 160 = £ 1, £ 30 = 90 Australian dollars, 10 Australian dollar = 250 Thai Bhat, how

much Bhat will be equal to Rs. 27,200?

c. Sadan buys goods in Bangkok for Thai Bhat 1000. Find the value of goods in NC
when the exchange rates are Rs. 3.08 = Rs. 1.5 Indian rupee and 1000 Thai Bhat = Rs.
500 Indian rupee.

5. a. Find the cost to galvanize a tank in pound (£) given that the exchange rate $3 = Rs.
250 and $6 = £ 4, if the cost is Rs. 7,392.

b. If Rs. 20,000 = $250, $50 = 60 Euro, 20 Euro, 30 Singapore dollers, express 900
Singapore doller in Nepali currency.

c. If $5 = £ 4, Rs. 225 = $3, how many pound (£) will be equal to Rs. 5,625?

d. If Rs. 30 = 100 yen, £ 4 = 1000 yen and $110 = £ 50, how many rupees are equivalent
to $2640?

22 GREEN Mathematics Book-10

3

Compound Interest

Population Growth and Compund Depreciation

Estimated Teaching Periods : 10

Albert Einstein was a German-born theoretical physicist. He developed
the theory of relativity, one of the two pillars of modern physics.
Einstein's work is also known for its influence on the philosophy of
science. “Compound interest is the eighth wonder of the world. He who
understands it, earns it ... he who doesn't ... pays it.”

Contents

3.1 Compound Interest
3.2 Population growth and compound depreciation

Objectives

At the end of this unit, students will able to :
calculate the compound interest, simple interest annually.
find the difference between compound interest semi annully and annually.
know about the depreciation value of any types of good.
find the population after increase by certain rate calculate time, rate of interst.

Materials

Chart papers, card boards, chart of formulae

GREEN Mathematics Book-10 23

3.1 Compound Interest

Let's study the given example.

Worked Out EXAMPLES

EXAMPLE 1 Sadan borrowed Rs. 10,000 from Sabin for 3 years at the rate of 15%
Solution : yearly interest then how much compound interest does Sadan pay at
the end of 3 years?

Now,

For first year

Principal (P) = Rs. 10000

Time (T) = 3 years

Rate (R) = 15% yearly

Using simple interest form

SI1 = PTR = 10,000 × 1 × 15
100 100

= Rs. 1500

For second year

Principal (P) = Rs. 10,000 + Rs. 1500

= Rs. 11,500

Time (T) = 1 year

Rate (R) = 15%

SI2 = PTR = 11,500 × 1 × 15
100 100

= Rs. 1725

And

For final year

Principal (P) = Rs. 11,500 + Rs. 1725

= Rs. 13225

Time (T) = 1 year

Rate (R) = 15%

SI3 = PTR = 13225 × 1 × 15
100 100

= Rs. 1983.75

24 GREEN Mathematics Book-10

Finally,
Total interest for 3 years.
I = SI1 + SI2 + SI3
= Rs. 15000 + Rs. 1725 + Rs. 1983.75
= Rs. 5208.75

Again,

From the above example, we can see principal remain unchanged but interest is changed
after the completion of the time in simple interest rule.

So, a fixed sum of money is known to be lent at compound interest when the starting sum
of money does not remain same for the period of time and the interest at the end of each
fixed period of time like half yearly or yearly is added to the original sum and total sum
at the end of each year or period is considered as the new sum of money or principal for
the another coming year. The phenomenon is called compound interest.

EXAMPLE 2 Calculate compound interest if principal (P) = P, Time (T) = T, Rate (R)
= R%.

Solution : Let Principal (P) = P

Rate of interest = R%

and Time = T

Interest for the first year.

SI1 = PTR
100

= PR for 1 year
100

Again principal for second year

Principal = oP + PR p
100

SI2 = P1T00R = oP + PR p× 1 × R
100 100

= oP + PR p × R
100 100

= P qo1 + R pr R
100 100

Total amount = P qo1 + R pr + P qo1 + R pr R
100 100 100

GREEN Mathematics Book-10 25

= P o1 + R p o1 + R p
100 100

For second year

Amount = P o1 + R 2
100
p

Similarly,

Amount for third year = P o1 + R 3
100
p

A mount for fourth year = P o1 + R 4
100
p

Amount for T year = P o1 + R T
100
p

So, its interest,

I = A – P, where A = Total amount

P = Principal (P)

or, I = P qo1 + R pTr – P
100

or, Iy = P qo1 + R pT – 1r
100

Finally, CI = P qo1 + R pT – 1r Where CI is called the
100 compond interest.

Note: R
2
I. If the compound interest is payable half yearly then rate of interest = %

Time = 2T half year (6 month each)

Now,

(i) CIh = P qo1 + R 2T – 1r
200
p

(ii) CAh = P qo1 + R 2T r
200
p

2. (i) CAy = P qo1 + R T r
100
p

for calculating compound amount yearly.

II. CA = P + CI

or, P = CA – CI

or, CI = CA – P

26 GREEN Mathematics Book-10

EXAMPLE 3 Calculate the compound interest and amount when the principal
(P) = Rs. 10,000, time (T) = 2 years at the rate of interest (R) = 12%.

Solution : Here,

Principal (P) = Rs. 10,000

Time (T) = 2 years If the time period in year and month

Rate of interest (R) = 12% interest is payable half yearly then
rate of interest = R%, Time = T

We have, For yearly,

CI = P qo1 + 12 pT – 1r CI = P qo1 + R T o1 + mR p–1r
100 100 1200
p

= 10,000 qo1 + R p2 – 1r CA = P qo1 + R T o1 + mR pr
100 100 1200
p

For half yearly,

= 10,000 qo 112 p2 – 1r CI = P qo1 + R 2T o1 + mR p–1r
100 200 1200
p

= 10000 × 0.2544 CA = P qo1 + R 2T o1 + mR pr
= Rs. 2544 200 1200
p

∴ CA = P + CI

= Rs. 10000 + Rs. 2544

= Rs. 12544

Alternate Method :
We can calculate CA by using the formula :

CA = P qo1 + R pT r
100

= 10,000 qo1 + 12 p2 r
100

= 10,000 qo 112 p2r
100

= 10000 × 1.2544

= Rs. 12544
C.I. = C.A. – P
= 12544 – 10000
= Rs. 2544

GREEN Mathematics Book-10 27

EXAMPLE 4 Sadan borrowed Rs. 1,50,000 from Sita at the rate of 21% per annum. At
the end of nine months how much compound interest should he pay if
the interest is compounded half yearly?

Solution : Here,

Rate (R) = 21%

Principal (P) = Rs. 1,50,000

Time (T) = 9 months

We have, For half year = 1 year + 3 months
2

CI = P qo1 + R p2× T – 1r
200

= 1,50,000 qo1 + 21 p2 1 – 1r
200 2

= 1,50,000 qo 221 p1 – 1r
200

= 1,50,000 (1.105 – 1)

= Rs. 150000 × 0.05

= Rs. 15750

Now,

Total principal (P) = Rs. 1,50,000 + Rs. 15750

= Rs. 165750

For 3 months, simple interest = PTR
100

= 165710500 × 3× 21
× 12

= Rs. 10442.25
1200

= Rs. 8701.875

Finally,

Interest for 9 months = Rs. 15750 + Rs. 8701.875

= Rs. 24451.875

28 GREEN Mathematics Book-10

EXAMPLE 5 The difference between the compound interest and simple interest on
a sum of money for 2 years at 20% p.a. is Rs. 400. Find the sum.

Solution : Principal (P) = Rs. x (suppose)

Rate of interest (R) = 20%

Time (T) = 2 years

Again SI = PTR = Rs. x ×2× 20 = Rs. 2x
100 100 5

CI = P qo1 + R pT – 1r
100

= Rs. x qo1 + 20 p2 – 1r
100

= Rs. x qo 120 p2 – 1r
100

= Rs. 44 = Rs. 11x
Now, 100 25

According to the question

CI – SI = Rs. 400

or, 11x – 2x = 400
25 5

or, 11x – 10x = Rs. 400
25

or, x = Rs. 400 × 25

or, ∴ x = Rs. 10,000

∴ the sum = Rs. 10000

GREEN Mathematics Book-10 29

EXAMPLE 6 The compound amount of a certain sum of money in 2 years and 3
years becomes Rs. 8820 and Rs. 9261 respectively. Find the sum and the
rate of interest.

Solution : Here,

CA1 = Rs. 8820 CA2 = Rs. 9261
Time = 2 years Time = 3 years

Let principal (P) = Rs. x

We have, We have,

CA1 = P o1 + R p T CA2 = P o1 + R p T
100 100

or, 8820 = x o1 + R p 2 9261 = x o1 + R p 3
100 100

8820 9261

or, o1 + R p 2 = x ..... (i) or, o1 + R p 3 = x ........... (ii)
100 100

Now, from (i) and (ii)

8820 9261

o1 + R p 2 = o1 + R p 3
100 100

or, 8820 o1 + R p = 9261
100

or, 1+ R = 9261
100 8820

or, R = 9261 – 1
100 8820

R = 5%
Again, putting the value of R in equation (i), we get

∴ Rs. 8820 = x o1 + 5 p 2
100

8820

or, o1 + 5 p 2 = x
100

or, x = 8820
1.1025

= Rs. 8000

∴ The sum is Rs. 8000 and rate of interest is 5%

30 GREEN Mathematics Book-10

EXAMPLE 7 In how many years will Rs. 8000 amount to Rs. 13824 at 20% p.a.
Solution : interest compounded yearly?

Here,

Rate of interest (R) = 20%

Principal (P) = Rs. 8000

Compound amount (CA) = Rs. 13824

We have, CA = P o1 + R p T
100

or, Rs. 13824 = Rs. 8000 o1 + 20 pT
100

Rs. 13824 = o 120 pT
Rs.8000 100

or, 1.728 = (1.2)T

or, (1.2)³ = (1.2)T

∴ T = 3 years.

∴ The required time is 3 years.

EXERCISE 3.1

A. Very Short Questions

1. Define compound interest.

2. Fill in the gaps.

a. .................... = P + I

b. P= 100 A × 100
+ [..............]

c. If principal = P, Time = T, R = Rate of interest then compound interest (CI) = .........

d. If principal = P, Time = T, rate of interest = R then simple interest (SI) = .................

e. R= I × 100
[..............]

f. If principal (P) = P, Time = T and rate of interest = R compound Amount (CA) = ....................

GREEN Mathematics Book-10 31

B. Short Questions

3. The compound interest and the compound amount are CI and CA respectively, write
down the relation among:

i. P, T, R and CI (yearly) ii. P, T, R, CA (half yearly)

4. At what rate percent per annum compound interest will Rs. 625 amount to Rs. 729 in
2 years?

5. At what rate percent per annum compound interest will Rs. 3000 amount to Rs.
3307.5 in 2 years?

6. Sabin deposited Rs. 5000 at 8% yearly compound interest in NMB Bank. Find the
difference between the amount compounded yearly and half yearly in 2 years.

7. The compound interest on a certain sum for 2 years at 8% per annum compounded
yearly is Rs. 312. Find the simple interest for the same time and same rate.

8. The compound interest on a certain sum of money for 2 years at 10% per annum is
Rs. 420. What would be the simple interest on the same sum at the same rate for the
same time?

9. In what time will Rs. 8000 amount to Rs. 10648 at 10% compound rate of interest?

10. Find the rate of interest if the compound interest on Rs. 15625 for 3 years is Rs. 1951.

11. Find the amount and interest on Rs. 4000 at 16% per annum for 3 years compounded
yearly.

12. Find the compound amount if principal (P) = Rs.500, R (Rate of interest) = 12%, Time
(T) = 2 years.

C. Long Questions

13. Ramita borrowed Rs. 1,70,000 from Sadan at the rate of 21% per annum. How much
compound interest and simple interest will she have to pay at the end of 1 year 6
months?

14. Sabin borrowed Rs. 1,30,000 from Sadan at the rate of 21% per annum. Find simple
interest and the compound interest at the end of 2 years compounded yearly.

15. A trader borrows Rs. 3000 at 8% per annum simple interest and invests the same
amount at 10% compound interest for two years. Find their difference.

16. Find the difference between simple interest and the compound interest on Rs. 18000
for 2 years at the rate of 15% per annum.

32 GREEN Mathematics Book-10

17. The difference between the compound interest and simple interest on a sum of money
for 2 years at the interest rate of 10% yearly is Rs. 240, find the sum.

18. Find the sum where the difference between simple and compound interest on a sum
of money for 2 years at 5% yearly is Rs. 15.

19. The simple interest on a sum of money in 2 years is Rs. 36 less than the compound
interest compounded yearly. If the rate of interest is 12%, find the sum.

20. The compound interest on a sum of money in 2 years at 15% yearly will be Rs. 180
more than simple interest, find the sum.

21. The difference between the compound interest and simple interest on a sum of money
for 2 years at 5% yearly is Rs. 30. Find the sum.

22. At the rate of 12% yearly the annual compound interest of a sum of money in 2 years
is Rs. 2019.24 less than the semi-yearly compound interest, find the sum.

23. The sum of simple and compound interest in 2 years is Rs. 202.50 and the rate of
interest is 5% yearly, find the sum.

24. The sum of simple and compound interest in 3 years is Rs. 504.80 and the rate of
interest is 10% yearly. Find the principal.

25. The compound amount of a certain sum of money in 2 years and 3 years become Rs.
7260 and Rs. 7986 respectively. Find the sum and rate of interest.

26. The interest is compounded yearly. If the amount of the sum of money becomes Rs.
12,100 at the end of two years and Rs. 13,310 at the end at 3 years, find the rate of
compound interest and the sum.

27. A certain sum of money amounts to Rs. 1323 in 2 years and to Rs. 1389 in 3 years. Find
the compound rate of interest and the sum.

28. The compounded amount of a sum in 2 years and 3 years are Rs. 10580 and Rs. 12167
respectively. Find the rate and the sum.

29. In how many years will Rs. 8000 amounts to Rs. 11520 at 20% per annum compounded
yearly?

30. According to the yearly compounded interest in what time will the compound
interest on Rs. 400000 at the rate of interest 6.5% yearly be Rs. 83179.85?

31. A person took a loan of Rs. 46875 at the rate of the compound interest 4 paisa per
rupee per year. In how many years will the compound interest be Rs. 5,853?

32. If the compound interest compounded half yearly on Rs. 15625 for 1 year and 6
months is Rs. 1951, find the rate of interest.

GREEN Mathematics Book-10 33

33. At what rate of compound interest per year will the compound interest on Rs. 343000
amount to Rs. 456533 in 3 years?

34. According to the system of semi annual compound interest of a sum of money
amounts to Rs. 400 in 1 year and Rs. 441 in 2 years, find the rate of interest.

35. The compound amount on a sum of money in 2 years and 3 years are Rs. 10,800 and
Rs. 12960 respectively. Find the sum and the rate of interest per annum.

36. According to the system of yearly compound interest, a sum of money amounts to
Rs. 14520 in 2 years and amounts to Rs. 15,972 in 3 years, find the sum.

3.2 Population growth and Compound depreciation

Population growth

Generally total number of living people in a country in particular time is the
population of the country. The population of country does not remain constant. It
generally increases. If it is increased of given population per unit on certain time, it
is known as the rate.

If P0 is the initial (starting) population of a particular place, R% yearly growth rate,
T is the particular time in years and PT is the population of that place at the end of T
year then

PT = Pn = P0 qo1 + R p T r
100

Note:

a. If R1 %, R2% and R3% be the rates of population growth of 3 successive
years then population after 3 years

PT / Pn = P0 o1 + R1 p × o1 + R2 p × o1 + R3 p .
100 100 100

b. If R1% be the yearly birth rate, R2% be yearly death rate and P0 be
starting population then population after T years, PT will be.

PT = Pn = P0 qo1 + R1 – R2 T
100
pr

c. PT = Pn = P0 qo1 + R T + (in-migrant) – (out–migrant) – death
100
pr

34 GREEN Mathematics Book-10

Compound depreciation

Depreciation : Any fixed asset like building machinery, other goods etc. after a
constant use for certain time causes the reduction in their original cost gently. This
reduction in the original value is known as depreciation. The following causes to
depreciation.

i. fall in the face value ii. efflux of a time

iii. less efficient in the working value iv. wear and tear

Generally, there are two types of depreciation.

a. Simple depreciation b. Compound depreciation

Compound depreciation : The amount of depreciation in each year is not the same
previous value; the residual value of the fixed assets to which it is reduced with the
time interval as the depreciated value and in general the difference between original
and depreciated value is known as depreciated amount.

Note: If D0 is the initial (original) value, R% is the rate of depreciation, T is
a. particular time, DT, Dn is the depreciated value after T years, then

DT = Dn = D0 qo1 – R T Depreciated Amount = D0 – DT
100
pr

Worked Out EXAMPLES

EXAMPLE 1 The population of ward no. 20 of Kumaripati increases every year
at the rate of 2% yearly. The present population at the ward no 20 is
20,000. What was the population of the city before 2 years.

Solution : Population before 2 years (P0) = x say
Growth rate (R%) = 2%

Present population = PT = Pn = 20,000

We have or Pn = P0 qo1 + R T
PT 100
pr

or, 20,000 = P0 qo1 + 2 2
100
pr

20,000

o 102 2 = P0
100
p

or, 20,000 = P0
(1.02)²

∴ P0 = 19223
The population of that word no. 20, Kumaripati before 2 year was 19,223.

GREEN Mathematics Book-10 35

EXAMPLE 2 The present population of a town is 40,000. If the yearly birth rate is
Solution : 5% and death rate is 3%, what will be the population of town after 2
years?

Yearly birth rate (R1%) = 5%
Yearly death rate (R2%) = 3%
Time period (T) = 2 years

Present population (P0) = 40,000
Using formula,

PT or Pn = P0 qo1 + R1 – R2 T
100
pr

or, = 40,000 qo1 + 5–3 2
100
pr

= 40,000 qo101000+ 2p2r

= 40,0001×0010002 × 102
∴ The population of the town after 2 years will be 41616.

EXAMPLE 3 The population of Biratnagar city was 1,00,000 before 2 years and it is
Solution : 104040 at present. What will be the population of that city after 3 years
with the same growth rate?

For this question, we have 2 cases

Case = I

Starting population (P0) = 1,00,000
Time (T) = 2 years

Present population (PT) = 104040
Growth rate (R%) = x % say

We know,
RT

PT or Pn = P0 qo1 + 100p r

or, 104040 = 1,00,000 qo1 + 1x00p2r

or, 11,0004,004000 = o 100 + x 2
100
p

36 GREEN Mathematics Book-10

or, 1.0404= o 100 + x 2
100
p

or, 1.0404 × (100)² = (100 + x)²
or, 10404 = (100 + x)²
or, (102)² = (100 + x)²
or, 102 – 100 = x
∴ x = 2%
Again,
Case II
Initial population (P0) = 104040
Time (T) = 3 years
Growth rate (R) = 2%
Present population (P) = ?

2T
PT or Pn = P0 qo1 + 100p r
or, Pn or PT = 104040 qo1 + 1x00p3r
= 104040qo110002p3r
= 104040 × (1.02)³
Pn = 1,10,408

EXAMPLE 4 The yearly rate of compound depreciation of a mobile was 10%. If
Solution : the present value of a mobile is Rs. 5000 at what price was it bought 3
years ago?

Original value (P0) = x (say)
Rate of depreciation (R) = 10%

Depreciated value (DT) = Rs. 5000

Using formula,

DT or Dn = D0 qo1 – R T
100
pr

or, 5000 = x qo1 – 10 3
100
pr

GREEN Mathematics Book-10 37

5000

or, o1 – 10 3 = x
100
p

5000

or, o 90 3 = x
100
p

or, x = 5000
0.729

x = Rs. 6858.71

EXAMPLE 5 The original value of motorbike is Rs. 450000. If its value depreciates
Solution : by 20% at the begining of each year, after how many years will its
value reduce to Rs. 2,88,000.

Original value (P0) = Rs. 450000
Rate of depreciation (R) = 20%

Time (T) = x Say

Depreciated value (DT) = Rs. 2,88,000

Using formula,

DT = Dn = D0 qo1 – R T
100
pr

or, 2,88,000 = 4,50,000 qo1 – 20 T
100
pr

or, 2,88,000 = qo 100 – 20 T
4,50,000 100
pr

or, 2,88 = o 80 T
4,50 100
p

or, 0.64 = (0.8)T

or, (0.8)² = (0.8)T

∴ T = 2 years.

38 GREEN Mathematics Book-10

EXAMPLE 6 After the continued devaluation of US dollar for 2 years at 5% every
Solution : year the present exchange rate is $ 1 = Rs. 100. What was the exchange
rate 2 years ago?

Present exchange rate of dollar = (Dn = Dt) = Rs. 68
Devaluation rate (R) = 5%

Time (T) = 2 years

Using formula, DT = Dn = D0 qo1 – R T
100
pr

or, = D0 qo1 – 5 2
100
pr

or, 100 = D0 o 95 2
100
p

or, 100 = D0
0.9025

or, D0 = Rs. 110.80

The exchange rate of $1 (dollar) was Rs. 110.80 'two' years ago.

EXERCISE 3.2

A. Very Short Questions

1. Fill in the gaps.

a. If PT be population after T years, P0 be starting population, R be growth rate and
Time be T, then PT = P0 [(1 + ...............)T]

b. The compound depreciation on P in T years at R% rate of depreciation is
....................

c. Compound depreciation (CD) = D0 – [....................]
B. Short Questions

2. a. The population of a town increases at the rate of 5% yearly. The present
population of town is 67200. What will be the population of the town after 2
years?

b. The population of a city at present is 170,000 and it grows at the rate of 2%
yearly. What will be the population after 2 years?

c. A village has population 17640. If its population growth rate is 5%, find the

GREEN Mathematics Book-10 39

population before 2 years.

d. The present population of a country is 2,65302. What was its population one
years ago if it is increased by 2% yearly?

e. In how many years will the population of a city be 66550 from 50,000 if the
population growth rate is 10% yearly?

f. In how many years will the population of a village be 46305 from 40,000 if the
growth rate is 5% yearly?

g. The population of a town is 96000. In how many years later would it be 1,05,840,
if the population increased at the rate of 5% every year?

h. In how many years will the population of a town be 209475 from 1,90,000 at the
growth rate of 5% yearly?

3. a. The present price of a scooter is Rs. 95000. If it is depreciated at 6% per year,
what will be the price of the scooter after 2 years?

b. A LED TV costing Rs. 11600 is depreciated at the rate of 15% yearly, what will
be the cost of TV after 2 years?

c. The present price of bike is, 140,000. If it is depreciated at 7% yearly, what will
be the price of the bike after 2 years?

d. If the cost is depreciated at the rate of 10% yearly, the cost of a bike after 3 years
becomes Rs. 87480. Calculate the original price of the bike.

e. A scooter which is supposed to be depreciated at the rate of 15% yearly was sold
at Rs. 72250 after a year. At what price was it bought?

f. A bike which was purchased at Rs. 90,000 was sold for Rs. 81,000 after one year.
At what percentage was it depreciated?

g. A radio costing Rs. 6000 is depreciated per year and 2 years later its price
becomes Rs. 5414. Find the rate of depreciation.

h. A tempo which was bought for Rs. 2,00,000 was sold at Rs. 1,79,000 after one
year. At what percent was it depreciated?

C. Long Questions

4. a. The population of a town increases every year by 10%. At the end of two years
the total population of the town was 30,000. If 5800 people were added by
migration, what was the population of the town in the beginning?

b. The population of Ramkot village increases every year by 2.5%. At the end of 2
years the total population of the village was 24895. If 320 people migrated to the
other village, what was the population of the village in the starting?

c. The population of a village increase every year by 5%. At the end of two years,
the total population of the village was 10,000. If 1025 migrated to other places,
what was the population of a village in the beginning?

40 GREEN Mathematics Book-10

d. In the beginning of 2065 BS, the population of a town was 10,0000 and the rate
of population growth was 2% every year. If in the beginning of 2066 BS, 8,000
people migrated there from different places, what will be the population of the
town in the beginning of 2068?

e. Two years ago the population of a village was 40,000. If the present population
of the village is 44100, find the yearly growth rate of population.

5. a. Sabin bought a motorcycle for Rs. 96100 and using it for 2 years he sold it for Rs.
90,000. Find the rate of compound depreciation of the motorbike.

b. Sadan bought a computer for Rs. 44100 and after using it for 2 years sold it for
Rs. 40,000. Find the rate of compound depreciation of the computer.

c. 4 ropanies of land was bought for Rs. 4,250,000 before 3 years. If the same land
is sold for Rs. 34,00,000 at present, what percentage of its value is depreciated?

d. A motorbike was purchased 2 years ago and it was depreciaed at the rate of 12%
yearly. If it is sold at Rs. 332992, what was its cost 2 years ago?

e. A car purchased a year ago was depreciated at the rate of 8% yearly. If it is sold
for Rs. 14,26,000 now, what was its cost a year ago?

f. After the continued devaluation at US dollars for 2 years at 10% yearly the
present exchange rate is 1 dollar = Rs. 107 then

i. what was the exchange rate 2 years ago?

ii. what will be the exchange rate 3 years hence with the same rate of
depreciation?

g. After the continuous devaluation of a piece of land for 3 years at the rate 4%
yearly its price is reduced to Rs. 90,00,000 at present.

i. What was its price 3 years ago?

ii. What will its price be 2 years hence with the same rate of depreciation?

6. a. The population of a village was 10,000 few years ago. The present population is
13,310. If the annual growth rate is 10%, how many years ago was the population
10,000?

b. The population of a town increases every year by 10%. At the end of two years,
the total population of the town was 30,000. If 5,800 people were added by
migration, what was the population of the town in the beginning?

c. At the end of 2055 B.S., the population of Kathmandu was 10 lakh. At the end
of 2056 B.S. 25,000 people migrated here from other parts of the country. If the
population growth rate is 4.5%, find its population at the end of 2058 B.S.

d. A man bought a motor cycle for Rs. 96,100 and after using it for 2 years, he sold
it for Rs. 90,000. Find the rate of compound depreciation of the motorcycle.

GREEN Mathematics Book-10 41

4

Mensuration

Estimated Teaching Periods : 24

Archimedes (287-212 BC) of Syracuse was a Greek mathematician,
physicist, engineer, inventor, and astronomer. Although few details of his
life are known, he is regarded as one of the leading scientists in classical
antiquity.

Contents

4.1 Introduction
4.2 Prism
4.3 Area and volume of cylinder
4.4 Pyramid : Introduction
4.5 Cone : Introduction

Objectives

At the end of this unit, students will be able to:
solve the problems related to area of plane surface, prism, pyramid.
find the total surface area volume of sphere and cylinder.
find the area of cross section, curved surface, area, total surface area and volume of
prism.
estimate the number of bricks required for making a wall.

Materials

Models of prism, nets of prism, core, pyramid, solid prism.

42 GREEN Mathematics Book-10

4.1 Introduction

The branch of mathematics which deals with the measurement of length, area and
volumes of plane and solid figures is called mensuration. In our day - to - day life we
come across different activities like - to find area of room, four walls and plot of land etc.
Similarly we have other activities such as to find the length of wire to fence a ground, to
find the capacity of a container, etc. All these activities are related to measure of length,
perimeter, area and volume.

In this lesson, we learn to find the area of triangle by using special formulae.

Area of triangle A
a. Area of triangle when base and height are given

In the given ∆ABC, AM⊥BC. Thus AM is the height
(altitude) and BC is the base of the triangle.

Area of ∆ABC = 1 base × height BC
2 M

= 21 × b × h Base (b)

b. Area of a right-angled triangle A

In right angled ∆ABC, AB⊥BC. So AB is the height and BC is

the base of the triangle. p

Area of ∆ABC = 1 base × height B b C
2

= 1 × b × p
2

c. Area of an equilateral triangle

In the given figure, ∆ABC is an equilateral triangle and AM⊥BC.

If AB = BC = AC = a then BM = CM = a A
2 a

In right-angled ∆AMB, AM = AB² – BM²

= a² – a² B aMa C
4 22 43

= 3 a a

2

= 1 base × height
2

Now, area of ∆ABC = 1 × a × 3 a
2
2

= 3 × a²

4

GREEN Mathematics Book-10

d. Area of an isosceles triangle A
In isosceles ∆ABC, AB = AC = b and base BC = a.

Then AM⊥BC is drawn. Thus BM = CM = a b hb
2

and AM = AB² – BM² B aM a C
2a 2
a² 4b² – a2
∴h = b² – 4 = 2

Area of ∆ABC = 1 . b × h
2

= 1 .a. 4b² – a²
2
2

= a 4b² – a²
4

e. Area of scalene triangle when measure of three sides are given

Let a, b and c be the three sides BC, AC and AB respectively of scalene triangle ABC.

AD is the height which is denoted by h. A

When BD = x then DC = a – x

Here, perimeter of ∆ABC = a + b + c c b C
2s = a + b + c where s = semi - perimeter Bx h
and area of ∆ABC = 1 a × h ................. (i)
D a–x
2 a

In right-angled ∆ADC,

h² = b² – (a – x)² ....................... (ii)

In the right-angled triangle ADB, h2 = c2 – x2 ...................... (iii)

From (i) and (ii), c² – x² = b² – (a – x)²

or, c² – x² = b² – a² + 2ax – x²

or, c² + a² – b² = 2ax

∴ x = c² + a² – b²
2a

Substituting the value of x in equation (iii), we get

h² = c² – o c² + a² – b² 2
2a
p

= qc + c² + a² – b² rqc – c² + a² – b² r
2a 2a

44 GREEN Mathematics Book-10