Prime Mathematics 5

Prime

Mathematics
Series

123 - 4 ÷ 5
5

9 - 8 x6+7
0

Raj Kumar Mathema

Dirgha Raj Mishra Bhakta Bahadur Bholan

Uma Raj Acharya Yam Bahadur Poudel

Naryan Prasad Shrestha Bindu Kumar Shrestha

Prime

Mathematics
Series

5

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

learning materials.

Raj Kumar Mathema Authors
Bhakta Bahadur Bholan Dirgha Raj Mishra
Yam Bahadur Poudel Uma Raj Acharya
Bindu Kumar Shrestha Naryan Prasad Shrestha

Editors
Anil Kumar Jha
Dhurba Narayan Chaudhary
Hari Krishna Shrestha

Language Editor
Mrs. Tara Pradhan

Pragya Books & Distributors Pvt. Ltd.

Printing history
First Edition 2074 B.S.
Second Edition 2077 B.S.

Authors
Raj Kumar Mathema
Dirgha Raj Mishra
Bhakta Bahadur Bholan
Uma Raj Acharya
Yam Bahadur Poudel
Naryan Prasad Shrestha
Bindu Kumar Shrestha

Editors
Anil Kumar Jha
Dhurba Narayan Chaudhary
Hari Krishna Shrestha

Layout and design
Rabi Man Shrestha

© Publisher
All rights reserved. No part of this book, or designs and illustrations here within, may be
reproduced or transmitted in any form by any means without prior written permission.
ISBN : 978-9937-0-2222-4
Printed in Nepal

Published by
Pragya Books & Distributors Pvt. Ltd.

Lalitpur, Nepal
Tel : 5200575
email : [email protected]

Preface

Prime Mathematics Series is a distinctly outstanding mathematics series
designed in compliance with Curriculum Development Centre (CDC) to meet
international standards. The innovative, lucid and logical arrangement of the
content makes each book in the series coherent. The presentation of ideas
in each volume makes the series not only unique, but also a pioneer in the
evolution of mathematics teaching.

The subject matter is set in an easy and child-friendly structure so that
students will discover learning mathematics a fun thing to do.A lot of research,
experimentation and careful gradation have gone into the making of the series
to ensure that the selection and presentation is systematic, innovative and both
horizontally and vertically integrated.

Prime Mathematics Series is based on child-centered teaching
and learning methodologies, so the teachers will find teaching this series
equally enjoyable. We are optimistic that this series shall bridge the existing
inconsistencies between the cognitive capacity of children and the course matter.

We owe an immense debt of gratitude to the publishers for their creative,
thoughtful and inspirational support in bringing about the series. Similarly, we
would like to acknowledge the tremendous support of teachers, educationists
and well-wishers for their contribution, assistance and encouragement in making
this series a success.

We hope the series will be another milestone in the advancement of teaching
and learning mathematics in Nepal. We solicit feedback and suggestions from
teachers, students and guardians alike so that we can refine and improvise the
series in the future editions.

Our team would like to express our special thanks to Mr. Nara Bahadur
Gurung, Mr. Ram Narayan Shah, Mr.Tulsi Kharel, Mr. Mani Ram Khabas, Mr. Umesh
Acharya, Mr., J. Phuldel, Mr. Kamal Raj Tripathee, Mr. Rudra Prasad Pokharel, Mr.
Uttam Prasad Panta, Mr. L.N. Upadhyaya, Mr. Shakti Prasad Acharya, Mr. Upendra
Subedi, Mr. Kul Narayan Chaudhary, Mr. Bishonath Lamichhane, Mr. Harilal
Lamichhane, Mr. Govinda Paudel, Mr. Krishna Aryal, Mr. Nine Bhujel, Mr. Santosh
Simkhada, Mr. Pashupati Upadhyaya, Mr. Dipak Adhikari, Mr. Mukti Adhikari, Mr.
Dipendra Upreti, Mr. Dipak Khatiwada, Mr. Narayan Nepal, Mr. Raj Kumar Dahal,
Mr. Bhim Raj Kandel, Mr. Prem Giri,Mr. Iswor Khanal, Mr. Balram Ghimire, Mr. Om
Kumar Chhetri, Mr. Ram Hari Bhandar, Mr. Krishna Kandel, Mr. Madhav Atreya,
Deep Raj Nigam Unai, Sangita Thapa, Shiva Devkota, Harihar Adhikari, Chandra
Dev Tiwari, Chura Gurung, Sagar Dhakal, Baikuntha Marahatha, Subash Bidari
Raghu Kandel, Sudip Poudel, Roshan Sapkota Sujan Dhungana, Tara Bahadur
Bhandari and Jiwan K.C. for their Painstaking effort in peer reviewing of this
book.

Contents

Unit Topic Page

1. Geometry 1

2. Numeration System 23

3. Concept of Numbers 33

4. Fundamental Operations 49

5. Measurements 67

6. Mensuration 97

7. Fractions, Decimals and Percentage 113

8. Unitary Method, Simple Interest, Profit and Loss 153

9. Charts, Bills, Bar-graph and Co-ordinates 163

10. Sets 173

11. Algebra 181

Model Question 217

U1nit Geometry

Estimated periods − 17

Objectives

At the end of this unit, the students will be able to:
• measure and draw angles using protractor .
• draw perpendicular and parallel line using set square.
• understand complementary and supplementary angles .
• classify the triangles according to sides and angles .
• know the properties of some special quadrilaterals .
• find unknown angles of triangle and quadrilaterals using their properties.

Teaching Materials
geometry set ( geometrical tools ), geo-board, models of triangles and quadrilaterals.

Activities
It is better to:
• discuss about different geometrical tools and their uses and perform activities related
to measuring and drawing angles.
• demonstrate the construction of perpendicular and parallel lines to the given line.
• discuss about the type of triangles and classify them.
• demonstrate to clarify the sum of the interior angles of triangles and quadrilateral .

Prime Mathematics Book - 5 1

Geometry

A geometry box (instrument box) consists –
1 Ruler: A ruler is also called a straight edge or scale. It is used to measure

length of lines and to draw straight lines.

mm 1 cm 2 3 4 5 6 7 8 9 10 11 12 13 14 15

centimeters

0 Inches 1 2 3 4 5 6

2. Protractor: It is a semicircular
transparent device used to measure and
draw angles.

3. Set squares: It is a set of two triangular
devices. It is used to measure some
special angles and to draw perpendicular
and parallel lines.

4. Pencil compass: It is a forked metallic
device with needle at one arm and pencil
adjusted at other arm. It is used to draw
angle and circles.

5. Divider: It is a forked metallic device with O A
needles at both arms. It is used to measure
the lengths of straight line. angle
B
Angles
When two straight lines meet at a point, the space
between them is called an angle. In the given figure,
two straight lines AO and BO meet at the point O
forming an angle where point O is the vertex and
two lines OA and OB are called arms. This angle
contains two arms and a vertex. We write the

2 Prime Mathematics Book - 5

angle AOB as ∠AOB or ∠BOA. Let's revolve the O B
line OB about point O keeping O and the line OA O
fixed, the size of ∠AOB go on increasing. When 360O A
OB makes a complete revolution, completes an BA
angle of 360O. Generally we measure angles in
degrees.

Measuring angles:
A protractor is used to measure angles.
• Given angle, ∠AOB
B 130 120 110 100 90 80 70 B
60 A
OA
140 50 60 70 80 90 100110 120 50
150 40
130 40
140 30

160 30 150 20
20 160
170 10 170 10

180 0 O 180 0

• Put the center of the protractor at O. Adjust the zero line on OA and
read the measure of the angle shown by OB. It is 40o so ∠AOB = 40o.

Types of angles:

• Acute angle: An angle whose size is less than 90o is called an acute
angle.
B

45o 0 A

O
acute angle

• Right angle: An angle of size 90o is called a right angle.

90o

O 0
right angle

• Obtuse angle: An angle whose size is greater than 90o and less than 180o
is called an obtuse angle.

B

120o A

O 0
Obtuse angle

Prime Mathematics Book - 5 3

• Straight angle: An angle whose size is exactly 180o i.e. two right angles
is called a straight angle.
180o
B OA

• Reflex angle: An angle whose size is greater than 180o is called a reflex
angle.

reflex angle O acute angle A

OA reflex angle
B obtuse angle
B

To measure the size of a reflex angle: Measure corresponding acute
or obtuse angle and subtract it from 360o. It gives the measure of the
reflex angle. Here in the figure given along side, obtuse angle ∠AOB =
120o.

∴ Reflex ∠AOB = 360o - 120o = 240o 360o - 120o = 240o A

O

120o

B

• Constructing angles: Angles of different sizes can
be constructed by using protractor and ruler.
110 100 90 80
For example 120 70 60
To construct an angle of 30o follow the
steps below : C 130 50 60 70 80 90 100110 120 50
140 40
150 130 40
140 30

160 30 150 20
20
A 170 10 160
180 0 170 10

B 180 0

– Draw a line segment AB. C
– Put the centre O of the protractor at B and 30o

the base line or zero line on the line BA. AB
– Mark 300 at C.

– Now, remove the protractor and join BC with
the help of ruler.

Thus ∠ABC = 30o is constructed.

4 Prime Mathematics Book - 5

Exercise 1.1

1 Write the types of the angles.
2 Write the name of angle, vertex and arms of the following figure.

(a) C (b) R

BA QP

(c) M (d) B C

ON A

3. Classify the following angles (acute or right or obtuse or straight or
reflex angle):
(a) 100o (b) 45o (c) 175o (d) 90o (e) 181o

(f) 180o (g) 300o (h) 89o (i) 15o (j) 111o

4 Write the types of the following angles without measuring them:

(a) P (b) W X Y
QR O Q
(d)
(c) B C P
A

(e) Y Z

X

5. Measure the following angles:

(a) A (b) R (c)
Q KLM
B CP
Prime Mathematics Book - 5 5

(d) R (e) Y X (f) M
Q N O
P
Z

6. Construct the following angles using protractor:

(a) 15 (b) 30o (c) 45o (d) 60o

(e) 75o (f) 90o (g) 120o (h) 135o

(i) 150o (j) 165o (k) 180o (l) 105o

7. Construct the following angles using compass: (e) 75o
(a) 60o (b) 30o (c) 90o (d) 120o

Perpendicular and parallel lines
Perpendicular lines: If the angle made by two straight lines is 90o, the lines
are said to be perpendicular to each other.
Here, in the given figure, ∠ABC = ∠ABD = 90o. So, the straight line AB is
perpendicular to the line CD. We also write AB⊥CD.

A

90o 90o D

CB

To construct a line perpendicular to the given line at a given point on it:

A. Using protractor Q Q
APB P
110 100 90 80 70 60
120
130
50 60 70 80 90 100110 120 50
140 40 130
150 140 40
30

160 30 150 20
20
170 10 160 10
170
180 0
A P 180 0 B A B

Let AB is the given line and we need to construct a line perpendicular to AB
at p on the line.
• Put center of the protractor at P and base line (zero line) along AB.
• Mark the point corresponding to 90o , say Q.

6 Prime Mathematics Book - 5

• Remove the protractor and by the help of ruler and pencil, join PQ.
• Thus, PQ⊥AB is drawn at p.

B. Using set square

O O

APB APB APB

Let's construct a line perpendicular to the line AB at P.
• Adjust the ruler along the line AB.
• Adjust one of the perpendicular edge of a set square on the ruler such

that line can be drawn through P along the other perpendicular edge
of the set square.
• Fix the set square in its position and draw line through P say PQ
Thus, PQ⊥AB is constructed.

To construct a line perpendicular to the given line through a point out
side it.

PP P

A B A QB A QB

Let AB be the line and P be a point outside it through which we need to draw
line perpendicular to AB.
• Adjust the edge of ruler along AB.
• Adjust one of the perpendicular edge of the set square on the ruler and

run the set square along the ruler such that we can draw line through P
along the other perpendicular edge.
• Drawn line through P, say PQ.
• Thus, PQ⊥AB is constructed.

Prime Mathematics Book - 5 7

Parallel lines:
If two straight lines are drawn at equal distance, they are said to be parallel
to each other. Two parallel lines never meet each other on extending either
sides.
A B

dd

CD

Here, the straight lines AB and CD are parallel to each other. We write AB//CD.
To construct a line parallel to the given line through the given point.
Let AB be the line and P be the given point through which line parallel to
AB is to be drawn.
• Adjust one of the edge of a set square to the line AB.
• Set the ruler to one of the other edge of the set square.
• Now keeping the ruler fixed, slide the set square over the ruler till the

edge set to the line AB is over the point P and draw the line CD through
P along the edge.
• Thus, CD//AB is drawn through P.

BD

P

AC

Exercise 1.2

1. Identify whether the following pairs of lines are perpendicular or
parallel lines:
(a) C P R
(b)

AB Q

DS

8 Prime Mathematics Book - 5

(c) N (d) W X
L Y Z
M
K

2. Use protractor to construct a line perpendicular to the given lines
through the given point P:

(a) P B (b) C PD
A

3. Draw a line perpendicular to the given line through the given point
out side it:

(a) P B (b) M

A QN

4. Construct a line parallel to the given line through the given point:

(a) A P (b) X

Q

BY

(c) P Q (d) M T
OS

Prime Mathematics Book - 5 9

5. In the given figure AB⊥BC, draw a line perpendicular to AB through
A and another line perpendicular to BC through C. If these two lines
intersect at D, name the figure ABCD.
A

BC

6. Copy the figure where PQ=QR and PQ⊥QR. Construct a line parallel
to PQ through R and line parallel to QR through P. If these two lines
intersect at S, name the figure PQRS so formed.
P

QR

Pair of angles: O A
Adjacent angles: R B
In the figure given alongside, two angles ∠AOB C
and ∠BOC have a common vertex O and a common O
arm OB. Such angles are called adjacent angles. Q
If the sum of two adjacent angles is 90o, they C
are called complementary adjacent angles. P
Here, ∠POQ + ∠QOR = 90o B
So, ∠POQ and ∠QOR are complimentary OA
adjacent angles.
If the sum of the two adjacent angles is 180o,
they are called supplementary adjacent angles
or linear pair.
Here, ∠AOC = 180o

∠AOB + ∠BOC = ∠AOC = 180o
So, ∠AOB and ∠BOC are supplementary
adjacent angles or linear pair.

10 Prime Mathematics Book - 5

Note:

If the sum of two angles is 90o (not necessarily adjacent), they are simply complementary angles
and if the sum of two angles is 180o (not necessarily adjacent), they are called simply supplementary
angles:

If ∠ABC = 60o, then angle complementary to ∠ABC is 90o - ∠ABC = 90o - 60o =
30o. Similarly, if ∠ABC = 60o, angle supplementary to ∠ABC is 180o - ∠ABC =
180o – 60o = 120o.

Vertically opposite angles:

When two straight lines intersect each other, the C A

opposite angles so formed are called vertically
opposite angles, hence ∠AOC and ∠BOD, similarly
∠AOD and ∠BOC are vertically opposite angles. O

Vertically opposite angles are equal. BD
∠AOC = ∠BOD and ∠AOD = ∠BOC

In the given figure, pair of angles ∠AOC + ∠BOC = 180o
∠COB + ∠BOD = 180o, ∠BOD + ∠AOD = 180o and ∠AOD + ∠AOC = 180o are
supplementary adjacent angles.

Sum of complementary angle = 90o,
Sum of complementary adjacent angles = 90o,

Sum of supplementary angles =180o,
Sum of supplementary adjacent angles = 180o,

Vertically opposite angles are equal.

Worked out Examples

Example 1: Find the complementary angle of the angle 50o.

Solution: Let x be the complementary angle of 50o, then
x + 50o =90o
x = 90o - 50o
∴ x = 40o
Alternately

Complementary angle of 50o = 90o - 50o

= 40o

Prime Mathematics Book - 5 11

Example 2: Find the supplementary angle of 60o.
Solution: Let x be the supplementary angle of 60o, then

x + 60o = 180o
or, x = 180o - 60o = 120o
∴ The supplementary angle of 60o is 120o
Alternately
Complementary angle of 60o = 180o - 60o = 120o

Example 3: Two straight lines AB and CD intersect at A O 50o D
O. If ∠BOD = 50o, find the size of angles C B
∠AOD, ∠AOC and ∠BOC.
Here, ∠BOD = 50o
Being supplementary adjacent angles
∠BOD + ∠AOD = 180o
or, 50o + ∠AOD = 180o
or, ∠AOD = 180o – 50o
∴ ∠AOD = 130o

Being vertically opposite angles
∠AOC = ∠BOD = 50o
And ∠BOC = ∠AOD = 130o

Exercise 1.3

1. Write whether the given pairs of angles are adjacent angles or

complementary adjacent, complementary, supplementary adjacent

or supplementary angles:

(a) A CQ (b) C B

x=40 y x A
y=50 O

BP R

12 Prime Mathematics Book - 5

(c) C A (d) R Q
B yx P
y
O Ox

(c) (d)
xy
110o y 70o
x

2. Name the pair of adjacent angles:

(a) Y (b) R (c) A
X Q C O
ZO B
P

3. Find the complementary angle of: (e) 35.5o
(a) 75o (b) 20o (c) 50o (d) 61o (e) 80.25o (f) 45
2

4. Find the supplementary angle of: (f) 12 1
(a) 40o (b) 100o (c) 90o (d) 135o 2

5. Find the value of unknown angle 'x' in the following figures.

(a) (b) 10o (c) x
x 65o x 45o

(d) (e) (f) 80o x

25o 50o x
x

Prime Mathematics Book - 5 13

(g) (h)
90o x

x 45o

6. Find the size of unknown angles x, y and z from the given figures.

(a) (b)

y x 55o 120o z
z xy

7. x and 50o are a part of complementary angles. Make

an equation and solve it to find the size of x. 50o x

8. m and 130o are a pair of suppplementary angles. 130o Y
m
Make and equation and solve it to find the size of m X Z

Triangles A C

A triangle is a plane figure bounded by three straight B
line segments. A triangle has three sides, three angles
and three vertices. If the vertices of a triangle are A, B
and C, then the triangle is named as ∆ABC.

The line segments AB, BC and AC are three sides and
∠BAC, ∠ABC, ∠ACB are the three angles. A set square
is an example of a triangular object.

Types of triangles:

A. According to the sides, there are three types of triangles

(i) Equilateral triangle: If all the three sides are A

equal, the triangle is called an equilateral
triangle. In the given figure, AB = BC = CA. So,
∆ABC is an equilateral triangle. All the angles of
an equilateral triangle are also equal. B C

14 Prime Mathematics Book - 5

(ii) Isosceles triangle: If any two sides of a triangle B A C
are equal, the triangle is called an isosceles A C
triangle. In the given figure, AB = AC so, it is an
isosceles triangle. Angles opposite to the equal
sides of an isosceles triangle are also equal. They
are called base angles of an isosceles triangle.

(iii) Scalene triangle: If all the three sides of a triangle
are unequal, it is called a scalene triangle.

In the given triangle ABC, AB ≠ BC ≠ AC, so it is a
scalene triangle.
The angle opposite to the longest side of a B
scalene triangle is the largest and that opposite
to the shortest side is the smallest.

B. According to the size of angles, there are three type of triangles
(i) Right angled triangle: If one of the angles of the triangle A

is right angle (90o), it is known as a right angled triangle.

In the given triangle ABC, ∠ABC = 90o. So, it is a right B C
angled triangle.
There can not be more than one right angle in a triangle.

(ii) Acute angled triangle: If all the angles of a A

triangle are acute angles (<90o), it is known as 70o

acute angled triangle. In the given ∆ABC, ∠BAC B 60o 50o C

= 70o, ∠ABC = 60o and ∠ACB = 50o, all are acute

angles. So it is an acute angled triangle.

(iii) Obtuse angled triangle: If one of the angles C
of a triangle is obtuse (>90o and <180o), it is 35o
known as obtuse angled triangle. In the given

∆ABC, ∠ABC = 120o is obtuse angle. So, ∆ABC A 25o 120o
is an obtuse angled triangle. In a triangle there B

can not be more than one obtuse angle.

Note:

Sum of all three angles of any kind of triangle is two right angles that is 180o.

Prime Mathematics Book - 5 15

Exercise 1.4

1. Define:

(a) Triangle (b) Equilateral triangle

(c) Isosceles triangle (d) Scalene triangle

(e) Right angled triangle (f) Acute angled triangle

(g) Obtuse angled triangle

2. Measure the sides and angles of the following triangles State your
remarks and conclusion:
(a) A

B BC C ∠BAC ∠ACB ∠ACB Remarks
AB AC

Conclusion:………………………………….. P
(b)

∠QPR ∠PQR ∠PRQ Remarks

R

Conclusion:………………………………….. Q

(c) A

AB BC AC ∠ABC ∠ACB Remarks

Conclusion:………………………………….. BC
(d) X

Y Z
∠XYZ
∠XZY Remarks

Conclusion:…………………………………..

16 Prime Mathematics Book - 5

3. Classify the following triangles:

(a) A (b) D (c) G (d) J

B CE P F 120o IK L
(e) M (f) (g) (h) X 60o 70o Z
60o H
N RM L
(i) W O Q 60o 60o N 50o
(j) F 50o 50o Y

6cm 3cm

Y 100o H
X 5cm
G

4. Find the unknown angles in the following triangles:

(a) x (b) 60o a 45o (c) b (d) 85o
yz b 45o 65o 95o x
a
35o y

(e) 105o a 25o (f) (g)
ab
80o 40o
x
b

(h) a (i) a

x
2a

40o

Sum of angles of a triangle:
(Teachers are suggested to demonstrate the following): A

a. Measure the angles of the given triangles B C
and complete the following: Remarks

∠BAC ∠ABC ∠ACB ∠BAC + ∠ABC + ∠ACB

Conclusion: .................................

Prime Mathematics Book - 5 17

b. Take a triangular piece of chart paper and name it triangle ABC. Cut its
corners and join the corners A,B and C.
∠A + ∠B + ∠C =A............C..................
ABC

B A
c. Draw a triangle ABC on a chart paper. Draw
B AC
AD⊥BC Cut the triangle along the outlines.
Fold the corners B and C such that B and C D
meet at a point D along BC. Fold the corner
A such that A also meets at this point D.

∠A + ∠B + ∠C = ∠BDC = 180o B C
Thus, the sum of the angles of a triangle is
180o (or two right angles)

Exercise 1.5

1. Measure the angles of the given triangle and complete the
following:
(a) A

B ∠ABC C ∠BAC + ∠ABC + ∠ACB Remarks
∠BAC ∠ACB

Conclusion: .................................

(b) A C

B ∠ABC ∠ACB ∠BAC + ∠ABC + ∠ACB Remarks
∠BAC

Conclusion: .................................

18 Prime Mathematics Book - 5

(c) A ∠BAC ∠ABC ∠BAC + ∠ABC Remarks

B C Conclusion: .................................

2. Find the size of the unknown angles in the following triangles:
A
(a) a (b) P (c) (d) x

50o 40o

B 60o 30o C Q xR 110o a 45o 100o

3. Find the unknown side or angles in the following triangles :

(a) c (b) (c) (d)
x 60o y x
6cm

ab 60o 60o 70o a 50o 50o

4cm

(e) (f) (g) a (h) x
y
45o 45o
a 60o

x 45o

5.2cm

Quadrilaterals D
Given figure has four sides, four vertices and four angles. It A C

is a quadrilateral. D

A quadrilateral is a plane figure bounded by four sides. B C
D
Two special quadrilaterals
C
a. A square: It is a quadrilateral having all four sides A
equal and all the angles 90o. The given figure, ABCD

is a square where sides AB = BC = CD = DA and B
∠A = ∠B = ∠C = ∠D = 90o

b. A rectangle: A rectangle is a quadrilateral A

having opposite sides equal and all the angles

90o. Given figure is a rectangle where AB = CD,

AD = BC and ∠A = ∠B = ∠C = ∠D = 90o. B

Prime Mathematics Book - 5 19

The sum of angles of a quadrilateral A D
a. Measure the angles of the given quadrilateral and B C

complete the following:

∠A ∠B ∠C ∠D ∠A + ∠B + ∠C + ∠D Remarks

∠A + ∠B + ∠C + ∠D =

Conclusion: the sum of the angles of a quadrilateral is ……………….

b. Draw a quadrilateral ABCD in a chart paper, cut along its outlines, cut
its corners and arrange the corner pieces such that the vertices meet
a point. The corners form a circular angles.

Thus, ∠A + ∠B + ∠C + ∠D = 360o

AD

AD
BC
BC
Name each of the quadrilateral, also name its sides, vertices and angles.

Exercise 1.6

1. Name each of the quadrilateral, also name its sides, vertices and angles.

2. Find the size of the unknown angles from the following:

(a) 130o 80o (b) 90o 90o (c) 90o 70o

90o a 90o
x
90o y

(d) x 100o (e) 60o 95o (f) 120o b

2x 80o y 120o 70o 120o

3. Find the size of the unknown angles of the following quadrilaterals:

(a) 60o 80o (b) a 50o (c) 140o y (d) b 125o
a 115o
y 100o 60o

x 85o 90o b x 110o 75o

4. a. If W, X, Y, Z are the angles of a quadrilateral where W = 100o,
X = 80o, Y = 95o, find Z.

20 Prime Mathematics Book - 5

b. If x, 2x, 3x and 4x are the angles of a quadrilateral, find the size

of each angle.

c. In a quadrilateral ABCD, if ∠A = x, ∠B = 2x, ∠C = 3x and ∠D = 90o

find the size of the angles ∠A, ∠B and ∠C.

d. If the four angles of a quadrilateral are x, x + 15o, x + 25o and
x + 40o, find the value of x.

5. Measure the angles of the following and work out the sum of the four

angles:

(a) A D (b) P S

B CQ R

Unit Revision Test
1. Measure the following angles:

(a) (b)

2. Construct the following angles using protractor:
(a) 50o (b) 120o

3. Construct a line perpendicular to the line AB through the given point P:

(a) P (b) B
A B A P

4. Find the complementary angle of:
(a) 35o (b) 72o

5. Find the value of unknown angle x from the following figures:
(a) (b)

x 50o x
63o
Prime Mathematics Book - 5 21

6. Find the size of the unknown angles from the given figures:

(a) 110o (b) 105o 100o

40o x x 90o

7. Measure the angles of the given triangle and complete the following:
A

B C ∠ACB ∠BAC + ∠ABC + ∠ACB Remarks
∠BAC ∠ABC

Conclusion: .................................
8. If the four angles of a quadrilateral are x, x+10o, x+30o and 80o, find

value of x.

Answers:

Show the answer to your teacher.Exercise: 1.1

Show the answer to your teacher.Exercise: 1.2

Exercise: 1.3
1. Show the answer to your teacher 2. Show the answer to your teacher
3. (a) 15o (b) 70o (c) 40o (d) 29o
4. (a) 140o (b) 80o (c) 90o (d) 45o (e) 10o
5. (a) 25o (b) 80o (c) 45o (d) 65o (e) 130o (f) 100o (g) 90o (h) 135o
6. (a) x = 125o , y = 55o, z = 125o (b) x = 60o , y = 120o , z = 60o

Exercise: 1.4
1, 2, 3, show the answer to your teacher
4. (a) x = y = z = 60o (b) a = 75o, b = 45o (c) a = 25o, b = 25o (d) x = 85o, y = 60o
(e) a = 50o, b = 50o (f) a = 60o, b = 120o (g) x = 70o (h) a = 100o
(i) a = 30o, x = 120o

Exercise: 1.5 (c) 30o (d) 55o
1. Show the answer to your teacher. (c) a = 70o (d) x = 6cm
2. (a) a = 90o (b) x = 40o (g) a = 30o (h) x = 45o
3. (a) a = b = c = 60o (b) x = y = 4cm
(f) a = 5.2 cm
(e) x = y = 45o

Exercise: 1.6
1. (a) x = 60o (b) y= 90o (c) a= 110o (d) x = 60o, 2x = 120o (e) y = 85o (f) b = 50o
2. (a) x = 95o, y = 125o (b) a = 50o, b = 120o (c) x = 30o, y = 80o (d) a = 65o, b = 95o
3. (a) z = 85o (b) x = 36o, 2x = 72o, 3x = 108o, 4x = 144o
(c) ∠A = 45o, ∠B = 90o, ∠C = 135o (d) 70o
4. Show the answer to your teacher

22 Prime Mathematics Book - 5

Un2it

Numeration System

Estimated periods − 20

Objectives

At the end of this unit, the students will be able to:
• read and write the numbers upto 10 digits in local place value system of number.
• read and write the numbers upto 12 digits in International place value system.
• form the greatest and smallest number formed by the given digits.
• use commas according to the periods of the Numeration System.
• to know the face value and place value of digits of a number.

Teaching Materials
Place value table, abacus, tens charts and blocks for big numbers.

Activities
It is better to:
• involve the students while representing big numbers in place value table.
• discuss how to put the commas in local and international place value system.
• discuss how to form biggest and smaller numbers using the given digits.
• organize on activity for students using abacus about place values.
• use play cards and conduct activity about number system and place value.

Prime Mathematics Book - 5 23

Crores Place Value System of Number
Ten lakhsLocal Place Value System of Number:
LakhsDigits used for Hindi - Arabic number system are
thousaTnedns0 1 2 3 4 5 6 7 8 and 9
Thousandsin Devanagari number system, we use our own symbols for these digits such
Hundredsas.
0 1 2 3 4 5 6 7 8 and 9
TensRepresent five crore twenty four lakh eighty seven thousand and ninety one
Onesin a place value table and write the number.

52487091
Thus, number is 5,24,87,091
In Local Place Value System, first comma is put before the last three digits
and then after every two digits.
Represent 35, 24, 89, 167 in a place value table and write in words.

352489167
It is thirty five crore twenty four lakh eighty nine thousand one hundred sixty
seven.
Represent two arab forty crore eighty one lakh fifty three thousand nine
hundred seventy three in a place value table and write the number. Also put
the commas.

2408153973
Number is 2,40,81,53,973.

24 Prime Mathematics Book - 5
Ten crores
Crores
Ten lakhs
Lakhs

thousTaennds
Thousands
Hundreds

Tens
Ones

Arabs
Ten crores

Crores
Ten lakhs

Lakhs
thousTaennds
Thousands
Hundreds

Tens
Ones

Face value and place value of digits of a number
Let's take an example.

Ten crores
Crores

Ten lakhs
Lakhs

thousTaennds
Thousands
Hundreds

Tens
Ones

194532786

In the above number 4 is in ten lakhs position so it’s value is 4×10,00,000 =
40,00,000.

Therefore, the place value of 4 in the given number is 40,00,000.

Similarly, 9 is in the crores position, so it's value is 9×1,00,00,000 = 9,00,00,000.
Thus, the place value of 9 in the number 19,45,32,786 is 9,00,00,000. Thus, the
position of a digit in the number determines the value of the digit. This value is
called the place value of the digit. But, the place value of 0 is always 0 whatever
it occurs.

The face value of digit 4 and 3 in the number 19,45,32,786 is 4 and 3 itself.
Thus, the face value of a digit in a number remains as it is.

Let’s take another example.

Place Face value Place value
0
Ones 258714690 Ones 0 9×10 = 90
Tens 6×100 = 600
Hundreds Tens 9

Hundreds 6

Thousands Thousands 4 4×1000 = 4000
Ten 1×10000 = 10,000
Ten thousands 1 7×100000 = 7,00,000
thousand
Lakhs Lakhs 7

Ten lakhs Ten lakhs 8 8×1000000 = 80,00,000

Crores Crores 5 5×10000000 = 5,00,00,000

Ten crores Ten crores 2 2×100000000 = 20,00,00,000

Prime Mathematics Book - 5 25

Hundred BillionsInternational Place Value System of Number:
Ten BillionsLet's take a number 153,467,982,576 and represent it in a place value table of
BillionsInternational place value system.
Hundred
millions153467982576
Ten millionsThe number is one hundred fifty three billion four hundred sixty seven million
Millionsnine hundred eighty two thousand five hundred seventy six.
Hundred
thousandsIn the international place value system, first comma is placed before first
Ten thousandsthree digits and then after every three digits.
ThousandsRepresent two hundred four billion six hundred fifty seven million nine hundred
Hundreds eighty four thousand and six hundred eighty nine in a place value table and
Tenswrite the numeral.
Ones
204657984689
The number is 204,657,984,689.
Local and International Number System:
Locally : We have ones, tens, hundreds, thousand, the thousands, lakhs, ten
lakhs, crores, ten crores, arabs, ten arabs & so on
Internationally: We have one, tens, hundreds, ten thousands, hundred
thousands, millions, ten millions, hundred millions, billion, ten billion, hundred
billion, etc.

26 Prime Mathematics Book - 5
Hundred Billions
Ten Billions
Billions
Hundred
millions
Ten millions
Millions
Hundred
thousands
Ten thousands
Thousands
Hundreds
Tens
Ones

Local and International Period chart

Local Periods Numeration Interna- Numeration system
system tional
Ones Ones -Ps_

Ones Tens Ones Tens -bz_
Hundreds Hundreds -;o_

Thousands Thousands Thousands Thousands -xhf/_
La khs Ten thousands Ten thousands -b; xhf/_
Lakhs Hundred thousand -nfv_
Ten lakhs Millions -bz nfv_

Crores Millions Ten millions -s/f]8_
Ten crores Hundred millions -bz s/f8] _
Crores

Arab Arabs Billions Billions -c/a_
Kharab Ten arabs Trillions Ten billions -bz c/a_
Kharabs Hundred billions -v/a_
Ten kharabs Trillions -bz v/a_

Prime Mathematics Book - 5 27

Conclusion from the above chart Conclusion from the above chart
1 Lakh = 100 Thousand
10 Lakh = 1 Million 100 Thousand = 1 Lakh
1 Crore = 10 Million
10 Crore = 100 Million 1 Million = 10 Lakh
1 Arab = 1 Billion
10 Million = 1 Crore

100 Million = 10 Crore

1 Billion = 1 Arab

Nepali number system International number system
Ones Ones
Tens Tens
Hunderds Hundreds
Thousands Thousands
Ten- Thousands Ten Thousands
Lakhs Hundred - Thousands
Ten - Lakhs Millions
Crores Ten - Millions
Ten - Crores Hundred - Millions
Arab Billion
Ten Arab Ten Billions
Kharab Hundred Billions

Exercise: 2.1

1. Write the numerals in words in local place value system by
representing them in a place value table:

a. 17,04,789 b. 82, 14, 082 c. 2,14,57,609
d. 4,54,67,810 e. 13,52,68,123 f. 24,67,89,563
g. 2,14,62,89,410 h. 4,65,78,92,605 i. 16,25,67,94,695

2. Write the given numerals in words in international place value system
by representing them in a place value table:

(a) 254,198 (b) 156,274

(c) 3,194,607 (d) 4,098,956

(e) 54,196,289 (f) 69,781,254

(g) 297,007,148 (h) 354,980,659

(i) 2,197,250,678 (j) 3,208,598,423

28 Prime Mathematics Book - 5

3. Write the given number names in numbers by representing them in a place
value table in local place value system. Also, put commas:
a. Five lakh thirty six thousand nine hundred and eighty seven.
b. Fourteen lakh four thousand eight hundred and ninety two.
c. Six lakh forty four thousand seven hundred and eighty.
d. Twenty five lakh thirty eight thousand nine hundred and forty seven.
e. Three crore forty eight lakh ninety two thousand seven hundred and four.
f. Nine crore fifty six lakh eighty two thousand and nine hundred and fifty six.
g. Twenty five crore five lakh ninety seven thousand two hundred and four.
h. Two arab five crore two lakh nine thousand four hundred and eighty five.

4. Write the given number names in numerals in the international place
value system by representing them in a place value table and also
put commas:
a. Five hundred two thousand nine hundred and sixty eight.
b. Eight hundred fifty two thousand four hundred and seventeen.
c. Four millon six hundred fifty two thousand eight hundred and one.
d. Eight million seven hundred thousand two hundred and fifty four.
e. Fourteen million five hundred sixty seven thousand two hundred
and fifteen.
f. Twenty five million four hundred eight thousand nine hundred and
five.
g. Three hundred fourteen million five hundred six thousand nine
hundred and fifty nine.
h. Eight hundred eighteen million two hundred twenty two thousand
five hundred and ninety eight.
i. Nine billion four hundred two million six hundred forty three
thousand eight hundred and four.
j. Seven billion five hundred eighty million two hundred four thousand
six hundred and eighty.

Prime Mathematics Book - 5 29

5. Write the following numbers in expanded form
a. 732015
b. 9257183
c. 35792146

6. Write the number names of the following in Devanagari numbers
a. 5286
b. 17349
c. 68941

7. Write the Devanagari numbers in Numerals.

s_ Ps s/f8] krf; nfv bz xhf/ tLg ;o c7f/ Ö

v_ b'O{ s/f]8 krf; nfv bz xhf/ kfFr ;o Ö

u_ tLg c/a ;ft nfv bz xhf/ kfrF ;o Ö

3_ Ps c/a krf; s/f]8 bz xhf/ kfFr ;o Ö

The greatest and the least numbers:
The greatest number of one digit is 9.
The greatest number of two digits is 99.
The greatest number of three digits is 999.
The greatest number of four digits is 9999.
The greatest number of five digits is 99,999 and so on.

The smallest number of one digit is 1.
The smallest number of two digits is 10.
The smallest number of three digits is 100.
The smallest number of four digits is 1000.
The smallest number of five digits is 10,000 and so on.

30 Prime Mathematics Book - 5

The greatest and the least numbers formed by the digits.

Let's take three digits 1,4 and 7.
The numbers formed by these three digits are 147 ,417,471,741,174,714.
Among these, the greatest number is 741. So, it is clear that to form the greatest
number of same digits, the digits should be arranged in descending order.
Similarly, the smallest number is 147. So, to form the smallest number the
digits should be arranged in ascending order.
Let's take four digits 1,4,0 and 7.
The greatest number formed by these four digits is 7410.
The smallest number formed by these four digits is 1047 but not 0147 because
0147 is simply 147 which is a number containing three digits 1, 4 and 7.

Exercise: 2.2

1. Write the given numbers in numerals, in local place value system and
international place value system and also in words:
(a) The greatest number of six digits.
(b) The smallest number of seven digits.
(c) The greatest number of eight digits.
(d) The smallest number of nine digits.
(e) The greatest number of ten digits.

2. Write the greatest and smallest numbers formed by the following
digits without repetition of any digit. Write them in local place value
system and international place value system and also in words:

(a) 2, 0, 4, 7,9 (b) 3, 4, 9, 8, 7, 5

(c) 1, 2, 4, 0, 7, 8, 9 (d) 1, 3, 5, 7, 9, 8, 4, 6

(e) 2, 4, 6, 8, 0, 3, 5, 7, 1

3. Solve :-
(a) Find the sum of the greatest and the smallest number of 5 digits.

(b) Find the difference of greatest and smallest number of 9 digits.

Prime Mathematics Book - 5 31

(c) Find the difference of the greatest and smallest numbers formed by
using the digits 2, 5, 1, 6, 3, 8 and 4.
(d) Find the sum of the greatest and smallest number formed by using
the digits 2, 0, 4, 5, 7 and 9.

Unit Revision Test

1. Write the following numerals in words in local numeration system by
representing them in a place value table:

a. 3,52,67,49,601 b. 24,57,60,913

c. 28,69,45,83,758 d. 5,67,42,68,498

2. Write the following numerals in words in international numeration
system by representing them in a place value table:

a. 68,120,695 b. 276,581,409

c. 62,107,405,679 d. 387,452,698,410

3. Write the given number names in numerals by representing them in
a place value table in local numeration system. Also, put commas:
a. Sixty seven lakh forty two thousand one hundred and ninety
eight.
b. Twenty six arab five lakh two thousand nine hundred and four.
c. Forty eight crore thirty six lakh five thousand nine hundred and
five.
d. Two arab fifty crore three lakh nine hundred and eight.

4. Write the given number names in numerals by representing them
in a place value table in International numeration system. Also, put
commas:
a. Twenty five million five thousand two hundred and eight.
b. Three thousand ninety eight million eighteen thousand two hundred
and ten.
c. Ninety five billion six million forty seven thousand five hundred
and twelve.
d. Five thousand seventeen billion twenty five million four thousand
eight hundred and seventy five.

32 Prime Mathematics Book - 5

U3nit Concept of Numbers

Estimated periods − 18

Objectives
At the end of this unit, the students will be able to:
• know about the natural numbers, whole numbers, odd numbers and even numbers.
• know about the prime numbers and composite numbers.
• find the factors and multiples of given numbers.
• find the prime factors of a number.
• know the divisibility tests of 2, 3, 5, 6, 8, 9, 10 and 11.
• find the square and square root, cube and cube root.
• find the H.C.F and L.C.M.

Teaching Materials
Multiplication and division table, models of factor tree, divisibility test charts, etc.

Activities
It is better to:
• differentiate between natural number and whole number, odd number and even
number.
• differentiate between prime number and composite number.
• discuss with the students to find the square and square root of a number, cube and
cube root of number.
• discuss all the methods to find the L.C.M. and H.C.F. of the given numbers.

Prime Mathematics Book - 5 33

Introduction: Numbers

The concept of number and the process of counting developed long before
the time of recorded history. Humans even in most primitive times had some
number sense. With the gradual evolution of society, simple counting became
imperative and thus for keeping count, counting or natural numbers gradually
developed. A German mathematician Leopold Kronecker (1823-1897) once
remarked, "God created natural number and the rest is the work of man".

Natural number (N) : 1, 2, 3, 4, ....................

Whole number (W) : 0, 1, 2, 3, .....................

Integers (Z) : ....,...., -3, -2, -1, 0, 1, 2, 3, ......,.....

The symbol Z for integers
comes from the German word

Zahlen means to count.

Even numbers: 2 ) 18 ( 9 2 ) 26 ( 13
-18 -2
2)6(3 0 6
-6 -6
0 0

Here 6, 18, 26 are all exactly divisible by 2. Such numbers are called even
numbers.
Thus, the numbers which are exactly divisible by 2 are called even numbers.
Whether small or big, the numbers ending with any of the digits 0, 2, 4, 6 and
8 are even numbers. Even numbers may be positive as well as negative.
Let's check whether 5,20,984 is an even number or not. Let's see the digit at
ones place. It is 4 and it is an even number. So, 5,20,984 is an even number.

34 Prime Mathematics Book - 5

Odd numbers: 2 ) 25 ( 12 2 ) 93 ( 46
-2 -8
2)9(4 5 13
-8 -4 -12
1 1 1

The above numbers 9, 25 and 93 are not exactly divisible by 2. In each case, the
remainder is 1. Such numbers are called odd numbers.

Thus, the numbers which are not exactly divisible by 2 are called odd numbers
whether small or big, an odd number ends with any of the digits 1, 3, 5, 7 and
9. Odd numbers may be positive as well as negative.

Let's check whether 6,25,497 is an odd or an even number. See the digit in
ones place. It is 7 and 7 is an odd number. So, the number 6,25,497 is an odd
number.

Prime Numbers

2÷2=1 3÷3=1 7÷7=1
7÷1=7
2÷1=2 3÷1=3

The numbers such as 2, 3, and 7 are exactly divisible by 1 and by itself only.

Such numbers are called prime numbers. Thus, the numbers which are exactly
divisible only by 1 and by itself only are called prime numbers. For example, 2,
3, 5, 7, 11 …... etc.

Composite Numbers

4÷4=1 6÷6=1 9÷9=1

4÷1=4 6÷1=6 9÷1=9

4÷2=2 6÷2=3 9÷3=3

6÷3=2

Prime Mathematics Book - 5 35

The numbers like 4, 6, 9 are exactly divisible by 1, by itself and by other numbers
as well. Such numbers are called composite numbers .

Thus, the numbers which are exactly divisible by 1, by itself and by other
numbers are called composite numbers. For example : 8, 10, 12 etc.

Keep in mind that 1 is
neither prime nor composite
number. 2 is the only even prime

number.

Exercise: 3.1

1. a) Write the first five natural numbers.

b) Write the first five whole numbers.

c) Mention the first five even numbers.

d) Mention the first five odd numbers.

2. a) What are the first five prime numbers?

b) What are the first five composite numbers? -6,

c) Write all the prime numbers from 10 to 50. 1, 10, 8,

3. From the given set of numbers, write the set of: -2, 27, 11,
7, 6,
a) natural numbers b) whole numbers -11, 0, 3, 2,

c) integers d) even numbers 5, 9, 15, 12,
-5, 20
e) odd numbers f) prime numbers

Factors and Multiples

Observe:

12 × 1 = 12 12 ÷ 1 = 12

2 × 6 = 12 or, 12 ÷ 2 = 6

3 × 4 = 12 12 ÷ 3 = 4

6 × 2 = 12 12 ÷ 4 = 3

4 × 3 = 12 12 ÷ 6 = 2

12 ÷ 12 = 1

36 Prime Mathematics Book - 5

Here, 1, 2, 3, 4, 6 and 12 are factors of 12. They divide 12 exactly leaving no
remainder.
Multiples

4x1=4

4x2=8

4 x 3 = 12

4 x 4 = 16

4 x 5 = 20 and so on.
Here, 4, 8, 12, 16, 20 and so on are multiples of 4.

Exercise: 3.2

1. Find the factors of the following numbers:

a. 15 b. 20 c. 45 d. 25
h. 95
e. 125 f. 250 g. 81 l. 639

i. 105 j. 385 k. 425

2. Find the first five multiples of the given numbers:

a. 11 b. 24 c. 25 d. 8

e. 12 f. 15 g. 13 h. 18

i. 20 j. 31 k. 16 l. 19

Test of divisibility:

Divisibility test of 2.
Look at the numbers 480, 672, 9454, 10586. The digits in ones place are either
even or zero. Such numbers are exactly divisible by 2.

Divisibility test of 3.

If the sum of the digits used in the number are exactly divisible by 3, then the
given numbers are also exactly divisible by 3.
Look at the numbers 123, 6945, 91577
1 + 2 + 3 = 6 is divisible by 3 so 123 is exactly divisible by 3
6 + 9 + 4 + 5 = 24 is divisible by 3 so 6946 is exactly divisible by 3
9 + 1 + 5 + 7 + 7 = 29 is not divisible by 3 so 91577 is not divisible by 3

Prime Mathematics Book - 5 37

Divisibility test of 4.
If the number formed by the last two digits of number is exactly divisible by 4,
then the given numbers is exactly divisible by 4.
Observe the numbers 748, 2076, 98528.
In 748, number formed by last two digits that is 48 is divisible by 4. So, 748 is
exactly divisible by 4.
Similarly, in 2076 and 98528 number formed by the last two digits are divisible
by 4. So, the numbers are also divisible by 4.

Divisibility test of 5.
Look at the numbers, 205, 1950, 21895.
These numbers are exactly divisible by 5 because the digits in the ones place
are either 5 or 0.

Divisibility test of 6.
The numbers which are even and exactly divisible by 3 are divisible by 6.
See the numbers 924, 1944 and 85914.
The numbers are even and also divisible by 3.
So, these numbers are exactly divisible by 6.

Divisibility test of 7
In a given number, it the difference between the double of units digit and
remaining digits is either zero or multipal of 7, the number is divisible by 7. for
example : in 294 make double of 4, that is 8 and remaining is 29. The difference
is 29-8=21 Which is multiple of 7. so, 294 is divisible by 7.

Divisibility test of 8
If the number formed by the last three digits its divisible by 6, the number is
exactly divisible by 8.
Observe the numbers 448, 2048 and 9568.
These numbers are exactly divisible by 8 because the numbers formed by the
last three digits are exactly divisible by 8.

38 Prime Mathematics Book - 5

Divisibility test of 9

If the sum of the digits of a number is divisible by 9, the number is also divisible
by 9.
Observe the numbers 297, 19872 and 20331.
2 + 9 + 7 = 18
1 + 9 + 8 + 7 + 2 = 27
2+0+3+3+1=9
These numbers are exactly divisible by 9 because the sum of the digits are
exactly divisible by 9.

Divisibility test of 10

If the numbers with ones digit is 0 (zero) are divisible by 10
Look at the numbers 980, 2570 and 67540.
All the given numbers are exactly divisible by 10 because the digit in ones
place is 0.

Divisibility test of 11

If the difference of the sum of alternate digits of a number is 0 or multiple of
11. Then the number is exactly divisible by 11.

Look at the numbers 1,34,596 and 20,510.

13 45 96 Sum of up-cup digits = 1 + 4 + 9 = 14

Sum of down-cup digits = 3 + 5 + 6 = 14

Difference = 14 - 14 = 0.

20 5 1 5 Sum of up-cup digits = 2 + 5 + 5 = 12

Sum of down-cup digits = 0 + 1 = 1

Difference = 12 - 1 = 11

The above given numbers are exactly divisible by 11 because the difference of
the sum of up-cup digits and the sum of down-cup digits is either 0 or multiple
of 11.

Prime Mathematics Book - 5 39

Exercise: 3.3

1. Which of the following numbers are exactly divisible by 2? Mention
with reason.

a. 1,258 b. 35,427 c. 5,27,490 d. 67,951

2. Which of the following numbers are exactly divisible by 3? Mention
with reason.

a. 27,501 b. 18,952 c. 3,27,459 d. 4,12,308

3. Which of the following numbers are exactly divisible by 4? Write the
reason.

a. 23,748 b. 85,471 c. 94,642 d. 1,23,452

4. Which of the following numbers are exactly divisible by 5? Write the
reason.

a. 24,507 b. 85,605 c. 94,804 d. 1,23,620

5. Which of the following numbers are exactly divisible by 6? Write the
reason.

a. 24,602 b. 54,906 c. 68,424 d. 1,25,642

6. Which of the following numbers are exactly divisible by 7? Write the
reason.

a. 8,085 b. 22,792 c. 30,275 d. 38,731

7. Which of the following numbers are exactly divisible by 8? Write the
reason.

a. 25,472 b. 84,985 c. 95,784 d. 1,42,186

8. Which of the following numbers are exactly divisible by 9? Write the
reason.

a. 14,922 b. 25,803 c. 45,917 d. 1,26,346

9. Which of the following numbers are exactly divisible by 10? Write
the reason.

a. 12,940 b. 24,815 c. 35,210 d. 1,93,894

10. Which of the following numbers are exactly divisible by 11? Write
the reason.

a. 15,489 b. 33,759 c. 45,942 d. 2,74,890

Prime Factorisation.

20 ÷ 1 = 20 20 ÷ 2 = 10 20 ÷ 4 = 5

20 ÷ 5 = 4 20 ÷ 10 = 2 20 ÷ 20 = 1

Here 1, 2, 4, 5, 10, 20 are factors of 20. But among them only 2 and 5 are prime
factors of 20.

40 Prime Mathematics Book - 5

Finding prime factors of numbers
A. Continuous division method
Let's find the prime factors of 30.

2 30 ∴[ 30 is an even number which is divisible by 2.]
3 15 ∴[ 15 is an odd number, the sum of whose digits 1 & 5 is 6, which is divisible by 3]
[Stop dividing because 5 is a prime number.]
5

∴ 30 = 2 x 3 x 5
B. Factor tree method .

30

2 × 15 ∴ 30 = 2 × 3 × 5

2× 3 × 5
Exercise: 3.4

1. Find the prime factors of the following numbers using continuous
division method:

a. 128 b. 330 c. 540 d. 200

e. 630 f. 900 g. 112 h. 800

i. 144 j. 216 k. 108 l. 924

2. Using the factor tree diagram, find the prime factors of the following
numbers:

a. 625 b. 1000 c. 64 d. 72

e. 420 f. 80 g. 375 h. 230

i. 243 j. 2000 k. 180 l. 720

Square number and square root of a number.

Let's observe some examples:

The square of 1 is 12 = 1 × 1 = 1. The square of 2 is 22 = 2 x 2 = 4.

The square of 3 is 32 = 3 x 3 = 9. The square of 4 is 42 = 4 x 4 = 16.

The square of 5 is 52 = 5 x 5 = 25.

Here, 1, 4, 9, 16, 25 are called square numbers. Thus, a square number is the
product of two identical numbers.

When square numbers are expressed in figures, they form square shapes.

Prime Mathematics Book - 5 41

14 9 16 25

From the above examples it is clear that the square of 1 is 1. 1 is called the
square root of 1.

The square of 2 is 4. 2 is called the square root of 4.

The square of 3 is 9. 3 is called the square root of 9.

The square of 4 is 16. 4 is called the square root of 16.

The square of 5 is 25. 5 is called the square root of 25.

Worked out Examples
Example 1: Find the square number of 15.
Solution: The square number of 15 is 152

= 15 × 15
= 225

Example 2: The square root of 441.

Solution: 3 441 First arrange the identical
3 147 prime factors into groups and take
7 49
a prime factor from a group of
77 two identical prime factor.

∴ 441 = 3 × 3 × 7 × 7
Square root of 441 = 3 × 7
= 21

Exercise: 3.6

1. Find the square of each of the following numbers:

a. 9 b. 12 c. 14 d. 25

e. 34 f. 48 g. 15 h. 11

i. 75 j. 100 k. 215 l. 405

m. 1000

42 Prime Mathematics Book - 5

2. Find the square root of each of the following numbers:

a. 64 b. 144 c. 196 d. 121

e. 100 f. 81 g. 324 h. 225

i. 36 j. 400 k. 900 l. 441

3. A chess board is square shaped board containing 8 squares along the
rows and columns. How many unit squares are there?

4. In a horticulture farm, 576 orange trees are planted in such a way
that there are equal number of plants along the rows and columns.
How many trees are there in a row?

5. A PT teacher wants to arrange some students in a square for a drill
with 16 students in a row. How many students are required for the
drill?

Cube and Cube root

The cube number of 1 is 13 = 1 × 1 × 1 = 1.
The cube number of 2 is 23 = 2 × 2 × 2 = 8.
The cube number of 3 is 33 = 3 × 3 × 3 = 27.
The cube number of 4 is 44 = 4 × 4 × 4 = 64.
The cube number of 5 is 55 = 5 × 5 × 5 = 125 and so on.
Thus, a cube number is the product of three identical numbers.
From above examples it is clear that 1 × 1 × 1 is a cube number of 1, so 1 is
cube root of 1.
(2 × 2 × 2) i.e. 8 is a cube number of 2, so 2 is cube root of 8.
(3 × 3 × 3) i.e. 27 is a cube number of 3, so 3 is cube root of 27.
(4 × 4 × 4) i.e. 64 is a cube number of 4, so 4 is cube root of 64.
(5 × 5 × 5) i.e. 125 is a cube number of 5, so 5 is cube root of 125.

Workout Examples

Example 1: Find the cube of 13.

Solution: The cube of 13 = 133

= 13 × 13 × 13

= 169 × 13

= 2197

Prime Mathematics Book - 5 43

Example 2: Find the cube root of 512.

Solution: 2 512
2 256
2 128 Like prime factors are
2 64 grouped and a prime factor is
2 32 taken from a group of three
2 16
like prime factors.
28
24

2

∴ 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

Cube root of 512 = 2 × 2 × 2
=8

Exercise: 3.7

1. Find the cube of the following numbers: d. 12
a. 6 b. 9 c. 10 h. 26
e. 15 f. 18 g. 24 l. 100
i. 30 j. 35 k. 72

2. Find the cube root of the following numbers:
a. 1728 b. 125 c. 343 d. 1000
e. 1331 f. 216 g. 729 h. 4096
i. 5832 j. 2744 k. 9261 l. 4913

Highest common factor (H.C.F.)

Let’s find the prime factors of 24 and 18.

2 24 2 18
2 12 39
26
23 3

∴ 24 = 2 × 2 × 2 × 3
18 = 2 × 3 × 3

44 Prime Mathematics Book - 5

Their common prime factors are 2 and 3. Their product is 6. So, 6 is the highest
common factor of 24 and 18. This method is called prime factorisation method
of H.C.F.

∴ H.C.F. of 24 and 18 is 6. It is clear that the
H.C.F. of two numbers is the
highest number which divides

them exactly.

We can also determine the H.C.F. of 24 and 18 by division method as:

18 ) 24 ( 1 This method is

-18 known as division
6 ) 18 ( 3 algorithm for H.C.F.
-18

0 H.C.F of two prime
numbers is always 1.

∴ H.C.F. of 24 and 18 is 6.

Exercise: 3.8

1. Find the H.C.F of the following number by factor method.

a. 12 and 15 b. 15 and 24 c. 10 and 25

d. 15 and 42 e. 64 and 84 f. 50 and 80

2. Find the H.C.F. of the following numbers by prime factorisation
method:

a. 60, 75 and 90 b. 144, 252 and 324 c. 75,125 and 150

c. 30, 45 and 70 e. 160 , 170 and 200 f. 40, 60, and 80

2. Find the H.C.F. of the following numbers by division method:

a. 75 and 120 b. 18 and 30

c. 63 and 105 d. 36 and 54

e. 80 and 90 f. 70 and 84

g. 96, 256 and 288 h. 75, 90 and 120

i. 75, 100 and 125 j. 48, 60 and 72

k. 275, 325 and 400 l. 192, 256 and 320

Prime Mathematics Book - 5 45