Prime Mathematics 7

Rotation about origin through +180o or -180o (half (x,y)
turn)
Consider a point P(x,y)
Join O and P and rotate OP through +180o or - 180o to
the position OP´

Draw PM and P´M´^ to XX´,
We get POM = P´OM´ and DOP´M´@DOPM

Where OM´ = -OM = -x
P´M´ = -PM = -y

\Co-ordinates of P´is (-x,-y)
\ Under rotation about through +180o or - 180o

P(x,y)→P´(-x,-y)

Example 5. Find the images of the points A(-3,5) and B(2,6) under rotation about origin.
(a) through 90o (b) through -90o (c) through 180o

Solution:

(a) Given points A(-3,5) and B(2,6)

We have,
Under rotation about origin through 90o

P(x,y) → P´(-y,x) \ A(-3,5) → A´(-5,-3) B(2,6) → B´(-6,2)

(b) We have,
Under rotation about origin through -90o

P(x,y) → P´(y,-x) \ A(-3,5) → A´(5,-(-3) = A´(5,3)

B(2.6) → B´(6,-2)

(c) We have,
Under rotation about origin through 180o

P(x,y) → P´(-x,-y) \ A(-3,5) → A´(3,-5) B(2.6) → B´(-2,-6)

Example 6: Rotate DABC with vertices A(2,1), B(4,-1) and C(5,3) about origin through
-90o. Plot the DABC and its image DA´B´C´ on the same graph.
Solution :

Vertices of DABC are A(2,1), B(4,-1) and C(5,3)

Centre of rotation is origin and angle of rotation is -90o
We have,
Under rotation about origin through -90o

P(x,y) → P´(y,-x)
A(2,1) → A´(1,-2)
B(4,-1) → B´(-1,-4)
C(5,3) → C´(3,-5)

94 Transformation

Exercise 4.1

1 (a) Name the fundamental transformations
(b) What do you mean by isometric transformations
(c) Define invariant point.
(d) Define identity transformation.

2. Draw the following figures approximately and translate by the given translation
vector.

(a) (b) (c)

(3) Copy the following figures approximately on your copy and draw the image of each
figure under reflection on the given line MN.

A (b)

(a)

BM

N
C

(c) (d)

Prime Mathematics Book - 7 95

4 (a) Rotate DABC about the point O through positive quarter turn.
(b) Draw the image of the figure under rotation about the point O through -60o.

(c) Rotate the figure ABCD about O through -90o

O

5. Find the image of the points A(1,4) and B(-3,4) under the following transformation.
(a) Reflection on X-axis (b) Reflection on Y-axis
(c) Rotation about origin through +90o (d) Rotation about origin through 180o

6. Find the coordinates of the vertices of image of DABC with vertices A(-1,4), B(2,-2)

C(5,2) under reflection on
(a) x-axis (b) y axis

Plot the DABC and DA´B´C´ on the same graph.

7 (a) Find the coordinates of the vertices of image of DABC with vertices A(-1.5),
B(4,-1)and C(2,3) under rotation about origin through +90o. Plot the DABC and
DA´B´C on the same graph

(b) Find the co-ordinates of the vertices of DABC under rotation about origin through
90o anti clockwise direction. Vertices of DABC are A(-3,-2), B(0,-1) and C(-1,-5).

(c) Under rotation about origin thought 180o, DABC→DA´B´C´. If co-ordinates of

A.B and C are respectively (2,5), (-1,7) and (-3,3). Find the co-ordinates of
A´, B´ and C´.

96 Transformation

4.2 Symmetry, Tessellation and Design of Polygons

4.2.1 Symmetry

Appearance of body

Bilaterally Symmetrical Radially Symmetrical

Symmetry is the correspondence of parts of a figure with reference to a point or line
or plane.

Line symmetry
Look at the following figures:

Each dotted line divides the figure into two identical parts. The figures are said to be
symmetrical about the line and such symmetry is called line symmetry. The line about
which the figure is symmetrical is called the line of symmetry (axis of Symmetry) and
the figure is said to have bilateral symmetry. In the above figure, the heart has only
one line of symmetry, the rectangle has two and the equilateral triangle has three
lines of symmetry. Symmetrical figure may have different number of lines (axes).
A circle has infinite numbers of lines (axes) of symmetry.

Prime Mathematics Book - 7 97

Most of the animals have their bodies externally (bilaterally) symmetrical. Leaves of
many plants are almost line symmetrical about their mid ribs.

Rotational Symmetry
The letter M is symmetrical about the dotted line. The
letter N seems symmetrical but not bilateral symmetry. If
we rotate about its mid point about 180o the corresponding
parts coincide. Such figures are said to have rotational
symmetry.
Thus, if corresponding parts of a figure, fit to each other
by a rotation less than and complete turn 360o about a
point, the shape is said to have rotational symmetry.

Order of rotational symmetry O
On completing a complete rotation
about O, the rectangle ABCD comes to
its original position twice. It is said to
have rotational symmetry of order
two.

On making a complete rotation about the centre O, the O
equilateral triangle comes to its usual position thrice and
an equilateral triangle is said to have rational symmetry of
order 3.

98 Transformation

Similarly, on making a complete rotation about centre O, The
square ABCD comes to its usual position 4 times. So, a square is
said to have rotational symmetry of order 4.

The order of rotation of any rotational symmetric figure is
defined as the number of times the figure is rotated by the
minimum angle to bring each of its corresponding point (part)
in position of exact fit as it was before rotation.

The order of rotation = 360o

The minimum angle of rotation (q)

Here, minimum angle of rotation = Angle between the lines joining adjacent
corresponding point to the centre.

Consider, the letter S. Join corresponding adjacent points A and B to the centre of
rotation O. Here AOB = 180o

\Minimum angle of rotation (q) = 180o

Now, order of rotation = 360o

q

= 360o

180o

=2.

In an equilateral triangle ABC, the points A, B and C are corresponding points. Joining
A and B to centre O.
We get

AOB = 120o = q

\ Order of rotation = 360o

q O

= 360o
120o

= 3.

Prime Mathematics Book - 7 99

Exercise 4.2.1
1. Draw possible lines of symmetry of the following figures (Use dotted line).

(a) (b) (c)

(d) (e) (f)

(g) (h) (i)

2. State whether the following letters have line symmetry or rotational symmetry or
no symmetry.

A, B, to Z

3. Find the minimum angle of rotation and order of rotation of the following
figures.

(a) (b)

100 Transformation

(c) (d)

(e) (f)

4.2.2 Tessellation

Tessellation is the process of covering a
plane with congruent geometrical shapes in a
repeating pattern with no space between
them and without overlapping.

In tesselation, regular polygon or other
shapes are used.
Tessellation is used in paving roads or
floors, tiling the walls, carpet and textile
designs.

Prime Mathematics Book - 7 101

Type of tessellation.
1. Regular tessellation: A tessellation using congruent regular polygon of same type

that meet so that no vertex of one polygon lies on the side of another is referred
to a regular tessellation . The pattern of equilateral triangles, squares as in chess
board pattern and regula r hexagonal pattern are the only tessellations of regular
polygons that exit.

2 Semi-regular tessellation: If two or more than two regular polygons are used, the
tessellation is called semi regular tessellation.

102 Transformation

3. Non-regular tessellation: In non-regular tessellations, non-regular polygons are
used, Tessellation of curved shapes are also non-regular tessellations.

Exercise 4.2.2
1. Complete the following regular tessellations.
(a) (b)

(c) (d)

2. Complete the following semi-regular tessellations.
(a) (b)

Prime Mathematics Book - 7 103

4.3 Bearing and Scale drawing

N

Dawa, can you explain Pemba’s house
the location of Pemba’s

and Norbu’s house ?

Yes, Mam, Pemba’s house 100 m Norbu’s
house
is 100 m to the North and 200 m
Norbu’s houe is 200 m to the

North East from my house.

Dawa ‘s house

To give the location of a place we need three things:

1. Reference point (place from where observation is made)

2. Distance (length of straight line joining two places)

3. Bearing (direction to which the place is from the reference point).

Representation of bearing :
1. In vedic vastu, the reference point is called Brahma Sthana and eight
directions are mentioned
Purba (East) (Direction of Surya),
Paschim (West) (Direction of Varuna), V Ishana (NE)

Uttar (North)(Direction of Kubera)

Dakshin (South) (Direction of Yama),

Ishana (NE, direction of Mahadeva)

Agneya (SE, direction of Agni),

Nairitya(SW, direction of Rahu)

Vayabya(NW, direction of Vayu).

2. Compass bearing: After invention of magnetic compass
people measured the directions with respect to N-S
dircetion as; N, S, E, W, NE, SE, NE, SW etc. The direction just
between N and NE was denoted by NNE, Similarly between
NE and E as NEE, between SE and E as SEE etc. But this way
of representing bearing is not convenient .

104 Transformation

Second Method. Second phase of representing R
compass bearing is use of angle with respect to N or S
like compass bearing of P is N 80o E, that of Q is S 30o W
and that of R is N 40o W.

3. Three figure bearing : Q
Three figure bearing is another way of locating the
direction of a place. In this system we explain the
direction of a point or place from a point or place in term
of angle in three digits taken in clockwise direction from
the North Line.

To measure the bearing of P from A , AN is North line and
NAP = 70o. So bearing of P from A is 070o . Thus in three figure
bearing, direction (bearing) of E is 090o, SE is 135o, S is 180o, W is
270o and NW is 315o. 070o

In a radar screen three figure bearing is used where the centre is the
position of airport, 0 or 0o is the North line. The dots show different
places and the circles show the distances in Km or aeronautical miles. The revolving
bright line shows the position of aeroplane.

Scale drawing:

The drawing represents a football ground
90m × 60m. Drawing of actual size of the
ground is not possible so it is to be reduced
to a drawing of desired size 9cm × 6cm.

Here, ratio of length of the drawing to the

length of actual ground

= 9cm = 9cm = 1
90m 9000cm 1000

Prime Mathematics Book - 7 105

Similarly, ratio of breadth of the drawing to the breadth of the actual ground

= 60cm = 60cm = 1
60m 6000cm 1000

The ratio of distance between any two points in the drawing to the distance
between corresponding points in real field is always same and is called scale of drawing
(map).
map distance
\Scale of drawing = actual distance

Scale of above drawing is 1:1000
Which means distance 1cm in map = distance 1000 cm in actual field i.e. map

distance 1cm is same as 10 m in actual field.

If map distance 1cm = Actual distance
100 km scale = 1cm:100 km
= 1:10000000 which seems clumsy, so in
such case scale is represented as
0 100 200km where length of the
line segment 2cm (by actually
measuring with scale) in map = 200 km
in actual distance and half of the
segment 1 cm in map = 100 km in actual field.

In the given map distance between Kathmandu and Birgunj is 2 cm and scale provided
is 1 cm = 100 km.

Now, to calculate actual distance, we have.

map distance = 1cm
actual distance 100km

2cm = 1cm
actual distance 100km

\Actual distance = 200km

Example 1. Express the following compass bearing in three figure bearing.
(b) S30o E
(a) NNE

Solution
(a) Given compass bearing = NNE.
which is a Point between N and NE
45o 1o
= 2 = 22 2

106 Transformation

(b) Compass bearing = S 30o E
Here, S 30o E means
SOP = 30o

\ Three figure bearing of P is

NOP = NOS - SOP
= 180o-30o
= 150o

Example 2. Draw and write the compass bearing of the following 3 figure bearings.
(a) 230o (b) 345o

Solution:

(a) Given 3 figure bearing = 230o

Here, SOP = 230o-180o = 50o

\Compass bearing of P is S 50o W.

(b) Given 3 figure bearing = 345o
Here, NOP = 360o- 345o = 15o

\Compass bearing of P is N 15o W.

Example 3. Figure given along side shows a drawing of a rectangular bed room of a
house. If the scale of the drawing is 1 :200, find the areas of the room in m2.

Solution:

Given scale 1:200

⇒ 1 cm in drawing = 200 cm in actual room

\ 2.5 cm in drawing = 200 × 2.5 cm in actual room

= 500 cm in actual room

= 5 m in actual room

And 2 cm in drawing = 200 × 2 cm in actual room

= 400 cm = 4 m

\Length (l) = 5m and breadth (b) = 4m
\Areas of room = l × b = 5m × 4m = 20 m2

Prime Mathematics Book - 7 107

Example 4: From the given map, find the distance and bearing of the places B and C

from A.

Solution:

From A a north line AN is drawn joining A and B ; A and C,
we get NAB = 70o and AB = 3.8cm

\Bearing of B from A is 070o, as scale is 1:10000

⇒1cm of map distance = 10000cm in real field.

3.8 cm of map distance = 10000 × 3.8 cm in real

field.

= 38000cm = 380m.
Again AC = 2.6cm and NAC = 154o
= 154o and as scale is
\Bearing of C from A

1:10000

⇒1 cm of map distance = 10000 cm of real field. scale
= 10000 × 2.6 cm of real field. 1:10000
2.6 cm of map distance = 26000 cm = 260 m

\Distance of C from A is 260 m

Exercise 4.3

1. Write down the following compass bearings in 3 figures bearing.

(a) E (b) S (c) W (d) N (e) NW (f) SW
(g) SE (h) NE (i) NNE (j) SEE (k) SSW (l) NEE

2. Write down the following compass beings in 3 figures. Show also in diagram.

(a) N 40° E (b) S 15°E (c) S 45°W (d) S 90° W (e) N 45°W

(f) N 90° E (g) S 15°W

3. Write the compass bearing of the following. (c)
(a) (b)

(d) (e) (f)

108 Transformation

4. Study the map and write the bearing of the following from Kathmandu.
(a) Pokhara
(b) Mt. Everest
(c) Birgunj
(d) Biratnagar
(e) Nepalgunj
(f) Janakpur.

5. Study the map of part of Kathmandu N Tokha Chunikhel
district shown alongside and find the 0 1 2 3 4 Km
distance between. Kapan
(a) Kathmandu to Kapan Gonggabu Jorpati
(b) Kathmandu to Gonggabu
(c) Chhunikhel to Jorpati Kathmandu
(d) Tokha to Kapan

6. The drawing shown alongside is a rectangular
room of a home. Using the given measures and
scale, find the area of the room in m2.

Scale 1 : 200

Unit Revision Test.

1. Name four fundamental transformations.

2. Draw the given figure approximately and translate the
DABC by the given vector.

3. Copy the given figure approximately and
reflect DABC on the line XY.

o

4. Copy the given figure approximately in your copy and

rotate the DPQR about the point ‘O’ though -90°.

Prime Mathematics Book - 7 109

5. Write the following compass bearing in three figure bearing
a) S 40o E. (b) NE.

6. Write the compass bearing of the point P.

o

7. Find the coordinates of the image of DABC with vertices A(2,3),

B(5,-2) and C(7,4) under reflection on X-axis.
8. ABC is a triangle with vertices A(-1,-3) , B (2,-2) and C(5,-4). Find the coordinates

of the image of DA´B´C´ after rotation about origin through positive quarter turn.
Plot the DABC and DA´B´C´ on the same graph paper.
9. Drawing of a rectangular room has length 2.75 cm and breadth
2 cm. If the scale of the drawing is 1:200, find the area of the
room.

Answers

Unit: 4 Transformation Exercise: 4.1

1. Show to your teacher. 2. Show to your teacher. 3. Show to your teacher.

4. Show to your teacher.

5. a) A´(1,-4), B´(-3,-4) b) A´(-1,4), B´(3,4)

c) A´(-4,1), B´(-4,-3) d) A´(-1,-4), B´(3,-4)

6. a) A´(-1,-4), B´(2,2), C´(5,-2) b) A´(1,4), B´(-2,-2), C´(-5,2)

Graph: Show to your teacher.

7. a) A´(-5,-1), B´(1,4), C´(-3,2) Graph: Show to your teacher.

b) A´(-2,3), B´(-1,0), C´(-5,1) c) A´(-2,-5), B´(1,-7), C´(3,-3)
Exercise: 4.2.1
1. Show to your teacher. 2. Show to your teacher.
3. a) 180°, order 2 b) 90°, order 4 c) 120°, order 3 d) 60°, order 6
e) 60°, order 6 f) 180°, order 2

Show to your teacher. Exercise: 4.2.2

Exercise: 4.3
1. a) 90° b) 180° c) 270° d) 0°
e) 315° f) 225° g) 135° h) 045°
i) 22.5° j) 112.5° k) 202.5° l) 67.5°
2. a) 040° b) 165° c) 225° d) 270°
e) 315° f) 090° g) 195°
3. a) N 80°E b) S 40°E c) N 90° W d) S 45° W
e) N 30°W f) N 80°W
4. a) 290° b) 008° c) 206° d) 123° e) 276° f) 151°
5. a) 2.7km b) 2.2km c) 2km d) 2.15 km 6. 12m2

110 Transformation

UNIT Objectives:

5 SETS

Estimated periods 12

At the end of this unit students will be able to

● get introduction of sets, universal sets, proper and improper subsets.
● illustrate sets in Venn-diagram.
● illustrate different set operations (union, intersection, difference and

complements) in Venn Diagram

AB

Teaching Materials:
Flash cards, number cards, charts of different things.

Activities:

It is better to
● explain about universal sets, subsets.
● let the students form proper and improper subsets from the given set.
● involve the students to draw Venn - Diagram and illustrate the different operations.

Historical fact:

The set theory was developed by a German mathematician Georg Cantor
(Georg Ferdinand Ludwig Philip Cantor) (1845 – 1918).

Introduction

A collection or aggregate of well defined objects is called a set.
For example:

A = Set of days of a week = {Sunday, Monday, Tuesday, Wednesday, Thurs-
day, Friday, Saturday}

N = Set of natural number = {1, 2, 3, 4, …}
Sets are denoted by capital letters A,B,C, …., Z etc. Each object of a set is called an
element or a member of the set.

Let V = {a,e,i,o,u} then ‘a’ is an element of set V and we write a∈V. The symbol ∈
stands for 'is member of 'or 'belongs to.' As b is not a member of V, we write b∉V
5.1 Describing a set

We describe a set by the following methods.

1. Description Method:

In this method we describe the common property/ properties of the members of the
set in words inside the braces.
For example:

A = {days of a week}
B = {whole numbers less than 12}

2. Listing Method (Roster Form) :

In this method we enumerate or list all the members of a set inside the braces.
For example:

A = {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}
B = {0,1,2,3,4,5,6,7,8,9,10,11}

3. Set-builder Method (Rule Method):

In this method the members of a set are represented by a variable and specify the
'defining property'.
Like, A = {x: x, has property p}
This is read as “A is the set of elements 'x' such that (where) x has the
property p”.
For example:

W = {x:x, is a whole number}
N = {x:x, is a natural number}
A = {x:(x-2)(x-3) = 0}

112 Sets

Various Sets

On the basis of number of elements, sets may be
1. The empty (or null or void) set: The set containing no elements is called the null set

or the empty set or the void set. It is denoted by { } or ∅ (phi)

For example:
A = {x: x is a whole number between 1 and 2}

B = {x: x+1 = 0, x∈N}

Note: ∅ ≠ {0}, since {0} is a set whose element is 0.

∅ ≠ {∅}, since {∅} is a set whose element is ∅.
Since there is one and only one empty set with different descriptions, we say ‘the
empty set’ instead of ‘an empty set’.
2. Singleton set or unit set: A set containing only one element is called a singleton set
or unit set.
For example:

A = {0},
B = {1}
C = {x: x is a planet of solar system having life in it}

3. Finite set: A set with finite number of elements is called a finite set.
For example:

A = { 1,2,3,4,…………….,50}
B = {odd numbers between 10 and 100}

4. Infinite set: A set which contains infinite number of elements is called an infinite
set.
For example:

N = {1,2,3,……………..}
W = {0,1,2,3,……………}

Cardinal numbers:
Let's take a set A = {a,e,i,o,u}. Here , the number of elements of the set A is 5. We
say cardinal number of the set A is 5. Mathematically we write n (A) = 5.

Note: For the empty set ∅, n (∅) = 0
For the set A = {0}, n(A) = 1.

Set Relations:
On the basis of the types and number of elements contained in two or more than two
sets the relationships can be defined.

Prime Mathematics Book - 7 113

1. Equal sets: Let's consider two sets A = { e,a,t} and B = {t,e,a}. Here the sets
A and B have same elements and also n(A) = n(B) = 3. Such sets
are called equal sets. Thus the two sets A and B are equal if they
contain exactly the same elements. i.e. every element of A is an
element B and every element of B is also an element of A.

2. Equivalent sets: Let’s consider two sets A = {1,2,3,4,5} and B = {a,e,i,o,u}. Here
n(A) = n(B) but the elements of the sets A are not contained in
the set B and the elements of the set B are not contained in the
set A. The two set are said to be equivalent. Thus two sets A and
B are said to be equivalent and written as A~B if they have equal
cardinal numbers but not have exactly same elements.

Note: If two sets are equal then they must be equivalent.
However two equivalent sets need not be equal.

3. Overlapping Sets:Let’s consider two sets A = {a,e,i,o,u} and B = {a,b,c,d,e}. Here,
the elements ‘a’ and ‘e’ are common to both the sets A and B .
So the set A and B are overlapping sets. Thus two sets are said
to be overlapping or intersecting sets if they have one or more
common elements.

4. Disjoint sets: Two sets are said to be disjoint sets if they have no element in
common.

For example: i) If A = {1,2,3} and B = {4,5,6} then A and B are disjoint sets since

there is no elements common in A and B.

ii) If A = {x:x ∈ N}, B = {x:x ∈ Z, x<0} then A and B are disjoint.

Exercise 5.1

1. Define sets. Which of the following collections are sets?
a) A collection of tall men.
b) A collection of planets of solar system.
c) A collection of beautiful girls.
d) A collection of districts of Nepal.

2. Represent the given sets as stated in the brackets.
a) A = {Natural numbers less than 7} [listing /Tabular form]
b) B = {Sun, Mon, Tues, Wed, Thur, Fri, Sat} [Descriptive form]
c) C = {0,1,2,3,4,5} [Set builder form]

d) D = {x:x = 2n+1, n ∈ N} [Tabular form]
[Set builder form]
e) E = {2,6,10,14,18,22,26}

3. Identify whether the following sets are null or singleton.
a) {x:x is an even prime number} b) {0}

c) {x:x+1 = 5, x ∈ N} d) {Cube number between 20 and 25}

4. List the elements of these pairs of sets and state with reasons whether they are
equal or equivalent sets.
a) A = {1,2,5,10} and B = {factors of 10}

114 Sets

b) P = {x:x ≤ 5, x ∈ N} and Q = {y:y < 5 ∈ W}

c) C = {x:x is a letter of the word TALENT} and D = { x:x is a letter of the word
LATENT}

d) R = {x:x < 5,x ∈ N} and S = {x:x is a letter of the word FOUR}

5. List the elements of these pairs of sets and state with reason whether they are
overlapping or disjoint sets.
a) A = {1,2,3,4} and B = {odd number less than 10}
b) C = {x:x is a factor of 12} and D = {y:y is a factor of 18}
c) P = {x:x is an even number; 5<x<12} and Q = {x:x is a prime number;5<x<12}
d) A = {a, b, c, d} and B = {composite number less than 10}

6. List the elements and write the cardinal number of the following sets.

a) A = {x:x is a prime number, x ≤ 13} b) B = {x:x ∈ Z, -2 ≤ x ≤ 3}

c) C = {x:x is a natural number, x < 1} d) D = {all the factors of 12}

5.2 Set Operation

Subset and Superset:

Let's consider two sets A = {a,b,c} and B = {a,b,c,d,e}. Here, each element of set A is
also an element of set B and we say set A is sub set of set B. Mathematically, we write

A ⊆ B and the set B is called super set of A. We write it on B ⊇ A

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

of set B and at least one element of B is not an element of A. We write A ⊂ B.

If N is the set of natural numbers, W is the set of whole number; Z is the set of
integers, Q is the set of rational numbers and R is the set of real numbers, then

N ⊂ W ⊂ Z ⊂ Q ⊂ R.

Let A = {a,b,c},using the elements of A the following sub set can be formed

Taking none element { } or ∅
Taking single element
Taking two elements {a}, {b}, {c}
Taking all three elements {a,b},{a,c} , {b,c}
{a,b,c}

The set of all the subsets of A = {∅},{a},{b},{c},{a,b},{a,c},{b,c},{a,b,c}}

Which is called the power set of A.
If a subset is formed by taking some elements from the given set, then the subset is

said to be a proper subset of the given set. It is denoted by ⊂. If P = {a, b, c, d} and Q

= {a, c} then Q is a proper subset of P i.e. Q ⊂ P.

If a subset is formed by taking all the elements from the given subset of given set, it

is denoted by ⊆. If P = {a,b, c, d} and B = {a, b, c, d} then P ⊆ B.

Note: 1. ∅ is a sub set of every set.
2. Number of sub sets of a set having 'n' elements is 2n
3. Number of proper sub sets of a set having 'n' elements is 2n-1
4. A set is not a proper sub set of itself.
5. ∅ is not a proper subset of itself.

Prime Mathematics Book - 7 115

Universal set: A set which contains all the sets under consideration as sub sets is called
the universal set. It is denoted by U or [ ] or ξ (pxi)
For the sets A = {1,2,3} and B = {2,3,4} the universal set might be {1,2,3,4,5,6} or {x:x

∈ N}or {x:x ∈ W} or {x:x ∈ N,x ≤ 5} etc.

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

Venn Diagram

A Venn diagram is merely a closed figure used to denote the set of all points within
the figure.
Venn diagram is used to represent sets and relation between the sets. A universal
set is denoted by a rectangle and its subsets by circles or ovals within it showing the
elements within their respective parts.
If A = {2,3,5,7} and B = {1,3,5,7,9}, then the sets can be represented in Venn diagram as

AB

If the universal set is considered to the set of digits of Hindu Arabic numerals then
U = {0,1,2,3,4,5,6,7,8,9},and the Venn diagram showing A,B and U is

AB

Venn diagrams showing other relations:

U U U
B
AB A AB

Set operations: In number system, four fundamental operations are addition,
subtraction, multiplication and division. In an operation between two numbers they
are replaced by a third number.

116 Sets

For example:

Addition of 1 and 2 ⇒ 1 and 2 is replaced by 3.

Multiplication of 5 and 2 ⇒ 5 and 2 is replaced by 10.

Similarly there are operations that replace two sets by a third set. These operations
are
1. Union of sets
2. Intersection of sets
3. Difference of sets
4. Complement of sets

Union of sets: Union of two sets A and B is the set of all the elements that are in A or
in B or in both the sets A and B. Union of two sets A and B is denoted by A ∪ B (read as
"A union B" or "A cup B")

Showing the union of two sets A and B in Venn diagram:

UU U
B
AB AB
A

Example 1: If A = {1,2,3,4} and B = {1,2,5} and C = {1, 2, 3} then which set is the
subset of A.
Solution:
Here : All elements of C are in the set A, It has at least 1 element less then A. So
that C is a proper subset of Set A.

Example 2: If A = {a,b,c} then write all subset of A.
Solution:
Here : A = {a, b, c}

Total number of elements (n) = 3
Total number of subset = 2n = 23 = 8
So, all possible subset of A are A1 = {a, b, c} A2 = {∅} A3 = {a } A4 = {b}
A5 = {c} A6 = {a, b} A7 = {a, c} A8 = {b, c}

Prime Mathematics Book - 7 117

Example 3: If A = {2,3,5,7} and B = {1,3,5,7,9}, list the elements of A ∪ B and show
the operation in a Venn diagram.
Solution:
Here A = {2,3,5,7}, AB

B = {1,3,5,7,9}

\ A∪B = {2,3,5,7}∪{1,3,5,7,9}

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

The operation is shown in Venn diagram where the shaded region represents A∪ B

Example 4: A = {0,2,4,6,8} and B = {1,3,5,7,9}. List the elements of the set A ∪ B and

Solution: represent in a Venn diagram. U
Here A = {0,2,4,6,8} AB
B = {1,3,5,7,9}

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

The operation is shown in Venn diagram as

Example 5: If A = {a,b,c,} and B = {a,b,c,d,e}, list the elements of the set A ∪ B and

Solution: represent it in a Venn diagram. U
B
Here A = {a,b,c}, B = {a,b,c,d,e}
A

\ A ∪ B = { a,b,c } ∪ { a,b,c,d,e }
= {a,b,c,d,e}

The operation is shown in Venn diagram as:

(Two sets being complete overlapping i.e. A being proper subset of B)

Intersection of two sets: Intersection of two sets A and B is the set of all the elements
which are in both A and B. Intersection of two sets A and B is denoted by A ∩B. (read
as "A intersection B" or "A cap B")
Intersection of two sets A and B can be shown in Venn diagram as:
U
UU

AB AB A
B

118 Sets

Example 6: If A = {a,e,i,o,u} and B = {a,b,c,d,e},list the elements of the set A ∩ B and
represent it in a Venn diagram.
Solution:
Here A = {a,e,i,o,u}, B = {a,b,c,d,e}

\ A ∩ B = {a,e,i,o,u } ∩ {a,b,c,d,e }
= {a,e}

The operation A ∩ B is shown in the Venn diagram as:

U

AB

Example 7: If A = {x:x is a factor of 6} and B = {x:x is a factor of 18}, write A ∩ B in set
builder form and show it in a Venn diagram.
Solution:
Here A = {x:x is a factor of 6} = {1,2,3,6}
B = {x:x is a factor of 18} = {1,2,3,6,9,18}

\ A ∩ B = {1,2,3,6} ∩ {1,2,3,6,9,18}
= {1,2,3,6}
= { x:x is a factor of 6}

The shaded region in the Venn diagram shows A ∩ B

U
B

A

⊆
Difference of two sets: Difference of two sets A and B is the set of all the elements
that are in A but not in B. It is denoted by A-B. Similarly the difference of two sets B
and A is the set of all the elements that are in B but not in A and it is denoted by B-A.

UU U
AB
AB AB

Prime Mathematics Book - 7 119

Example 8: If A = {a,e,i,o,u} and B = {a,b,c,d,e}, list the elements of the set A-B
U
and represent it in a Venn diagram.
AB
Solution:

Here A = {a,e,i,o,u} and B = {a,b,c,d,e}

\ A- B = { a,e,i,o,u } - { a,b,c,d,e }
= {i,o,u}

A-B is showing in the adjoining Venn Diagram. A-B

Example 9: If A = {1,2,3,4,5} and B = {6,7,8,9},list the elements of the set B-A and show

it in a Venn diagram. U
AB
Solution:

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

\ B-A = {6,7,8,9} - {1,2,3,4,5}
= {6,7,8,9}

This operation is shown in Venn diagram as:

(Since A and B are disjoint)

Example 10 : A = {0,1,2,3,4,5,6,7,8,9} and B = {2,3,5,7}, find the set A-B and represent

it by a venn diagram. U
A
Solution:
B
Here A = {0,1,2,3,4,5,6,7,8,9}
\ B = {2,3,5,7}
A-B = {0,1,2,3,4,5,6,7,8,9} - {2,3,5,7}

= {0,1,4,6,8,9}
It is illustrated in a Venn diagram as
Complement of a set: If A is a subset of a universal set U, complement of the set A
denoted by Ac or A´ or A is the set of all the elements which are in U but not in A.
Complements of various sets are shown below
UU U

A AB AB

A BU U
AB

120 Sets

Complement of a set A is the difference of the universal set and the set A.

i.e. A = U-A.

Complement of the complement of a set is the set itself. i.e. A = A

It can be shown as A = U-A = U-(U-A) = U-U+A = A

Example 11: If U = {0,1,2,3,4,5,6,7,8,9} and A = {0,2,4,6,8},list the elements of

Solution: set A and illustrate it in a venn diagram. U
Here U = {0,1,2,3,4,5,6,7,8,9} and A = {0,2,4,6,8} A

\ A = U-A

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

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

It is illustrated in Venn diagram as.

Example 12: If U = {1,2,3,…………,10}, A = {2,4,6,8}and B = {1,2,3,4}, list the elements

of following and show them in Venn diagram.

(a) A∪B (b) A∩B (c) A-B U
AB
Solution:

Here, U = {1,2,3,…………,10},

A = {2,4,6,8} and

B = {1,2,3,4}

(a) \ A∪B = {2,4,6,8} ∪ {1,2,3,4}
= {1,2,3,4,6,8}

Now A∪B = U - (A∪B)

= {1,2,3,4,5,6,7,8,9,10} - {1,2,3,4,6,8}
= {5,7,9,10}
It can be represented in Venn diagram as.

(b) \ A∩B = {2,4,6,8} ∩ {1,2,3,4} U
AB
= {2,4}

Now A∩B = U - (A∩B)

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

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

It can be illustrated in venn diagram as

(c) \ A-B = {2,4,6,8}-{1,2,3,4} = {6,8}

Now A-B = U-(A-B)

Prime Mathematics Book - 7 121

= {1,2,3,4,5,6,7,8,9,10} - {6,8} U
= {1,2,3,4,5,7,9,10} AB
It can be illustrated in Venn diagram as.

Example 11: If U = {1,2,3,4,5,6,7} and A = {2,4,6}, show that
A = A.

Solution:
Here U = {1,2,3,4,5,6,7} and A = {2,4,6}

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

Now A = U - A d) Power set e) Universal set.
= {1,2,3,4,5,6,7} - {1,3,5,7}
= {2,4,6} = A

\ A = A

proved.

Exercise 5.2
1. Define the following:

a) Sub-set b) Proper sub-set c) Super-set

2. Define the following sets.

a) A ∪ B b) A ∩ B c) A - B d) A’

3. write the all possible subset of following sets .

a) A = {a} b) B = {a,b} c) B = {1,2}

d) C = {a,b,c} e) A = {1,2,3}

4. a) If A ={a,b,c,d,e}, P = {a,b,e}, Q={c,d,f} then which set is the sub-set of set A

b. If P={1,2,3,4}, A={4,2,3}, Q= {3,6,9} then which set is the sub-set of set P.

c. If A ={a,e,i,o,u} , B={a,e,i} and C= {a,b,c}which set is proper sub-set of set A.

d. If A={1,2,3,4},B= {3,4}, C={3,4,2,1} which set is proper sub-set of set A.

e. If A= {p,r,i,n,h} , B={p,r,a,n,i,s,h}, C={p,r,i,n,h} then which set is an improper
sub-set of set A.

122 Sets

5. a) If A ={1,2,3} then write down the two proper sub sets of set A.
b) If A= {q,b,c,d} then write down a improper sub-set of set A.

6. a) Which is the proper sub-set of set A in the given A cU
venn-diagram? B

b. which is the proper sub set of set B in the A U
B
c
given venn diagram?

7. Find the union of the following sets and also illustrate the operation in Venn
diagram.
a) A = {3,2,4,6,8,} and B = {1,3,5,7,9}
b) P = {1,2,3,4,5,} and Q = {2,4,6,8,10}
c) X = {a,b,c,d} and Y = {w,x,y.z}
d) M = {x:x is multiple of 2,x ∈ N ≤12} and N = {y:y is multiple of 3, x ∈ N ≤15}

8. Find the intersection of the following sets and illustrate them in Venn diagrams.

a) A = {a,b,c,d,e,} and B = {a,e,i,o,u}

b) P = {x:x ∈ N, 2 < x < 6} and Q = {y:y ∈ N, y < 10}

c) X = {factors of 8} and Y = {factors of 12}

d) M = {x:x is a multiple of 3, x ≤18} and N = {x:x is an odd integer , x ≤ 21}

9. If A and B are subsets of the set U and U = {1,2,3,4,5}, A = {1,2,3} and B = {2,3,5},
find the following sets and illustrate them in Venn diagram.

a) A-B b) B-A c) A d) B

e) A-B f) A∩B g) A∪B h) B-A

10. From the adjoining diagram, list the elements of the following sets. U

a) A∪B b) A∩B c) A-B d) B-A A B

e)A f) A-B g) A∩B h)A∪B

11. Write the set represented by the shaded part of the following in set notation.

a) U b) U
B B
A A

Prime Mathematics Book - 7 123

c) U d) U

B AB

A

e) U f) U
B
A AB

g) BU h) A B U

A

Unit Revision Test

1. Define
(a) A set (b) The empty set.

2. Represent the given sets as stated in the brackets.

(a) A= { x : x = 3x +1, x ∈N} [Tabular form]

(b) B ={ 3, 7, 11, 15, 19, 23, 27} [Set builder form]

124 Sets

3. List the elements and write the cardinal number of the sets.
(a) A = { x : x is a prime number, x ≤ 13}

(b) B = { x : x is a natural number, x < 1}

4. Define. (b) Universal set.
(a) Proper subset

5. Define (b) A.
(a) A ∪ B

6. List the intersection of the sets A = {factors of 8}, B= {factors of 12} and illustrate
in a venn diagram.

7. Write the set represented by the shaded part of the following sets in set notation.
a) b)

8. Find the following sets and illustrate in Venn diagram.

(a) A ∩ B (b) A-B

Where A = {1, 2, 3, 4, 5} and B = {2, 4, 6, 8, 10}

9. Find the proper sub-set of the set A = {1,3,5}

10. a. If A = {1,2,3,4}, B = 1,3,4}, C = {1,3,5} which set is proper sub-set of Set A.

b. If P = {a,b,c}, Q = {a}, R = {c,b,a} then which set is an imprper sub set of set P.

Prime Mathematics Book - 7 125

Answers

1. Show to your teacher. Unit:5 Sets
2. Show to your teacher.
Exercise: 5.1

3. a) Singleton b) Singleton c) Singleton d) Null

4. a) A=B b) P~Q c) C=D d) R~S

5. a) overlapping b) overlapping c) overlapping d) disjoint

6. a) n(A) = 5 b) n(B) = 6 c) n(C) = 0 d) n(D) = 6

Exercise: 5.2

1. Show to your teacher.

2. Show to your teacher.

3. a) {a}, ø b) {a,b}, {a}, {b} and ø c) {1,2}, {1}, {2}, ø

d) {a,b,c}, {a,b}, {b,c}, {a,c}, {a}, {b}, {c}, ø

e) {1,2,3}, {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, ø

4. a) P b) A c) B d) B e) C

5. Show to your teacher.

6. Show to your teacher.

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

c) {a,b,c,d,w,x,y,z} d) {2,3,4,6,8,9,10,12,15}

8. a) {a,e} b) {3,4,5} c) {1,2,4} d) {3,9,15}

9. a) {1} b) {5} c) {4,5} d) {1,4} e) {2,3,4,5}

f) {1,4,5} g) {4} h) {1,2,3,4}

10. a) {1,3,5,6,7,9,11,12} b) {3,9} c) {1,5,7,11}

d) {6,12} e) {0,2,4,6,8,10,12}

f) {0,2,3,4,6,8,10,12} g) {0,1,2,4,5,6,7,8,10,11,12}

h) {0,2,4,8,10}

11. a) A∩B b) AUB c) B-A d) AUB e) A-B
f) A∩B g) B-A h) B

126 Sets

6UNIT 6 WHOWLEHNOULMEBNERUSMBERS
Estimated periods 15 Objectives:

At the end of this unit students will be able to

● find the square of numbers and square roots of perfect squares by
factorization method.

● find the square roots of perfect as well as other numbers by using division
algorithm.

● cubes of numbers and cube roots of perfect cube numbers by prime
factorization method.

● find H.C.F and L.C.M of numbers by prime factors and division method.
● relate two numbers and their H.C.F and L.C.M.
● introduce number system, binary and quinary number system.

1=! 7 =Q 1=I 5 =V
2=L 8 =( 10 = X 50 = L
3=# 9 =) 100 = C 500 = D
4=$ 10 = * 1000 = M
5=% 102 = =
6=& 103 = ?

Teaching Materials:

Tables of squares and square roots, tables of cube and cube roots, graph of
square and cubic functions. List of prime numbers, List of ancient numerals,
conversion table of numerals.

Activities:

It is better to

● show and let the students to find squares and square roots, cube and
cube roots using tables, graph.

● discuss processes of finding square roots and cube roots.
● discuss about process of factorization and use to find H.C.F. and L.C.M.

of numbers.
● discuss about old number system like Egyptian, Greek, Roman, Chinese,

Hindu number systems.
● discuss about decimal binary and quinary number system.

6.1 Squares and square roots:

We know 1×1 = 12 = 1

2×2 = 22 = 4

3×3 = 32 = 9 and so on.

Here, the products 1,4,9 etc. are perfect squares and the number 1 is the square root
of 1,2 is the square root of 4 and 3 is the square root of 9. A perfect square number is
the product of two identical numbers and one of the identical numbers is the square
root of the square number.

Thus, square of 5 is 25 i.e. 52 = 25.

and square root of 25 is 5 i.e 25 = 5

The radical sign (root sign) is used to denote the square root of a number.

Methods to find the square root of numbers:

i. Factorization method

For example, to find the square root of 225.

Solution: 3 225
225 = 3×3×5×5 3 75
5 25
=32×52
5
\ 225 = (32 × 52 )

= 3×5 = 15

Process
Find the prime factors of the number
Make the pairs of identical factors
For the square root, take the product of one of the factors taken from each pairs

Note:
1. Method of factorization is useful for finding the square roots of perfect squares only.
2. In case of very big numbers, the process is not so convenient.

ii. Division method (division algorithm)

For example, to find the square root of 69169
2 69169 263

+2 4-
46 291

+ 6 2- 76
523 1569
+ 3 - 1569
526 0

128 Whole Number

Process

From ones place, group the number in pairs. Like, pair 69, pair 91, pair 06.(you can
put bars over them)
The last pair is 06 i.e. 6. Think the largest square less than or equal to 6. It is 4.
Think the square root of 4. It is 2.
Write 2 to the left as divisor and quotient to the right.
Multiply 2 x 2 = 4 and put 4 just below the last pair 6 and subtract. The
remainder is 2. Bring down (drop) next pair 91 thus new dividend is 291.
In the divisor column add 2 + 2 = 4
To determine the next digit in the new dividend 291, 29 is 7 times 4.So trial with
divisor 47 but 47x7 = 329 exceed 291 so trial with 6 i.e. trial with divisor 46. Since
46 x 6 = 276 which doesn’t exceed 291. Hence the next digit is 6.
Put 6 next to 4 in the divisor side and next to 2 in quotient side. And since
46 × 6 = 276, put it just below 291 and subtract. The remainder is 15 and bring
down next pair 69 forming new dividend 1569.
Now add 6 to new divisor 46. The new trial divisor is 52.
In the dividend 1569,156 is 3 times the new trial divisor 52 i.e. 52 × 3 = 156.
Put 3 next to 52 thus new divisor is 523. And we get 523 × 3 = 1569 which equal
to the new dividend. Now put 3 next to 6 in quotient side. Put 3 next to 52 in
divisor side and 1569 just below 1569(divisor) remainder is 0.
Add 3 with divisor 523 we get 523 + 3 = 526, Which should be double of the
quotient 263

Thus 69169 = 263

Note: We can find the square root of any number by this method.

iii. Guess, divide and average method

For example, to find the square root of 24.
Since, 16 < 24 <25

4 < 24 < 5

\ Square root of 24 must lie between 4 and 5

\ Our first guess is (4+5) = 4.5
2

Now divide 24 by 4.5 we get 24 = 5.3
4.5

\ Square root of 24 lie between 4.5 and 5.3
(5.3+4.5) 9.8
\ Our second guess is 2 = 2 = 4.9

\ 24 = 4.9

Prime Mathematics Book - 7 129

Alternately: To find the value of 24 . We can also start with an arbitrary guess.
Since 4 < 24 <5
Let our first guess = 4
24
Dividing 24 by 4 = 4 =6

\ Our new guess = 4+6 =5
4
24
Dividing 24 by 5 we get = 5 = 4.8

\ Our new guess = 5+4.8 = 4.9
4
\ 24 = 4.9

Note:
i) If our guess is average of square roots of the nearest small and bigger square,

it is not necessary to take average more than twice or thrice.
ii) With arbitrary guess, it may take more steps.
iii) Guess, divide and average method is not necessary for this class, you can learn

for your extra knowledge.

Example 1: Find the square of 34. = 34×34 = 1156
Solution:

we know the square of 34 = 342

\ Square of 34 = 1156

Example 2: Find the square root of 1764 by using factorization method.
Solution:
Here, 2 1764

\ 1764 = 2×2×3×3×7×7 2 882

= 22×32×72 3 441
3 147

\ 1764 = (22 × 32 ×72) = 2×3×7 = 42 7 49
7

Example 3: Find the square root of 2916 by division algorithm.
Solution: 5 2916 54

+5 -25
104 416

+4 - 416
108 0

\ 2916 = 54.

130 Whole Number

Example 4: Simplify: 12 + 147 2 12 3 147
Solution: 26 7 49
Here,
3 7

12 + 147 = 2 3 +7 3 =9 3
= (2×2×3) + (3×7×7) = (22×3) + (3×72)

Example 5. Find the square root of 1.96
Solution:
196
Here, 1.96 = 100 Alternately
using division algorithm
\ 1.96
196 1 1.96 1.4
= 100 +1 -1
24 96
= (22×72) = 2×7 = 14 = 1.4 +4 - 96
(22×52) 2×5 10 28 0
(Pair from the position of
\ Square root of 1.96 is 1.4 decimals to left & right)

Example 6: Find the square root of 3 correct to 2 decimal places.
Solution:
Here, using division algorithm
1 3.000000 1.732
+1 1-

27 200
+7 -189
343 1100
+3 1- 029
3462 7100
+2 6- 924
3464 017600

\ 3 = 1.732 = 1.73 up to 2 decimal places.

Alternately: (if you are interested)
Since 1<3 <4
1 < 3< 4
1< 3 <2

\ Square root of 3 lies between 1 and 2
1+2
\ Our first estimate = 2 = 1.5

Prime Mathematics Book - 7 131

Dividing 3 by 1.5 we get 3 =2
1.5
And since 1.5 × 2 = 3
∴ Square root of 3 lies between 1.5 and 2
1.5+2 3.5
∴ Our second estimate = 2 =2 = 1.75

Dividing 3 by 1.75 we get 3 = 1.71
1.75

Again Square root of 3 lies between 1.75 and 1.71

∴ Our third estimate = 1.71+1.75 = 3.46 = 1.73
2 2

∴ 3 = 1.73 (correct to 2 decimal places.)

Example 7: Find the least number by which when 252 is multiplied, it becomes a

perfect square. = 2×2×3×3×7 2 252
Solution: = 22×32×7 2 126
3 63
Here, 252 3 21

7

Here, 7 is not in pair so, the given number must be multiplied by 7 to make it a perfect
square.

Example 8: Find the least number by which when 121275 is divided, it becomes a

perfect square. 3 121275

Solution: 3 40425

Here, 121275 = 3×3×5×5×7×7×11 5 13475
= 32×52×72×11 5 2695
7 539
Where 11 is not in pair. If 11 is removed, the number formed will be 7 77
perfect square. So the number should be divided by 11 to make it a
perfect square. 11

Example 9: Find the least number which should be subtracted from 200 to make it a

perfect square. 1 200 14
Solution: +1 -1
Finding square root of 200 by division method. 24 100
+4 - 96
In this process, the last remainder is 4. If there is Zero (0) instead of 28 4
4, the number would be a perfect square. So 4 should be subtracted
from 200 to make it a perfect square.

132 Whole Number

Example 10: Find the least number which should be added to 7102 to make it a perfect square.

Solution:

Finding the square root of 7102 using division method. 8 7102 84
+8 -64
The last remainder is 46 which shows that 7102 is not a 164 702
perfect square. The whole number part of its square root +4 - 656
is 84. So, the square number just greater than 7102 is (85)2 168 46
= 7225.

∴ The required least number to be added is 7225 - 7102 = 123.

Example 11: What is the length of a square plot of area 2809 m2 ?
Solution:

We know that the length of a square is the square root of its area.
Here, finding 2809

5 2809 53
+5 -25
103 309
+3 -× 3 0 9 ∴∴
2809 = 53 the square plot is 53m.
The length of

Example 12: Find the least number which is a perfect square exactly divisible by 10,

12 and 15.

Solution:

The least number exactly divisible by 10, 12 and 15 is their L.C.M.

L.C.M of 10, 12 and 15

= 2×3×5×2 2 10, 12, 15

= 60 3 5, 6, 15
Here, 60 = 22 × 3 × 5
5 5, 2, 5

Where 3 and 5 are not in pair. 1, 2, 1
∴ The least number by which 60 is to be multiplied to make it a perfect square is
3x5 = 15.

∴ The required perfect square exactly divisible by 10, 12 and 15 is 60×15 = 900

(∵ L.C.M. is exactly divisible by the numbers and multiple of L.C.M are also
divisible by the numbers.)

Prime Mathematics Book - 7 133

Example 13: Dristant made a chess board pattern but
it contained 7 rows and 7 columns only. How many unit
squares should be added to make it a perfect chess board
with 8 rows and 8 columns.
Solution:
No. of unit square in the given square pattern = 72 = 49

No. of unit in a chess board = 82 = 64

\ No. of unit square to be added = 64 – 49 = 15

1. Find the square of : Exercise 6.1
c) 108
a) 23 b) 81 d) 153

2. Find the square root of the following perfect squares using factorization method.

a) 16 b) 144 c) 196 d) 576

e) 7056 f) 63504 g) 11664 h) 11025

3. Find the square roots of the following by using division algorithm

a) 169 b) 225 c) 1369 d) 2809

e) 6889 f) 12769 g) 23409 h) 54756

4. Simplify: b) 396 c) 3 5 + 180 d) 108 + 2 3
a) 45 f) 49 × 144 g) 72 × 18
e) 2205 - 605

h) 18 × 48 × 50 i) 27 × 21 × 63 j) 10 × 6 × 5

5. Simplify:

a) 4 × 25 b) 162 c) 153 ÷ 3 17
5 × 20 480

d) 875 e) 22×32×72 42×92×72
1125 16 f) 62

6. Find the square roots of :

a ) 0.36 b)1.44 c) 2.25 d) 0.2304

134 Whole Number

7. Find the square roots correct to two decimal places of the following.
a) 3 b) 24 c) 72 d) 125

8. Find the length of the squares having the following area.
a) 156 cm2 b) 1296 cm2 c) 4096 m2 d) 12544 m2

9. By what least number should each of the following numbers be multiplied to make
them perfect squares?
a) 72 b) 90 c) 432 d) 4375

10. Find the least number by which each of the following number should be divided to
make them perfect square.
a) 675 b) 768 c) 968 d) 1620

11. a) Find the least number which is a perfect square exactly divisible by 16, 24 and 32.
b) Find the least number to be subtracted from 1234 to make it a perfect square.
c) Find the least number to be added to 18325 to make it a perfect square.

6.2 Cubes and Cube roots.

Historical Fact:
Diophantus (about 250 A.D.) used ∆γ for unknown square (taken from first two

letters of Greek word dunamis (∆gNAMI∑).

ky for unknown cubes (taken from first two letters of Greek word Kubos (KgBO∑).

Johann Widman (born 1460, Garmany) used x ce for unknown square i.e. x2 (ce
from censo). x cu for unknown cube i.e. x3 (cu from cuba)

Thomas harriot (1560-1621) used today’s notation for power.

x.x = x2
x.x.x = x3

We Know
3×3×3 = 33 = 27
A cube number is the product of three identical (same) numbers.
Thus 1, 8, 27, 64 etc. are cube numbers.
Here, 27 is called the cubes of 3 and 3 is called cube root of 27.
Symbolically, we write

33 = 27 and 3 27 = 3

Prime Mathematics Book - 7 135

To find the cube root of perfect cube numbers.
e.g. to find the cube root of 216.
216 = 2×2×2×3×3×3

= 23×33

∴ Cube root of 216 i.e. 3 216 = 3 23 × 33 = 2×3 = 6

Find the factors
Group, three identical number
To find the cube root, remove power 3 and take the product of single factors.

Example 1: Find the cube of a) 8 b) 12
Solution : b) 3375
a) 83 = 8×8×8 = 512
b) 123 = 12×12×12 = 1728

Example 2: Find the cube root of a) 64
Solution:
(a) Here, 64 = 2×2×2×2×2×2 = 23 × 23

∴ 3 64 = 3 23 × 23 = 2 × 2 = 4

(b) 3375 = 3×3×3×5×5×5 = 33 × 53

∴ 3 3375 = 3 33 × 35 = 3×5 = 15

Example 3: Find the value of:

a) 3 125 b) 3 0.064
343
Solution:

a)3 125 = 3 5×5×5 =3 53 = 5
343 7×7×7 73 7

b) 3 0.064 =3 0.064 = 3 64 = 3 2×2×2×2×2×2 = 3 23×23 = 2×2
1.000 1000 2×2×2×5×5×5 23×53 2×5

= 2
5
Example 4: Simplify: a) 3 54 + 3 16 b) 3 (3×22) × 3 (32×2)

Solution:

(a) 3 54 + 3 16 = 3 3 × 3 × 3 × 2 + 3 2 × 2 × 2 × 2

= 3 33 × 2 + 3 23 × 2
= 3 3 2 +2 3 2 = 5 3 2

136 Whole Number

(b) 3 3 × 22 × 3 32 × 2 = 3 3 × 22 × 32 × 2
= 3 33 × 23 = 3×2 = 6

Example 5: Find the volume of a solid cube whose one side is 15 cm.
Solution:
Here, length of a side of a cube (L) = 15 cm
We know,
Volume of a cube (V) = L3 = (15 cm)3

= 15cm × 15cm ×15cm
= 3375cm3

\ The volume of the solid cube is 3375cm3.

Example 6: The volume of a cube is 216000cm3. Find the length of its side.

Solution:

Here, volume of the cube (V) = 216000cm3

= 216 × 1000cm3

= 63 ×103cm3

\ Side of the cube (L) =3v

= 3 63×103 cm3

= 6×10cm

= 60cm

\ The length of the side of the cube is 60cm.

Example 7: Check whether 1944 is perfect cube or not. If not by what least number
must it be multiplied to make it a perfect cube?

Solution:
Here, 1944 = 2×2×2×3×3×3×3×3 = 23 × 33 × 32
Where 23 and 33 both are cube but 32 is not a cube. So 1944 is not a perfect cube.
If 23 × 33 × 32 is multiplied by 3, it will be a perfect cube. So the least number required
to make 1944 a perfect cube is 3.

Exercise 6.2

1. Find the cube of the following numbers.

a) 4 b) 12 c) 26 d) 50

2. Find the cube root of the following numbers.

a) 27 b) 125 c) 216 d) 343 e) 512

Prime Mathematics Book - 7 137

f) 729 g) 2197 h) 1331 i) 1728 j) 4096

3. Find the cube roots of.

a) 8 b) 27 c)76249 d) 343
27 512 1000

e) 0.008 f) 0.125 b) 2 3 54 + 3 375 - 3 192 + 3 432
4. Simplify: d) 3 10000 - 3 7290
f) 3 4 × 3 × 3 9 × 2
a) 3 24 + 2 3 81
c) 5 3 320 - 3 1715 + 3 1024
e) 3 27×125
g) 3 2662 ÷ 3 3456

5. a) A solid cubical block is 24cm long. Find its volume.
b) A cubical water tank is 1.2m long. Find the capacity of the tank in liters.
c) If the volume of a cube is 729cm3, find its length.
d) The capacity of a cubical cistern is 216000m3. Find its length.

6. Test whether the following are perfect cube or not. If not, find the smallest

number with which it should be multiplied to make it a perfect cube.

a) 500 b) 343 c) 108 d) 392

6.3 Highest Common Factor (H.C.F.)

Dolma wants to divide 36 pencils and 48 chocolates among some students. Among
maximum of how many students can she divide the things equally?
Let’s observe in this way.

No of students No of pencils in each share No of chocolate in each share
1 36 48
2 18 24
3 12 16
4 9 12
6 6 8
12 3 4

138 Whole Number

We see that the maximum number of students is 12 and each will get 3 pencils and 4 chocolates.
It is the case of common factors.
Factor of 36 are 1,2,3,4,6,9,12,18,36
Factor of 48 are 1,2,3,4,6,8,12,16,24,48
The common factors are 1,2,3,4,6,12.

All these factors divide the numbers 36 and 48 exactly and among these 12 is the
greatest which is known as the highest common factor (H.C.F.) or the greatest common
divisor(G.C.D). H.C.F. or G.C.D. of the given numbers is the greatest number that
divides the number exactly.

Finding H.C.F of numbers

1. Prime Factorization Method.
In ths method we find the prime factors of the numbers and the product of the prime
factors common to all the numbers in the H.C.F.
e.g to find the H.C.F. of 24, 36 and 72

24 = 2 × 2× 2 × 3
36 = 2 × 2 × 3 × 3
48 = 2 × 2 × 2 × 2 × 3

\ H.C.F. = product of the common prime factors

= 2 × 2 × 3 = 12
Note: If the numbers have no common factors, H.C.F. is 1. In such case the numbers
are said to be the numbers prime to each other.

2. Common Divisor Method.
In this method, we divide each number by divisors common to all. The product of the
common divisors is the H.C.F.
e.g to find the H.C.F. of 24, 36 and 48

2 24, 36, 48 divide each by 2
2 12, 18, 24 divide each by 2
3 6, 9, 12 divide each by 3
3, 4 no common from all, so stop
2,

\ H.C.F. = product of the common divisors.

= 2 × 2 × 3 =12

3. Euclidean Algorithm.
In this method, we divide the larger number by smaller. If remainder is zero, H.C.F. is
the divisor (first divisor). If the remainder is not zero, then we divide the first divisor
by the remainder (first remainder). If the remainder is zero the divisor of second step
tsr1hteeispl(atrhs2)tebdHyiv.tCihs.oeF.rrgaeinmvdianiignf drteehmrerar3i.enmWdeaerinczodenertroin(isou)enioustnttzhieelrwoH,e.Cwa.rFer.ivdeivaitdtehtehreemreaminadinedrezreroof(ose).cAonndd

Prime Mathematics Book - 7 139

e.g. to find H.C.F. of 64 and 112
64)112 (1

-64
48)64 (1

- 48
16 )48(3
-48
0

The last divisor in the algorithm is 16

\ H.C.F = 16

To find the H.C.F. of more than two numbers, we find the H.C.F. of the first two by
using the Euclidean algorithm and repeat the process with the H.C.F. of the first two
and next number.
The final H.C.F. is the actual H.C.F.
e.g. to find the H.C.F. of 120, 420 and 510.
Firstly to find the H.C.F. of 120 and 420
120)420(3

-360
60)120 (2
-120
0

\H.C.F of 120 and 420 is 60.

Now, finding H.C.F of 60 and 510
60)510(8
-480
30)60 (2
-60
0

\H.C.F. of 60 and 510 is 30
\ H.C.F. of 120, 420 and 510 is 30.

Example 1. Find the H.C.F. of 12 and 18 by finding set of possible factors.
Solution.
Here, factors of 12 are 1, 2, 3, 4, 6, 12

factors of 18 are 1, 2, 3, 6, 9, 18

where the greatest among the factors is 6.

\ H.C.F = 6

140 Whole Number

Example 2. Find the H.C.F. of 42, 56, 70 by factorization method.
Solution:
42 = 2 × 3 × 7
56 = 2 × 2 × 2 × 7
70 = 2 × 5 × 7

\ H.C.F = product of the common factors

=2×7
= 14

Example 3. Find the H.C.F. of 112 and 128 by using Euclidean algorithm.
Solution:
Here, given numbers are 112 and 128.

Now using Euclidean algorithm.
112)128(1
-112
16)112(7
-112
0

\ H.C.F = 16.

Example 4. Find the H.C.F. of 54, 90, 126

Solution:

Firslty finding H.C.F of 54 and 90
54) 90 ( 1
-54
36)54 (1
-36
18) 36 (2
-36
0

Again finding H.C.F of 18 and 90

18) 90 (5
-90
0

\H.C.F. = 18

Example 5. Find the greatest number that divides 168, 224 and 308 without leaving

any remainder. 2 168 2 224 2 308
Solution: 2 84 2 112 2 154
The number is the H.C.F. of 168, 224 and 308
Now, to find their H.C.F. 2 42 2 56 7 77
3 21 2 28 11
168 = 2 × 2 × 2 × 3 × 7 2 14
224 = 2 × 2 × 2 × 2 × 2 × 7 7
7

Prime Mathematics Book - 7 141

308 = 2 × 2 × 7 × 11

\ H.C.F = 2 × 2 × 7 = 28

\ The greatest number that divides 168, 224 and 308 exactly is 28.

Example 6: Find the greatest number that divides 131, 215 and 740 leaving remainder

5 in each case.

Solution:
Here,

131 - 5 = 126
215 - 5 = 210
740 - 5 = 735
H.C.F. of 126, 210, 735 should be the required number.

Now, 126 = 2 × 3 × 3 × 7 2 126 2 210 3 735
210 = 2 × 3 × 5 × 7 3 63 3 105 5 245
735 = 3 × 5 × 7 × 7
3 21 5 35 7 49
\ H.C.F. = 3 × 7 7 7 7

= 21

Therefore, the required number is 21

Exercise 6.3

1. Find the H.C.F. of the following, finding all the possible factors.

(a) 12, 18 (b) 60, 84

(c) 32, 40, 56 (d) 28, 42, 70
2. Find the H.C.F. of the following by factorization method.

(a) 16, 24 (b) 18, 27

(c) 18, 24, 30 (d) 24, 48, 56

(e) 30, 45, 60 (f) 24, 36, 60
3. Find the H.C.F. of the following numbers using division method (Euclidean algorithm).

(a) 24, 40 (b) 42, 70

(c) 36, 60 (d) 18, 45

(e) 60, 75, 90 (f) 60, 100, 180

(g) 125, 175, 225 (h) 156, 180, 204
4 (a) Find the greatest number that divides 36, 54 and 90 without leaving any
reminder.

(b) There are three containers containing 50 ltr, 70 ltr and 90 ltr of kerosene.
Find greatest capacity of bucket which can empty each container with exact
number of feelings.

142 Whole Number

(c) Find the greatest number of students to whom 72 oranges, 108 bananas and
180 litchis can be equally distributed. Also find the number of each item each
student will get.

(d) Find the greatest number that divides 101, 137 and 197 leaving remainder 5 in
each case.

(e) Find the greatest number that divides 85, 72 and 101 leaving remainders
respectively 1, 2 and 3.

6.4 Lowest Common Multiple (L.C.M.)

Consider two numbers 9 and 12.
Multiples of 9 are 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99, 108
Multiples of 12 are 12, 24, 36, 48, 60, 72, 84, 96, 108 ........
The common multiples are 36, 72, 108, ... which are exactly divisible by the
numbers 9 and 12.
Where the smallest common multiple is 36 which is the lowest common multiple
(L.C.M.) of 9 and 12. Thus L.C.M. is the smallest number which is divided exactly by
the given numbers.

Finding the L.C.M. of the numbers.

1. Finding the multiples: In this method we find some multiples of each,
number and find the smallest common multiple which is the L.C.M.
e.g. to find the L.C.M. of 6 and 8.
Multiples of 6 are 6, 12, 18, 24, 30, 36, 42, 48, ........
Multiples of 8 are 8, 16, 24, 32, 40, 48, ..........

\ L.C.M. = 24

2. Factorization method: In this method we find the prime factors of each numbers.
Then the product of the common factors as well as non common prime factors is the
L.C.M. of the given numbers.
e.g. To find the L.C.M. of 20, 24 and 30.
20 = 2 ×2 ×5
24 = 2 ×2×2×3
30 = 2×3×5

\ L.C.M. = 2×2×2×3×5 = 120
\L.C.M. of 20, 24 and 30 is 120

Prime Mathematics Book - 7 143