Math 7

Properties

(i) Object and image are equal in area.

(ii) The line segments joining the object points and its corresponding image points

are parallel and equal in length. P'(x + a, y + b)

Translation using co–ordinates: Any point P(x, y) P(x, y) b
moves h units parallel to x–axis and b units parallel a
to y–axis. Then new co-ordinates of P will be (x + a,
y + b). We mark the point P' on the plane whose co–
ordinates are (x + a, y + b).

( ) a , the point P (x, y) is translated to P'(x + a, y + b)
Thus, if the translation vector T = b

Symbolically, P(x, y) T (ba( P' (x + a, y + b)

Worked Out Examples

Example: 1 a

Translate ∆ABC given in the figure by the given vector.
Here, ∆ABC is displacing by vector as
• From A, draw AA' || a→ such that AA'= a→

• From B, draw BB' || a→ such that BB' = a→

• From C, draw CC' || a→ such that CC' = a→

• Join A', B' and C'

Hence, ∆A'B'C' is the displaced image of ∆ABC. O

Example: 2

Find the translation vector which displaces a point 3 units right and 4 units downward.

Solution:

Since a translation vector displaces a point 3 units right, X-component of the vector is 3.

Again, since it displaces the point 4 units downward, Y-component = –4.

( (Hence the required vector = 3
-4

Example: 3

ABC is a triangle with vertices A(–2, 5), B(–2, 3) and C(2, 3). Translate the ∆ABC by

translation vector T = ( 4 ( . Present the ∆ABC and its image on the graph.
3

Oasis School Mathematics – 7 95

Solution: Y A' C'
We have, P(x, y) T (ba( P' (x + a, y + b) X
A(–2, 5) T (34( A'(–2 + 4, 5 + 3) = A' (2, 8) B'
B(–2, 3) T (34( B'(–2 + 4, 3 + 3) = B' (2, 6) A
and C(2, 3) T (43( C'(2 + 4, 3 + 3) = C' (6, 6) BC

X'

Hence, ∆A'B'C' is the translated image of ∆ABC. Y'

Exercise 7.3

1. Copy the figures given below in your copy and transform them by the translation
vector in the direction and magnitude of it. Shade the image so formed.

(a) a→ (b) A (c) P R
B a→
A O a→

B CS

2. (a) Find the translation vector if it displaces a point 3 units right and 4 units
upward.

(b) Find the translation vector if it displaces a point 5 units left and 3 units
downward.

3. (a) Find the image of the following points (i) A(2, 3), (ii) B(6, –3), (iii) C(3, 0),

( ((iv) D(–4, –6) when they are translated by the vector T = 4
3

( ( 1
(b) Translate the following points by the translation vector: T = -2

(i) P (2, -3) (ii) Q (1, 4) (iii) R (0, 3) (iv) S (-2, 5)

4. (a) A point P(4, 5) is translated to the point P(6, 8). Find the translation vector by
which P is translated to P'.

(b) Find the translation vector which transforms A (2, 1) to A' (3,3).

5. (a) ABC is a triangle with vertices A(3, 7), B(2, 2) and C(6, 1). Translate the ∆ABC
( (by the vector T =2
3 . Present the ∆ABC and its image on the graph.

(b) Vertices of ∆PQR are P (2,1), Q (5, 2) and R (0, -1). Translate ∆PQR by the

( (translation vector-1 . Draw both figures on the graph paper.
2

Answer

Consult your teacher

96 Oasis School Mathematics – 7

Unit Symmetry Tessellation

8 and Designs

8.1 Symmetry

a. Line Symmetry
As we have already studied about line of symmetry, a line divides the given figure

into two parts such that one part is the reflection of another part about that line. This
line is called line of symmetry or axis of symmetry.

Line of symmetry

Line of symmetry

Line of symmetry

b. Rotational symmetry:

Figures which can be made to coincide with themselves by rotation through an an-
gle about a point are called the rotational symmetry.

Look at this figure: D BA C

AC

Rotation through Rotation through O
O 180º about O. O 180º about O.

B DC A B D

Fig. (i) Fig. (ii) Fig. (iii)

If the first figure is rotated through 180° about O, we will get the second figure. If the
second figure is rotated through 180° about O, we will get the third figure which is
the original position of the figure. Such symmetry is called rotational symmetry.

Order of rotational symmetry

Let us look some figures.

(i)

This figure has only one line of symmetry. If we rotate it
about a point we will get the same shape only after a rota-
tion of 360°. Therefore, order of this rotational symmetry
is 1.

Oasis School Mathematics – 7 97

(ii) A Rotation throu gh B Rotation through A

180º about O. 180º about O. O

B O O

A B

If we rotate the given figure through 180°, it can be fitted over the original figure.
Again if it is rotated through another 180° we will get the original position of the
figure.

Therefore order of this rotational symmetry is 2. B

C

(iii)

Rotation through
120º about O.
120º
O 120ºO
AB CA
120º
120º
AC

Rotation through Rotation through
120º about O. 120º about O.

O O
BC AB

If we rotate the given figure through 120°, it can be fitted over the original figure. To
bring in the original position it has to rotate three times.

Therefore order of this rotational symmetry is 3. Similarly, order of rotational
symmetry of a square is 4.

Exercise 8.1

1. Draw the given figures in your copy and draw the possible lines of symmetry in
each of the following figures.

(a) (b) (c)

(d) (e) (f)



98 Oasis School Mathematics – 7

2. Find the order of symmetry of the following figures.

O O
O

3. Indicate the point of rotation and order of rotational symmetry of the following
figures.

Answer
Consult your teacher.

8.2 Tessellation

A tessellation is a method which is used to cover surface with congruent
geometrical shapes in a repeating pattern without leaving any gaps and without
overlapping each other.

Tessellation is done on the surface of floor or wall or carpet to make them more
attractive.

Oasis School Mathematics – 7 99

Types of Tessellations
Depending upon the tiles used in tessellation, there are three types of tessellations.

They are:
(a) Regular tessellation
(b) Semi regular tessellation

(c) Non regular tessellation

a. Regular tessellations

Regular tessellations are those tessellations in which regular polygons of only one
type are used. Some examples of regular tessellations are given below:



b. Semi-regular tessellations
Semi-regular tessellations are those tessellations in which two or more than two-reg-

ular polygons are used. Some examples of semi-regular polygons are given below:



Tessellation having square, Tessellation having square,
triangle and hexagon and hexagon

c. Non-regular tessellations

Non-regular tessellations are those tessellations in which non-regular polygons are
used. Some examples of non-regular polygons are given below:


100 Oasis School Mathematics – 7

Exercise 8.2

1. Complete the following regular tessellations and expand them in full page.

2. Complete the following semi regular tessellations and expand them in full page.

3. Complete the following non–regular tessellations and expand them in full page.

4. Complete the following tessellations and also mention their types.

5. Take the following figures and tessellate them.
(a) Triangle
(b) Regular hexagon



Answer
Consult your teacher.

Oasis School Mathematics – 7 101

8.3 Designs Using Circles and Polygons

Different designs can be made of circles, polygons, arc, etc. They can be used to
decorate walls or floors of rooms or halls.

Look at the following designs, study them and try to draw them.

(a) Design using circle (b) Design using circle and arc


(c) Design using squares (d) Design using circles and triangles

(e) Design using pentagon

Exercise 8.3

1. Draw a circle of radius 4 cm and design as in (a).
2. Construct a square of side 4 cm and make a design as in (c).
3. Draw a regular pentagon of 5 cm and design it as in (e).

Answer 
Consult your teacher.

102 Oasis School Mathematics – 7

Unit Bearing and Scale
9 Drawing

9.1 Bearing

Bearing is the method of finding the distance between the two places in degree.
There are two types of bearing

(a) The compass bearing
(b) The three figure bearing

a. The compass bearing

The compass bearing is the old method of locating the di- NW N
rection. In the given figure, NOS and EOW represent north– NE

south and east west direction respectively. O is the point of W OE
reference for the direction. We use line ON as the base line.
SE
Now, we have simple compass card having 8 directions as SN S
shown in the figure alongside.

Where, N
NE – North East direction

SE – South East direction W E
SW – South West direction

NW – North West direction S

Look at these two examples: N

NP

30º

WO E WO E

20º

S Q
S

The direction of P is N 30° E. The direction of Q is S 20° E.

Note: The bearing is always measured from N or S but not from E and W.

Oasis School Mathematics – 7 103

b. The three-digit bearing N30º 60º
The direction of any point in terms of an angle ex- P

pressed in three digits as measured in clockwise di W O 150º E
rection with reference to the north line is called the Q
three-digit bearing.
Here, the bearing of P from O is 060°. S
The bearing of Q from O is 150°. (1800 - 300)
N N'
B
Note:
∠NAB is the bearing of B from A. A

Reflex ∠N'BA is the bearing of A from B.



Worked Out Examples

Example: 1

From the given figure, write down the compass bearing of P, NE and SE.

Solution: NW N P NE
Bearing of point P is N 15° E. 15º

Bearing of NE is N 45°E. W OE
SE
Bearing of SE is S 45° E.
S
SW

Example: 2

Write down the three digit bearing of the given points from O.

NN N

(a) A (b) (c) C

60º O 140º 25º
O
O
Solution:
B

Here, (a) The bearing of A from O is 060°.
(b) The bearing of B from O is 140°.
(c) The bearing of C from O is (360° – 25°) = 335°.

104 Oasis School Mathematics – 7

Example: 3 N

Write in three digit bearing. O

(a) N 60° E N (b) S 30° W 30º

Solution: A S

From the figure, 60º S30°W = (180° + 30°)
N 60° E = 060° O = 210°

A

Example: 4

If the bearing of B from A is 060°, what is the bearing of A from B?
Solution:

Here, the bearing of B from A is 060°.

i.e. ∠NAB = 60°

Since, NA||N'B ∠NAB + ∠N'BA = 180° [∵ Sum of co–interior angles is 180°]

60° + ∠N'BA = 180° N'
or, ∠N'BA = 180° – 60°

= 120° N B
∴ The bearing of A from B = reflex ∠N'BA
= (360° – 120°) 60°
= 240° A

Exercise 9.1

1. Write down the compass bearing of the following figures.

(a) N (b) (c) (d) N (e) N

A 60º W NW
O SW
O O EW E
30º 45º S S
AB
S S

2. Draw the diagrams to show each of the following bearings.

(a) N50°E (b) S 30°E (c) S 60°W (d) N 80°W

N NE

3. From the given figure write down the three-digit bearing of NW

NE, E, SE, S, SW, W, and NW. W OE

SW SE

S

Oasis School Mathematics – 7 105

4. Write the three-digit bearing of given points from O.

(a) N (b) N (c) N (d) N (e) N

C

80º 160º 40º D O 150º
O A O O

B E

5. Sketch the following bearing.

(a) 075° (b) 145° (c) 215° (d) 305°

6. Write the three digit bearing of:

(a) N60°E (b) S70°W (c) S50°E (d) N30°W

7. In each of the given figures, bearing of B from A is given. Find the bearing of A from B.

N N' (b) N' N' N N'

(a) N (c) (d) A
310º
A 120º N A B
230º
A 40º B
B
B

8. From the given map of Nepal, using three-digit bearing write down the bearing of:



(b) Gorkha from Nepalganj.
(d) Biratnagar from Janakpur.
(a) Kathmandu from Pokhara.
(c) Jumla from Dhangadi.

Answer

1. a. S30°E b. S 45°W c. N60° W d. S 45°W e. N 45° W 2. Consult your teacher

3. 045°, 090°, 135°, 180°, 225°, 270°, 315° 4. a. 080° b. 200° c. 320° d. 270°

e. 150° 5. Consult your teacher. 6. a. 060° b. 250° c. 130° d. 330°

7. a. 300° b. 220° c. 050° d. 130° 8. Consult your teacher.

106 Oasis School Mathematics – 7

9.2 Scale Drawing

Introduction:

The figures or maps can be drawn by using different scales. It is impossible to show
lengths and distance which are too great or too small in a sheet of paper. In such
case, we have to enlarge smaller figure and reduce the larger figure by taking a
suitable scale.

Scale factor:

In the given fig. (i) and (ii) we can see that each side of the second figure is twice that
of the first figure.

A A'

6 cm
3 cm
3 cm
6 cm
Fig. (i) Fig. (ii)

B 3 cm C B' 6 cm C'

i.e. each side of the first figure is enlarged twice.

∴ AB = 3 cm = 1
A'B' 6 cm 2

BC = 3 cm = 1
B'C' 6 cm 2

AC = 3 cm = 1
A'C' 6 cm 2

Thus the first figure is the scale drawing of the second figure to the scale 1 : 2.
Thus the ratio 1:2 is called the scale factor.

Note: The scale factor 1:100 means 1 unit length in drawing is 100 units of length
in actual size.

i.e. length in drawing = 1 = scale factor.
actual length 100

Worked Out Examples

Example: 1

What is the actual distance between two places which is represented by 1.5 cm on a
map which is drawn to the scale 1: 6000000?
Solution:

Here, Scale = 1: 6000000

i.e. 1 cm on a map represents 6000000 cm.

Oasis School Mathematics – 7 107

= 6010000000m = 610000000km. = 60 km.

∴ 1.5 cm on a map represents = 60 × 1.5 km

= 90 km.

∴ Distance between the two places = 90 km

Example: 2

The actual length and breadth of a rectangular field are 100 m and 80 m respectively.
This is drawn on the scale 1:800. Find the length and breadth of the field on the drawing.

Solution:

Here, 800 cm is represented by 1 cm Alternative method:

i.e. 800 m is represented by 1 cm We have, scale = Length in map
100 Actual length

or, 8 m is represented by 1 cm 8010 length in map
length of the field = 100 m = 100 m
8 m is represented by 1 cm
or, length in map = 180000 m = 1 m = 1 ×100 cm
8 8

1 m is represented by 1 cm = 12.5 cm.
8
Again, scale = length in map
Actual length
1
80 m is represented by 8 × 80 cm 1 = length in map
800 80 m
= 10 cm
80 m = 110
100 m is represented by 1 × 100 cm ∴ breadth in map = 800 m
8
= 1 ×100 cm = 10 cm.
= 12.5 cm 10

Exercise 9.2

1. The given figure shows the position of students in a class. If the scale is 1:150, find

the actual distance between. Ram Sita
(a) Ram and Shyam. (b) Sita and Gita

(c) Ram and Sita (d) Shyam and Gita Gita Shyam

2. If the scale in a map is 1:1500, find the actual distance between any two places if

the distance in a map is:

(a) 4 cm (b) 8 cm (c) 6 cm (d) 15 cm (e) 29cm

3. The scale in a map is 1:5000. Find the distance between two places in a map if the
actual distance between the two places is:

(a) 0.8 km (b) 600 m (c) 1500 m (d) 3 km (e) 4 km

108 Oasis School Mathematics – 7

4. (a) Find the actual height of the tree whose height in a map is 5 cm and is repre-
sented by a scale 1: 600.

(b) What is the actual distance between two places which are represented by 25

mm if the scale on the map is 1 mm: 2 km? 3cm 1cm 2cm 4 cm 3 cm 3 cm

5. The figure alongside shows the outline Kitchen Common
sketch of a house plan. Toilet Room

Draw a scale of 1 cm to represent 1 m.

(a) Find the actual length and breadth of Bed Room Garage 3 cm
bedroom, kitchen and toilet.

(b) Find the area of garage. 4 cm 3 cm

6. Find the actual distance between the given places from the given scale.

Scale: 1:1560000

Dhankuta

(a) Kathmandu and Dhankuta. (b) Pokhara and Jumla.
(c) Gorkha and Janakpur. (d) Dhangadh and Biratnagar.



Answer

1. Consult your teacher. 2. a. 60 m b. 120 m c. 90 m d. 225 m e. 435 m

3. a. 16 cm b. 12 cm c. 30 cm d. 60 cm e. 80 cm 4. a. 30 m b. 50 km

5. a. Bedroom - 4m; 3 m. Kitchen - 4m, 2 m, Toilet - 4 m, 1 m b. 9 m² 6. Consult your teacher.

Oasis School Mathematics – 7 109

Objective Questions

Choose the correct alternatives.

1. Reflection of point (-2, 3) on X-axis is

(i) (2, -3) (ii) (-2, -3) (iii) (2, 3)

2. Quarter turn clockwise direction about origin is

(i) R [O, +900] (ii) R [O, –900] (iii) R [O, + 1800]

3. R [0 + 2700] is equivalent to

(i) R [O, – 900] (ii) R [O, + 900] (iii) R [O, –2700]

4. If P(3,4) P'(-4, 3) then the transformation is

(i) reflection on X-axis (ii) reflection on Y-axis

(iii) rotation through + 900 about origin.

5. If an object is displaced 4 units right and 3 units upward then the translation vector
is

( ( 4 ( ( (ii) 3 ( ((iii) -4
(i) 3 4 3

6. Three digit bearing of S 500 W is

(i) 2300 (ii) 3100 (iii) 0500

7. An object is displaced 2 units left and 3 units upward. Then the translation vector is

( ( 2 ( ( (ii) -2 ( ((iii) -2
(i) 3 3 -3

8. The image of a point (–2, 3) after the rotation about origin through 3600 is

(i) (–3, –2) (ii) (–2, –3) (iii) (–2, 3) N'
(iii) 1400 B
9. In the given figure, bearing of B from A is N

(i) 0800 (ii) 2600 80°
A
10. The order of symmetry of the given figure is

(i) 2 (ii) 3 (iii) 4

11. How many lines of symmetry does the given figure have?

(i) 1 (ii) 2 (iii) 3

110 Oasis School Mathematics – 7

Assessment Test Paper

Attempt all the questions. Full Marks : 26

Group 'A' [10×1=10]

1. (a) Reflect a point A(2, 3) on X-axis.

(b) Rotate a point P(–5, 4) through + 90° about origin.
( )2. 1
(a) Translate a point Q(–3, 2) by the translation vector 2 .

(b) Reflect a point R(–4, 5) on Y-axis.

3. Draw the line of symmetry in the given figure.

(a) (b)

4. Write the comp ass bearing of following figures.

(a) N (b) N

O 30°
O
45°

SS

5. Draw the diagram to show the following bearing.

(a) N60°E (b) N50°W

6. (a) Group 'B' [4 × 2 = 8]

Co-ordinates of the point P and Q are P(2, –3) and Q(5, 6). Reflect the points
P(2, –3) and Q(5, 6) on Y - axis.

(b) Co-ordinates of the points A and B are A(1, 2) and B(–2, – 3). Rotate the points
7. (a)
(b) A and B through –90° about origin. N

In the given figure, bearing of B from A is given, find the bear- N1

ing of A from B. 120º

The scale of a map is 1:1000. Find the actual distance between A
two places if the distance in the map is 6cm.

Group 'C' [2×4=8] B

8. Vertices of ∆ABC are A(2, 5) , B(–3 , 2) and C(–1, 3). Reflect ∆ABC pm X- axis. Plot both
figure on the graph paper.

9. Vertices of ∆PQR be (P( –2, 3), Q( 1, 2) and R(2, – 4). Rotate ∆PQR through + 90° about
origin. Draw both the triangles on the graph sheet.

Oasis School Mathematics – 7 111

•

Sets

12Estimated Teaching Hours

Contents

• Definition of set
• Set notation
• Representation of sets
• Types of sets
• Set relation
• Subsets

Expected Learning Outcomes

At the end of this unit, students will be able to develop the
following competencies:
• Identify whether the given collections are set or not
• Use different notations in set
• Represent the set by various method
• To find the cardinal number of given set and identify its type
• Relate two sets and identify their relation
• Find the subsets of given set
• Find union, intersection, difference and complement of given

sets

Teaching Materials

• Model of Venn diagram

112 Oasis School Mathematics – 7

Unit

10 Sets

10.1 Definition of sets

Set: The word 'set' simply gives the meaning as a collection of objects, where a
collection of objects must be well defined i.e. the elements present in the set must be
distinguishable.

Elements/Members of set: The objects of a set are called the elements or members of
the given set. For instance, V = {a, e, i, o, u} is a set which contains the objects a, e, i, o,
u called the elements of set V. The membership of an element to a set is denoted by
a symbol ∈ (Greek letter Epsillon) '∈'. We can write a ∈V, e ∈ V, or a ∈ { a, e, i, o, u},
e ∈ {a, e, i, o, u} and so on , it can be read as 'a' is an element of set V. Similarly, ∉
stands for the words "not an element of" or "doesn't belong to".

Example, b ∉ V and d ∉ V. Write Say

Let's see some examples: a∈v a belongs to V
• Collection of odd numbers less than 10 b∉V or, a is an element of V
b does not belong to V

• Collection of vowels of English alphabet or b is not an element of V

• Collection of English months starting from J

In each of the above examples, objects are well defined:

1 3 5 7 9 Collection of odd numbers less than 10

a e i o u Collection of vowels of English alphabet

January June

July Collection of English months starting from J


Again, let's see some other examples,
• Collection of intelligent students

• Collection of good people in a community

In each of the above examples, objects are not well defined. Hence, above collections
do not represent set.

Oasis School Mathematics – 7 113

• Set notation

A set (collection of elements) is generally written inside a pair of curly brackets
separated by a comma. A set is represented by capital letter of English or
represented by capital letter of Greek alphabet.

Example:

C = {b, c, d, f, g, h}, P = {2, 3, 5, 7}.

∆ = { 1, 2, 3,}, γ = {w, x, y, z}.

Here, C and P are English letters whereas ∆ (Delta) and γ (Gamma) are Greek letters.

Note: the elements/members of a set can be written in any order.

Example: A = {2, 4, 6, 8} or A = {8, 6, 4, 2}

• Representation (Method of describing) of set

A given set may be described (represented) in the following ways:
(i) Description method: In this method, we describe the property of elements

(members) in a sensible sentence.
Example V = a set of vowels of English alphabet
W = a set of whole numbers less than 20
N = a set of letters of the word "school"
(ii) Roster method (Tabular/Listing method)
In this method, we list out the non–repeated elements (members) of a set enclosing

in a pair of curly braces. The elements are separated by a comma.
Example P = {2, 3, 5, 7, 11, 13}
N = {s, c, h, o, l}
V = {a, e, i, o, u}
(iii) Rule method (Set–builders): In this method, the elements (members) of a set are

represented by a sensible statement or a formula using some variables like x, y, z.
e.g. P = {x : x is a prime number less than 15}
V = {y : y is a vowel of English alphabet}
W = {z : z is a whole number, 10 < z< 15 }

Remember !

• ∈ → belongs to
• ∉ → does not belong to
• Collection of objects having words intelligent, honest, etc. are not well defined.
• Elements of sets should not be repeated.
• Elements of set can be arranged in any order .

114 Oasis School Mathematics – 7

Note:

Description Method Listing method Set-builder method

A set of prime numbers {2, 3, 5, 7} {x : x is a prime number
less than 10 less than 10}

Exercise 10.1

1. State whether the following sets are well defined or not.

(a) A collection of intelligent girls of standard VII

(b) A set of animals in a zoo
(c) A set of all the teachers of your school
(d) A set of ripen fruits
(e) A set of even numbers less than 20
(f) A set easy problems of mathematics

2. If P = {1, 2, 3, ..., 10} and Q = {2, 4, 6, 8, 10, 12, 14}, state whether the following
statements are true or false.

(a) 11 ∈ P (b) 1 ∉ Q (c) 7∈ Q

(d) 9 ∈ P (e) 12 ∉ Q (f) 14 ∈ P

3. Insert the appropriate symbols ∈ or ∉ in the given blanks if A = {0, 1, 2, 3, 4}

and B = {2, 3, 5, 7}. (d) 7 ....... A
(a) 0..... A (b) 6 ....... B (c) 5 ....... B

4. Write the following sets in description method.

(a) P = {w, x, y, z} (b) C = {1, 2, 3, 4, 6, 12}

(c) Q = {2, 3, 5,7,11 } (d) V = {3, 6, 9, 12, 15}

5. Write the following sets by roster method.
(a) A = letters of the word "NARAYANI".
(b) B = first five letters of English alphabet
(c) P = x:x is a day of the week
(d) Q = y: y < 10, y ∈ Even numbers

6. Write the following sets by rule method.
(a) O = {1, 3, 5, 7, 9}
(b) Q = { January, February, March, April}

(c) R = {Sugarcane, Tobacco, Jute, Mustard} (d) A = {2, 4, 6, 8}



Answer
Consult your teacher

Oasis School Mathematics – 7 115

• Cardinal number of set

Let A = {1, 2}, B = {3, 5, 7}
Set A has 2 elements and set B has 3 elements
∴ Cardinal number of A = 2 and cardinal number of B = 3.
Hence the total number of non–repeated elements of a set is called its cardinal number.

For any set A, its cardinal number is denoted by n(A) or #(A). {# → stands for hash}.

Example, If A = { a set of English alphabet}, then n(A) = 26;

If T = a set of the letters of the word 'mathematics', T = {m, a, t, h, e, i, c, s}, n(T) = 8
For U = {2, 3, 5, 7, 2, 5, 3, 2, 6, 4}, n(U) = 6.

Note: Cardinal number of a set is always a whole number.
• Types of set
Sets can be categorized in the following ways on the basis of the number of elements

contained. They are described as follows.
(i) Empty/Null/Void set: A set is called a void set if it contains no element. A void set

can be denoted by a Greek letter φ (Phi) or a pair of empty braces { }.
e.g. A = {counting numbers between 9 and 10}
B = {a set of men with three eyes}
O = { a set of countries having no capital}
Cardinal number of Null set is 0.
(ii) Singleton/Unit set: A set is called a singleton set if it contains only one element.
Example: C = the set of smallest natural number.
D = { φ }
E = a set of the deepest river of Nepal
Cardinal number of singleton set is 1.
(iii) Finite set: A set is called a finite set if it contains countable collection (limited)

number of elements.
Example: A = counting numbers less than 100
B = {2, 4, 6, 8, ...., 20}
(iv) Infinite set: A set is called infinite set if it contains uncountable (unlimited) number

of elements.
Example : P = a set of natural numbers, Q = {1, 2, 3, ………...}.



116 Oasis School Mathematics – 7

Remember !

• Cardinal number of a set is the number of its elements.
• Cardinal number is always a whole number .
• A null set has no element, its cardinal number is zero.
• A unit set has only one element, its cardinal number is 1.
• A finite set has finite number of elements whereas an infinite set has infinite

number of elements.

10.2 Set Relations

The relations of sets can be defined on the basis of the types and number of elements
contained.

(i) Overlapping Sets

Two or more sets are called overlapping sets if they contain at least one common
element.

Example, for sets A = {a, b, c, d, e} and B = {a, e, r, o} two elements a and e are
common. So, they (A and B) are overlapping sets. They are also called joint sets.

(ii) Disjoint Sets

Two or more sets are called disjoint sets if they contain no common elements.

Example, for sets P = {1, 3, 5, 7, 11, 13} and E = {2, 4, 6, 8, 10}

There is no common element, so they are disjoint sets. They are also called non
overlapping sets.

(iii) Equal Sets

Two or more sets are called equal sets if they contain same elements. The equality of
sets is denoted by '=' sign.

Example, if A = {2, 4, 6, 8} and B = {the first four multiples of 2}, then A = B.

If two sets are not equal, then we use '≠' sign to represent their relation.

e.g. if P = {1, 2, 3, 4} and Q = {a, e, i, o, u} then, P ≠ Q.

(iv) Equivalent Sets

Two or more sets are called equivalent sets if they contain the same number of
elements . The equivalent sets are denoted by "~" sign.

Example, i f X = {1, 3, 5, 7, 9, 11} and Y = {b, c, d, f, g, h}, both the sets contain equal
number of elements.

∴ X ~ Y [∵ n(X) = n(Y) ]

Note: Every equal set is equivalent but every equivalent set is not equal.

Oasis School Mathematics – 7 117

Exercise 10.2

1. Write the cardinal number of the following sets.
a. A = {1,3,5,7,9}
b. B = A set of prime numbers from 10 to 20
c. C = A set of the letters of word RAJSHREE
d. P = {x:x is a vowel letter}
e. Q = φ
f. M = {x:x is an English month starting from M}
g. N = A set of the highest peak in the world

2. State whether the following set are unit set or null set.
a. P = a set of even number between 3 and 4
b. Q = {x:x is an even number which is prime}
c. R = A set of the longest river of Nepal
d. A = A set of male students in Padma Kanya College
e. B = {x : x is a set of people having 30 fingers}
f. Q = A set of integers which are neither negative nor positive

3. Identify whether the given sets are finite set or infinite set
a. W = A set of whole numbers
b. N = A set of natural number less than 20.
c. R = {x : x is a mountain in Nepal whose height is more than 8000 m}
d. P = {x : x is a prime number between 20 to 30}
e. Q = A set of integers

4. State whether the given pairs of sets are overlapping or disjoint.
(a) C = {a, b, c, d, e}
D = {e, f, g, h, i}
(b) A = letters of the word "fastest"
B = letters of the word "apple"
(c) M = whole numbers less than 5
N = prime numbers greater than 4 but less than 15
(d) P = {x : x is a multiple of 3 less than 20}
Q = {y : y is one of the first five square natural numbers}
(e) X = {paddy, potato, pineapple}
Y = {pomegranate, peach, papaya}

118 Oasis School Mathematics – 7

5. State whether the given pairs of sets are equal or equivalent.

(a) C = {letters of the word 'eat'}, D = {letters of the word 'tea'}
(b) G = {1, 3, 5, 7, 9, 11, 13, 15}, H = {2, 4, 6, 8, 10, 12, 14, 16}
(c) K = {Raju, Raaj, Shyam, Vim}, L = {Kalpana, Bidhya, Junu, Sheela}
(d) O = {y : 10 < y <2 0, y is a prime number}
P = {x: x is a prime number between 9 and 20}.
(e) M = {a set English months stating from 'J'}
N = {a set of even numbers lessthan 8}

Answer
Consult your teacher

10.3 Subset∪∪

Let A= {1, 2, 3}, B = {1, 2}, C = {1, 3}
Let's compare set A and B.
Every element of B is also the element of 'A'
Then B is the subset of A.
It is written as B A.
Every element of C is also the element of A.
Then C is the subset of A.
It is written as C A.
Hence a set B is said to be the subset of A , if every element of B is also the element of A.

Note: Every set is the subset of itself.
Null set is the subset of every set .

Number of subsets of given set

Let, A = {1} Its subset are φ and {1}
when , n = 1, number of subsets = 2 = 21
Let , A = {a, b}. Its subsets are φ, {a}, {b}, {a,b}
When n = 2, number of subsets of A = 4 = 22
If the Super set : number of elements = n, the number of subsets : 2n
If Q is a subset of P then P is called a super set of Q. It is denoted as P ⊇ Q. It is read

as "P is a super set of Q".
For example, let A = {1, 2, 3, 4, 5,}, B = {2, 4} then A ⊇ B.
i.e. A is the superset of B.

Oasis School Mathematics – 7 119

Proper subset:

Let A and B are any two sets. B is called the proper subset of A if every element of B
belongs to A and B contains at least one element less than A. It is denoted as B ⊂ A,
read as "B is a proper subset of A."

For example, A = {1, 3, 5, 7, 9}, B = {1, 3, 5, 9},
C = {5, 7, 9}, D = {1, 5},
B ⊂ A, C ⊂ A, D ⊂ A.

Note: Empty set is proper subset of every set:
Number of proper subsets of a given set can be determined by a
relation 2n – 1.

For example, If P = {2, 4, 6}, number of proper subsets of P is,
2n – 1 = 23 – 1 = 8 – 1 = 7.
i.e. { }, {2}, {4}, {6}, {2, 4}, {2, 6}, {4, 6} are the proper subsets of P.

Universal set:

A set which contains all elements of other sets under consideration is called a
universal set. It is denoted by U of English alphabet.

For example, If A = {0, 1, 2}, B = {2, 3, 5, 7} and C = {2, 4, 6, 8} then we can consider
U = {0, 1, 2, ..., 8} as a universal set of A, B and C.

Note: Every set is the subset of universal set.


Exercise 10.3

1. State whether the given statements are true or false.
(a) Every set is the subset of itself.
(b) If B= {1, 2}, C = {1, 2, 3}, then B is the proper subset of C.
(c) If A= B, A is the proper subset of B.
(d) Null set is the subset of every set.
(e) Number of subsets of a set is 2n-1.

2. If A = {1, 2, 3, 4}, B = {2, 3, 4,}, C = ({3, 4, 5}, state whether the given statements
are true of false.

(a) B ∉ A (b) C ⊇ A (c) B = C (d) A ⊇ B

120 Oasis School Mathematics – 7

3. Write the possible subsets of the following sets. (c) C = {bat, ball, wicket}
(a) A = {k} (b) B = {r, s}

4. Write all the possible proper subsets of the following sets.

(a) P = {14, 1 , 12} (b) Q = {–2, –1, 0} (c) R ={a, b}
3

5. Write the number of possible subsets of the following sets.

(a) {1, 2} (b) {a, b, c} (c) {p, a} (d) {1}

6. Write the number of possible proper subsets of the following sets.

(a) {1} (b) {a, b} (c) {3, 5, 7}

7. Suggest a universal set for the pairs of given sets.
(a) A = {a, e, i, o, u} and B = {b, c, d, f, g, h, ...., x, y, z}
(b) P = {girls of standard VII}and Q = {boys of standard VII}
(c) R = {goalkeeper, defender}and S = {midfielder, striker}
(d) T = {x : x is an odd number} and V = {y : y is an even number}



Answer

1. (a) Consult your teacher. 2. Consult your teacher 3. (a) φ, {k} (b) φ, {r}, {s}, {r, s}

(c) φ, {bat}, {ball}, {wicket}, {bat, ball}, {bat, wicket}, {ball, wicket}, {bat, ball, wicket}

4. (a) φ, { 1 }, {13 }, { 21 }, { 14 , 1 }, { 1 , 1 }, { 1 , 1 }
4 3 4 2 3 2

(b) φ, {–2}, {–1}, { 0 }, {–2, –1}, {–2, 0}, {–1, 0}. (c) φ, {a}, {b}

5. (a) 4 (b) 8 (c) 4 (d) 2 6. (a) 1 (b) 3 (c) 7

7. (a) a set of English alphabet (b) a set of students of class VII (c) a set of football players

(d) a set of natural numbers

10.4 Set operations

There are four major operations on sets. They are:

(i) Union (ii) Intersection

(iii) Difference (iv) Complement

(a) Union of Sets

A collection of all the non–repeated elements from given sets (two or more sets) is
called the union of given sets. The union of sets is denoted by '∪'. The union of any
two sets A and B is written as A ∪ B and read as A union B or 'A cup B'.

"or" "A ∪ B" = {x : x ∈ A or x ∈ B}

For example:

Oasis School Mathematics – 7 121

If A = {1, 2, 3, 4} and B = {3, 4, 5, 6, 7, 8}
then, A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8} [write repeated elements only once]
If P = {ball, bat, wicket} and Q = {umpire, batsman, bowler, wicketkeeper},
then, P ∪ Q = {ball, bat, wicket, umpire, batsman, bowler, wicketkeeper}

(b) Intersection of Sets
A collection of all the repeated elements from the given sets (two or more sets) is

called the intersection of given sets. The intersection of sets is denoted by '∩'. The
intersection of any two sets P and Q is written as P ∩ Q and read as 'P intersection 'Q'.
P ∩Q = {x: x ∈ P and x ∈ Q}
For example:
(i) If A = {2, 3, 5, 7, 11} and B = {prime numbers less than 5},
⇒ A ∩ B = {2, 3} [ ∵ {2, 3} are repeated common elements]
(ii) Again, if X = {volleyball, basketball, football} and
Y = {court, ground, diamond}
⇒ X ∩Y = { } [∵ none of the element is common in both sets]

(c) Difference of Sets
Let A and B be any two sets. The difference of A and B denoted as A – B is

defined as the collection of all the elements of A only, which are not in B.
∴ A – B = {x: x∈A and x∉B}
Example:
(i) If M = {a, e, i, o, u} and N = {letters of the word 'pea'} ={p, e, a}
then, M – N = {i, o, u} and N – M = {p}
(ii) If P = {2, 4, 6} and Q = {2, 4, 6, 8, 10, 12}, then P – Q = { } and Q – P = {8, 10, 12}

Note: A ∆ B = (A – B ) ∪ (B – A), union of symmetric difference. Read as 'A delta B'

(d) Complement of Set
Let '∪' be a universal set and A be any subset of U. Then complement of A denoted

by A or A' is defined as the collection of all elements of U only which are not in A.
i.e. A = U – A

Similarly, (A ∪ B) = U – (A ∪ B)
(A ∩ B) = U – (A ∩ B)
(A - B) = U – (A – B)

122 Oasis School Mathematics – 7

Example:
If U = {0, 1, 2, ...., 11, 12}, P = {1, 3, 5, 7, 9, 11} and Q = {2, 7, 4, 6, 8, 10}, then
P = U – P = {0, 2, 4, 6, 8, 10, 12} and, Q = U – Q = {0, 1, 3, 5, 9, 11, 12}.

Note: A= A, (A ∪ B) = (A ∪ B) and so on]

10.5 Venn–diagrams

Sets can be represented by using diagrams called Venn–diagrams. Generally, a
rectangle represents a universal set and circle represents the subsets. The diagram
was developed by a British Mathematician John Venn.

Some set relations and their operations are described here. U
(i) (ii) (iii)

UU

A AB AB
A
A∪B A∪B

(iv) [For overlapping sets] [For disjoint sets]
(v) (vi)

U U

AB AB AB
A–B
A∩B A∩B
[For overlapping sets]
[For overlapping sets] [For disjoint sets]
(ix)
(vii) (viii) U
U
U

AB A
AB

A–B (A–B) ∪ (B–A) A

[For disjoint setsU] A = (U–A)

(x) U (xi) U (xii) U

AB AB AB

(A∩B) = U–(A∩B) (A–B) = U–(A–B)

(A∪B) = U–(A∪B)

Oasis School Mathematics – 7 123

Worked Out Examples

Example: 1

For A = {1, 2, 3, 4, 5}, B = {4, 5, 6, 7, 8, 9}, find

(a) A ∪ B (b) A ∩ B (c) A – B (d) B – A

Represent each of the relation is Venn–diagrams.
Solution:

(a) A = {1, 2, 3, 4, 5} and (b) A ∩ B = {4, 5}

B = {4, 5, 6, 7, 8, 9},

∴ A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8, 9}

U U

1 46 1 4 6
2 678 5
3 2 78
9 3 9

A B (A ∪ B) AB (A ∩ B)

(c) A – B = { 1, 2, 3} (d) B – A = {6, 7, 8, 9}

U A–B U

14 6 14 6

2 1 78 2 5 78
3 9 3 9

AB AB B–A



Example: 2

If U = {1, 2, ....., 12}, A = {1,3, 5, 7, 9} and B = {1, 2, 4, 6, 8, 10}, then find

(a) A (b) A∪B (c) A∩B (d) A - B

Also represent the relations in Venn diagrams.
Solution:

∴ A ∪ B = ∪ − (A ∪B) = {11,12}

U U

A3 2B A3 2B

57 1 46 57 1 46
9 8 10 9 8 10

11 12 11 12

124 Oasis School Mathematics – 7

(d) A - B = { 3, 5, 7, 9}
(c) A ∩ B = { 1 } ∴ A – B = U – (A–B) = {1,2,4,6,8,10,11,12}

∴ A ∩ B

= U – (A ∩B) = {2,3,4,5,6,7,8,9,10,11,12}

U U

A5 2 B A3 2B

37 146 57 1 46
8 9 8 10
8 10

11 12 11 12

Example: 3

From the given Venn diagram, list the elements of the following sets:

( i) A (ii) B (iii) (A ∩ B) (iv) A ∪ B

(v) A ∪ B (vi) A – B (vii) B – A (viii) A ∩ B

(ix) A – B (x) B – A

Solution:

From the Venn diagram, U
(i) A = {1, 2, 3, 4, 8} AB
(ii) B = {4, 5, 6, 7, 8, 9} 14 5
(iii (A ∩B) = {4, 8} 67 11
9
2 38 12

10

(iv) A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8, 9}

(v) A ∪ B = {10, 11, 12} (vi) A – B = {1, 2, 3}
(vii)
(ix) B – A = {5, 6, 7, 9} (viii) A ∩ B = {1, 2, 3, 5, 6, 7, 9, 10, 11, 12}
(x)
A – B = {4, 8, 5, 6, 7, 9, 10, 11, 12}

B – A = {1, 2, 3, 4, 8, 10, 11, 12}

Example: 4

If U = {1, 2, 3…………. 15}, A = {2, 4, 6, 8, 10, 12, 14}

B = {3, 6, 9, 12, 15} and C = {4, 8, 12}, find

(i) A ∪ B ∪ C (ii) A ∩ B ∩ C (iii) (A ∪ B) ∩ C
Solution: Here,
(i) A ∪ B = {2, 4, 6, 8, 10, 12, 14} ∪ {3, 6, 9, 12, 15}

= {2, 3, 4, 6, 8, 9, 10, 12, 14, 15}

∴ A ∪ B ∪ C = (A∪ B) ∪ C

= {2, 3, 4, 6, 8, 9, 10, 12, 14, 15} ∪ {4, 8, 12}

= {2, 3, 4, 6, 8, 9, 10 12, 14, 15}

Oasis School Mathematics – 7 125

(ii) Here, A ∩ B = {2, 4, 6, 8, 10, 12, 14} ∩ {3, 6, 9, 12, 15}
= {6, 12}
∴ A ∩ B ∩ C = {6, 12} ∩ { 4, 8, 12} = {12}
(iii) (A ∪ B) = {2, 4, 6, 8, 10, 12, 14} ∪ {3, 6, 9, 12, 14, 15}
= {2, 3, 4, 6, 8, 9, 10, 12, 15}

(A ∪ B) ∩ C = {2, 3, 4, 6, 8, 9, 10, 12, 15} ∩ {4, 8, 12} = {4, 8, 12}

Exercise 10. 4

1. List out the elements of the union of the following sets.
(a) A = {a, e, i, o, u} and B = {a, b, c, d}
(b) P = {2, 3, 5, 7, 11} and Q = {1, 3, 5, 7, 9}
(c) M = { 10, 20, 30, 40, 50, 60, 70} and N = {φ}
(d) X = {x : x is a multiple of 3 less than 15} and
Y = {x: x is a multiple of 6 less than 20}

2. List out the elements of the intersection of the following sets.
(a) A = {1, 2, 3, 4, 5, 6} and B = {2, 4, 6, 8, 10}
(b) C = {letters of the word 'fastest'} and D = {letters of the word 'staff'}
(c) E = {factors of 10} and F = {factors of 15}
(d) G = {Romi, Rozy, Mary, Krisha}, H = {Krish, Rose, Harry}
(e) I = {Maths, Science, Social, English}, J = { }

3. Find the differences A – B and B – A from the given sets.
(a) A = {5, 10, 15, 20, 25} and B = {multiples of 10 less than 40}
(b) A = {a, b, c, d, e, f, g, h, i, j} and B = {a, e, i}
(c) A = {–1, 0, 1} and B = { }
(d) A = {2, 4, 6, 8, 10, 12, 14}, B = {3, 6, 9, 12, 15}.

4. Let U = {a, b, c, ..... k}, A = {b, d, f, h, j, k} and B = {c, d, e, f, g, h, i, j, k}, then find the
following:

(a) A (b) B (c) (A ∪ B)

(d) (A ∩ B) (e) (A - B) (f) (B - A)

5. What does the shaded part of each of the given Venn diagrams represent?

(a) U (b) A BU (c) U
B B
A A

126 Oasis School Mathematics – 7

(d) (e) (f)

A U U A BU
BA B

6. (a) Copy the given Venn diagram in your copy and shade the following relations.

(i) A∩B (ii) A∪B (iii) A – B A U
B

(iv) B – A (v) A ∪ B (vi) A

(b) Copy the given Venn diagram, shade the following set relations. BU

(i) A∩B (ii) A∪B A

(iii) A – B (iv) A ∪ B

7. (a) If A = {a, e, i, o, u} and B = {a, b, c, d, e}, find

(i) A ∪ B (ii) A ∩ B (iii) A – B

(iv) B – A and show them in Venn diagram.

(b) If M = {2, 3, 5, 7, 11}, N = {1, 3, 5, 7, 9, 11}, find

(i) M ∪ N (ii) M ∩ N (iii) M – N

(iv) N – M and show them in Venn diagram.

8. If U = {1, 2, 3 …… 15}, A = {2, 4, 6, 9, 10, 11, 12} and B = {1, 3, 5, 7, 9, 11, 13}, find

(i) A (ii) B (iii) A ∪ B (iv) A ∩ B

(v) A – B and show them in Venn diagram.

9. From the given Venn diagram, list out the given sets. U

(i) A (ii) B (iii) A ∪ B Ad a j B

e b kg
i fcl h
(iv) A ∩ B (v) A ∪ B (vi) A ∩ B

10. If U = {1, 2, 3, 4………………13}, A = {2, 3, 5, 7, 11}, B = {2, 4, 6, 8, 10, 12} and

C = {1, 2, 3, 5, 7, 9, 11}, find:
(i) A ∪ B ∪ C (ii) A ∩ B ∩ C
(iii) (A ∪ B) ∩ C (iv) (A ∩ B) ∪ C

Do you know!
The modern study of set theory was initiated by Georg Cantor and Richard
Dedekind in the 1870s.

Oasis School Mathematics – 7 127

Answer (b) P ∪ Q = {1, 2, 3, 5, 7, 9, 11}

1. (a) A ∪ B = {a, b, c, d, e, i, o, u}

(c) M ∪ N = {10, 20, 30, 40, 50, 60, 70, φ} (d) X ∪ Y = {3, 6, 9, 12, 18}

2. (a) A∩B = {2, 4, 6} (b) C ∩ D = { f, a, s, t} (c) E ∩ F = {1, 5}

(d) G ∩ H = { } or φ (e) I ∩ J = φ

3. (a) A - B = {5, 15, 25} and B - A = {30} (b) A - B = {b, c, d, f, g, h, j} and B - A = { }

(c) A - B = { -1, 0, 1} and B - A = { } (d) A – B = {2, 4, 8, 10, 14} and B – A = {3, 9, 15}

4. (a) A = {a, c, e, g, i} (b) B ={a, b} (c) (A∪B) = {a} (d) ( A∩B ) = {a, b, c, e, g, i}

(e) ( A–B ) = {a, c, d, e, f, g, h, i, j, k} (f) (B – A) = {a, b, d, f, h, j, k}

5. (a) A ∩ B (b) A ∪ B (c) A – B (d) B – A (e) A∪B (f)( A ∩ B) 6. Consult your teacher.

7. (a) (i) A ∪ B = {a, b, c, d, e, i, o, u} (ii) A ∩ B = {a, e} (iii) A –B = (i, o, u}
U U
U A bB A bB
A bB
a i a c i a c
i e c e d e d
d o o
o u u
u

(iv) B – A = (b, c, d} (b) (i) M ∪ N = {1, 2, 3, 5, 7, 9, 11) (ii) M ∩ N= {3, 5, 7, 11}

A bB U M U M U
N N
i a c 2 31 2 31
e d 57 57
o 11 9 11 9
u

(iii) M – N = {2} (iv) N – M = {1, 9} 8. (i) A = {1, 3, 5, 7, 8, 13, 14, 15}

M U U A U
N N B
2 31 M 31 8 2 1 3
57 57 46 9 5 7 15
11 9 2 11 9 10 12 11 14
13

(ii) B = {2, 4, 6, 8, 10, 12, 14, 15} (iii) A∪B = {8, 14, 15}

A2 U A U
B B
46 9 1 3 8 2 11 33
5 15 46 9 55 77 15
8 10 12 11 13 7 14 10 12 11 14
1133

(iv) A∩B = {1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 13, 14, 15} (v) A – B = {1, 3, 5, 7, 9, 11, 13, 14, 15}
U
U A B
A B
2 1 8 21 3
8 46 5 3 15 46 9 5 7 15
10 12 9 7 14 11
11 13 10 12 13 14

9. (i) A = {a, b, c, d, e, f} (ii) B = {a, b, c, j, k, l } (iii) A∪B = {a, b, c, d, e, f, j, k, l}
(iv) A ∩ B = {a, b, c} (v) A∪B = {i, g, h} (vi) A∩B = {d, e, f, g, h, i, j, k, l}

10. (i) A ∪ B ∪ C = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} (ii) A ∩ B ∩ C= {2}

(iii) (A ∪ B) ∩ C = {2, 3, 5, 7, 11} (iv) (A ∩ B) ∪ C= {1, 2, 3, 5, 7, 9, 11}

128 Oasis School Mathematics – 7

Objective Questions

Choose the correct alternatives.
1. Symbol ∈ (Greek letter Epsilon) represents

(i) belongs to (ii) does not belong to (iii) is a subset of

2. Which one of the following is not a set

(i) collection of prime number from 1 to 20.

(ii) collection of honest students from class VII.

(iii) collection of the letters of word SCIENCE.

3. If A = a set of stars in the sky then the set A is

(i) finite set (ii) infinite set (iii) universal set

4. Cardinal number of the set { φ } is

(i) 0 (ii) 1 (iii) 2

5. If n(A) = n(B) then 'A' and 'B' are

(i) equal sets (ii) overlapping sets (iii) equivalent sets

6. If A = {1, 2, 3, 4}, then the number of subsets of A is

(i) 8 (ii) 16 (iii) 12

7. Which set is the subset of every set?

(i) Unit set (ii) Null set (iii) Universal set

8. Which one of the following statement is not true?

(i) Every set is the subset of itself.

(ii) Unit set is the subset of every set.

(iii) Null set is the subset of every set.

9. Shaded part of the given Venn diagram represents P U
Q
(i) P–Q (ii) Q–P (iii) P∩Q U
B
10. If 'A' and 'B' are disjoint sets, then A∩B is equal to

(i) φ (ii) [ φ } (iii) 0

11. The shaded part of the given figure represents A

(i) A ∩ B (ii) A ∪ B (iii) A – B

Oasis School Mathematics – 7 129

Assessment Test Paper

Attempt all the questions. Full Marks : 26

Group 'A' [5 × 1 = 5]

1. (a) Express the set of letters of the word AADHYA by listing method.

(b) Copy the given Venn diagram and shade the set which represents A∩B.

U
AB

(c) Find the number of sub sets of the set A = {p, q, r].
(d) A = a set of highest peak in the world, identify whether 'A' is a null set or a

unit set.
(e) A = {a, e, i, o, u}, B = a set of vowels of English alphabet. What is the set relation

between 'A' and 'B'?
Group 'B' [4 × 2 = 8]

2. (a) If A = a set of the letters of the word APPLE and
B = a set of the letters of word BAGMATI, list A and B. Identify whether they

are disjoint or overlapping.
(b) If A = {2, 3, 5, 7, 1}, represent set A by set builder method.

3. (a) If A = {p, q, r}, write all the possible subsets of A.

(b) If M = {2, 3, 5, 7, 11} and N = {1, 2, 3, 4, 5}, find
(i) M∪N and (ii) M∩N

Group 'C' [2 × 3 = 6] BU

4. From the given Venn diagram, find A •7 •9

(i) A∩B (ii) B–A (iii) A–B •5 •8 •11
•6
•10

5. If M = {1, 2, 3, ..........10}, N = a set of even numbers less than 20.

Find M–N and M – N and show these relations in Venn diagram.

130 Oasis School Mathematics – 7

Arithmetic

Contents 49Estimated Teaching Hours

• Binary and Quinary Numbers

• Whole numbers

- H.C.F. and L.C.M - Square and Square root

- Cube and Cube roots

• Integers

- Integers on the number line - Operations on integers-simplification

- Rational Numbers and Irrational Numbers - Fraction and Decimal

• Fundamental Operations on Fraction

• Simplification of Fractions

• Simplification of Simple Verbal Problems on Fraction

• Four Fundamental Operations on Decimal

- Commercial Arithmetic

- Ratio and Proportion

- Percentage - Unitary method - Simple Interest - Profit and Loss

Expected Learning Outcomes
At the end of this unit, students will be able to develop the following
competencies:

• Convert the given decimal number into binary and quinary numbers
• Convert the given binary and quinary numbers into decimal number
• Calculate the H.C.F and L.C.M of given numbers
• Apply the relation of H.C.F and L.C.M with given number
• Calculate the square root of given number using factorisation and division

method
• Calculate the cube and cube root of given number
• Show the integers on the number line
• Add, subtract, multiply and divide integers on the number line
• Simplify the integers
• Identify whether the given numbers are rational number or an irrational number
• Solve the simple problems related to direct and indirect proportion
• Solve the simple problems on unitary method
• Calculate the simple interest of given sum with given rate and time

• Solve simple problems on profit and loss

Teaching Materials

• Graph paper, A4 Size paper, Glue, Scissors, etc.

Oasis School Mathematics – 7 131

Unit Operation on
11 Whole Number

11.1 Binary Number System (Base two system)

In Hindu Arabic system of numeration the ten base symbols used are 0, 1, 2, 3,
4, 5, 6, 7, 8, 9 which are called digits. In this system every whole number can be
expressed as the power of 10.

For example: 327 = 3 × 102 + 2 × 101 + 7 × 100

In this system, moving from right to left, value of each place is ten times the value
of its right. So this number system is called 10–base system or the decimal number
system or the denary system.

Binary Number system

In binary number system, we use only the two digits 0 and 1. In this number
system the numbers are represented by the power of 2. So it is also called the base
two system. Hence the system in which 0 and 1 are used to represent the number by
the power of two is called binary number system.

I n this system value of each place is two times the value of the place at the right.
Binary number system is widely used in computer system. Computers consider
electric circuit on as 1 and off as 0. Binary numbers are written with their base 2 on
the suffix.

Example 102, 112, 1002 etc. Look at the following table.

Decimal 0 1 2 3 4 5 6 7 8 9 10 11

(Denary)

Binary 0 12 102 112 1002 1012 1102 1112 10002 10012 10102 10112
System

Place value chart of binary number Binary number I understand !
20, 21, ..... are the
27 26 25 24 23 22 21 20 12 places of binary
1 102 number.
1011012
10
101101

132 Oasis School Mathematics – 7

I. Conversion of Binary Numbers into Decimal Numbers

To convert a binary number into decimal number, expand the given binary number
by giving the place value of each digit in power of 2.

For example:

1002 = 1×22 + 0×21 + 0×20
=4+0+0=4

111012 = 1 × 24 + 1 × 23 + 1 × 22 + 0 × 21 + 1 × 20

= 16 + 8 + 4 + 0 + 1 = 29

II. Conversion of Decimal Numbers into Binary numbers

Let's be clear to convert decimal number into binary number with the help of given

example. Steps
• Divide the given number successively by 2
For example:
until the quotient is zero.
Convert 30 into a binary number. • List the remainder obtained in each successive

2 30 Remaider division in a column.
2 15 • Write all the remainders from the bottom to the
27 0
23 1 top.
21 1 Which is the required binary number.
1
0 1 Remember!



Writing all the remainders from the • Digits used in Decimal number
bottom to the top 11110. system are 0, 1, 2, 3, .......... 9

∴ 30 = 111102 • Digits used in binary number
system are 0 and 1.

Worked Out Examples

Example: 1

Display 1012 and 1011012 in place value chart and write in expanded form.
Solution:

Index notation 2² 2¹ 2°

Number 10 1

Expanded form = 1×2² + 0×2¹ + 1×2°.

Index notation 25 24 23 22 21 20
1 10 1
Number 10

Expanded form : 1×25 + 0×24 + 1 × 23 + 1×22 + 0×21 + 1 × 20.

Oasis School Mathematics – 7 133

Example: 2

Convert the following binary numbers into denary (decimal) number.

(a) 110112 (b) 1000112
Solutions:

(a) 110112 = 1×24 + 1×23 + 0×22 × 1×21 + 1×20.
= 16 + 8 + 0 + 2 + 1 = 27

(b) 1000112 = 1×25 + 0×24 + 0×23 + 0×22 + 1×21 + 1×20.
= 32 + 0 + 0 + 0 + 2 + 1 = 35

Example: 3

Convert the following decimal number into binary numbers.

(a) 46 (a) (b) 132 (b)
Solution:
2 46 Remainder 2 132 Remainder
2 23 2 66
2 11 0 0
2 5 1
2 2 1 2 33 0
2 1 1
0 0 2 16 1
1
2 8 0

2 4 0

2 2 0

2 1 0

0 1

Writing the remainders from the Writing the remainders from the

bottom to the top 101110. bottom to the top: 10000100

∴ 46 = 1011102 ∴ 132 = 100001002

Exercise 11.1

1. Display the following numbers in place value chart and write them in expanded
form.

(a) 11101102 (b) 1000102 (c) 11012

2. Convert the following binary numbers into decimal numbers.

(a) 112 (b) 1102 (c) 1112
(d) 10012 (e) 11012 (f) 100112
(g) 111002 (h) 1110012 (i) 101101012

134 Oasis School Mathematics – 7

3. Convert the following decimal numbers into binary numbers.

(a) 7 (b) 24 (c) 49 (d) 142 (e) 175
(j) 101
(f) 274 (g) 312 (h) 400 (i) 97

Answer 2. (a) 3 (b) 6 (c) 7 (d) 9
1. Consult your teacher.

(e) 13 (f) 19 (g) 28 (h) 57 (i) 181

3. (a) 1112 (b) 110002 (c) 1100012 (d) 100011102 (e) 101011112

(f) 1000100102 (g) 1001110002 (h) 1100100002 (i) 11000012 (j) 11001012

11.2 Quinary number system

We have already learnt that decimal number system is the number system having
base 10, binary system is the number system having base 2. If the number system
has base 5 such number system is Quinary number system.

The number system having base five and in which the numbers are expressed in the
power of 5 is called Quinary number system.

Five basic symbols used in quinary number system are 0, 1, 2, 3, and 4. Quinary
numbers are written with base 5 in the suffix.

Decimal 0 1 2 3 4 5 6 7 8 9 10 11 12 13
Quinary 0 1 2 3 4 0 11 12 13 14 20 21 22 23

Place value chart of quinary number

57 56 55 54 53 52 51 50 Quinary number
231
231 1430

1430 123241

12 3 2 4 1

I. Conversion of quinary numbers into decimal numbers

To convert a quinary numbers into decimal numbers we should expand the given
number by giving the place value of each digit in the power of 5.

For examples:

13245 = 1× 53 + 3×52 + 2×51 + 4×50
= 125 + 75 + 10 + 4

= 214

2045 = 2×52 + 0×51 + 4×50

= 50 + 4 = 54

Oasis School Mathematics – 7 135

II. Conversion of decimal numbers into quinary numbers:

Let's be clear to convert Decimal number in to Quinary number with the help of
given example

For example:

Convert 258 into quinary number system.

Here, 5 258 Remainder

5 51 3 Steps

5 10 1 • Divide the given number successively
by 5 until the quotient is zero.

52 0 • List the remainder obtained in each

02 successive division in a column.

Written the remainder from the • Write all the remainders from the

bottom to the top 2013. bottom to the top.

∴ 258 = (2013)5

Worked Out Examples

Example: 1

Display 132045 and 21105 in place value chart and write them in the expanded form.
Solution:

Index notation 54 53 52 51 50
Number 1320 4

Expanded form = 1 × 54 + 3×53 + 2 × 52+ 0 ×51 + 4 ×50

Again,

Index notation 53 52 51 50

Number 21 1 0

Expanded form = 2×5³ + 1×52 + 1×51 + 0×50.

Example: 3

Convert the following quinary numbers into decimal numbers.

(a) 12315 (b) 13041235
Solution:

(a) 12315 = 1×53 + 2×52 + 3×51 + 1×50
= 125 + 50 + 15 + 1 = 191

(b) 13041235 = 1×56 + 3×55 + 0×54 + 4×53 + 1×52 + 2×51 + 3×50.
= 15625 + 9375 + 500 + 25 + 10 + 3 = 25538

136 Oasis School Mathematics – 7

Example: 3

Convert the following decimal numbers into quinary numbers.

(a) 2171 (b) 6750
Solution:

(a) Remainder (b) Remainder

5 2171 1 5 6750 0
5 434 4 5 1350 0
5 86 1 5 270 0
5 17 2 5 54 4
5 3 3 5 10 0
0 5 2 2
0
Written the remainder from the
bottom to the top. 32141. Written the remainder from the bottom to the
top 204000.
∴ 2171 = 321415
∴ 6750 = 2040005

Exercise 11.2

1. Display the following quinary numbers in place value chart and write each of
them in expanded form.

(a) 20135 (b) 123405 (c) 3421015

2. Convert the following quinary numbers into decimal numbers.

(a) 325 (b) 125 (c) 1425 (d) 2125

(e) 3215 (f) 10005 (g) 314205 (h) 21345

3. Convert the following decimal numbers into quinary numbers.

(a) 6 (b) 9 (c) 27 (d) 85

(e) 123 (f) 567 (g) 874 (h) 1000

1A.n swCeornsult your teacher. 2. (a) 17 (b) 7 (c) 47 (d) 57 (e) 86
(f) 125 (g) 2110 (h) 294 (c) 1025 (d) 3205
(e) 4435 (f) 42325 (g) 114445 3. (a) 115 (b) 145
(h) 130005

Oasis School Mathematics – 7 137

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

L et's take two numbers 12 and 18

Let A = Set of factors of 12

= {1, 2, 3, 4, 6, 12}

B = Set of factors of 18

= {1, 2, 3, 6, 9, 18}

Common factor of A and B = A ∩ B = {1, 2, 3, 6}

Among these common factors, the greatest factor is 6.

∴ H.C.F. (G.C.D.) = 6

Hence, the highest common factor (H.C.F.) of two or more than two numbers is the
highest number that divides each of them exactly. It is also called the greatest com-
mon divisor (G.C.D.). The greatest common factor (G.C.F.).

• Methods of finding H.C.F.

The methods of finding H.C.F. of two or more than two numbers are:
Set of factors method

Prime factor method

Division method

Set of factors method

Let's be clear to find the H.C.F. of two numbers with the help of given example.

For example: Steps

To find the H.C.F. of 15 and 25. • Find the factors of each of the given
The set of factors of 15 = {1, 3, 5, 15}. numbers.

The set of factors of 25 = {1, 5, 25} • Express the factors in set notation.

• Find the set of common factors.

The common factors of both sets = {1, 5} • Select the greatest factor.

∴ The greatest common factor is 5. The greatest factor thus obtained is H.C.F.

Prime factorisation method
Look at this example properly and get the idea to find H.C.F. Using the method:

For example: To find the H.C.F. of 18 and 24.

Here, Steps
• Break down each of the numbers into
∴ Prime factors of 24 = 2 × 2 × 2 × 3 its prime factors.

Again, 2 18 2 24 • Take out the common prime factors.
3 9 2 12
• Multiply all the common factors.
3 2 6
Which is the H.C.F. of given numbers.
Prime factors of 18 = 2 × 3 × 3 3

138 Oasis School Mathematics – 7

Here,
24 = 2 × 2 × 2 × 3

18 = 2 × 3 × 3
C ommon factors of 24 and 18 are 2 and 3

∴ H.C.F. = 2 × 3 = 6.

Note: If two numbers have only 1 as a common factor, these numbers are
called co–prime number.

4 = 2 × 2 × 1

9 = 3 × 3 × 1

∴ 4 and 9 are co–prime numbers.

(c) Division Method

Look at this example properly and get the idea to find H.C.F. of two numbers, using
this method:

For example:

To find the H.C.F. of 90 and 100.

Here, Steps

90)100 )1 • Divide the greater number by the smaller number.
• Divide the first divisor by the remainder.
–90 • Continue this process till the remainder is zero.

10 ) 90 )9

–90
× • The last divisor is the H.C.F.


Here, the last divisor is 10

∴ H.C.F. = 10

While finding the H.C.F. of three numbers.


Steps
• Find the H.C.F. of first two numbers.

• Take H.C.F. of previous two numbers as divisor of third number.

• Repeat the process till the remainder is zero.

Note: This method is suitable, while finding the H.C.F. of large numbers.

Oasis School Mathematics – 7 139

Worked Out Examples

Example: 1

Using set of factors method, find the H.C.F. of 36 and 48.
Solution:

Here, Set of factors of 36 = {1, 2, 3, 4, 6, 9, 12, 18, 36}

Set of factors of 48 = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}

Common factors of 36 and 48 = {1, 2, 3, 4, 6, 12}

Largest common factor = 12

∴ H.C.F. = 12

Example: 2

Using prime factorisation method, find the H.C.F. of 24, 36 and 30.

Solution:

Here, 2 24 2 36 2 30
2 18 3 15
2 12 3 9 5
3
2 6

3

Here, 24 = 2 × 2 × 2 × 3

36 = 2 × 2 × 3 × 3

30 = 2 × 3 × 5


∴ H.C.F. = 2 × 3 = 6

Example: 3

Using division method, find the H.C.F. of

(a) 125 and 900 (b) 135, 180 and 300.

Solution:

(a) Here, )900 )7

125 –875

25 ) 125 ) 5
–125
× ∴ H.C.F. = 25

140 Oasis School Mathematics – 7

b) 135, 180, 300

Arranging them in descending order

300, 180, 135

Now, )300 )1 Again,

180 –180 60 )135 )2

120) 180 )1 –120
–16200)120 )2
–120 15 )60 )4

–60
×

∴ H.C.F. = 15 ×

Example: 4

Find the greatest number which will divide 125 and 238 leaving the remainders 5 and
6 respectively.

Solution:

Here, 5 is the remainder in the first case

∴ (125 – 5) = 120 is exactly divisible.

Again, 6 is the remainder in the second case

∴ (238 – 6) = 232 is exactly divisible.

Now,

120 )232 )1 Here, H.C.F. of 120 and 232 = 8.
∴ Required number is 8.
–120

112 ) 120 ) 1

–112

8) 112 )14

–8
32
–32
×

Exercise 11.3

1. Find the H.C.F. of each of the following sets of numbers by set of factors method:
(a) 4, 6 (b) 18, 24 (c) 35, 50

(d) 12, 18, 30 (e) 16, 24, 32

2. Find the H.C.F of the following using prime factor method:

(a) 12, 18 (b) 15, 24 (c) 24, 36 (d) 60, 84

(e) 120, 192 (f) 24, 36, 60 (g) 84, 120, 156 (h) 135, 180, 300

(i) 60, 90, 120 (j) 32, 48, 64

Oasis School Mathematics – 7 141

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

(a) 20, 48 (b) 36, 60 (c) 39, 91

(d) 60, 75, 90 (e) 54, 90, 126 (f) 96, 240, 400

(g) 360, 600, 504 (h) 132, 220, 330 (i) 1176, 3240, 3816

4. (a) Find the greatest number which exactly divides 36 and 48.

(b) Find the greatest number which exactly divides 120 and 144.

(c) Find the greatest number which divides 60, 84 and 144 exactly.

5. (a) What is the greatest number which will divide 55 and 146 leaving remainder 3
in each case?

(b) Find the greatest number which will divide 125 and 238 leaving remainders 5
and 6 respectively.

(c) Find the greatest number that will divide 640, 710, 1526 so as to leave 11, 7,
9 as remainders respectively.

6. (a) Find the maximum number of students among which 144 apples, 192 bananas
and 216 oranges may be divided equally. Also find how many apples, bananas
and oranges will each get?

(b) Two buckets contain 540 litres and 720 litres of milk. Find the bucket of
greatest capacity that will empty both buckets.

(c) Find the length of the longest stick which can measure 120 m and 180 m
exactly.

Answer (c) 5 (d) 6  2. (a) 6 (b) 3 (c) 12 (d) 12
1. (a) 2 (b) 6 (g) 12 (h) 15 (j) 16 3.(a) 4 (b) 12 (c) 13
(e) 24 (f) 12 (e) 8
(i) 30

(d) 15 (e) 18 (f) 16 (g) 24 (h) 22 (i) 24
4. (a) 12 (b) 24 (c) 12 5. (a) 13 (b) 8 (c) 37 6. (a) 24 6 apples, 8 bananas, 9 oranges

(b) 180 litres (c) 60 m

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

Let us take numbers 4 and 6.

Now, set of the multiples of 4 = {4, 8, 12, 16, 20, 24, ...........}

Set of the multiples of 6 = {6, 12, 18, 24, ................}

Set of common multiples = {12, 24, ....................}

Smallest common multiple = 12

∴ L.C.M. = 12

Lowest common multiple of two or more than two numbers is the smallest number
which is exactly divisible by each of the given numbers.

142 Oasis School Mathematics – 7

Methods of Finding L.C.M.

L.C.M. of two or more numbers can be obtained by the following methods.
(a) Set of multiples method
(b) Prime factor method
(c) Common multiple method

(a) Set of multiples method:

Let's be clear to find the L.C.M. with the help of given example.

For example: Steps

To find the L.C.M. of 12 and 8 • Write the set of multiples
of given number.

The set of multiples of 12 = {12, 24, 36, 48, ...............} • Make the set of common

The set of multiples of 8 = {8, 16, 24, 32, 40, 48, .......} multiples from sets.

The common multiple of 12 and 8 = {24, 48, ............} • Select the smallest
multiple which is
The smallest common multiple is 24.
the L.C.M. of given

∴ L.C.M. = 24. numbers.

(b) Prime factor method

For example:

While finding the L.C.M. of 18, 24, and 30. Steps

Here, 2 24 2 30 • Resolve each of the given numbers
2 12 3 15 into their prime factors.
2 18 26
39 5 • Take out the common prime factors.

3 • Take out the remaining prime factors
which are not common.

3 • Multiply all the factors from step (ii)
and (iii) which is the required L.C.M.
Now,

18 = 2 × 3 × 3

24 = 2 × 2 × 2 × 3 L.C. M. = Common factors from all
30 = 2 × 3 × 5 × Common factors from two numbers
L.C.M. = 2 × 3 × 3 × 2 × 2 × 5 = 360 × Remaining factors from all

Note: L.C.M. can also be obtained by multiplying all the prime factors having
highest power.

Example:
18 = 2 × 3 × 3 = 2 × 32
24 = 2 × 2 × 2 × 3 = 23 × 3

30 = 2 × 3 × 5

Oasis School Mathematics – 7 143

Product of prime factors having highest power is

32 × 23 × 5 = 360.

∴ L.C.M. of 18, 24 and 30 is 360.

Remember !

L.C.M. of two co–prime numbers is their product.


(c) Common division method:

Let's be clear to find the L.C.M. using this method with the help of given example.

Example: Steps
• Arrange the given numbers in array.
Find the L.C.M. of 20, 30, 36
Solution: • Divide them by the least common factors.

Here, • Set down the quotients thus obtained and
undivided numbers side by side in the next
2 20, 30, 36 line.
2 10, 15, 18
3 5, 15, 9 • Proceed the method until we get numbers
5 5, 5, 3 which are prime to one another.
1, 1, 3
• Find the product of all divisors and the
numbers on the last line,

Which is the L.C.M. of given numbers.

∴ L.C.M. = 2 × 2 × 3 × 5 × 1 × 1 × 3 = 180

Note: This method of finding L.C.M. is suitable if the given numbers are
large.

Worked Out Examples

Example: 1

Using set of multiples method, find the L.C.M. of 3, 6, and 9.

Solution:

The set of multiples of 3 = {3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36........}

The set of multiples of 6 = {6, 12, 18, 24, 30, 36, ................}

The set of multiples of 9 = {9, 18, 27, 36, ...................}

The set of common multiples = {18, 36, ..............}

The smallest common multiple = 18

∴ L.C.M. = 18.

144 Oasis School Mathematics – 7