Math 8

(iii) R(–9, –3) Reflection R'(9, –3) (iv) M(4, 7) Reflection M'(–4, 7)
y-axis y-axis

Example: 4

Find the reflection image of ∆ABC with vertices A(3, 3), B(–3, –2) and C(5, 1) on (i) x–axis
(i.e. y = 0) (ii) y–axis (i.e. x = 0). Also, show ∆ABC and ∆A'B'C' on the same graph paper.

Solution: Y

(i) x–axis (.i.e. y = 0)

We have, P(x, y) Reflection P'(x, –y) B' A
x-axis X' C
X
∴ A(3, 3) Reflection A'(3, –3) B C'
x-axis
A'
∴ B(–3, –2) Reflection B'(–3, 2)
x-axis

∴ C(5, 1) Reflection C'(5, –1) Y'
x-axis

Hence, ∆ABC and its image ∆A'B'C' are shown on the graph.

(ii) Y–axis (i.e. x = 0) Y
Y'
We have, P(x, y) Reflection P'(–x, y) A' A
y-axis

∴ A(3, 3) Reflection A'(–3, 3) C' C X
y-axis X' B'

∴ B(–3, –2) Reflection B'(3, –2) B
y-axis

∴ C(5, 1) Reflection C'(–5, 1)
y-axis

Hence, ∆ABC and its image ∆A'B'C' are shown on the graph.

Exercise 6.1

1. Copy the following figures in your copy and find the images of the following

figures under the reflection on the line 'M'.

(a) (b) (c)



(d) (e)


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2. Find the images of the following points under the reflection on x–axis (i.e. on y = 0).

(a) A(5, 3) (b) B(7, –4) (c) C(–2, –3) (d) D(–5, 4)

3. Find the images of the following points under the reflection on y–axis i.e. on x = 0).

(a) N(3, 4) (b) Q(–4, 5) (c) R(–4, –9) (d) S(6, –5)

4. Find the axis of reflection in each of the following cases:

(a) A(4, 6) → A'(4, –6) (b) M(6, 5) → M'(–6, 5)

5. Find the reflection image of ∆ABC with vertices A(3, 2), B(5, 6) and C(8, 1) on

(a) x–axis (i.e. y = 0) (b) y–axis (i.e. x = 0).

Also show ∆ABC and ∆A'B'C' on the same graph paper.

6. Find the reflection image of ∆PQR with vertices P(4, –2), Q(8, –2) and R(8, 2) on

(a) y–axis (i.e. x = 0) (b) x–axis (i.e y = 0)

Also show ∆PQR and ∆P'Q'R' on the same graph paper.

7. Draw a parallelogram on graph paper where A(–2, 3), B(–1, 1), C(4, 1) and D(3,3).
Find the image of the parallelogram and plot on graph paper after reflecting on

(a) x–axis and (b) y–axis.

8. Find the image of the quadrilateral ABCD whose vertices are A(–5, 4), B(–3, 1),

C(–1, 3) and D(–2, 6) under reflection on y–axis. Present the quadrilateral ABCD
and its image on the graph.



Answer
1. Consult your teacher. 2. (a) (5, –3) (b) (7, 4) (c) (–2, 3) (d) (–5, –4)
3. (a) (–3, 4) (b) (4, 5) (c) (4, –9) (d) (–6, –5)
4. (a) X-axis (b) Y-axis 5. (a) A'(3, -2), B' (5, -6) C' (8, -1)
(b) A'(-3, 2), B' (-5, 6) C' (-8, 1) 6. (a) P'(-4, -2), Q' (-8, -2) R' (-8, 2)
(b) P'(4, 2), Q' (8, 2) R' (8, -2) 7. (a) A'(-2, -3), B' (-1, -1) C' (4, -1), D' (3, -3)
(b) A'(2, 3), B' (1, 1) C' (-4, 1), D' (-3, 3)
8. A'(5, 4), B' (3, 1) C' (1, 3), D' (2, 6)

96 Oasis School Mathematics-8

6.2 Rotation–Review

Study the following illustrations and get the idea of rotation of a point. Now, we
take a point P, that rotates about a point O as shown in figures.

In figure (i), ∠POP'= 90° and OP = OP' i.e. the point Figure (i)
P rotates about O through angle 90° in the anti- Figure (ii)
clockwise direction.



In figure (ii), ∠POP' = 90° and OP = OP' i.e. the point P
rotates about O through an angle 90° in the clockwise
direction.

In figure (iii), OP = OP' and ∠POP'= 180° i.e. the point
P rotates about O through a 180° in both direction
producing the same image.

Figure (iii)

In figure, (iv), OP = OP' and ∠POP' = 270° i.e. the
point P rotates about O through an angle 270° in anti-
clockwise direction.

Figure (iv)

In figure (v), OP = OP', ∠POP' = 360°, the point P
rotates about O through an angle of 360° in both
direction producing the same image.

Figure (v)

Thus, a rotation is defined when its centre, the angle of rotation and the direction of
the rotation are given. The direction of rotation can be clockwise or anticlockwise
direction.

Rotation is a rule which shifts each point of an object in the same direction
through a certain angle about a fixed point.

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I. Rotation of a Line PQ, about the Point O Through 90° in Anticlockwise
Direction

Steps:
• Join OP and OQ.
• Taking O as centre and OP as radius,

draw an arc in anticlockwise direction.
• At O, draw ∠P'OP = 90° and OP = OP'

meeting the arc through P at P'.
• Similarly, taking O as centre and OQ

as radius draw an arc in anticlockwise
direction.
• At O, draw ∠Q'OQ = 90° and OQ = OQ',
meeting the arc drawn through Q at Q'.
• Join P'Q'. Hence, P'Q' is the image of PQ.

II. Rotation of a ∆ABC about the Point O Through 60° in Clockwise
Direction

Let ABC be a triangle and O is the point.

Steps:
• Join AO, BO and CO.
• Taking O as centre and OA as radius, draw

an arc in clockwise direction.
• At O, draw ∠A'OA = 60° and OA = OA'

meeting the arc drawn through A at A'.
• Similarly, taking O as centre and OB and OC

as radius, draw arcs in clockwise direction
respectively.
• At O, draw ∠B'OB = 60° and OB = OB' and
∠C'OC = 60° and OC = OC' meeting the
arcs drawn through B and C at B' and C'
respectively.
• Join A'B', B'C' and A'C'.
Hence, ∆A'B'C' is the image of ∆ABC.

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6.3 Properties of Rotation

The following are the properties of rotation.
(i) Area of object and its image are equal.
(ii) The centre of rotation is the invariant point.
(iii) Each point of an object turns through an equal angular displacement in the same

direction.
(iv) The perpendicular bisector of the line segment joining a point of the object and its

corresponding image passes through the centre of rotation.

6.4 Rotation Using Co-ordinates

Just as in reflection, the images of the objects can easily be obtained by the rotation
through a given angle using co-ordinates. Look and learn the following illustrations.

(a) Rotation through 90° (or –270°) (b) Rotation through 90° in clockwise
in anticlockwise direction about direction about origin.
origin.

Anticlockwise

Clockwise

P(3,4) +900 P'(–4, 3) P(3, 4) – 900 P' (4, –3)

Symbolically, Symbolically,
P(x, y) R(O, –900) P' (y, –x)
P(x, y) R(O,+900) P'(–y, x)


(c) Rotation through 180° in anti-clockwise and clockwise direction.

Anticlockwise

Clockwise P(3, 4) R(O,±1800) P'(–3, –4)
Symbolically,
P(x, y) R(O,±1800) P'(–x, –y)

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Remember !

• R(O, + 90°) is equivalent to R (O, – 270°)
• R(O, –90°) is equivalent to R(O, +270°)
• R (O, + 180°) is equivalent to R(O, – 180°)

Worked Out Examples



Example: 1

Find the image of the point A(–4, 6) under the rotation about the origin through the
angle
(i) +90° (ii) –90° (iii) 180° (iv) 270°
Solution:
We have,
(i) P(x, y) R(O,+900) P' (–y, x)
∴ A(–4, 6) R(O,+900) A'(–6,–4)

(ii) P(x, y) R(O,–900) P'(y , –x)
∴ A(–4, 6) R(O,–900) A'(6, 4)

(iii) P(x, y) R(0,1800) P'(–x, –y)
∴ A (–4, 6) R(O, 1800) A'(4, –6)
(iv) P(x, y) R(O,2700) P'(y, –x)
∴ A(–4, 6) R(O,2700) A(6, 4)

Example: 2

A(1, 3), B(6, 4), C(7, 2) are the vertices of a triangle
ABC. Rotate the ∆ABC about origin through 90°
in anticlockwise direction. Present the ∆ABC and
its image in the graph.
Solution:
We have, R(O, + 90°) : P(x, y) → P'(–y, x)
∴ A(1, 3) R(O,+900) A'(–3, 1)
B(6, 4) R(O,+900) B'(–4, 6)
and C(7, 2) R(O,+900) C'(–2, 7)
Hence, ∆ABC and ∆A'B'C' are presented on the graph.

100 Oasis School Mathematics-8

Exercise 6.2

1. Copy the figures given below in your exercise book. Rotate them about O through

the given angle in the given direction. Draw the image formed by rotation.
P A
(a) A (b)
(c)

B Q 3 R B5 C
O• Rotation O • Rotation
O • Rotation C
through +90 through –600 through 1800

2. Draw a ∆ABC in which AB = BC = AC = 4.5 cm. Take any point O outside of
∆ABC and rotate it about O through 60° in anticlockwise direction.

3. (a) Find the image of the following points under the rotation through +900 about origin.

(i) (4, 3) (ii) (2, –3) (iii) (-5, -4) (iv) (-3, 2)

(b) Find the image of the following points under the rotation through –90º about the
origin.

(i) (–2, 4) (ii) (1, 3) (iii) (5, –3) (iv) (-1, -4)

(c) Find the image of the following points under the rotation through 180° about the
origin.

(i) (1, –5) (ii) (–3, –2) (iii) (2, 6) (iv) (–1, 2)

4. (a) Rotate the ∆ABC with verities. A(1, 2), B(4, 5) and C(6, 1) through +90° about
origin. Draw both the figure on the graph paper.

(b) ABC is a triangle with vertices A(–1, 4), B(5, 2) and C(3, 7). Find the images of the
vertices A, B and C of the triangle ABC under the rotation about the origin through

(i) 180° (ii) 270°. Plot both the triangles on the same graph paper.

(c) ABCD is a quadrilateral with vertices A(2, 4), B(–1, 2), C(3, –1) and D(4, 1). Find
the images of the vertices A, B, C and D of the quadrilateral ABCD under the
rotation about the origin through

(i) +90° (ii) –90°. Present the quadrilaterals ABCD and A'B'C'D' on the same
graph paper.

Answer

1. Consult your teacher. 2. Consult your teacher.
3. (a) (i) (–3, 4) (ii) (3, 2) (iii) (4, –5) (iv) (–2, –3)
(b) (i) (4, 2) (ii) (3, –1) (iii) (–3 –5) (iv) (–4, 1)
(c) (i) (–1, 5) (ii) (3, 2) (iii) (–2, –6) (iv) (1, –2)
4. (a) A'(–2, 1), B'(–5, 4) C'(–1, 6) (b) (i) A' (1, -4) B' (-5, -2), C' (–3, –7)
(ii) A' (4,1), B' (2, -5), C'(7, -3) (c) (i) A' (–4, 2), B' (–2, –1), C' (1, 3), D' (–1, 4)
(ii) A' (4, -2), B' (2,1), C' (-1,-3), D' (1,–4)

Oasis School Mathematics-8 110011

6.5 Translation or Displacement

Each point of an object (or a geometrical figure) shifted along the same distance and
same direction is called translation or displacement.

The displacement of an object (or geometrical figure) has magnitude as well as

direction. Hence it is a vector quantity. A A'

Example: Let AB be given line and a→ be translation Translaa→tion vector Image
vector. Now, A and B are shifted to A' and B' such that AA'
= a→ . Then AB is said to be translated to A'B' by translation

vector a→ . B B'

Translation vector: The fixed value by which the geometrical

figure will be displaced in definite direction of translation is

called translation vector.

Properties

(i) Object and image are congruent.

(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 a units parallel to x–axis and b units parallel to a
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

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

Worked Out Examples



Example: 1

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

102 Oasis School Mathematics-8

Example: 2

ABC is a triangle with vertices A(–2, 3), B(–2, 5) and C(2, 3). Translate the ∆ABC by
( (translation vector T =4
3 . Present the ∆ABC and its image on the graph.

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

and C(2, 3) T (34( C'(2 + 4, 3 + 3) = C' (6, 6) X' O X

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

Y'

Exercise 6.3

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. 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 vectors (a) T =4 -3 .
3 (b) T = 5

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

4. ABC is a triangle with vertices A(3, 7), B(2, 2) and C(6, 1). Translate the ∆ABC by the

( ( ( (vector (i) T =2 -3
3 (ii) T = -4 . Present the ∆ABC and its image on the graph.


Answer
1. Consult your teacher.
2. a) (i) A' (6,6) (ii) B' (10,0) (iii) C' (7, 3) (iv) D' (0,–3) b) A' (-1, 8), B' (3, 2), C' (0, 5), D' (-7, -1)

( )3. 2 4. (i) A'(5,10), B' (4,5) C' (8,4) (ii) A' (0,3), B' (-1,-2) C' (3, –3)
3

Oasis School Mathematics-8 110033

Unit Bearing and Scale

7 Drawing

7.1 Bearing

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

I. The compass bearing

II. The three figure bearing

I. The compass bearing

The compass bearing is the old method of locating the direction. NW N NE

In the given figure NOS and EOW represent north-south and east OE
west direction respectively. O is the point of reference for the di- W

rection. We use line ON as the base line. Now, we have simple SW SE

compass card having 8 directions as shown in the figure alongside. S

Where,

NE - North East direction

SE - South East direction

SW - South West direction

NW - North West direction

Look at these two examples:

N N
P

30 E WO E

WO

S 200
The direction of P is N 30° E.
Q
S

The direction of Q is S 20° E.

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

104 Oasis School Mathematics-8

II. The three digits bearing

The direction of any point in terms of an angle expressed in N P
O 600 E
three digits as measured in clockwise direction with reference

to the north line is called the three digit bearing. W

Here, the bearing of P from N is 060°. 300

The bearing of Q from N is 150° SQ

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

Worked Out Examples



Example: 1

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

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

Bearing of NE is N 45°E. WO E

Bearing of SW is S 45° W. SW SE

S

Example: 2

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

(a) N (b) N (c) N
C
A

550 1400 250
O O O

Solution: B

Here,

(a) The bearing of A from O is 055°.

(b) The bearing of B from O is 140°.

(c) The bearing of C from O is (360° - 25°)

= 335°.

Oasis School Mathematics-8 110055

Example: 3 N

Write in three digits bearing.

(a) N60° E N (b) S 30° W

Solution: AO
From the figure, 600
N 60° E = 060° O S30°W = (180°+ 30°)

Example: 4 = 210° 300
WS
N'

If the bearing of B from A is 60°, what is the bearing of A from B? N

Solution: 600 B

Here, A

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°

or, ∠N'BA = 180° – 60°

= 120°

∴ The bearing of A from B = reflex ∠N'BA

= (360° – 120°)

= 240°

Exercise 7.1

1. Write down the compass bearing of the following figure:

(a) N (b) A (c) N (d) (e) N
O
O C 600 N NW S

O 450 O WO EW E
SW
300 A B
S S
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

106 Oasis School Mathematics-8

3. From the given figure, write down the three digit bearing of NE, NW N

NE

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

4. Write the three digit bearing of given points from O. SW SE
S

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

800 A 1600 O 400 O 1500
E
O O DO

B (d) 305°

5. Sketch the following bearing.

(a) 075° (b) 145° (c) 215°

6. Write the three digit bearing of: (d) N30°W
(a) N60°E (b) S70°W (c) S50°E

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

(a) N (b) N' (c) N N' N
NN N' (d) A 3100
1150
400 B B
A
BA B A

2300

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

Kathmandu

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


Oasis School Mathematics-8 110077

Answer
1. (a) S300E (b) S450W (c) N600W (d) S450W (e) N450W
2. Consult your teacher.
3. 0450, 0900, 1350, 1800, 2250, 2700, 3150
4. (a) 0800, (b) 2000, (c) 3200 (d) 2700 (e) 1500
5. Consult your teacher. 6. (a) 0600, (b) 2500 (c) 1300, (d) 3300
7. (a) 2950 (b) 2200 (c) 500 (d) 1300 8. Consult your teacher



7.2 Scale Drawing

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 this 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 second figure is twice that of

first figure.
A'

A

6 cm

3 cm
3 cm

6 cm

A 3 cm C

Figure (i) B' 6 cm C'

Figure (ii)

Here each side of first figure is enlarged twice.

∴ AB = 3 cm = 1 or, BC = 3 cm = 1 or, AC = 3 cm = 1
A'B' 6 cm 2 B'C' 6 cm 2 A'C' 6 cm 2

Thus, first figure is the scale drawing of 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. LenAgctthuainl ldenragwthing = 1 = scale factor.
100

108 Oasis School Mathematics-8

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 1:60,00,000?

Solution:

Here, Scale = 1: 6000000

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

= 6000000 m
1000

= 60000 m.
100

= 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 is 100 m and 80 m respectively, which
is drawn on the scale 1:800. Find the length and breadth of the field on the drawing.

Solution: Alternative method

Here, 800 cm is represented by 1 cm We have, scale = length in map
actual length
i.e. 180000 m is represented by 1 cm
or, 8 m is represented by 1 cm 1:800 = length in map/100 m

length of the field = 100 m or, length in map = 100 m
800

8 m is represented by 1 cm = 1 m
8

1 m is represented by 1 cm = 1 × 100 cm
8 8

80 m is represented by 1 × 80 cm = 12.5 cm.
8
Again, Scale = Length in map
= 10 cm Actual length

100 m is represented by 1 × 100 cm 1:800 = Length in map
8
80 m

= 12.5 cm. ∴ Length in map = 80 m
800

= 1 m
10

= 1 ×100 cm
10

= 10 cm.

Oasis School Mathematics-8 110099

Example: 3

If an aeroplane flies 600 km in the bearing of 030° and 800 km in the bearing of 120°,
find the distance between the two places and the bearing of the starting place from the
last place using the scale 1 cm = 100 km.

Solution: N1 N2
B
In the given figure, A is the starting place, last place is 1200

B, AB = 10 cm. (By measurement) therefore, distance

between the places (AB)=10 cm = 1000 km. N 8 cm

[∵ 1 cm = 100 km] From the given figure ∠ABN2 = 113° (By 6 cm

measurement). 300

Hence, bearing of the starting place from the last A
place = 360° – 113° = 247°.

Exercise 7.2

1. Given figure shows the position of students in a class. If the scale is 1:150. Find
the actual distance between:

(a) Ramesh and Shyam. Goma

(b) Goma and Bidhya. Ramesh
(c) Ramesh and Goma.
(d) Shyam and Bidhya. Shyam Bidhya

2. If the scale in a map is 1:1500. Find the actual distance between the given places

if the distance in the map is:

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

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

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

(b) What is the actual distance between two places which is represented by 25 mm.
on a map which is drawn to the scale of 1 mm = 2 km?

110 Oasis School Mathematics-8

5. The figure alongside shows the outline sketch of a 3 cm 4 cm 3 cm 3 cm
house plan. Draw a scale of 1 cm to represent 1 m.
Kitchen 1cm Common
(a) Find the actual length and breadth of bedroom, Toilet Room
kitchen and toilet. 3 cm 3 cm
Bedroom Garage
(b) Find the area of the garage.
4 cm 3 cm

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


(a) Kathmandu and Dhangadi
(b) Pokhara and Jumla
(c) Gorkha and Janakpur

7. The figure or map of Nepal is drawn on the scale of 1 cm is equal to 40 km showing
the major trade centers. Find out the actual distance between:

(a) Dhangadhi and Nepalgunj.
(b) Bhaktapur and Birgunj.
(c) Jiri and Biratnagar.
(d) Birgunj and Biratnagar.

8. If an aeroplane flies 400 km in the bearing of 030° and then 300 km in the bearing
of 120°, find the distance between the two places and the bearings of the starting
place from the last place using the scale 1 cm = 100 km. N

9. A motor is at S, a place 200 km east to T. Travelling a

distance of 300 km to the due west of S, it reached at U.

Express this information by the drawing and find T

(a) the distance from T to U. U
(b) the bearing of T from U. 2cm S

3cm

Using scale : 1cm = 100km

Oasis School Mathematics-8 111111

Answer 

1. Consult your teacher.

2. (a) 60m (b) 120m (c) 90m (d) 225m (e) 435m

3. (a) 16cm (b) 12cm (c) 30cm (d) 60cm (e) 80cm

4. (a) 30m (b) 50km 5. (a) Bedroom 4m, 3m Kitchen - 4m, 2m, Toilet-4m, 1m.

(b) 9m2 6. Consult your teacher 7. (a) 152km (b) 88km (c) 152 km.

(d) 72km 8. 500km and 2470 9. (a) 100km (b) 0900

Assessment Test Paper

Full marks : 12

Attempt all the questions:

Group A [4 × 1 = 4] N

1. (a) Draw a compass bearing of 0450. W 45° E
(b) In the given figure, write the compass bearing of SE N1 SE
S
B
N 120°

(c) Find the bearing of B from A in the given figure.

A

(d) Find the actual height of a tower whose height in a map is 3.5cm and represented

by scale 1:800.

Group B [4×2=8]
2. Draw a triangle ABC with the vertices A (3,1), B (5,2) and C (2, 5) on graph paper.

Reflect ∆ABC on Y-axis. Also draw ∆A'B'C' on the same graph paper.

3. Plot ∆PQR with the vertices P (-4, -1), Q(2,1) and R (-1, 5) on a graph paper then plot

the image ∆P'Q'R' after rotating through +900 about origin.

112 Oasis School Mathematics-8

Sets

5Estimated Teaching Hours

Contents

• Operations on Sets
- Union of Sets
- Intersection of Sets
- Differences of Sets
- Complement of Sets
• Cardinality of Sets

Expected Learning Outcomes

At the end of this unit, students will be able to develop the following competencies:
• Find the union and intersection of two sets and show these relations

in the Venn diagram.
• Find the difference of two sets and complement of a set and show

these relations in Venn diagram.
• Find the cardinal number of sets and use the cardinality relations to

solve problems related to sets.

Teaching Materials

• Tracing paper, flash card, chart paper, etc.

Oasis School Mathematics-8 111133

Unit

8 Sets

8.1 Review

A collection of well-defined objects is called a set. The term well-defined means that
we must be able to tell whether a particular object belongs to the set or not. There
must be any repeated objects in a set.

(i) The set of boys of class VIII in a school

(ii) The set of vowels in the English alphabet

The objects of a set are called its members or elements.

Set notation: The sets are usually denoted by capital letters A, B, C etc. Elements are
enclosed inside the braces { } and separated with comma (,).

Example: A = {2, 3, 5, 7, 11}

2 is an element of A, then 2∈A. 4 is not an element of A, then 4 ∉A.

Note: Write
2 ∈ A.
Say 4 ∉ A.
2 belongs to A
4 does not belong to A

Specification of Sets

The common method to represent the set are-

(i) Listing method (ii) Description method (iii) Set builder method

Let us represent the set of prime numbers less than 15 by three different methods:

Method Way to specify Example
Listing method
All the elements are listed A = { a, e, i, o, u}
Description method within curly braces separating
each element by comma.
Set builder method
The common property of ele- A set of vowel letters.
ments are described by words
with suitable sentence.

A rule or a statement or a for- A = {x : x is a vowel letter of
mula is written to represent English alphabet}
the sets.

114 Oasis School Mathematics-8

Types of Sets

Null set: A set having no element is called null set or an empty set or void set. It is
denoted by φ (phi) or { }.

Example: A set of triangle having 4 sides.

Singleton Set: A set having only one element as its member is called singleton set or
unit set.

Example: {1}, {0}, etc. are singleton sets.

Finite set: A set having finite number of elements is called a finite set.
Example: A = {x : x is a prime number less than 20} is a finite set.

Infinite set: A set having an infinite number of elements is called an infinite set.
Example: A = {1, 2, 3, 4, ……} is an infinite set.

Remember !

• A set of male students in Padmakanya college is a null set.
• A set of the highest peak of the world is a singleton set.
• A set of even numbers from 10 to 30 is a finite set.
• A set of natural numbers is an infinite set.
• φ is a null set and {φ} is a unit set.

Set Relations

The set relations between two sets can be understood with the help of the given
table.

Type Nature Example
Overlapping sets
Disjoint sets Sets having at least one A = {1, 2, 3, 4} and B = {3, 4,
Equal sets
common element 5, 6}
Equivalent sets
Sets having no common A = {p, q, r, s} and B = {x, y, z}
element

Sets containing same elements A = {a, b, c, d}
B = {a, b, c, d}
then A = B.

Sets containing equal number A = {1, 2, 3, 4}

of elements B = {w, x, y, z}

Subsets:

In this case, A may be equal to B. Let A = {1, 2, 3} and B = { 1, 2, 3, 4, 5}. Here every
element of set A is also the element of set B. A is subset of B.

Hence, set A is said to be a subset of set B, if every element of set A is also an
element of set B. It is denoted by A ⊂ B which is read as 'A is a subset of B or A is
contained in B'.

Oasis School Mathematics-8 111155

Note:
• If A = {a, b} and B = {a, b, c}, then A is the proper subset of B., i.e. A ⊂ B.
• Every set is a subset of itself. i.e. A ⊆ A, B ⊆ B.
• Null set is a subset of every set i.e. φ ⊂ A, φ ⊂ B.

Number of subsets of a given set

Let us observe the following examples,

• Consider the null set { } or φ. Since every set is a subset of itself, the subset of the null
set = { }.i.e. number of subsets of a null set = 1 = 2°.

• Consider the set A = {1}. Since the null set is a subset of every set and a set is also a
subset of itself, its subsets are { } and {1}. ∴ Number of subsets of this sets = 2 = 2¹

• Consider the set A = {1, 2} of two elements. Its subsets are { }, {1}, {2}, and {1, 2}.
So, the number of subsets = 4 = 2² .

• Consider the set A = {1, 2, 3}. Its subsets are { }, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}
and {1,2,3}.

∴ Number of subsets of this set = 8 = 2³.

The above information obtained can be written in the tabular as below.
In general, the number of subsets of set having 'n' elements = 2n.

Set No. of elements No. of subsets
{ } 0 1 = 20

{1} 1 2 = 21

{ 1, 2 } 2 4 = 22

{ 1, 2, 3 } 3 8 = 23

Exercise 8.1

1. Write the following sets in listing method
(a) A = a set of prime numbers less than 10.
(b) B = a set of first three days of the week.
(c) C = {x : x is an even number less than 10}.
2. Write the following sets in the set-builder form
(a) A = {2, 3, 5, 7}
(b) B = {1, 3, 5, 7, 9}
(c) W = {a set of whole numbers less than 5.}

116 Oasis School Mathematics-8

3. Write the following sets in description method

(a) A = {1, 3, 5, 7, 9, 11, 13}

(b) B = {x : x is a natural number less than 5}.

(c) C = { 5, 10, 15, 20}.

4. Determine whether the following sets are disjoint or overlapping.

(a) A = {1, 2, 3}, B = {a, b, c}

(b) C = {4, 5, 6, 7}, D = {6, 7, 8, 9}

(c) E = {4, 5, 6, 7, 8}, F = {7, 8, 9, 10, 11}

(d) P = {x, y}, Q = {a, b, c, d}

(e) F(8) = the set of factors of 8, F(7) = the set of factors of 7.

5. Identify whether the following sets are null set or unit set.

(a) { } (b) A = {men over 200 years old}

(c) D = {0} (d) E = the set of men with height more than 15 feet

(e) {φ}

6. Identify whether the following sets are finite or infinite.

(a) A set of prime numbers less than 20

(b) A set of natural numbers

(c) A set of odd numbers from 1 to 100

7. Which of the following pairs of sets are equal?

(a) A = {1, 2, 3, 4}, B = {4, 3, 2, 1}

(b) C = {4, 5, 6, 7}, D = {a, b, c, d}

(c) A = a set of prime numbers less than 8, B = {2, 3, 5, 7}

(d) P = a set of odd numbers less than 8, R = { 1, 3, 5, 7}

8. Which of the following pairs of sets are equivalent?

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

(b) C = {Umesh, Dolendra,Prabhat}, D={Manita, Alina, Gita}

(c) A = {1,2,3,4,5}, B = {4,5,6,7}

9. Let A = {0, 1, 2, 3, 4, 5}. State whether the following statements are true or false.

(a) {2} ⊂ A......... (b) {o} ⊆ A........ (c) φ ⊆A..... (d) {1, 3} ⊆ A...

(e) A ⊆ A.... (f) 2 ⊆ A....

10. State whether the following statements are true or false.

(a) No set is a proper subset of itself.

(b) The universal set is the subset of every set.

(c) Null set is subset of every set.



Oasis School Mathematics-8 111177

11. If A = {0, 1, 2}, which of the following are proper subsets of A?

(a) {0, 1} (b) {2, 1, 0} (c) {2} (d) {a, b, c} (e) {4, 5}

12. Write down all possible subsets of the following sets.

(a) {a} (b) {a, b} (c) {2, 4, 6}

13. Find the number of subsets of the following sets.

(a) { p } (b) {x, y, z} (c) { 1, 2}



Answer

1. (a) A = {2, 3, 5, 7} (b) B = {Sunday, Monday, Tuesday} (c) C= {2, 4, 6, 8}

2. (a) A = {x: x is a prime number less than 10} (b) B = {x : x is an odd number less then 10}

(c) W = {x : x is a whole numbers less then 5}

3. (a) A = a set of odd number less then 15 (b) B = a set of natural numbers less then 5

(c) C = a set of first 4 multiples of 5

4. (a) disjoint set (b) overlapping sets (c) overlapping sets (d) disjoint sets (e) overlapping sets

5. (a) null set (b) null set (c) unit set (d) null set (e) unit set

6. (a) finite set (b) infinite set (c) finite set

7. (a) equal sets (b) unequal sets (c) equal sets (d) equal sets

8. (a) equivalent sets (b) equivalent sets (c) not equivalent sets 9. (a) true (b) false

(c) true (d) true (e) true (f) false 10. (a) true (b) false (c) true

11. a and c are the proper subsets.

12. (a) φ, {a} (b) φ, {a}, {b}, {a, b} (c) φ, {2}, {4}, {6}, {2,4}, {2, 6}, {4, 6}, {2,4,6}

13. (a) 2 (b) 8 (c) 4

8.2 Operations on Sets

Four operations on sets are -

(i) Union of sets (ii) Intersection of sets

(iii) Difference of sets (iv) Complement of a set

Union of sets

The union of two sets A and B is a new set of whose elements are either in A or in B
or in both of them. It is denoted by A ∪ B which is read as 'A union B' or 'A cup B'
Symbolically, it is defined as A ∪ B = { x : x belongs to at least one of the sets A or B}
= {x : x ∈ A or x ∈ B}

To form A ∪ B from two given sets A and B, first take the elements of A and then
take elements of B which are not in A. For example,

118 Oasis School Mathematics-8

(i) If A = {a, b, c} and B = {b, c, d, e}, then AB U
A ∪ B = {a, b, c} ∪ {b, c, d, e} = {a, b, c, d, e}.
Here, b and c are common elements of A and B. Its Venn- a b d
c e
diagram is given alongside.
A∪B is shaded
(ii) If A = {a, b, c} and B = {d, e}, then
A ∪ B = {a,b,c} ∪ {d, e} A U
= {a, b, c, d, e} B
Here, A and B have no common element. Its Venn-diagram is a d
bc e
given alongside.
A∪B is shaded

(iii) If A = {1, 4, 5}, B = {1, 4, 5, 8, 10} and U

U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, then U B 7
A ∪ B = {1, 4, 5} ∪ {1, 4, 5, 8, 10} = {1, 4, 5, 8, 10} A
9 81

6 45

Here, A is subset of B. Thus, we have A ∪ B = B. Its 3 10 2
A∪B is shaded
Venn-diagram is given alongside.

Intersection of sets

The intersection of two sets A and B is the set of all elements common to both A
and B. The intersection of two sets A and B is denoted by A ∩ B which is read as
"A intersection B" or "A cap B".

Thus, A ∩ B = {x : x ∈ A and x ∈ B}

In order to form A ∩ B from the two given sets A and B, we take all those elements
which are common to A and B. If there is no common element between A and B,
then A ∩ B is the null set and A and B are called disjoint sets.

For example,

(i) If A = {a, b, c, d, e, f} and

B = {b, d, f, h, j}, then

A ∩ B = {a, b, c, d, e, f} ∩ {b, d, f, h, j}

= {b, d, f}

Here, b, d, f are common elements to both A and B.

U
AB

ab h
cd j

ef

A∩B is shaded

Oasis School Mathematics-8 111199

(ii) If A = {1, 3, 5, 7, 9} and

B = {2, 4, 6, 8} then

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

= φ

There is no common element between A and B. So A and B are disjoint sets.

A U
B
1
53 24
79 68



Difference of Sets

The difference of two sets A and B is the set of all elements of A only, which are
not in B. The difference of two sets A and B is denoted by A – B which is read as "A
difference B" or "A minus B". Symbolically, it is defined as,

A – B = {x : x ∈ A and x ∉ B} = elements of A, which are not in B

Similarly, B – A = {x : x ∈ B and x ∉ A} = elements of B, which are not in A

In order to form A – B, we take all those elements of A which are not in B. This means
that A – B is the new set which is formed by taking elements of A only.

For examples, U U
(i) If A = {b, c, d, e} and B ={a, b, c}, then AB AB
A–B = {b, c, d, e}–{a, b, c}={d, e}and d ba d ba
B – A = {a, b, c} – {b, c, d, e} = {a}. ec ec

Their Venn-diagrams are given alongside. A–B is shaded B–A is shaded

Complement of a Set

Let U = {x:x is an odd number less than 15} and A = {1, 3, 5, 7, 9},
then U = {1,3,5,7,9,11,13}.

Here, two elements 11 and 13 from U do not belong to set A.

∴ {11, 13} is the complement of set A.

The complement of a set A with respect to universal set U is the set of all those
elements of U only, which are not in A. The complement of a set A is denoted by
A' or A or Ac. Symbolically, it is defined as U – A = {x : x ∈ U and x ∉ A}. Its
Venn-diagram is given below. The shaded portion represents the complement of a
set A.



120 Oasis School Mathematics-8

For example, U
(i) If U = {1, 2, 3, 4, 5, 6, 7} and A = {1, 2, 3, 4}, then UA
= U – A = {1, 2, 3, 4, 5, 6, 7} – {1, 2, 3, 4}
= {5, 6, 7}. 6 17
Its Venn-diagram is given alongside. 23
Also, (A) = U – A = {1, 2, 3, 4, 5, 6, 7} – {5, 6, 7}
= {1, 2, 3, 4} = A A 45
∴ (A) = A.
This shows the complement of a complement set is given set. A is shaded
Its Venn-diagram is given alongside.
U

A

6 17

23

A 4
5

(A) is shaded

(ii) If U = {1, 2, 3, 4, 5, 6, 7}, A = {1, 2, 3} and B = {3, 4, 5}, then
A ∪ B = { 1, 2, 3} ∪ {3, 4, 5} = {1, 2, 3, 4, 5}.

Thus, A ∪ B is the set of all the elements which belong to either A or B or
both A and B.

∴ (A ∪ B) = U – (A ∪ B)

= {1, 2, 3, 4, 5, 6, 7} – {1, 2, 3, 4, 5}

= {6, 7}. A U
B

1 34 6
25

7

A∪B is shaded

Thus, it is the set of all those elements which belong to neither A nor B.

Worked Out Examples



Example: 1

If U = {x : x is a natural number less than 13}, A = {x : x is a prime number less than 13},
B = {x : x is an even number less than 13} and C = {x : x is a natural number less
than 6}, find; (i) A ∪ B (ii) A ∪ C. Show it in Venn-diagrams.

Solution:

Here, U = {x : x is a natural number less than 13}

= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

Oasis School Mathematics-8 112211

A = {x : x is a prime number less than 13}={2, 3, 5, 7, 11}
B = {x : x is an even number less than 13}={2, 4, 6, 8, 10, 12}
and C = {x : x is natural number less than 6} ={ 1, 2, 3, 4, 5}

(i) A ∪ B = {2, 3, 5, 7, 11} ∪ {2, 4, 6, 8, 10, 12} U
AB

= {2, 3, 4, 5, 6, 7, 8, 10, 11, 12} 11 2 12 6
7 53 8 10
Its Venn-diagram is given alongside.
4
19

(ii) A ∪ C = {2, 3, 5, 7, 11} ∪{1, 2, 3, 4, 5} A∪B is shaded

= {1, 2, 3, 4, 5, 7, 11}. U
AC

Its Venn-diagram is given alongside. 12 11 3 1
2
7 5 4

68 9

Example: 2 A∪C is shaded

If A = {factors of 24} and B = {factors of 36}, find U
(i) A ∪ B (ii) A ∩ B. Show it also in Venn-diagram. AB

Solution: A = factors of 24 = {1, 2, 3, 4, 6, 8, 12, 24} 8 213 9
Here, 24 4126 18

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

(i) A ∪ B = {1, 2, 3, 4, 6, 8, 12, 24} ∪{1, 2, 3, 4, 6, 9, 12, 18, 36} A∪B is shaded

= { 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36}.

Its Venn-diagram is given alongside. U
AB

(ii) A ∩ B = {1, 2, 3, 4, 6, 8, 12, 24} ∩ {1, 2, 3, 4, 6, 9, 12, 18, 36} 8 13
23
46 18

= {1, 2, 3, 4, 6,12}. 24 12 36

Its Venn-diagram is given alongside. A∩B is shaded

Example: 3

If U = set of natural numbers less than 10, A = {1, 2, 3, 4, 5}, B = set of odd numbers less
than 10, find- (i) A – B (ii) B–A (iii) A ∩ B. Also show them in the Venn diagram.

Solution: U = {1, 2, 3, 4, 5, 6, 7, 8, 9}


A = {1, 2, 3, 4, 5} A BU

B = {1, 3, 5, 7, 9} 2 17
3
(i) A–B = {1, 2, 3, 4, 5} – {1, 3, 5, 7, 9} 45 9
68
= {2, 4}
A–B is shaded


122 Oasis School Mathematics-8

Illustration in Venn diagram.

(ii) B–A = {1, 3, 5, 7, 9} – {1, 2, 3, 4, 5} A BU

= {7, 9} 2 17
3
Illustration in the Venn diagram. 45 9
86
(iii) A ∩ B = {1, 2, 3, 4, 5} ∩ {1, 3, 5, 7, 9}
B–A is shaded

= {1, 3, 5} A BU

A ∩ B = U – (A∩B)

= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 3, 5} 2 135135 7
4 9

= {2, 4, 6, 7, 8, 9} 86

Illustration in the Venn diagram. A∩B is shaded

Example: 4 U
AB
From the given Venn diagram, find

(i) A (ii) B (iii) A ∩ B (iv) A ∪ B (v) A–B pq
r
(vi) B–A (vii) A ∪ B a 153ou t
e
i
S olution : s

Here, from the Venn diagram,

(i) A = {a, e, i, o, u}

(ii) B = {p, q, r, o, u}

(iii) A ∩ B = {o, u}

(iv) A ∪ B = {a, e, i, o, u, p, q, r}

(v) A–B = {a, e, i}

(vi) B–A = {p, q, r}

(vii) A ∪ B = {s, t}.

Example: 5

Shade the following sets separately in the adjoining Venn-diagrams.

(i) A – B (ii) B – C (iii) A – (B ∩ C) (iv) A (v) A ∪ B

Solution: (i) A – B U (ii) B – C U (iii) A–(B∩C)
B B
A A U
AB

C C C
A–B is shaded B–C is shaded
A–(B∩C) is shaded

Oasis School Mathematics-8 112233

(iv) A U (v) A ∪ B U
B
A AB

C C
A∪B is shaded
A is shaded

Example: 6

If A = {1, 2, 3, 4, 5}, B = {4, 5, 6, 7, 8}, C = {4, 6, 9, 10}, verify that

(i) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

(ii) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
Solution:

(i) B ∩ C = {4, 5, 6, 7, 8} ∩ {4, 6, 9, 10} = {4, 6}

∴ A ∪ (B ∩ C) = {1, 2, 3, 4, 5} ∪ {4, 6} = {1, 2, 3, 4, 5, 6} .................. (i)
Again, A ∪ B = {1, 2, 3, 4, 5} ∪ {4, 5, 6, 7, 8} = {1, 2, 3, 4, 5, 6, 7, 8}

and A ∪ C = {1, 2, 3, 4, 5} ∪ {4, 6, 9 10} = {1, 2, 3, 4, 5, 6, 9, 10}

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

From (i) and (ii), we get,

A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C). Hence, verified.
(ii) B ∪ C = {4, 5, 6, 7, 8} ∪ {4, 6, 9, 10} = {4, 5, 6, 7, 8, 9, 10}

∴ A ∩ (B ∪ C) = {1, 2, 3, 4, 5} ∩ {4, 5, 6, 7, 8, 9, 10} = {4, 5}............ (iii)

Again, A ∩ B = {1, 2, 3, 4, 5} ∩ {4, 5, 6, 7, 8} = {4, 5}
And A ∩ C = {1, 2, 3, 4, 5} ∩ {4, 6, 9, 10} = {4}

∴ (A ∩ B) ∪ (A ∩ C) = {4, 5} ∪ {4} = {4, 5} ...................... (iv)

From (iii) and (iv), we get, A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C). Hence, verified.

124 Oasis School Mathematics-8

Exercise 8.2

1. What does the shaded portion represent in each of the given venn-diagrams?
(a) (b) (c)
U U
A BU AB AB

(d) (e) (f) U
AB
U U
AB AB

(g) U (h) U (i) U
AB
AB AB

If U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 3, 5, 7} and B = {2, 4, 6, 8}, find
2. (a)
(i) A ∪ B (ii) A ∩ B.

(b) (iii) A ∪ U (iv) B ∩ U.
(c)
If A = {a, e, i, o, u} and B = {r, a, e, s}, find (i) A ∩ B. (ii) A ∪ B.

If P = {1, 2, 3, 4, 5}, Q = {1, 2, 4, 6, 7}, find (i) P – Q. (ii) Q – P.

(d) If A = {factors of 8}, and B = {factors of 12}, find (i) A ∪ B (ii) A ∩ B.

(e) If U = {1, 2, 3, 4, 5, ………….., 10}, A = {factors of 2}, B = {factors of 3}, find

(i) A ∪ B (ii) A ∩ B (iii) A ∪ B (iv) A ∩ B . U
AB
3. From the given Venn diagram, find the
following sets (i) A (ii) B (iii) A ∪ B 37 2 4
(iv) A ∩ B (v) A–B (vi) B–A (vii) A (viii) B 5 6

8
91

4. (a) If U = {x : x is a positive integer less than 20}; A = {x : x is a prime number less
than 20} and B = {x : x is divisible by 3, x ≤ 15}, find

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

(v) A ∪ B (vi) A – B (vii) B – A.

Show all these relations in the Venn diagram.

Oasis School Mathematics-8 112255

(b) If U = {x : x is a positive integer less than 10}, A = {x : x is a prime number less
than 10} and B = {x : x is an odd number less than 10}, find

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

Also draw Venn-diagram to represent the above sets.

5. Show the following set relations in the Venn diagram.

(a) A–B (b) B–A (c) A (d) B (e) A – B

(f) A ∪ B (g) A ∩ B (h) A–(A ∩ B) (i) (A ∪ B) – A

6. What does the shaded part of each figure represent?
(a) (b) (c)

(d) (e) (f)



7. Draw Venn-diagram separately like the adjoining figure and shade the portion of

the following sets. (b) A ∩ B (c) A ∩ C U
(a) A ∪ B AB

(d) A ∪ (B ∪ C) (e) (A∩B)∩C (f) U ∩ C

C

8. From the adjoining Venn-diagram, write down the elements of the following sets.

(a) U (b) A (c) C (d) A ∪ B A BU

(e) B ∩ C (f) A ∪ (B ∪ C) (g) (A ∩ B) ∩ C 3 46
1
9 528 11

7 10 12 C

9. If A = {2, 3, 5, 7, 11}, B = {3, 6, 9, 11, 15, 17} and C = {5, 6, 7, 11}, then verify
that-
(a) A ∪ (B∩C) = (A ∪ B) ∩ (A ∪ C)
(b) A∩(B ∪ C) = (A∩B) ∪ (A∩C)



126 Oasis School Mathematics-8

Answer
1. (a) A∪B (b) A∩B (c) A – B (d) B – A (e) A ∪ B (f) A ∩ B (g) A – B (h) A (i) B – A
2. (a) (i) {2, 3, 4, 5, 6, 7, 8} (ii) {2} (iii) {1, 2, 3, ..... 9} (iv) {2, 4, 6, 8}
(b) (i) {a, e} (ii) {a, e, i, o, u, r, s} (c) (i) {3, 5} (ii) {6, 7}
(d) (i) {1, 2, 3, 4, 6, 8, 12} (ii) {1, 2, 4} (e) (i) {1, 2, 3} (ii) {1} (iii) {4,5,6,7,8,9,10}

(iv) {2, 3, 4, 5, 6, 7, 8, 9, 10} 3. (i) {2, 3, 5, 7} (ii) {2, 4, 6, 8} (iii) {2, 3, 4, 5, 6, 7, 8}
(iv) {2} (v) {3, 5, 7} (vi) {4, 6, 8} (vii) {1, 4, 6, 8, 9} (viii) {1, 3, 5, 7, 9}
4. (a) (i) {2, 3, 5, 6, 7, 9, 11, 12, 13, 15, 17, 19}
(ii) {1, 4, 8, 10, 14, 16, 18} (iii) {3} (iv) {1, 2, 4, 5, 6, 7, 8, 9, 10 .... 19}
(v) {1, 2, 3, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 18, 19}
(vi) (2, 5, 7, 11, 13, 17, 19} (vii) {6, 9, 12, 15}
(b) (i) {1, 4, 6, 8, 9} (ii) {2, 3, 5, 7} (iii) {2, 4, 6, 8} (iv) {1, 3, 5, 7, 9}
(v) {4, 6, 8} (vi) {1, 2, 4, 6, 8, 9}

5.

(g) A U (h) A U (i) A U
B B B

A∩B A–(A ∩ B) (A ∪ B) – A

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

7. (a) U (b) U (c) U
B A B B
A A

C C C
A∩C
A ∪ B A ∩ B

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(d) A U (e) A U U
B B (f) A B

CC C

A ∪ B ∪ C (A ∩ B) ∩ C U∩C

8. (a) {1, 2, 3, ...12} (b) {1, 2, 3, 4, 5} (c) {1, 2, 5, 8, 10, 12} (d) {1, 2, 3, 4, 5, 6, 8}

(e) {1, 2, 8} (f) {1, 2, 3, 4, 5, 6, 8, 10, 12} (g) {1, 2}

8.3 Cardinality of Two Sets

Let A = {a, b, c, d}, B = { 1, 2 }, C = { 0 } and φ are some sets. There are four elements
in set A, two elements in set B:

In short, we write n(A) = 4, n(B) = 2, n(C) = 1 and n(φ) = 0, which are called cardinal
numbers of sets.

Hence, the number of elements present in the set is called cardinal number of sets.

Let 'A' and 'B' be any two overlapping sets, such that n(A) = a, n(B) = b and
n(A∩B) = x.

Then, from the Venn–diagram, U
AB
n(A∪B) = a–x+x+b–x

n(A∪B) = a+b–x (a-x) x (b-x)
n(A∪B) = n(A) + n(B) –n(A∩B)

If two sets are disjoint, then n(A∩B) = 0

n(A∪B) = n(A) + n(B)

Some important results on cardinality of two sets

(i) n(A∪B) = n(A) + n(B) – n(A∩B)

(ii) n(A∪B) = n(A) + n(B) if A and B are disjoint sets.

(iii) n0(A) = n(A) – n(A∩B) U
(iv) n0(B) AB
(v) n(A∪B) = n(B) – n(A∩B) n(A∪B)
= n0(A) + n0(B) + n(A∩B) n(A) n(B)
(vi) n(A∪B) = n(U) – n(A∪B) n0(A) n0(B)

(vii) n0(A) = n(A∪B) – n(B) n(A∩B)

(viii) n0(B) = n(A∪B) – n(A)

Note: n0(A) = n(A–B) and n0(B) = n(B – A)

128 Oasis School Mathematics-8

Worked Out Examples

Example: 1

If A = {1, 2, 3, 4, 5} and B = {4, 5, 6, 7, 8}, find

(i) n(A) (ii) n(B) (iii) n(A∩B)
(v) n(A–B) (vi) n(B–A)
(iv) n(A∪B)

Solution:

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

(i) n(A) = number of elements in set A = 5

(ii) n(B) = number of elements in set B = 5

(iii) A ∩ B = {1, 2, 3, 4, 5} ∩ {4, 5, 6, 7, 8} = {4, 5}

∴ n(A ∩ B) = number of elements common to both sets A and B = 2

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

∴ n(A ∪ B) = number of elements in set A ∪ B = 8

(v) A – B = {1, 2, 3, 4, 5} – {4, 5, 6, 7, 8} = {1, 2, 3}

∴ n(A – B) = number of elements in set A only = 3

(vi) B – A = {4, 5, 6, 7, 8} – {1, 2, 3, 4, 5} = {6, 7, 8}

∴ n(B – A) = number of elements in set B only = 3.

Example: 2

In the given Venn diagram "A" and "B" are the subsets of universal set U. Then find

(i) n(A ∩ B) (ii) n(A) (iii) n(B) (iv) n(A–B) U
(v) n(B–A) (vi) n(A ∪ B) (vii) n(A ∪ B) AB

Solution: 12 5 9
From the Venn diagram 6

(i) n(A ∩ B) = 5 (ii) n(A) = 12 + 5 = 17

(iii) n(B) = 5 + 9 = 14 (iv) n(A–B) = 12

(v) n(B – A) = 9 (vi) n(A ∪ B) = 12 + 5 + 9 = 26

(vii) n(A ∪ B) = 6

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Example: 3

In the given Venn diagram, if n(U) = 200, find
(i) n(A ∩ B) (ii) no(A) (iii) no(B) (iv) n(A ∪ B).

U
AB
Solution:
120-x x 100-x
40

Given, n(U) = 200

From the Venn diagram

n(U) = 120 – x + x + 100 – x + 40

or, 200 = 260 – x

or, x = 260 – 200

or, x = 60

∴ (i) n(A ∩ B) = x = 60

(ii) no(A) = 120 – x = 120 – 60 = 60

(iii) no(B) = 100–x = 100–60 = 40

(iv) n(A ∪ B) = n(U)–40

= 200–40

= 160.

Example: 4

If n(A) = 40, n(B) = 60 and n(A ∪ B) = 80, using Venn diagram, find the value of
n(A ∩ B).

Solution: n(A) = 40 U
Given, n(B) = 60 AB
40–x x 60–x
n(A ∪ B) = 80


n(A ∩ B) = ?
Let, n(A ∩ B) = x

From the Venn diagram, n(A ∪ B) = 40 – x + x + 60 – x
80 = 100 – x
x = 100 – 80
or,
or, n(A ∩ B) = 20

∴

130 Oasis School Mathematics-8

Example: 5

In a class of 150 students, 70 study English, 90 study Mathematics and 50 study both.
Find the number of students who study neither of them.

Solution:
Let 'E' and 'M' be the set of students who study English and Mathematics respectively.

Here, n(U) = 150 Alternative method
n(E) = 70

n(M) = 90 Let, n(E ∪ M) = x

n(E ∩ M) = 50 U
EM
n(E ∪ M) = ?

We have, 70–50 50 90–50
= 20 = 40

n(E ∪ M) = n(E)+n(M)–n(E ∩ M) x

= 70 + 90 – 50 From the Venn diagram,
n(U) = 20+50+40+x
= 110. 150 = 110 + x
or, x = 150 – 110
Again we have, or, x = 40

n(E ∪ M) = n(U)–n(E ∪ M)

= 150–110 = 40

Hence, 40 students study neither subjects.

Example: 6

In a group of students 45 speak either Nepali or English or both, 22 can speak Nepali
only and 12 can speak English only. How many students can speak both Nepali and
English?

Solution: Alternative method
Here,
Let, 'N' and 'E' be the set of students who speak Nepali and n0(N) = 22, n0(E) = 12,
English respectively. N E
n(A∪B) = 45
Here, n0 (N) = 22 no(N) no(E) Let, n(N∩E) = x
n0(E) = 12 = 22 = 12
n(N∪E) = 45 NE
n(N∩E) = 11
n(N∩E) = ? 22 x 12

We have, n(N∪E) = n0 (N) + n0 (E) + n(N∩E) From the Venn diagram,
22 + x + 12 = 45
45 = 22 + 12 + n(N∩E) or, 34 + x = 45
or, x = 45 – 34
or, 45 = 34 + n(N∩E) Hence, 11 students can speak
both Nepali and English.
or, n(N∩E) = 45 – 34

= 11

∴ 11 students can speak both Nepali and English.

Oasis School Mathematics-8 113311

Example: 7

In a group of 50 students, 25 play cricket, 30 play football and 8 play neither game. Find
the number of students who play both cricket and football.

Solution:

Let, 'C' and 'F' be the sets of students who play cricket and

football respectively.

Here, n(U) = 50 Alternative method

n(C) = 25 Let, n(C∩F) = x

n(F) = 30 U
CF
n(C∪F) = 8
25 – x x 30 – x 8
n(C∩F) = ?

We have, n(C∪F) = n(U) – n(C∪F) From, the Venn diagram,
= 50 – 8

= 42 25 – x+ x + 30 – x + 8 = 50

Again, We have, n(C∪F) = n(C) + n(F) – n(C∩F) or, 63 – x = 50

42 = 25 + 30 – n(C∩F) or, x = 63 – 50

or, n(C ∩ F) = 25 + 30 – 42 or, x = 13

= 13 ∴ 13 students play both cricket
and football.
∴ 13 students play both cricket and football.

Exercise 8.3

1. (a) If A = {w, x, y, z}, find n(A). (b) If B = {a, b, c, d, e}, find n(B).

2. From the given Venn diagram, find the cardinal number of following sets.

(i) n(A) (ii) n(B) (iii) n0(A) A B U

(iv) n0(B) (v) n(A ∩ B) (vi) n(A ∪ B) ab jk p
cd l
xf
(vii) n(A∪B) (viii) n (A) (ix) n (B) hi
em

no

3. (a) If A = {2,3,5,7} and B={1,2,3,4,6} be two subsets of the universal set

U={1, 2, 3, 4, 5, 6, 7, 8}, find (i) n(A) (ii) n(B) (iii) n(A – B) (iv) n(B – A)

(b) If P = {2, 4, 6, 8} and Q = {a set of prime numbers less than 10} be the subsets of
universal set U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, find

(i) n(P) (ii) n(Q) (iii) n(P∩Q) (iv) no(P)
(v) no(Q) (vi) n(P∪Q) (vii) n(P ∪ Q).

132 Oasis School Mathematics-8

4. If n(A) = 12, n(B) = 14, find the maximum value of n(A∪B) and n(A∩B)
5. (a) If n(A) = 40, n(B) = 60 and n(A∩B) = 10, find the value of n(A∪B).

(b) n(A) = 80, n(B) = 70, n(A∪B) = 100, find n(A∩B).

(c) If n0(A) = 35, n0(B) = 45, n(A∩B) = 20, find (i) n(A) (ii) n(B) (iii) n(A∪B).
6. (a) If n(U) = 50, n(A) = 15, n(B) = 30 and n(A∩B) = 7, find

(i) n(A∪B) (ii) n(A – B) (iii) n(B – A) (iv) n(A ∪ B) .

(b) If n(P) = 35, n(Q) = 45, n(P∪Q) = 60, find the n(P∩Q).
7. If n(U) = 200, n(M) = 160, n(N) = 150, n(M ∪ N) = 30, find the value of

(i) n(M∩N) (ii) n0(M) (iii) n0(N)

(iv) Show the above information in the Venn diagram.

8. (a) In a group of 90 people, 70 like mango, 50 like both mango and orange. Each
people like at least one of the two fruits,. Using Venn diagram find,

(i) the number of people who like orange.

(ii) the number of people who like only one fruit.

(b) In a survey conducted among 350 people in a community, 200 people are found
influenced by materialism and 120 people are found influenced by spiritualism.
If 120 persons are found influenced by both,

(i) Show the above information in the Venn diagram.

(ii) Find the number of people who are influenced by neither of them.

9. (a) In a school, all students play either football or volleyball or both of then 300
play football, 250 play volleyball and 110 play both the games. Draw a Venn
diagram to find,

(i) the number of students playing football only.
ii) the number of students playing volleyball only.
(iii) the total number of students.

(b) In a class of 50 students, 20 students play football and 16 students play hockey.
It is found that 10 students play both the games. Find the number of students
who play (i) at least one game (ii) neither of them.

(c) Out of 30 students of class VIII, 15 like to play cricket, 12 like to play football, 6
do not like to play any game.

(i) Draw the Venn diagram to illustrate this information.
(ii) How many students like to play both the games?
(iii) How many students like to play cricket only?
(iv) How many students like to play football only?

Oasis School Mathematics-8 113333

10. 40 students of a class study either Mathematics or English or both. 12 students take
both subjects and 17 study English only. How many students study

(i) Mathematics only? (ii) English ?

Answer 

1. (a) 4 (b) 5 2. (i) 9 (ii) 8 (iii) 5 (iv) 4 (v) 4 (vi) 3 (vii) 13 (viii) 7 (ix) 8

3. (a) (i) 4 (ii) 5 (iii) 2 (iv) 3 (b) (i) 4 (ii) 4 (iii) 1 (iv) 3 (v) 3 (vi) 7 (viii) 2

4. 26 and 12 5. (a) 90 ((b) 50 (c) (i) 55 (ii) 65 (iii) 100

6. (a) (i) 38 (ii) 8 (iii) 23 (iv) 12 (b) 20 7. (i) 40 (ii) 120 (iii) 110 8. (a) (i) 70
(ii) 40 b. (ii) 150 9. (a) (i) 190 (ii) 140 (iii) 440 (b) (i) 26 (ii) 24 (c) (ii) 3 (iii)
12 (iv) 9 10. (i) 11 (ii) 29

Assessment Test Paper

Full marks : 20

Group - A [6 × 1 = 6]

1. (a) If A = a set of prime numbers less than 10 and B = {2, 3, 5, 7}, identify whether
A and B are equal sets or not.

(b) If A = {a, b, c}, find the number of subsets of A.

2. (a) What does the shaded part of given Venn diagram represent?

(a) AB U (b) U
AB

3. (a) If A = {1, 2, 3, 4, 5} and B = {4, 5, 6, 7, 8}, find n(A∪B).

(b) Write the formula that shows the relation among n(A), n(B), n(A∪B) and n(A∩B).

Group - B [ 3×2=6]

4. (a) Let A = {2, 3, 5, 7, 11, 13}, B = {2, 4, 6, 8, 10, 12, 14}, find (i) A∪B (ii) A∩B.

(b) Draw a Venn diagram to show the relation (A∩B)∩C.

(c) If A = {p, q, r}, B = { q, r, s}, find (i) A–B and B–A. U
Group- C [2 × 4 = 8] AB

5. (a) In the given Venn diagram, if n(∪) = 150, 70–x x 80–x
find the value of (i) x, (ii) no (A), (iii) no (B) 20

(b) In a group of 100 people, 70 like tea and 80 like coffee. If 60 like both, find the
number of people who do not like both.

134 Oasis School Mathematics-8

Arithmetic

49Estimated Teaching Hours

Contents
• Number System of Different Bases
• Integers
• Rational and Irrational Numbers
• Ratio and Proportion
• Percentage
• Unitary Method
• Simple Interest
Expected Learning Outcomes
At the end of this unit, students will be able to develop the following
competencies:

• Convert decimal number into binary and quinary numbers and
vice-versa.

• Add or subtract the binary numbers and quinary numbers.
• Use the four fundamental operations on integers.
• Simplify the problems on integers.
• Identify rational and irrational numbers.
• Operate on irrational numbers.

Teaching Materials:

• Chart paper, A4 size paper, flash cards, etc.

Oasis School Mathematics-8 113355

Unit

Number System of

9 Different Bases

9.1 Review

In Hindu-Arabic system of numeration, ten basic symbols are used to represent the
large numbers. They are 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9. The numbers are expressed in
powers of 10. This number system is 10 based number system or decimal number
system.

For example,

24 = 2 × 101 + 4 × 100

386 = 3 × 102 + 8 × 101 + 6 × 100

4512 = 4 × 103 + 5 × 102 + 1 × 101 + 2× 100

9.2 Binary number system

This is another system of numeration. This system uses only two digits 0 and 1. In
this system, numbers are expressed in terms of powers of 2. So it is base-two system.

For example,

10012 = 1 × 23 + 0 × 22 + 0 × 21 + 1 × 20
= 8 + 0 + 0 + 1

= 9

11102 = 1 × 24 + 1 × 23­ + 1 × 22 + 1 × 21 + 0 × 20
= 16 + 8 + 4 + 2 + 0

= 30.
∴ 10012 = 9 and 11102 = 30

binary decimal binary decimal
number number number number

136 Oasis School Mathematics-8

Place value chart

Decimal numeration Base two grouping Binary numeration

27 26 25 24 23 22 21 20
0 00

1 1 12

2 10 102

3 11 112

4 10 0 1002

5 10 1 1012

6 11 0 1102

7 11 1 1112

8 100 0 10002

9 100 1 10012

10 101 0 10102

In the binary system, we see from the above table that the even numbers have zero (0) in
ones digit and the odd numbers have one (1) in ones digit.

I. Conversion of Binary System (2 base system) into decimal number system (10
base system):

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

10012 = 1 × 23 + 0 × 22 + 0 × 21 + 1 × 20
= 1×8+0×4+0×2+1×1=8+0+0+1=9

1000012 = 1 × 25 + 0 × 24 + 0 × 23 + 0 × 22 + 0 × 21 + 1 × 20
= 1 × 32 + 0 × 16 + 0 × 8 + 0 × 4 + 0 × 2 + 1 × 1

= 32 + 0 + 0 + 0 + 0 + 1

= 33

II. Conversion of decimal numbers into binary numbers:

To convert a decimal number into a binary number system, follow the given steps.

For example:

Convert 30 into a binary number. • Divide the given number successively
by 2 until the quotient is zero.
2 30 Remainder
• List the remainder obtained in each
2 15 0 successive division in a column.

27 1 • Write all the remainders from the
23 1 bottom to the top.

21 1 Which is the required binary number.
0 1

Writing all the remainders from the bottom to the top 11110.

∴ 30 = 111102

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9.3 Quinary Number System

The number system which uses the number 5 as the base is called quinary number

system. Under the quinary number system, the numbers are expressed in powers of

5. The five basic symbols used in the quinary number system are 0, 1, 2, 3 and 4.

For example, 135 = 1 × 51 + 3 × 50 = 1 × 5 + 3 × 1 = 8


∴ 135 = 810

2335 = 2 × 5² + 3 × 5¹ + 3 × 5º = 2 × 25 + 3 × 5 + 3 × 1

= 50 + 15 + 3 = 68

∴ 2335 = 6810

Let's see the given table

Decimal numbers Quinary numbers
0
1 0
2
3 15
4 25
5 35
6 45
7 105
8 115
9 125
10 135
145
205

Place value chart

Base 5 system 54 53 52 51 50 Base-10 system

235 23 13
1425 142 47
12305 12 3 0 190

I. Conversion of quinary number system (base-5 system) into decimal number
system (base - 10 system)

To convert a quinary number system into decimal number system, we should
expand the given number by giving the place value of each digit in powers of 5.
Find the product and then the sum of all numbers.

For example,

1235 = 1 × 5² + 2 × 5¹ + 3 × 5°
= 1 × 25 + 2 × 5 + 3 × 1 = 25 + 10 + 3 = 38

∴ 1235 = 3810

138 Oasis School Mathematics-8

and 11435 = 1 × 5³ + 1 × 5² + 4 × 5¹ + 3 × 5°
= 1 × 125 + 1 × 25 + 4 × 5 + 3 × 1
= 125 + 25 + 20 + 3 = 173

∴ 11435 = 17310.

II. Conversion of decimal numbers into quinary numbers

To convert decimal numbers into quinary number system
• Divide the given number successively by 5 until the quotient is zero.
• List the remainder obtained in each successive division in a column.
• Write all the remainders from the bottom to the top.
Which is the required quinary number.

For example:
Convert 258 into quinary number system.

Here,

5 258 Remainder
5 51
5 10 3
52 1
0
0 2

Writing the remainder from the bottom to the top 2013.

∴ 258 = 20135

Worked Out Examples



Example: 1

Display the numbers 10012, 111012, 1000112 and 10000012 in place value chart in binary
system.
Solution:
Place value chart in binary system

26 25 24 23 22 21 20

10012 10 01
111012
1000112 1 1 1 01
10000012
1 0 0 0 11

1 0 0 0 0 01

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Example: 2

Convert the following binary numbers into decimal system.

(i) 10012 (ii) 10000012
Solution:

(i) 10012 = 1 × 23 + 0 × 22 + 0 × 21 + 1 × 20
= 1×8+0×4+0×2+1×1=8+0+0+1=9
(ii)
10000012 = 1 × 26 + 0 × 25 + 0 × 24 + 0 × 23 + 0 × 22 + 0 × 21 + 1 × 20
= 1 × 64 + 0 × 32 + 0 × 16 + 0 × 8 + 0 × 4 + 0 × 2 + 1 × 1
= 64 + 0 + 0 + 0 + 0 + 0 + 1 = 65

Example: 3

Convert 59 into binary (base two) number system.

Solution:

2 59 Remainder
2 29 1

2 14 1

27 0
23 1
21 1
0 1

∴ 59 = 1110112

Example: 4

Display the numbers 23015, 123145 and 2304105 in place value chart of quinary number
system.

Solution:

55 54 53 52 51 50

23015 23 0 1

12345 12 3 1 4

2304105 2 3 0 4 1 0

Example: 5

Convert the following quinary numbers into decimal numbers

(i) 1005 (ii) 23405 (iii) 11345

Solution:

(i) 1005 = 1 × 5² + 0 × 5¹ + 0 × 50
= 1 × 25 + 0 × 5 + 0 × 1 = 25 + 0 + 0 = 25

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(ii) 23405 = 2 × 53 + 3 × 52 + 4 × 51 + 0 × 50
= 2 × 125 + 3 × 25 + 4 × 5 + 0 × 1

= 250 + 75 + 20 + 0 = 345
iii) 11345 = 1 × 53 + 1 × 52 + 3 × 51 + 4 × 50

= 1 × 125 + 1 × 25 + 3 × 5 + 4 × 1

= 125 + 25 + 15 + 4 = 169

Example: 6

Convert the following decimal numbers into quinary numbers.

(i) 145 (ii) 9999

Solution:

(i) Remainder

5 145 0
5 29
55 4
51 0
1
0

Successive remainders are taken from bottom to top 1040.

∴ 145 = 10405

Checking : 10405 = 1 × 5³ + 0 × 5² + 4 × 5¹ + 0 × 50
= 125 + 0 + 20 + 0 = 145

(ii) 5 9999 Remainder
5 1999 4

5 399 4

5 79 4

5 15 4

53 0
0 3

∴ 9999 = 3044445

Example: 7

Convert 21345 into binary number system.
Solution:

Lets convert 21345 into decimal system.

Now, 21345 = 2 × 53 + 1 × 52 + 3 × 51 + 4 × 50

= 250 + 25 + 15 + 4

= 294

∴ 21345 = 29410 .....................(i)

Oasis School Mathematics-8 114411

Again, let's convert 29410 into binary number.

2 294 Remainder
2 147
2 73 0
2 36 1
1
2 18 0
29 0
24 1
22 0
21 0
0 1


∴ 29410 = 1001001102 .................(ii)
From (i) and (ii)

∴ 21345 = 1001001102 .....................(i)

Example: 8

Convert 110102 into quinary number system.

Solution:

Let's convert 110102 into decimal number system.
Now,

110102 = 1 × 24 + 1 × 2³ + 0 × 2² + 1 × 2¹ + 0× 20
= 1 × 16 + 1 × 8 + 0 × 4 + 1 × 2 + 0 × 1

= 16 + 8 + 0 + 2 + 0 = 26

∴ 110102 = 2610 ..........................(i)

Again, lets convert 2610 into quinary number system.

5 26 Remainder
55 1
51 0
01

∴ 2610 = 1015...........................(ii)
From (i) and (ii)

110102 = 1015.


Exercise 9.1

1. Display the following binary numbers in place value chart.

(a) 1101002 (b) 1000100112 (c) 11110012

142 Oasis School Mathematics-8

2. Convert each of the binary number (i.e. base 2 system) into decimal number
system.

(a) 102 (b) 1002 (c) 11112 (d) 10012

(e) 101002 (f) 101012 (g) 111102 (h) 11110102

(i) 10000012 (j) 11111112

3. Convert the following decimal numbers into binary numbers.

(a) 45 (b) 105 (c) 355 (d) 400 (e) 529

(f) 512 (g) 190 (h) 101 (i) 723

4. Display the following quinary numbers in place value chart.

(a) 423015 (b) 1230015 (c) 23101235

5. Convert the following quinary numbers into decimal number (i.e. base 10

number).

(a) 1005 (b) 2105 (c) 4235 (d) 200015 (e) 32345
6. Convert the following decimal numbers into quinary numbers.

(a) 150 (b) 500 (c) 1024 (d) 5054 (e) 8085

7. Express the following base two numbers into base five numbers.

(a) 111102 (b) 1111112 (c) 1011012
8. Convert the following base five numbers into base two numbers.

(a) 10125 (b) 44445 (c) 31045 (d) 400025



Answer
1. Consult your teacher
2. (a) 2 (b) 4 (c) 15 (d) 9 (e) 20 ( f) 21 (g) 30 (h) 122 (i) 65 (j) 127
3. (a) 1011012 (b) 11010012 (c) 1011000112 (d) 1100100002 (e) 10000100012
(f) 10000000002 (g) 101111102 (h) 11001012 (i) 10110100112
4. (a) Consult your teacher 5. (a) 25 (b) 55 (c) 113 (d) 1251 (e) 444
6. (a) 11005 (b) 40005 (c) 130445 (d) 1302045 (e) 2243205
7. (a) 1105 (b) 2235 (c) 1405
8. (a) 100001002 (b) 10011100002 (c) 1100101002 (d) 1001110001102

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9.4 Addition and Subtraction of Binary Numbers

a. Addition of Binary Numbers

In decimal system, if the sum of the digits becomes 10 or more, we start to take carry
over 1 ten, 2 tens, 3 tens, etc. to the higher places. But, in binary (base 2) system, if
the sum of the digits becomes 2 or more, we start to take carry 1 two, 2 twos, 3 twos,
etc. to the higher places.

Binary number system has only two digits 0 and 1.

Let's remember the basic rule

0+0=0
1+0=1
1 + 1 = 10 i.e. write 0 in its place and carry over 1

1 + 1 + 1 = 10 + 1 = 11, i.e write 1 in its place and carry over 1.

Example, Steps
Add : 1012 + 112
• 1 + 1 = 10, set down 0 in its place, carry over 1.
11 • 1+ 0 + 1 = 10, set down 0, carry over 1.
• 1 + 1 = 10
1 0 12
+0 1 1 2
10 0 0

b. Subtraction of binary numbers

Let's remember the basic rule for subtraction.

0–0=0
1–1=0
1–0=1
10–1 = 1

We can borrow from higher places if it is necessary. Let's be clear with the help of
given example.

Subtract : 11012 – 1112
Here,

0 10 10 Steps

1 1 0 12 • 1 – 1 = 0, set 0 down.
• Borrow 1. then 10–1 = 1, set 1 down
– 1 1 12 • Again, borrow 1, 10–1=1, set 1 down
110

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