Prime Mathematics 8

Prime Mathematics Book - 8

Unit - 8: RevAisrieona TReesvtision Test (Mensuration)

1. Find the area of the given figures. A B

(a) A (b)

10cm

8.5cm



B 14.6cm C D 15cm C

A

(c) (d) A 12cm B
D B

C D25cm 19cm C
AC=20cm, BD=32cm (b) 6cm20cm

2. Find the area of the following.

(a) 14cm

8cm

27cm

20cm 5cm
7cm
3. Find the area of shaded part of the given figure.



6cm

8cm
4. Volume of a cubical box is 729cm3. Find its total surface area.
5. Find the total surface area of the a cuboid having length, breadth and height

respectively 2.4m, 1.2m and 80cm.

96 3 Mensuration

Prime Mathematics Book - 8

Answers

Exercise - 1.1 (Perimeter, Area and Volume)

1. (a) 432cm2 (b) 256cm2 (c) 30.55cm2 (d) 160cm2 (e) 57.04cm2 (f) 192cm2

2. (a) 96cm2 (b) 1134cm2 (c) 250.8cm2 (d) 1219.05cm2 (e) 224cm2 (f) 327.25cm2

(g) 72cm2 (h) 220cm2

3. (a) 270cm2 (b) 154cm2 (c) 91.712cm2 (d) 240cm2

4. (a) 90cm2 (b) 216cm2 (c) 234.95cm2 (d) 99cm2 (e) 80cm2 (f) 60cm2

5. (a) 100cm2 (b) 156.25cm2 (c) Rs.8450 (d) Rs.19200

6. (a) 14cm (b) 2.8cm (c) 11cm

Exercise - 1.2 (Perimeter, Area and Volume)

1. (a) 192cm3 (b) 1728cm3 2. (a) 592cm2 (b) 384cm2

3. (a) 792cm2, 1440cm3 (b) 4.32m2, 0.576m3 (c) 12.24cm2, 2.592cm3

4. (a) 648.96cm2, 1124.864cm3 (b) 216cm2, 216cm3 (c) 277.44cm2, 314.432cm3 (d) 69.36cm2, 39.304cm3

(e) 384ft2, 512ft3

5. (a) 486cm2 (b) 26cm (c) 10584cm2 (d) 9240cm3

6. (a) 352 cm2 (b) 6480 litres (c) l = 18cm; h = 6cm (d) 60cm

Answers 97

Estimated periods - 10

At the end of this unit the student will be able to Objectives:

● draw the image of figures under translation, reflection and rotation on simple plane.
● find coordinator of the image of the given points under translation, reflection on x-axis,

y-axis, y = x and rotation about origin through ±900, ±1800.
● know the bearing and reversed bearing of a point from a given point.
● know and use the scale in maps.

Teaching Materials:

Models of different polygons, instrument box, graphs, different symmetrical figures and maps.

Activities:

It is better to
● demonstrate drawing image of figures under different transformations in simple plane.
● discuss about the formula to find image of a point under different transformations in co-

ordinate plane.
● discuss about uniting bearing in different forms.
● explain about revived bearing.
● discuss about scale and use of bearing of scale in maps.

Unit - 1 Transformation Prime Mathematics Book - 8
Estimated period : 5

Introduction

Observe the following activities.



Map of Nepal is shifted Different sized photos from the position A to B.


Image of Mt. Machhapuchhre on Phewa lake. Position of hands of clock
in different times. Transformation simply means change. Geometrical transformation
refers to the change in geometrical figures and the change may be in shape, size and
position. In transformation, the original (initial) point or figure is called object and the
corresponding point or figure formed latter is called its image.

If A' is the image of point A, we write A A' (read as A maps to A' or A' is the image
of A). If ∆A'B'C' is the image of ∆ABC, we write ∆ABC ∆A'B'C'

If a figure and its image after transformation are congruent, the transformation is
known as isometric. If size of the object and its image are not same, the transformation
is known as nonaromatic. During transformation some points may remain unchanged i.e.
points and their images coincide. Such points are called invariant points. If all the points
remain unchanged , i.e. figure and its image coincide, the transformation is called an
entity transformation.

Transformation 99

Prime Mathematics Book - 8

Types of Transformation:

There are different types of transformations but the following are the fundamental
transformations.
a) Translation b) Reflection c) Rotation d) Dilation

1.1. Translation :

It is the transformation in which each and every point of an object is shifted by a

vector. In the given figure, ∆ABC ∆A'B'C' where

AA' = BB' = CC' = ν' ν
A'
We write, under translation by vector ν' A
B B'
T
or ∆ABC ∆A'B'C' or ∆ABC ν' ∆A'B'C'

T ν' : ∆ABC ∆A'B'C' C C'

Characteristics of translation
(i) It is isometric transformation.
(ii) Each and every point of the figure get displaced so there exists no invariant point.

Translation in simple plane: P

Given a translation vector PQ and a figure ∆ABC. A A' Q

- Draw lines parallel to PQ from A, B and C B

using scale and set squares. B'

- Take AA' = BB' = CC'= PQ C
C'

- Join A', B' ; B', C' ; A' C' Thus, T PQ' : ∆ABC ∆A'B'C'


Translation in co-ordinate plane:
Let a point P(x, y) is to be translated by a vector a to P'. So PQ = a, QP' = b.
b
Draw the Perpendiculars PM and P'N on x-axis respectively. y
P'

We get, OM = x, PM = y

ON = OM + MN = OM + PQ = x + a (x, y) a b
NP' = NQ + QP' = PM +QP' = y +b. P Q
∴ Co-ordinates of P' is (x + a, y + b).

∴ T a : P(x, y) P' (x + a, y + b) OM x
N
b

100 4 Transformation

Prime Mathematics Book - 8

Example 1: Find the co-ordinates of image of point A(3, 2) under translation by vector 2 .
-3

Solution : Given point A(3, 2) 2
-3
Given translation vector

We have,

T a : P (x, y) P' (x + a, y + b)

b

∴ T 2 : A (3, 2) A' (3 + 2, 2 - 3) = A' (5, -1)

-3

∴ Co-ordinates of image of A(3, 2) is A' (5, -1).

Example 2: Find the coordinates of the image of the vertices of ∆ABC under translation

by vector -1 where the coordinates of the vertices of ∆ABC are A(3,3),
2

B(4,0) and C(1,-1). Also plot the ABC and A'B'C' on the same graph.

Solution: Given vector a = -1 Y
b 2 A'

We know,

T a : p (x, y) p' (x + a, y + b) A
b B'

∴ T -1 : A (3, 3) A' (3 - 1, 3 + 2) = A' (2, 5) BX
2

: B (4,0) B' (4 - 1, 0 + 2) = B' (3, 2) C'
X' O

: C (1,-1) C' (1 - 1, -1 + 2) = c' (0, 1) C
Thus, Y'

T -1 : ∆ABC ∆A'B'C' where the coordinates of vertices of image triangle
2

A'B'C' are A'(2, 5), B'(3, 2), c'(0, 1).

Finding translation Vector:

If a point p(x, y) p'(x', y') under translation by vector a then
b

x' = x + a y' = y + b
⇒ a = x' - x
⇒ b = y' - y

Translation vector a = x' - x
b
y' - y .

Transformation 101

Prime Mathematics Book - 8

Example 3: Find the vector which translates a point A(3, -1) to A'(1, 2) and hence find

the image of the point B(-4, 2).

Solution: here,

Given point = A (3, -1) = (x, y)

Image = A' (1, 2) = (x', y')

Let a be the vector, then we have
b

a = x' - x = 1- 3 = -2
b y' - y 2 - (-1) 3

∴ The required translation vector is -2
3

Now,

we know, T a = p(x, y) p'(x + a, y + b)
b

∴ T -2 : B(-4, 2) B' (-4 - 2, 2 + 3) = B'(-6, 5)
3

∴ Image of B(-4, 2) is B'(-6, 5).

Exercise 1.1

1. a) Define translation (displacement).
b) Write the characteristics of translation.
c) Write the names of fundamental transformations.

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

(a) ν b) A ν (c)

AB ν

C

D CB

3. a) Find the coordinates of the point A(-3, 5) under translation by vector; (i) 1 (ii) 3 .
-2 1

b) If a point A(4, 1) is transformed to the point A'(2, 4) under translation, find the

translation vector.

102 4 Transformation

Prime Mathematics Book - 8

c) Under translation by a vector 1 point P maps to P'(2, 4), find the co-ordinates
-2
of the point P.

d) Under translation a point A(0, 3) displaces to A'(4, 5), find the image of

B(-3, 2). Also plot the ∆ABC, ∆A'B'C', ∆A''B''C'' on the same graph.

4. a) ∆ABC with vertices A(-2, 2), B(2, 0) and C(-4, -3) is translated by vector 4 . Find
3

the co-ordinate of vertices of the image ∆A'B'C'. Also plot the ∆ABC and ∆A'B'C' on

the same graph.

b) Find the coordinates of the images of the points A(0,0), B(-3, 1), C(0, -2), D(3, 1)
-1
under translation by vector 3 . Plot the figures ABCD and A'B'C'D'.

c) T -4 : ∆ABC ∆A'B'C' where coordinates of A, B, C are A(1, 4), B(3, -1)

2

and C(0, -1). Find the coordinates of A'B'C' and plot the ∆ABC and ∆A'B'C' on the
same graph.

1.2 Reflection: A' C'
Reflection of a point P on a given fixed line MN is the M
A
transformation which maps P to P' such that MN is
perpendicular bisector of PP'. P' is the image of P and B B'
MN is called line of reflection or axis of reflection or
mirror. N
C
i) Characteristics of reflection :

(i) Reflection is an isometric transformation.

(ii) The point/points of the figure which lies

on the axis remain M
P
invariant.

(iii) Figure and its image form a symmetrical

pattern about the axis. O
N
(iv) The position of image is reversed. P'

Transformation 103

Prime Mathematics Book - 8

ii) Reflection on simple plane : AM

Given figure ∆ABC and axis of reflection MN. P

- Draw AP, BQ, CR perpendiculars to MN. B A'

Q B'

- Produce AP, BQ and CR to A', B' and C' respectively such

that AP = PA', C

BQ = QB' and CR = RC'. R

- Join A' and B' ; B' and C' ; C' and A' N
C'

Thus ∆ABC ∆A'B'C' under reflection on MN.

iii) Reflection on co-ordinate plane:

Reflection can be made on any line (general line) but here we will take only the

following special lines as axis of reflection. Y
(a) x-axis (b) y-axis P(x, y)

a. Reflection on x-axis (y = 0) : X' O MX
Consider a p(x, y) point Y' P'
The axis of reflection is XX'. So draw PM
⊥ XX' and produce P' such that PM = MP'.
Here, OM = x and PM = y MP' = -PM = -y
∴ Co-ordinates of P' is (x, -y)

Under reflection on x-axis Also we write as:
Rx-axis : P (x, y)
P (x, y) P' (x, -y) P' (x, -y)

b. Reflection on y-axis (x = 0) :

Consider a point P(x, y). Axis of reflection is YY'. So, draw PM ⊥ PP' and produce

to P' such that PM = MP'. Y

Here, P' M P(x,y)
OM = y

MP' = -MP = -x. X' o X

∴ Co-ordinates of P' is (-x, y) Y'

∴ Ry-axis : P (x, y) P' (-x, y)

104 4 Transformation

Prime Mathematics Book - 8

Example 1: Find the image of the point A(-3, 2) under relation on (i) x - axis (ii) y - axis.

Solution: Here, given point A(-3, 2).

R
We have, (i) P (x,y) x-axis P' (x, -y)

R
∴ A (-3, 2) x-axis A' (-3, -2)

(ii) P (x, y) R P' (-x, y)
y-axis

R
∴ A (-3, 2) y-axis A' (-(-3), 2) = A' (3, 2).

Example 2: Find the coordinates of the image of ∆ABC with vertices A(2, 0), B(-1, 2), C(4, 3)

under reflection on X-axis. Y

Solution: We have, C

Under reflection on X-axis B

P(x, y) P' (x, -y)

∴ A(2,0) A' (2, 0) X' O A X
B' (-1, -2) A'
B(-1, 2)

C(4, 3) C' (4, -3) B'
C'
∴ Co-ordinates of ∆ A'B'C' are

A'(2, 0), B'(-1, -2), C'(4, -3). Y'

Example 3: Find the co-ordinates of the image of ∆ABC with vertices A(1, 2), B(4, 4),

C(1, -2) under reflection on Y-axis. Also plot the ∆ABC and its image ∆A'B'C'

Solution: on the same graph. B' Y
Here, Vertices of given ∆ABC are B
A(1,2), B(4,4), C(1,-2)

Line of reflection is Y-axis. A'
we know, A

R y-axis : P (x, y) P' (-x, y) X' O X
∴ A (1, 2) A' (-1, 2)

B (4, 4) B' (-4, 4) C' C
C (1, -2) C' (-1, -2) Y'

Co-ordinates of image ∆A'B'C' are A'(-1, 2), B'(-4, 4), C'(-1, -2).

Transformation 105

Prime Mathematics Book - 8

Exercise 1.2

1. a) Define reflection.
b) Write the characteristics of reflection.
c) What do you mean by isometric transformation?
d) In which condition, a point is invariant under reflection ?

2. Copy the given figures approximately and draw their reflections on the line MN.

a) A M b) A
M


BB
C

C N N A
M M

A D

c) d)

B B
C C

NN

3. Find the co-ordinates of the following points after reflection on x-axis and y-axis
respectively;
a) P(2, -3) b) M(-3, -9) c) R(7, 8) d) K(-10, 14)

4. a) Find the co-ordinates of the images of the point A(3,4) and B(-5, -3) after reflection
on (i) x-axis (ii) y-axis

b) Find the co-ordinates of the image of ∆ABC under reflection on x = 0 where the
coordinates of the vertices of ∆ABC are A(2, 0), B(-4, -3) and c(-2, 2). Also plot the
triangle ABC and image ∆A'B'C' on the same graph.

c) Reflect the ∆ABC on x-axis where co-ordinates of the ∆ABC are A(6,-2), B(-2,-2) and
c(2,5) and plot the ∆ABC and ∆A'B'C' on the same graph.

d) Vertices of ∆ABC are A(-3,1), B(-4,-1) and c(-1,-4). Reflect ∆ABC on x-axis to ∆A'B'C'
and then reflect ∆A'B'C' on y-axis to ∆A''B''C''.

106 4 Transformation

Prime Mathematics Book - 8

1.3 Rotation : P'

Rotation of a point P about a fixed point O through an angle θ
is the transformation which maps the point P to P' such that
OP = OP' and ∠POP'= θ.

We write, under rotation about O through an angle θ. θ

P P' or, R[0, θ]: P P' O P

The fixed point O is called centre of rotation and ∠POP'= θ is called angle of rotation.
∠POP' measured counter clockwise direction is taken as positive and clockwise direction
is taken as negative.

i) Characteristics of rotation:

(a) Rotation is an isometric transformation.
(b) If the point of the figure lies at the centre, it remain invariant.
(c) A figure and its image after rotation form a radially symmetrical pattern.

ii) Rotation on simple plane:
(a) Given ∆ABC, to rotate about the point O through 900.

- Join OA, OB, OC.
- Draw ∠AOX = ∠BOY = ∠COZ = 900 in positive direction.

- Take OA' = OA along OX, OB' = OB along OY and OC' = OC along OZ using compass.

- Construct A'B', B'C' and C'A'

Thus ∆ABC ∆A'B'C'

(b) Given equilateral triangle ABC, to rotate about its centroid through -600.

- Draw medians from the vertices A, B and C which meet at centroid O.

- Draw ∠AOX = ∠BOY = ∠COZ = 600 in clockwise direction (-600)

- Take OA' = OA along OX, OB' = OB along OY A

and OC' = OC along OZ.

- Construct A'B', B'C' and C'A'. Z X
C' A'
Thus ∆ABC ∆A'B'C'. B

iii) Rotation on co-ordinate plane : 600
Rotation of a point or figure can be made about O

any point (centre) and through any angle but in C
this class we will discuss about rotation about

origin through angles 900, - 900, 1800, 3600 only. B'
Y

Transformation 107

Prime Mathematics Book - 8

a. Rotation about origin through 900(Positive quarter turn):
Let p(x, y) be a point on co-ordinate plane. Joining O and P, rotate OP about O

through + 900 to OP'. Y
Here, ∠POP' = 900.
P'(-y, x)

If ∠XOP = θ, then we get ∠YOP' = θ θ P(x, y)
Draw PM and P'M' perpendicular to XX'. θ X
From ∆OP'M ≅ ∆OPM, we get
OM' = - PM = -y P'M' = OM =x X' θ M
M' o

∴ Co-ordinates of P' is (-y, x)


∴ Under the rotation about 0(0, 0), through +900

P(x, y) P'(-y, x). Y'

or, R [(0, 0), 900]: P(x, y) P'(-y, x)

b. Rotation about origin through -90o (Negative quarter turn):

Let P(x,y) be a point on coordinate plane. Y

Joining O and P, rotate OP about O through an P
angle of -900 to OP'.

∴ ∠POP' = -900. X' θ M' X
o M
Draw PM and P'M' ⊥s to OX, if ∠XOP = θ, we get
θ

∠Y'OP' = θ and ∆OP'M' ≅ ∆OPM where

OM' = PM = y P'M' = -OM = -x P'
Co-ordinates of P' is (y, -x). Y'

c. Rotation through 1800 about origin (Half turn):
Let P(x, y) be a point on coordinate plane. Joining O
and P, rotate OP about O through 1800 or -1800 to Y OP.'

∴ ∠POP' = 1800

Draw PM and P'M' ⊥s to XX', if ∠XOP = θ, we get 1800 P(x,y)
θ
∠X'OP' = θ and ∆OP'M' ≅ ∆OPM X' M' X
θo M
where, OM' = OM = -X P'M' = -PM = -Y

∴ Co-ordinates of P' is (-x, -y). P'

Y'
Under the rotation origin through 1800

P(x, y) P'(-x, -y).

or R [(0, 0), 1800] : P(x, y) P'(-x, -y)

Note: Under rotation through 1800 or -1800, the position of image is same.

108 4 Transformation

Prime Mathematics Book - 8

d. Rotation about origin through 3600 (Full turn):

Let P(x, y) be a point.

First, turning P(x, y) about origin through + 1800 or -1800.

R[(0, 0), 1800] : P(x, y) P'(-x, -y) and

again turning P'(-x, -y) about origin through 1800 to the same direction

R[(0, 0), 1800] : P'(-x, -y) P''(x, y)

P''(x, y) is the point P(x, y) itself.

Hence R[(0, 0) , 3600] or full turn is an identity transformation.

Example 1: Find the image of the point A(4, -6) under rotation about origin through;

(i) 900 (ii) -900 (iii) 1800

Solution: Here, given point A(4, -6) P'(-y, x)
We have, A'(6, 4)
P'(y, -x)
(i) R[(0,0), 900] : P(x, y) A'(-6, -4)
∴ R([0,0), 900] : A(4, -6)
P'(-x, -y)
(ii) R([0,0), -900] : P(x, y) A'(-4, 6)
∴ R([0,0), -900] : A(4, -6)

(iii) R([0,0), 1800] : P(x, y)
∴ R[(0,0), 1800] : A(4, -6)

Example 2: Find the coordinates of the image of triangle ABC with vertices A(1, -2),
B(4, -3) and C(3, 2), under rotation about origin through quarter turn.
Plot the ∆ABC and its image ∆A'B'C' on the same graph.

Solution : Vertices of ∆ABC are A(1, -2), B(4, -3), C(3, 2) Y
We have, B'
Under rotation about origin through a
quarter turn (900)

P(x, y) P'(-y, x) C' C

∴ A(1, -2) A'(2, 1) A'

B(4, -3) B'(3, 4) X' O X

C(3, 2) C'(-2, 3)

The coordinates of vertices of image A B
∆A'B'C' are A'(2, 1), B'(3, 4) and C'(-2, 3). Y'

Transformation 109

Prime Mathematics Book - 8

Exercise 1.3

1. a) Define rotation. b) Write the characteristics of rotation.

c) What do you mean by invariant point ?

d) what do you mean by identity transformation ?

2. Copy the following figures and rotate them about the point O through 900
A b) A
a)

B O

O C

B
C

3. Rotate the given figure about the point O through -900.

a) b) B A

O CO

4. Draw the image of the figure after rotation abut the point O through 1800.

a) A b) A

C

O

CB

BO
5. a) Find the coordinates of the point A(4, -2) under rotation about O(0, 0) through

(i) Quarter turn (ii) Negative quarter turn (iii) Half turn

b) If R [(0,0), 900] : A(a, b) A'(-3, 5). Find the co-ordinates of A.

c) Co-ordinates of the vertices of ∆ABC are A(2,5), B(6,2), C(3,-2). Find the co-ordinates

of the vertices of image ∆A'B'C' after rotation about origin through 900.

d) Find the vertices of the image of ∆ABC with vertices A(4, 1), B(2, 5), C(-2, -1) under
rotation about origin through 1800. Also plot the ∆ABC and ∆A'B'C' on the same graph.

e) If R([0,0), -900] : ∆ABC ∆A'B'C', where coordinates of A, B and C are (1, 1),

(-3 ,-2) and (-1, 3) respectively. Find the co-ordinates of image ∆A'B'C' and plot the

∆ABC and ∆A'B'C' on the same graph.

110 4 Transformation

Prime Mathematics Book - 8

Unit - 2 Bearing and scale drawing Estimated period : 5

2.1 Bearing

Direction is always of great concern in our daily life. People in old days took the
reference of rising sun, setting sun, position of pole star, etc to find the direction.
After the discovery of magnetism, magnet or compass became the most convenient
way to find the direction.
Bearing of a point B from a fixed point A is the direction stated either in the form
of angle the line AB makes with the line due North and South through A or in term of
the angles the line AB makes with the line running due North through A in clock wise
direction. There are two ways of describing bearing.
(i) Compass bearing
(ii) Three figure bearing

Bearing and scale drawing 111

Prime Mathematics Book - 8

A. Compass bearing

There are four major directions North (N), W NW N NE
East (E), South(S) and West(W). In the old method SW SES NEE
of compass bearing, direction is located along E
these direction or direction between them. The
direction exactly between N and E is denoted by B
NE, between S and E by SE, between S and W by E
SW, between N and W by NW, further the direction
exactly between N and NE was denoted by NNE, C
between NE and E by NEE and so on.

How to denote the direction exactly between P N
NEE and E ? Of course NEEE but it is inconvenient.
These are only the fixed directions. 45o 50o

In the new version of compass bearing, W A
bearing of a point B from a point A is stated in
terms of angle the line AB makes with the line due 20o 75o
North and South through A . In the figure,bearing of
B from A is N50ºE (Staring from N line towards E), DS
bearing of C is S70º E, that of D is S20º W and that
of P is N45ºW.

B. Three figure bearing N B

It is the modern version of compass bearing. 120o Q
In this method, direction of a point B from a point
A is stated in terms of the angle the line AB makes A
with the line running due North through A (North
line) in clockwise direction using three figure NP
(digits).
25o
In the figure, A is the point (Place) of 90o
observation, AN is the North line and ∠ NAB=120º
in clockwise direction. Thus bearing of B from A A
is 120º, similarly in the second figure, bearing of
P,Q,R are respectively 025º, 090º and 240º from A. 240o

R

112 4 Transformation

Prime Mathematics Book - 8

C. Reversed bearing N1
Let B be the bearing of the point B from the point A
which means ∠ NAB = θ (in 3 figure) N
\∠ ABS = ∠ NAB = θ (alternate angles) B
\∠ N1BA (reflex) =∠ ABS + ∠N1BA
i.e. Bearing of A form B = θ +180º θ
Again ∠ NAB (reflex) = θ be the bearing of the point A (i)
B from the point A. (fig ii)
S

\∠ SAB = reflex ∠ NAB - ∠NBS

= θ -180º N1 N
\∠ N1BA =∠ SAB = θ -180º (alternate angles)
i.e. bearing of A from B = θ - 180º

Hence, if bearing of the point B from the point A is B
θ ≤ 180º, bearing of the point A from the point B i.e.
reversed bearing = θ + 180º A
θ

And if the bearing of the point B from the point A is (ii)
θ > 180º, then the bearing of the point A from the S
point B i.e. reversed bearing = θ - 180º

Note:
We can always use reversed bearing = θ + 1800. If θ is reflex reversed bearing θ + 1800
gives reversed bearing greater than 3600, so(θ + 1800) - 3600 gives the reversed bearing.

Example 1: Write the compass bearing of the following direction

(a) E (b) S (c) W (d) NE (e) SE

Solution:

S.N. Directions Compass bearing
(a) E N90ºE or S90ºE
(b) S S0ºS
(c) W N90ºW or S90ºW
(d) NE N45ºE
(e) SE S45ºE

Bearing and scale drawing 113

Prime Mathematics Book - 8

Example 2: Write the 3 figure bearing for the following compass bearings;

(a) N50ºE (b) S30ºE (c) S40ºW (d) N75ºW

Solution:

Given compass bearings are shown in the figure respectively by the points A, B,

C, and D. NA
(a) A is between N and E and ∠ NOE (in

clockwise direction) = 50º D 75o 50o
\The 3 - figure bearing of A is 050º B

(b) Point B is between S and E and ∠SOB=30º W O E
\ ∠ NOB = 180º - ∠ SOB = 180º - 30º = 150º
\The 3 - figure bearing of B is 150º 40o 30o

(c) The Point C lies between S and W direction C
and ∠SOC = 40º S

\ ∠ NOC (reflect angles, in clockwise
direction)

=180º + ∠ SOC = 180º + 40º = 220º

\The 3 - figure bearing of C is 220º

d) The Point D lies between N and W direction and ∠ NOD = 75º

\ ∠NOD(in clockwise direction from north line)

=360º - ∠ NOD = 360º - 75º = 285º

\The 3 - figure bearing of D is 285º.

Example 3: The bearing of a place B from the place A is 050º. Find the bearing of the
place A from the place B.

Solution: N1
Here, bearing of B from A ∠ NAB = 050º = θ N
bearing of A from B, ∠NBA = ?

We have,

Reversed bearing = θ +180º [θ ≤ 180º] 050o B?
\ Bearing of A from B = Bearing of B from A + 180º
= 050º + 180º A
= 230º

\ The bearing of the place A from the place B is 230º

114 4 Transformation

Prime Mathematics Book - 8

Exercise 2.1

1. Write the compass bearing of the following directions

(a) N (b) S (c) E (d) W
(h) NW
(e) NE (f) SE (g) SW (l) NWW

(i) NEE (j) SEE (k) SSE

2. Write the 3 - figure bearing for the following compass bearings

(a) N30ºE (b) S40ºE (c) S75ºW (d) N45ºW

(e) N0ºE (f) S10ºE

3. Write the compass bearings for the following 3 - figure bearings. Also show each in
figure.

(a) 060º (b) 090º (c) 140º (d) 225º
(e) 300º (f) 180º (g) 270º (h) 351º

4. Measure the angles and write the 3 - figures bearing of the point P from the point

O in each of the following figures.

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

P P

O O

ON PP O

N

(e) (f)

P
P OO

5. Measure the angles and write the compass bearings of the point P from the point

O in each of the following figures. N

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

P P

O O

O PP o

Bearing and scale drawing 115

Prime Mathematics Book - 8 (f) N

(e)
N

O PP O

6. (a) The bearing of a place Y from a place X is 055º, find the bearing of X from the
place Y.

(b) If the bearing of Pokhara from Kathmandu is 300º, find the bearing of Kathmandu
from Pokhara.

(c) The bearing of a point B from another point A is 090º, find the bearing of the
point A from the point B.

(d) If the bearing of the point Q from the P is 180º, what is the bearing of the point
P from the point Q ?

7. From the given map, find the bearings of the following airports from Tribhuvan
International Airport (T.I.A.).

(a) Simara
(b) Biratnagar
(c) Jomsom
(d) Pokhara
(e) Jumla
(f) Nepalgunj
(g) Phaplu
(h) Gautam Buddha

116 4 Transformation

Prime Mathematics Book - 8

2.2 Scale drawing:

Scale drawing is the representation of an actual object. It is either reduction or en-

largement of two dimensional picture. Standard maps are drawn to a certain scale or scale

factor which is the ratio that compares the map distance to the actual distance in real

field. = map distance
Thus scale (scale factor) actual distance

= size of drawing
actual size of object

The map of Nepal given below is drawn
to the scale 1cm to 40km. This means 1cm in map
Nepal

represents 40km in the real field.
Here:
Scale = 1 cm
40 km

1 cm 0 40 km 80 km
= 40,00,000 cm

= 1: 4000000
0 40 km 80 km

Which is given in the map as which means the distance of 1cm in the map is 40km
and 2cm is 80km in real field.

Example 1: A diagram is drawn to the scale of 1:250. Find the length in real field if the
length of the diagram is 5cm.

Solution:

Here, given scale = 1: 250

map distance =5cm

actual length (x)= ?

we have,

scale = map distance
actual distance

1 = 5cm
250 X

x = 1250cm

= 1250m
100

= 12.5m

\ The length in real field is 12.5 m.

Bearing and scale drawing 117

Prime Mathematics Book - 8

Example 2: Map of Bagmati zone is given below to the scale of 1:2000000. Find the

bearing and distance from Kathmandu to:

(a) Melamchi (b) Patan Darbar Square (c) Trisuli (d) Palanchok



Trishuli N
Melamchi

Kathmandu

0Km 20km 40km Patan Palanchowk
Purbar square

Solution: a) Kathmandu to Melamchi : bearing = 0750 and map distance = 2cm

∴ Actual distance = 2 × 20km = 40km

b) Kathmandu to Patan Durbar-Square : bearing = 1850 and map distance = 0.4cm

∴ Actual distance = 0.4 × 20km = 8km

c) Kathmandu to Trishuli : bearing = 3200 and map distance = 2.2cm

∴ Actual distance = 2.2×20km = 44km

d) Kathmandu to Palanchowk : bearing = 1200 and map distance = 3cm

∴ Actual distance = 3 × 20km = 60km

Example 3: Reduce the figure to the scale 1:3. C
Solution: Here, the D ABC is an equilateral triangle of side 6cm scale = 1:3 C'

length in scale drawing = 1
3
real length

X = 1 B A
6cm 3 B' A'

\ x = 2cm

Thus, for the scale drawing A' B'= B' C'= C' A'= 2cm taking
sides A' B'= B' C'= C' A' = 2cm, equilateral triangle A' B' C' is
completed.

118 4 Transformation

Prime Mathematics Book - 8

Example 4: An aeroplane flies 300 km from A to B at the bearing of 30º and then flies

400 km from B to C at the bearing of 120º. Find the distance and bearing of

C from A by drawing scale drawing of suitable scale.

Solution: N N1
Here, AB =300 km and

BC=400 km 120º
Let 1cm = 100km (HCF of 300 & 400) B

\ AB =300 km = 3cm and 30º 4cm
BC= 400 km = 4cm 3cm

Bearing of B from A= 30º and bearing of C A C

from B is 120º. From A, north line AN is 5cm

drawn and ∠NAB= 030º and AB =3cm are taken. Again at B, north line BN1 is
drawn and ∠N1BC= 120º and AC =4cm are taken.
AC is completed.

By measurement AC = 5cm and ∠NAC = 85º
∴ Distance of C from A = 5×100 km = 500 Km and bearing of C from A is 850.

Exercise 2.2

1. The given figure is the map of a rectangular plot drawn to the 4 cm
scale of is 1:200. Find the actual length and breadth of the plot.
6 cm
2. What is the actual distance between two places represented by
3.5cm on a map drawn to the scale 1:1000000 ?

3. What may be the distance between two places on map in centimetre between two
places which are 50km away in real field? The map is drawn to the scale 1:500000.

4. Actual distance between two places is 32km, what is the distance between the two
places on the map drawn to the scale 1:1000000 ?

5. Actual distance of 12km is represented on a map by 6cm, find the scale factor.

6. Reduce the figures to the scale 1:2 (b) A

(a) P Q 6cm

10cm

4cm

S 4cm R B 8cm C

Bearing and scale drawing 119

Prime Mathematics Book - 8 (b) 1.5cm
1.5cm
7. Enlarge the following figure to the scale 3:1 3cm 1.5cm
(a) A

1.5cm

B 20cm C

3cm

8. From a place A, a helicopter flies 300km in the bearing of 0400, then changes its
direction and flies 400km in the bearing of 1100. By scale drawing of suitable scale
find the distance and bearing of the final destination from the first point.

9. A boat moves from a island A, 60km in the bearing of 1200 to the island B. Then it

moves 30km in the bearing of 0300 to the island C. Construct a scale drawing to the
suitable scale and find the distance of the island A from the island C.

10. Study the map of a part of Nepal and find the distance and bearing of

a) Kathmandu to Mt. Everest b) Kathmandu to Pokhara

c) Kathmandu to Mustang d) Kathmandu to Kanchanjungha



0Km 40km 80km

120 4 Transformation

Applied Mathematics Book - 8 Prime Mathematics Book - 8

Area Revision Test (Transformation)

1. (a) write the names of fundamental transformations.

(b) What do you mean by isometric transformation? Which of the four fundamental
transformation is non-isometric ?

2. (a) Draw the image of ∆ ABC under translation by the vector

given by the given directed line segment. A

PM

(b) Copy the given figure approximately and draw the Q B C
reflection of ∆PQR on the line MN. N A

R

B
o

(c) Rotate the triangle ABC about the point o through 90°
C

3. (a) Under translation by vector(2, -3), P maps to P’(3, -1) find the co-ordinates of the point P.

(b) Find the image of point A(2, 5) under reflection on y-axis.

4. (a) Find the vertices of the image of ∆ABC with vertices A(4, 2), B(3, 6) and c(-2, -1) under
rotation about origin through 180°

(b) Reflect the ∆ABC on x-axis where co-ordinates of the vertices of ∆ABC are A(6, -2), B(-2,
-1) and C(2, 5) and plot the ∆ABC and ∆A'B'C' in the same graph.

5. (a) Write the compass bearing of : (i) SE (ii) NWW

(b) Write there figure bearing of the compass bearing : (i) S75°W (ii) N90°W

(c) Write the compass bearing of : (i) 060° (ii) 140°

(d) Write the bearing of the point P from the point O in each case.

(i) N (ii) N

PO

OP

6. (a) The bearing of the point B from the point A is 135°. Find the bearing of A from the point B.

(b) What is the actual distance between two places represented 4cm on a map drawn to the
scale 1 : 1000000 ?

7. (a) Reduce the given figure to the scale 1 : 2 A

Transformation (b) Ram moves 300m from A to B at the bearing of 30° and then moves
400m from B to C at the bearing of 120°. Find the distance and
bearing of C from A by drawing scale drawing of suitable scale.

B 5cm C

Bearing and scale drawing 121

Answers

Exercise 1.1 (Translation)

1. Show to your teacher. 2. Show to your teacher.
3. (a) (i)(-2,3) (ii) (0,6) ( ((b)
−2 (c) (1,6) (d) (1,4)
3

4. (a) A'(2, 5), B'(6, 3), C'(0, 0) (b) A'(−1, −3), B'(−4, −2), C'(−1,− 5), D'(2, −2) (c) A'(−3, 6), B'(−1, 1), C'(−4, 1)

Y A’ Y A’ Y
A
A B’ B A D
X o X C’ B’
X’ X’ D’
C’ B’ A’ C

B

C C’ X’ o X
Y’ Y’ CB

Y’

Exercise 1.2 (Reflection)

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

3. (a) (2, 3), (−2, -3) (b) (-3, 9), (3, -9) (c) (7, -8), (−7, 8) (d) (-10, -14), (10, 14)

4. (a) i. (3, -4), (-3, 4) ii. (-5, 3), (5, -3)

(b) A'(−2, 0), B'(4, −3), c'(2, 2) (c) A'(6, 2), B'(−2, 2), C'(2, −5)

Y C
C Y C ’
B’ A’
X’ o X

X’ A’ o A X

BA

B B’
Y’ C’

Y’

(d) A'(−3, −1), B'(−4, 1), C'(−1, 4) and A"(3, −1), B"(4, 1), C"(1, 3)

Exercise 1.3 (Rotation)

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

4. Show to your teacher 5. (a) (i) A'(2, 4) (ii) A'(−2, −4) (iii) A'(−4, 2) (b) A (5, 3)
e) A'(1, −1), B'(−2, 3), C'(3, 1)
(c) A'(−5, 2), B'(−2, 6), C'(2, 3) (d) A'(−4, −1), B'(−2, −5), C'(2, 1)
B’ C Y
YB

C’ A C’
AX o X

X’ o A’
A’ Y’

C X’
B
B’ Y’

122 Answers

Exercise 2.1 (Bearing)
1. (a) N 0°E (b) S 0°S (c) N 90°E (d) N 90°W (e) N 45°E (f) S 45°E (g) S 45°W (h) N 45°W (i) N 67.5°E
(j) S 67.5°E (k) S 22.5°W (l) N 67.5°W
2. (a) 030° (b) 140° (c) 255° (d) 315° (e) 000° (f) 170°
3. (a) N 60°E (b) N 90°E (c) S 40°E (d) S 45°W (e) N 60°W (f) S 0°E (g) N 90°W (h) N 9°W
4. (a) 060° (b) 125° (c) 260° (d) 290° (e) 270° (f) 090°
5. (a) N 60°E (b) S 65°E (c) S 50°W (d) N 55°W (e) N 90°E (f) N 90°W
6. (a) 235° (b) 120° (c) 270° (d) 360° 7. Show to your teacher.

Exercise 2.2 (Scale drawing)

1. l = 12m, b = 8m 2. 35km 3. 10cm 4. 3.2cm 5. 1:200000 4.5cm
9cm
6. (a) P' Q' (b) 7. (a) A' (b) 9cm
4.5cm
2cm 5cm
3cm 4.5cm

S' R' 4cm B' 60cm C'

2cm

8. N N 9. N

B 1100 0920 6.7cm C
1200 N
A

3cm 4cm 6cm 0300 3cm

0400 830 5.8cm B
Ac=580km, 0830
A C

10. (a) 136km, 0750 (b) 152km, 2850 (c) 228km, 3150 (d) 240km, 0850

Answers 123

Estimated periods - 10

At the end of this unit the student will be able to Objectives:

● know about proper sets improper sets, supper set and cardinalities of set.
● get the idea of set operation.
● demonstrate the sets and set operation in venn diagram.
● use the cardinality relations to solve simple word problems on sets.

Teaching Materials:

Flash cards, number cards, charts of different things.

Activities:

It is better to
● explain about subsets, proper subsets and super sets with examples.
● discuss about cardinalities of sets.
● explain about different set operation and use of Venn diagram.
● explain the cardinality values.

Prime Mathematics Book - 8

Unit - 1 Sets Estimated period : 10

Historical fact:

John Venn (1834-1923) an English Logician first illustrated "Classes of objects" or sets
diagrammatically in his book "Symbolic Logic" which we call today Venn diagram.
Leonard Euler (1707-1783, Switzerland) also used such diagrams to represent sets (but
not formally published). So Venn diagrams are sometimes called Venn-Euler diagram.

1.1 Introduction

Sets are collections of well defined objects. The " well defined objects" refers to those
objects which we can easily identify and say whether they belong to that set or not.

For example:

A = {Set of days of a week} = {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}
B = { Set of Vowels of English alphabets} = { a, e, i, o, u}

Each object of the set is called an element or a member of the set. Sets are denoted by
capital letters A, B, C, ... e.t.c. and the elements of sets are enclosed within braces { }
separated by commas. In the set B = {a, e, i, o, u}, a is an element of the set B and we
write a ∈ B (read as a belongs to B or a is a member of the set B). The symbol ∈ stands
for " is member of " or "belongs to ". As b is not an element of B, we write b ∉ B where
the symbol ∉ stands for "is not a member of" or "doesn't belong to".

Describing a set:

Sets are described by the following methods:
1. Description method: In this method we describe the common property/ properties
of the elements of the set in words inside the braces.
For example: A = {Country of SAARC}, B = {Whole number less than 20}

2. Listing method (Roster form): In this method we enumerate or list all the elements
of a set inside the braces.

For example:
A = {Nepal, Indian, Bhutan, Bangladesh, Pakistan, Sri-lank, Maldives, Afghanistan}
B = {0, 1, 2, 3, ...... , 19}

3. Set builder method (Rule method): In this method, the elements of a set are
represented by a variable specifying with defining property inside braces as.

W = {x:x is a whole number}, S = {x: x is a square number}

Sets 125

Prime Mathematics Book - 8

Various sets:

On the basis of number of elements, sets may be:
A. The empty (null or void) set: The set containing no elements is called the null set
or the empty set or the void set. It is denoted by { } or φ (phi).
For example:
A = {x:x is a natural number between 1 and 2}, B = {female president of Nepal}

B. Unit set or singleton set: A set containing only one element is called a singleton set
or unit set.

For example: A = {5} B = {set of planets of solar system having life in it}

C. Finite set: A set which contains a finite number of elements is called a finite set.
For example:

A = {set of digits of decimal numerals}, B = {x : x is a square number, x ∈ N, x ≤ 100}

D. Infinite set: A set containing infinite number of elements in called an infinite set.
For example: N = {1, 2, 3, .......}, M = {stars in the universe}

On the basis of relationship between two or more than two sets, sets may be

E. Equal sets: Two sets are equal if they contain exactly the same elements.
For example: A = {a, t, e} and B = {e, a, t} So, A = B

F. Equivalent sets: If the two sets have equal number of elements, they are said to be
equivalent.

Consider, A = {1, 2, 3, 4, 5} and B = {a, b, c, d, e}
Here A and B have not same elements but have same number of elements. So A and

B are equivalent and we unite A ~ B.

G. Overlapping sets: If two sets A and B have one or more common elements then A
and B are said to be overlapping or intersecting.

For example: A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, B = {2, 3, 5, 7, 11}

H. Disjoint sets: If two sets have no elements in common, the two sets are called
disjoint sets.

I. Sub set and Super set:
Sets consider two sets A = {1, 3, 5, 7, 9} and B = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. Here

each element of set A is also an element of set B and we say set A is subset of set B.
Mathematically we write A ⊆ B and B is called super set of A and we write B ⊇ A.

J. Proper Subset: If every element of set A is an element of set B and at least one one
element of B is not an element of A, then the set A is called proper subset of set B
and mathematically we write A ⊂ B.

126 5 Sets

Prime Mathematics Book - 8

Note: - φ is a subset of every set

- A set is not a proper subset of itself.

- φ is not a proper subset of itself.

K. Subsets and Power Set: Consider a set A = {a, b}. Taking the elements of the set the
following subsets can be formed.

Taking non element φ or { }

Taking single element {a}, {b}

Taking two elements {a, b}

The set of all the subsets of A = {φ , {a}, {b}, {a, b}} which is called the power

set of A. Similarly consider B = {1, 2, 3}. Following subsets of B can be formed.

φ, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}. The power set of B is { φ,
{1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}. The set of all the subsets that can be
formed from the elements of a set is called its power set. We observed in the above
examples that the number of subsets having n elements = 2n. And the number of
proper subsets of a set having "n" elements is 2n-1.

L. Universal set: A set which contains all the sets under consideration as subsets is
called the universal set. It is denoted by U or ∪ (psi). For the sets A = {1, 2, 3} and
B = {2, 3, 5, 7} the universal set might be {1, 2, 3, 4, 5, 6, 7, 8, 9} or {0, 1, 2, 3, 4, 5,
6, 7, 8, 9} or {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} e.t.c.

Note: The choice of a universal set is not unique

Exercise 1.1

1. Define: b) finite set c) sub set d) super set
a) sets f) power set g) universal set
e) proper sub set

2. Which of the following collections are sets?

a) A collection of strong men. b) A collection of countries of Asia.

c) A collection of beautiful girls. d) A collection of districts of Nepal.

3. If A = {p, q, r, s, t}, B = {1, 4, 9, 16, 25}, use ∈ or ∉ in the following.

a) o A b) 25 B c) 10 A d) r A

e) q B f) 1 B

4. Represent the given set as stated in the brackets.
a) A = {Whole numbers less than 10} [listing/tabular form]
b) B = { Jan, Feb, Mar, Apr, May, June, July, Aug, Sep, Oct, Nov, Dec} [descriptive form]

Sets 127

Prime Mathematics Book - 8 [set builder form]
[tabular form]
c) C = {0, 1, 4, 9, 16, 25} [rule method]
d) D = {x : x = 3n+1, n ∈ w} [descriptive form]
e) E = {0, 1, 8, 27, 64, 125}
f) F = {Sun, Mon, Tue, Wed, Thu, Fri, Sat}

5. Identify whether the following sets are null or finite or infinite singleton.

a) {0} b) {x : x + 3 = 7, x ∈ N}

c) {an even prime number} d) {natural numbers}

e) {male students of Padma Kanya College}

f) {planets bigger than Jupiter in Solar System}

g) {satellite of Venus}

h) {x : -10 ≤ x ≤ 10, x ∈ Z}

6. List the elements of the following pairs of sets and state with reasons whether they
are equal or equivalent.

a) A = {1, 3, 5, 15} and B = {factors of 15}
b) C = {x : x ≤ 5, x ∈ N} and D = {y : 0 < y ≤ 5, y ∈ W}
c) E = {x : x ≤ 5, x ∈ N} and F = {vowels of English alphabets}
d) G = {letters of the word tea } and H = {letters of the word eat}
e) I = {a, b, c, d, e} and J = {last five letter of English alphabets}

7. List the elements of the following pairs of sets and state whether they are overlapping
or disjoint.

a) A = {factors of 10} and B = {prime number less than 10}
b) P = {x : x is an even number 5 < x < 12} and Q = {x : x is a prime number 5 < x ≤ 12}
c) M = {x : x ∈ N} and N = {x : x ≤ 0, x ∈ Z}
d) R = {x : x is a prime number less than 20} and S = {x : x is an odd number, x ∈ N, x ≤ 20}

8. Form the subsets of:

a) A = {1} b) B = {a , b} c) C = {p, q, r} d) D = {a, b, c, d}

9. Write the proper subsets of:

a) A = {a} b) B = {1, 2} c) C = {a, b, c} d) D = {w, x, y, z}

10. Write the power sets of:

a) P = {a} b) Q = {a , b} c) R = {a, b, c}

128 5 Sets

1.2 Venn diagram: Prime Mathematics Book - 8

A Venn diagram is merely a closed A B
figure used to denote the set of all points 31
within the figure.
259
A Venn diagram is used to represent 7
sets and their relations or operations. On
representing sets in Venn diagram, universal A B U
set is represented by a rectangle and its 31
subsets by circles or ovals within it showing
the elements within their respective parts . If 259
A = {2, 3, 5, 7} and B = {1, 3, 5, 7, 9}, then the 7
sets can be represented in Venn diagram as in
the figure. If the universal set is considered to A B U
be U = {0, 1, 2, 4, 5, 6, 7, 8, 9}, than the Venn 04
diagram showing U, A and B is;
31
Venn diagram showing other relations: 259

A U 7
A 68

B UU
B AB

A and B are disjoint A and B are overlapping A ⊂ B (A is proper subset of B)

Set operations:

There are four operations related to sets.

1) Union of sets 2) Intersection of sets

3) Difference of two sets 4) Complement of a set

1. Union of sets:

Union of two sets A and B is the set of elements which are in A or in B or in both. It
is denoted by A ∪ B = {x : x ∈ A or x ∈ B}. Under different conditions, union of sets can
be illustrated in Venn diagrams as;

Sets 129

Prime Mathematics Book - 8

A U U U
A B AB

B

A and B are disjoint A and B are overlapping A⊂B

Union of three sets:

A U U A U
B A B B


C

A, B, C are disjoint sets A and B are overlapping and C A, B, C are overlapping

U is disjoint with A and B U
A B C A
U B
C
AB C

A and B are overlapping, B A ⊂ B but C is disjoint with A C⊂B⊂A
and C are overlapping But A and B

and C are disjoint

Example 1: If U = {x : x ∈ w, x < 10}, A = {x : x < 6, x ∈ N} and B = {x : x ∈ N, 3 < x < 8},

find A ∪ B and illustrate in Venn diagram. A B U

Solution:

Here, U = {x : x ∈ W, x < 10} 14 6
= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} 7
A = {x : x < 6, x ∈ N} = {1, 2, 3, 4, 5} 25
B = {x : x ∈ N, 3 < x < 8} = {4, 5, 6, 7} 9
3
08

∴ A ∪ B = {1, 2, 3, 4, 5} ∪ {4, 5, 6, 7} = {1, 2, 3, 4, 5, 6, 7}

Shaded region shows A ∪ B in the Venn diagram.

130 5 Sets

Prime Mathematics Book - 8

Example 2: If A = {x : x is an odd number, x ≤ 10 , x ∈N} and B = {y : y is an even number, y
≤ 10, y ∈ N}, list the elements of the set A ∈ B and represent in a Venn diagram.

Solution: Here, A = {x : x is an odd number, x ≤ 10, x ∈ N} = {1, 3, 5, 7, 9}

B = { y : y is an even number, y ≤ 10, y ∈ N} A 0 BU
= {0, 2, 4, 6, 8, 10} 4
13 8 2
∴ A ∪ B = {1, 3, 5, 7, 9} ∪ {0, 2, 4, 6, 8, 10} 57 6
= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} 10
9

The operation is shown in Venn diagram as; A∪B

Example 3: If A= {x, y, z} and B = {u, v, w, x, y, z}, list the U
elements of A ∪ B and represent it in Venn diagram. AB

Solution: Here, A = {x, y, z} and B = {u, v, w, x, y, z} u
xy v
∴ A ∪ B = {x, y, z} ∪ {u, v, w, x, y, z} zw

= {u, v, w, x, y, z} A∪B

The shaded part represent A ∪ B.

Example 4: If U = {1, 2, 3, ..., 10}, A = {2, 3, 4, 5} B = {3, 5, 7, 9} and C = {5, 6, 7, 8},
list the elements of A ∪ B ∪ C and represent it in a Venn diagram.

Solution: Here,
U = {1, 2, 3, ....., 10}, A = {2, 3, 4, 5}, B = {3, 5, 7, 9}, C = {5, 6, 7, 8}

∴ A ∪ B ∪ C = {2, 3, 4, 5} ∪ {3, 5, 7, 9} ∪ {5, 6, 7, 8}

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

The shaded part of the Venn diagram 2 39
4

57

represent the set A ∪ B ∪ C 68

2. Intersection of Sets: A∪B∪C

Intersection of two sets A and B is the set of all the elements which are in both sets

A and B. Intersection of two sets A and B is denoted by A ∩B (read as "A intersection B" or

"A cap B").

Sets 131

Prime Mathematics Book - 8 U
B
Intersection of sets can be shown in Venn diagram as.

U
A B A


A∩B A∩B=φ U
B
A and B are overlapping A and B are disjoint.

U A
AB

A∩B=A C

A⊂B

A∩B∩C

A, B and C are overlapping

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

and represent it in a Venn diagram. U

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

∴ A ∩ B = {a, e, i, o, u} ∩ {a, b, c, d, e} i ab
= {a, e} oe c
ud

A ∩ B is shown in Venn diagram with shadow.

A∩B

Example 6: If A = {x : x is a factor of 25} and B = {x : x is a factor of 50}, write A ∩ B in

set builder form and show it in Venn diagram. U
AB
Solution: Here, A= {x : x is a factor of 25} = {1, 5, 25}
2
B = {x : x is a factor of 50} = {1, 2, 5, 10, 25, 50} 1 5 10
25 50
Now, A ∩ B = {1, 5, 25} ∩ {1, 2, 5, 10, 25, 50}

= {1, 5, 25}

∴ A ∩ B = {x : x is a factor of 25} A∩B=A
The shaded region in the Venn diagram shows A ∩ B.

132 5 Sets

Prime Mathematics Book - 8

Example 7: If A = {x : x ∈ N, 6 ≤ x ≤ 12}, B = {x : x is a multiple of 3, x ∈ N, x ≤ 21} and

C = {6, 12, 18, 24}, list the elements of the A ∩ B ∩ C and illustrates it in a

Venn diagram. A U
B

solution: Here, A = {x : x ∈ N, 6 ≤ x ≤ 12} 8 3
7 10 21
= {6, 7, 8, 9, 10, 11, 12} 11
6 15
B = {x : x is a multiple of 3, x ∈ N, x ≤ 21} 12 18

= {3, 6, 9, 12, 15, 18, 21} 68 C
C = {6, 12, 18, 24} 24

A∪B∪C

∴ A ∩ B ∩ C = {6, 7, 8, 9, 10, 11, 12} ∩ {3, 6, 9, 12, 15, 18, 21} ∩

{6, 12, 18, 24} = {6, 12}

The shaded region in the Venn diagram represent A ∩ B ∩ C.

3. Difference of two Sets:

The difference of two sets A and B denoted by A−B is the set of all the elements
that are in A but not in B. Similarly the difference of two sets B and A denoted by B−A is the
set of all the elements that belong to B but do not belong to A. Under different conditions,
difference of two sets A and B are show in Venn diagram as.

When A and B are disjoint A U
U B

AB


A-B=A B-A=B

When A and B are overlapping U
U AB

AB

A-B B-A

Sets 133

Prime Mathematics Book - 8

When one set is proper subset of their.

U U
BA A B

A-B B-A

Example 8: If A = {a, e, i, o, u} and B = {a, b, c, d, e}, list the elements of the set A − B

and B − A and illustrate in Venn diagram.

Solution: Here, A = {a, e, i, o, u} and A BU A BU

B = {a, b, c, d, e} ib ib
o aa c o aa c
∴ A − B = {a, e, i, o, u} u ee d u ee d

− {a, b, c, d, e}

= {i, o, u} A-B B-A

And, B − A = {a, b, c, d, e} − {a, e, i, o, u}

= {b, c, d}

The differences are shown by shaded region in the Venn diagram.

Example 9: If A = {0, 2, 4, 6, 8} and B = {1, 3, 5, 7, 9}, list the elements of the sets A − B
and B − A and represent them in Venn diagram.

Solution: Here, A = {0, 2, 4, 6, 8} and B = {1, 3, 5, 7, 9}

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

= {0, 2, 4, 6, 8} = A [A and B being disjoint]

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

= {1, 3, 5, 7, 9} = B [A and B being disjoint]

Shaded parts in the Venn diagram represent the difference.

BU A B U

A

02 13 02 13
4 5 4 5

68 79 68 79

A-B=A B-A=B

134 5 Sets

Prime Mathematics Book - 8

Example 10: If A= {x : x is a digit of Hindu Arabic numerals} and B= {x : x is a prime
number, x < 9, x ∈ N}, list the elements of the set A − B and illustrate if in
Venn diagram.

Solution: Here, A = {x : x is a digit of Hindu Arabic numerals}

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

B = {x : x is a prime number, x < 9 , x ∈ N} = {2, 3, 5, 7}

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

A − B is shown by the shaded part in the Venn diagram.

4. Complement of sets:

Complement of a set A under universal set U is the set of all the elements of U which
do not belong to A and is denoted by Ac or A' or A.

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

i.e. A = U − A complements of various sets are shown in Venn diagram as:

U U U U
AA BA BA B

A' (A ∪ B)' (A ∩ B)' (A - B)

Facts about complement of sets.

1) Complement of the complement of a set is the set itself.

i.e. (A')' = A

It can be shown as (A')' = (U - B)' = U − (U − A) = A

2) De Morgan's law: (a) (A ∪ B) = A ∩ B , (b) (A ∩ B) = A ∪ B

Proof: (a) (A ∪ B) = {x : x ∉(A∪B)} (b) (A ∩ B)' = {x : x ∉ (A ∩ B)}
= {x : x ∉ A and x ∉ B}
= {x : x ∉ A and x ∉B}

= {x : x ∈ A and x ∈ B } = {x : x ∈ A' and x ∈ B'}

=A ∩ B = {x : x ∈ (A' ∪ B')}

∴ (A ∪ B) =A ∩ B =A' ∪ B'

∴ (A ∩ B)' = A' ∪ B'

Sets 135

Prime Mathematics Book - 8

Example 11: If U = {x : x ∈ W, x ≤ 12}, A = {x : x is a prime number, X < 12} and B = {x :

x ∈ W, x < 10}, list the following sets and illustrate them in Venn diagram.

(a) A' (b) (A ∪ B)' (c)(A ∩ B)' (d) (B - A)'
Solution: Here, U = {x : x ∈ W, x ≤ 12} = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

A = {x : x is a prime number, X ≤ 12} = {2, 3, 5, 7, 11}

B = {x : x ∈ W, x < 10} = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

(a) A' = U − A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} − {2, 3, 5, 7, 11}

= {0, 1, 4, 6, 8, 9, 10, 12} A B U

A' is shown in Venn diagram as. 01
23 4 6
11 57 8 9

(b) A ∪ B = {2, 3, 5, 7, 11} ∪ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} 12

10 A'
BU
= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11}

Now: (A ∪ B)' = U − (A ∪ B) A

= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} − {0, 1, 11 01
2, 3, 4, 5, 6, 7, 8, 9, 11} = {10, 12} 23 4 6
57 8 9
A ∪ B)' is represented by the shaded part in the 10
Venn diagram as. 12

(c) A ∩ B = {2, 3, 5, 7, 11} ∩ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} A (A ∪ B)'
= {2, 3, 5, 7}
BU
Now, (A ∩ B)' = U − (A ∩ B)
01
= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} 11 23 4 6
57 8 9
− {2, 3, 5, 7} 10
= {0, 1, 4, 6, 8, 9, 10, 11, 12} 12

(A ∩ B)'

(d) B − A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} − {2, 3, 5, 7, 11}

= {0, 1, 4, 6, 8, 9} A BU

Now, 01
23 4 6
(B - A)' = U − (B − A) 11 57 8 9 (B - A)'

= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 10 12
11, 12} − {0, 1, 4, 6, 8, 9}

= {2, 3, 5, 7, 10,11 12}

(B - A)' is illustrated in the adjoining Venn diagram.

136 5 Sets

Prime Mathematics Book - 8

Example 12: If U ={0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, A ={2, 3, 5, 7} and B ={2, 4, 5, 6}, verify
the following. (a) (A)' = A (b) (A ∪ B)' = A' ∩ B' (c) (A ∩ B)' = A' ∪ B'

Solution: Here, U = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} A = {2, 3, 5, 7} and B = {2, 4, 5, 6}

Now,

(a) A' = U − A

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

Again A = U − A'

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

=A

∴ (A')' = A Verified.

(b) A ∪ B = {2, 3, 5, 7} ∪ {2, 4, 5, 6} = {2, 3, 4, 5, 6, 7}

∴ (A ∪ B)' = U − (A ∪ B)

= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} − {2, 3, 4, 5, 6, 7} = {0, 1, 8, 9} __ (i)

Again, A' = U − A

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

B' = U − B = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}−{2, 4, 5, 6} = {0, 1, 3, 7, 8, 9}

∴ A' ∩ B' = {0, 1, 4, 6, 8, 9} ∩ {0, 1, 3, 7, 8, 9} = {0, 1, 8, 9} ______ (ii)

Hence, we get from the relation (i) and (ii)

∴ (A ∪ B)' = A' ∩ B' Verified.

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

∴ (A ∩ B)' = U − (A ∩ B)

= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} − {2, 5} = {0, 1, 3, 4, 6, 7, 8, 9} __ (i)

Again A' = U − A

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

B' = U − B

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

∴ A' ∪ B' = {0, 1, 4, 6, 8, 9} ∪ {0, 1, 3, 7, 8, 9} = {0, 1, 3, 4, 6, 7, 8, 9} ___ (ii)

Hence, we get from the relation (i) and (ii)

∴ (A ∩ B)' = A' ∪ B' Verified.

Exercise 1.2

1. If A and B are two overlapping sets under the universal set U. Draw the Venn diagrams

shading the following.

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

(e) A' (f) (A ∪ B)' (g) (A ∩ B)' (h) (A - B)'

Sets 137

Prime Mathematics Book - 8

2. If A ⊂ B and A and B are subsets of universal set U. Illustrate the following in Venn diagram.

(a) A ∪ B (b) A ∩ B (c) B − A

3. Write in set notation, the following shaded region in the diagram.

(a) A U (b) A U
B B

(c) A U (d) A U
B (f) A B

U
(e) A U B
B
C
(g) A U (h) A U
B B

C

4. Copy the Venn diagram. A B U
Shade and list the elements of the following.
6
(a) A − B (b) A ∪ B (c) A ∩ B 15 3 12
7 11 9

(d) (B - A)' (e) (A ∪ B)' (f) (A ∩ B)' 0 2 4 8 10
A B U
5. List the following sets.
Also copy the diagram and shade in each case. 5 36
4 21
(a) A ∪ B ∪ C (b) A ∩ B ∩ C
(c) (A ∪ B) − C (d) (A ∪ B ∪ C)' 8 C
(e) (A ∪ B) ∩ C (f) (A ∩ B) ∪ C

079

138 5 Sets

Prime Mathematics Book - 8

6. If A = {x : x is a factor of 6} and B = {x : x ∈ N, x ≤ 5}, list the following.

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

7. If A and B are subsets of the universal set U. Where, U = {x : x ∈ N, x ≤10}, A = {x : x
is a prime number} and B = {x : x is an odd number}, list and represent the following

sets in a Venn diagram.

(a) A ∪ B (b) A ∩ B (c) B − A (d) (A ∪ B)' (e) B' (f) (A ∩ B)'

8. A and B are subsets of universal set A. If U={x : x ∈ N, x ≤ 15}, A={x : x ∈ N, x ≤ 6},
B = {x : x is even number, x ≤ 12} and C = {x : x is multiple of 3 , x ≤ 15}, list the
elements of the following sets and also show them in a Venn diagram.

(a) A ∪ B ∪ C (b) A ∩ B ∩ C (c) (A ∪ B) − C
(d) (A ∪ B) ∩ C (e) (A ∪ B ∪ C)' (f) (A ∩ B ∩ C)'

9. if A = {b, c, d, e}, B = {a, c, e, g} and C = {d, e, f, g} are the subsets of universal

set U= {a, b, c, d, e, f, g, h}, verify the following

(a) A ∪ (B ∪ C) = (A ∪ B) ∪ C (b) (A ∩ B) ∩ C = A ∩ (B ∩ C)

(c) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) (d) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

(e) (A')' = A (f) (A ∪ B)' = A' ∩ B' (g) (A ∩ B)' = A' ∪ B'

1.3 Cardinal number and cardinality relation between sets :

Cardinal number of a set A denoted by n(A) is the number of elements of the set A.
Cardinal number is also called cardinality.

If A = {a, e, i, o, u}, then cardinal number of A or cardinality of A = n(A) = 5
Let U = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
A = {1, 2, 3, 4, 5} and B = {2, 3, 5, 7} the
∴ n(U) = 10, n(A) = 5, n(B) = 4
A ∪ B = A ∪ B = {1, 2, 3, 4, 5} ∪ {2, 3, 5, 7} = {1, 2, 3, 4, 5, 7}
∴ n(A∪B) = 6
A ∩ B = A ∩ B = {1, 2, 3, 4, 5} ∩ {2, 3, 5, 7} = {2, 3, 5}
∴n(A ∩ B) = 3
(A ∪ B)' = U − ( A ∩ B) = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}−{1, 2, 3, 4, 5, 7} = {0, 6, 8, 9}
∴ n( A ∪ B)' =4

Similarly A − B = {1, 2, 3, 4, 5} − {2, 3, 5, 7} = {1, 4}
∴ n(A − B) = 2

n(A − B) is the number of elements contained in the set A − B which is also denoted

by n0(A) read on 'n only A'.
Thus n(A−B)= n0(A)
Also we can take n0(A) = n(A) − n(A ∩ B).

Sets 139

Prime Mathematics Book - 8

Cardinality relation of union of two sets

A. When A and B are disjoint. A U
(i) n(A ∪ B) = n(A) + n(B) B n(U)

(ii) n(U) = n(A ∪ B) + n (A∪B)

(iii) n(A ∩ B) = 0 (Being disjoint sets)

(iv) n(only A ) i.e. n0(A) = n(A − B) = n(A) n(A ∪ B)

(v) n(only B) i.e. n0(B) = n(B − A) = n(B) n(A) n(B)

B. When A and B are overlapping.

Let, n0(A) = a, n0(B) = b and n(A ∩ B) = c, then obviously from the Venn diagram,
n(A) = a + c, n(B) = b + c and n(A ∪ B) = n0(A) + n0(B) + n(A ∩ B)
= a + b + c = (a + c) + (b + c) − c = n(A)+ n(B) − n(A ∩ B)

∴ n(A ∪ B) = n(A)+ n(B)− n(A ∩ B)

∴ n(A ∩ B) = n(A) + n(B)− n(A ∪ B) U
B
Also n(U) = n(A ∪ B) + n(A ∪ B) n(A) A n(B)
∴ n(A ∪ B) = n(U)− n(A ∪ B) n0(B) = b
n(A ∪ B)
and n(A ∪ B) = n(U) − n(A ∪ B)

Also,

n(only A )= n0(A) = n(A − B)= n(A) − n(A ∩ B) no(A) = a n(A ∩ B) = c
and n(only B )= n0(B) = n(B − A)

= n(B) − n(A ∩ B) If n(A ∪ B) = 0 then, n(A ∪ B) = n(U)

Example 13: If n(U) = 60, n(A) = 40, n(B) = 36 and n(A ∩ B) = 25, find

(a) n(A ∪ B) (b) n (A∪B) (c) n0(A) (d) n0(B)

Also illustrate the above information in a Venn diagram.

Solution: Here, we have n(U) = 60, n(A) = 40, n(B) = 36 and n(A ∩ B) = 25

(a) n(A∪B) = n(A) + n(B) − n(A ∩ B) Venn diagram
U
(b) n (A∪B) = 40+36−25 = 51
AB
= n(U)−n(A∪B) n(A) = 40 n(B) = 36
= 60−51 = 9 n0(B) = 11
n(A ∪ B) = 9
(c) n0(A) = n(A)−n(A∩B)

= 40−25 = 15

(d) n0(B) = n(B)−n(A∩B) = 36−25 = 11 n0(A) = 15 n(A ∩ B) = 25

Note: Showing the cardinality relation between sets in Venn diagram,
each loops (spaces) inside the rectangle should be filled.

140 5 Sets

Prime Mathematics Book - 8

Example 14: In a survey of 52 students of class VIII of a school, it was found that 35
like basket ball, 32 like cricket and 20 like both the games.

(a) Find the number of students who like either basket ball or cricket.
(b) Find the number of students who like neither basket ball nor cricket.
(c) Find the number of students who like only one game.
(d) Illustrate the above information in Venn diagram.

Solution:

Let the sets of those who like basket ball and cricket be B and C respectively.
Here, n(U) = 52, n(B) = 35, n(C) = 32, n(B ∩ C)=20
Now, we have;

(a) Number of students who like either basket ball or cricket n(B ∪ C) = n(B) + n (C) − n(B∩C)

= 35 + 32 − 20 = 47

(b) Number of students who like neither basket ball nor cricket

n (B∪C) = n(U) − n(B ∪ C) = 52 − 47 = 5

(c) Number of students who like only one game n0(B) + n0(C) = {n(B) − n(B ∩ C)}
+ {n(C) - n(B ∩ C)} = (35 − 20 )+ (32−20) = 15 + 12 = 27

(d) Venn diagram n(B) = 35 B U
C n(C) = 32

n0(C) = 12
n(B ∪ C)= 5

n0(B) = 15 n(B ∩ C) = 20

Example 15: There are 48 students in a class. If 32 students participate in sports and

28 participate in music. If every student has to participate at least one of

the activities sports or music,

(a) illustrate the above information in Venn diagram.
(b) how many students participate in sports as well as in music?

(c) how many students participate in music only?

Solution: Let the set of students who participate in sports be A and the set of those who

Participate in music be B.

Then, n(U) = 48, n(A) = 32, n(B) = 28

Let, n (A∩B) = x then n0(A) = n(A) − n(A ∩ B)

= 32 − x

n0(B) = n(B) − n(A ∩ B)

= 28 − x

Sets 141

Prime Mathematics Book - 8

Since every student has to participate at least one of the activities sports or music,

n (A∪B) = 0 . i.e. n(A∪B)=n(U) U n(U) = 48
(a) Venn diagram B n(B) = 28

A

n(A) = 32 n0(B)

n0(A) n(A ∩ B)
(b) From the Venn diagram

n(A∪B) = n0(A) + n0(B) + n(A ∩ B)
or, 48 = 32 − x + 28 − x + x
or, 48 = 60−x
or, x = 60−48
∴ x = 12
(c) No. of Students participated in music only
n0(B) = 28 - x = 28 - 12 = 16

Example 16: In an examination, 39% of the students passed in compulsory mathemat-
ics only and 30% passed in optional mathematics only. If 8% of the stu-
dents failed in both the subjects,

(a) find the percentage of students passed in both the subjects.
(b) find the percentage of students passed in compulsory mathematics.

(c) illustrate the above information in a Venn diagram.
Solution: Let the set of students passed in C.Maths = C and the set of those passed in Opt.

math = O
Here, n0(C) = 39%; no(O) = 30% ; n (C∪O) = 8%
Total percentage of students n(U) = 100%
Now,
(a) Percentage of students passed in both the subjects n(C∩O)=?
we have,
n(U) = n(C∪O) +n (C∪O)
or, n(U) = n0(C)+n0(O)+n(C∩O) +n (C∪O)
or, 100% = 39%+30%+n(C∩O) +8%
or, n(C∩O) = 100%−77% = 23%
∴23% of the students passed in both the subjects

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Prime Mathematics Book - 8

(b) Percentage of students passed in compulsory mathematics [n (C)]=?

Here, n(C) = n0 (C) + n(C ∩ O)
= 39% + 23% = 62%

∴62% of the students passed in Compulsory Maths.

(C) Venn diagram

Venn diagram n(o)
OU

C

n0(o) =30%
n(C ∪ O) =8%

n0(C) = 39% n(C ∩ O) = 23%

Exercise 1.3

1. (a) If n(U) = 50; n(A) = 27; n(B) = 21 and n(A ∩ B) = 12, where A and B are sub
sets of universal set U, find:

(i) n(A∪B) (ii) n (A∪B) (iii) n0(A) (iv) n0(B)

(b) If P and Q are the sets under universal set U where n(U) = 90 ; n(P) = 45;
n(Q) = 52 and n (P∪Q) = 15, find:

(i) n (P∩Q) (ii) n(P∪Q) (iii) n0(P) (iv) n0(Q)

(c) If A and B are two disjoint sets under universal set U where n(A) = 35; n(B)

= 27; and n (A∪B) = 18, find: (i) n(A∪B) (ii) n(U)

(d) If n(U) = 100%, n(A) = 50%; n(B) = 59% and n (A∪B) = 12%, find:

(i) n(A∪B) (ii) n(A∩B) (iii) n0(P) (iv) n(only B)

2. (a) In the given diagram, N and E denote the sets of students passed in Nepali

and English respectively. Answer the following. N EU

(i) How many passed in Nepali? n0(N) =33
(ii) How many passed in English?

(iii) How many students passed in Nepali or in English?

(iv) How many students failed in both Subjects? n (N∪E) = 5 n (N∩E) = 40 n0(E) =42
(v) How many students were there in total?

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(b) From the given diagram find. A BU
n0(B) = 20
(i) n(A∪B) (ii) n(A) n0(A) = 21
(iii) n(U) n (A∩B)=17
(iii) n(B) (iv) n(U) n(A ∪ B) =12
(c) Study the Venn diagram and find. U
T n(U)
(i) n(F∩T) (ii) n(N∪E)

F U N EU F
T

n0(N) =33

n (N∪E) = 5 n (N∩E) = 40 n0(E) =42 n(F) = 30 n(F ∪ T) = 25
n(T) = 15
n(F) = 30 n(T) = 25

(d) Study the Venn diagram given along side and find U n(B) = 34
AB n0(B) = ?
(i) n(A ∩ B) (ii) n(A ∪ B) n(A∪B) = 6

n(A)=20 n0(Q)=20

(iii) n0(B) (iv) n(U) n(P∪Q) = 25

3. If n(∪) = 90, find the value of x. P U n(B) = 75
n0(P)=40
Q

n (P∩Q) = x

4. If n(∪) = 120, find the value of x. A U
B
n(A) = 65

n(A∪B) = 10

5. If n(A) = 40, n(B) = 50, n(A∩B) = 20, find n(A∪B) andn (sA∩hBo) w= x the result in a venn-
diagram.

6. If n(A) = 60, n(B) = 40, n(A∪B) = 10 and n (∪) = 80, find n(A∩B) and show the
result in a Venn-diagram.

7. If n0(X) = 15, n0(Y) = 20, n(X∪Y) = 10 and n(∪) = 60, find the value of n(X∩Y).
8. Out of 100 people it was found that 40 liked mango only, 45 liked apple only and

10 liked both the fruits. Show the above information in a Venn-diagram and find
the people who didn't like these fruits.

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9. A survey was made among 110 children and found that 70 liked milk, 60 liked
curd and 10 like none. Find the number of children who liked both and show
the result in a Venn-diagram.

10. In a survey of 200 people, it was found that 130 liked folk song, 110
liked modern songs and 70 liked folk as well as modern songs.

(i) How many people liked either folk or modern songs?
(ii) How many people liked none of the songs?
(iii) How many people liked only folk songs?
(iv) How many people liked only modern songs?
(v) Illustrate the above information in a Venn diagram.
11. In a survey, it was found that every people liked at least one types of
movie, Hindi or Nepali. If 70% liked Hindi movies and 50% liked Nepali movies,
(i) how many liked both types of movies?
(ii) how many people liked Nepali movies only?
(iii) illustrate the information in a Venn diagram.
12. Out of 48 students of a class, the number of students who like coffee and tea are

in the ratio 3:2, 15 like both the beverages and 3 do not ike any of the beverages.
(i) Illustrate the above information in a Venn diagram.
(ii) Find the number of students who like coffee.
(iii) Find the number of students who like tea.

13. In a survey of 1000 tourists who visited Nepal, it was found that 600 visited
Pokhara, and 680 visited Lukla. If every tourist visited at least one of the place,

(i) how many of them visited both the places?
(ii) how many of them visited only one place?
(iii) represent the above information in a Venn diagram.

14 Among 120 students in a school, It was found that 70% liked maths, 40% liked sci-
ence and 20% liked both the subjects and some of the students did not like these
subjects. Show the above information in a venn diagram and the find the number
of students who did not like these two subjects?

15. In a survey of 60 householdes, 10 liked to watch television only, 12 liked both
watching television and listening Radio. The number of people who liked to listen
radio is twice the number o f people who liked to watch television . By drawing a
Venn diagram , find

a) the number of people who liked to listen radio.
b) the number of people who did not like both.

Sets 145