Vedanta Excel in Mathematics Book - 8 Final (2078)

Unit Area and Volume

20

20.1 Area of plane figures - Looking back
Classroom - Exercise

Let’s say and write the answers as quickly as possible.
1. a) If p cm is the length of a square, its area is ...................

b) If the length of a rectangle is x cm and breadth is y cm, its area is ............
c) If b is the base of a triangle with height h, its area is ...........
d) If the length of the side of a square is 7 cm, its area is ...................
e) If the length of a rectangle is 5 cm and the breadth is 4 cm, its area

is .............
f) The base of a triangle is 8 cm and height is 6 cm, its area is ...................
2. a) In the given figure, the area of the bigger rectangle

is 54 cm2 and that of the smaller rectangle is
28 cm2. The area of the shaded region is ...................

b) In the adjoining figure, the area of the bigger triangle is
40 cm2 and that of the smaller triangle is 24 cm2. The area of the
shaded region is ...................

The plane surface enclosed by the boundary line of a plane closed figure is known as

its area. Area is measured in square units. For example, sq. centimeters (cm2), sq. metres

(m2), etc. Let’s review the formulae to calculate areas of the following plane geometrical

figures.

Name of plane figure Shape of plane figure Formula of area

Area = 1 base u height
2
Triangle h
b A = 1 bh
2

Rectangle b Area = length u breadth
A=lub

l

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Area and Volume d Area = (side)2 or, Area = 1 (diagonal)2
Square A = l2 2
1
or, A = 2 d2

l

Parallelogram h Area = base u height
A =buh

b

Rhombus d1 Area = 1 d1 u d2
d2 2

d1 Area = 1 (Product of diagonals)
d2 2
Kite
Trapezium l2 A = 1 d1 u d2
Quadrilateral 2
h
l1 Area = 1 (Sum of parallel sides) u h
2
p1 d
p2 A = 1 h (l1 + l2)
2

Area = 1 diagonal (Sum of
2 perpendiculars)

A = 1 d (p1 + p2)
2

20.2 Area of triangle

Let's cut a cardboard paper to form a triangular shape. Say the triangle is ABC. First fold

a perpendicular from the vertex A to the base BC. Open and mark the perpendicular

with a pen. It is the height of the triangle and mark it as h. Fold the height in half so that
vertex A sits on the base BC. Mark half of the height as h2.

AA A

h

h P 2 Q F 1E
Xh
22

BD CB D CB D C

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Area and Volume

Now, cut the folded edges AX, PX, and QX. Place the edge AP along PB and the edge AQ
along QC. Thus, a rectangle BCEF is formed by this arrangement.

Here, area of 'ABC = area of rectangle BCEF
= length × breadth

= BC × CE

= base × height

= b × 1 h
2

= 1 bh
2

Thus, area of triangle = 1 base × height = 1 b × h.
2 2

Area of equilateral triangle A

Let the length of each side of an equilateral triangle be a. Let's draw a ha
AD A BC. Here, AD bisects the base BC at D.

By using Pythagorus theorem, B a D a C
2 2
AD = h = AB2 – BD2 = a2 – a 2 = 3 a
2 2

Now, area of ' ABC = 1 base × height = 1 × a × 3 a = 3 a2
2 2 2 4

Thus, area of an equilateral triangle = 3 a2.
4

20.3 Area of quadrilateral

In the adjoining figure, ABCD is a quadrilateral in which AC is D
p1
the diagonal. Let's draw BX A AC and DY A AC. Here, BX (p1) X C

and DY (p2) are the heights of 'ABC and 'ACD respectively. A Y
p2
Now, area of 'ABC = 1 b × h = 1 AC × p1
2 2 B

Also, area of 'ACD = 1 b × h = 1 AC × p2
2 2

? Area of quad. ABCD = area of ('ABC + 'ACD)

= 1 AC × p1 + 1 AC × p2
2 2

=12 AC (p1 + p2)

Thus, area of a quadrilateral = 1 diagonal (p1 + p2).
2

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Area and Volume D C
h BF
(i) Area of parallelogram
Eb
Let's draw a parallelogram ABCD in a cardboard paper. Fold
the base AB through D along DE such that DE A AB. Mark
DE as h which is the height of the parallelogram. Now, cut A
the paper along DE and place the edge AD along BC so that a
rectangle EFCD is formed.
Here, AB = EF and DE = CF = h
Now,area of parallelogram ABCD = area of rectangle EFCD

= length × breadth
= EF × FC
= AB × h
=b×h
Thus, area of parallelogram = base × height = b × h.

(ii) Area of rhombus C

D

The figure alongside is a rhombus ABCD in which AC and BD

are the diagonals. Let AC be d1 and BD be d2. The diagonals E B
of a rhombus bisect each other perpendicularly. So, BE A AC A
and DE A AC.

Now, area of rhombus ABCD = Area of ('ABC + 'ACD)

= 1 AC × BE + 1 AC × DE
2 2

= 1 AC (BE + DE)
2

= 1 AC × BD
2

= 1 d1 × d2
2

Thus, area of rhombus = 1 d1 × d2.
2

(iii) Area of kite

Also in the case of a kite, the diagonals intersect each other d1
d2
perpendicularly.

So, the area of the kite = 1 d1 × d2.
2

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(iv) Area of trapezium DE Area and Volume

The figure given alongside is a trapezium ABCD in which C
h
AB // CD and AC is the diagonal. Here, AE A CD. So, AE is the B

height (h) of the trapezium. A

Now, area of trapezium = Area of ('ABC + 'ACD)

= 1 AB × h + 1 CD × h
2 2

= 1 h (AB + CD)
2

Thus, area of trapezium = 1 height (Sum of the two parallel sides)
2

Worked-out examples

Example 1: Find the area of the following figures. c) 7 cm

a) b)

6cm
8 cm
8cm

6 cm

Solution: 7 cm 12 cm
a) Area of triangle
= 1 base u height
A 2
1
b) Area of rhombus = 2 u 7 cm u 6 cm = 21 cm2

= 1 u product of diagonals
2

= 1 u 8 cm u 6 cm = 24 cm2
2

c) Area of trapezium = 1 height (Sum of parallel sides)
2

= 1 u 8 cm (12 cm + 7 cm) = 4 cm u 19 cm = 76 cm2
2

Example 2: Find the area of the following plane figures.

a) b)

7 cm
6 cm

8 cm 24 cm 8 cm 20 cm

Solution:

a) Area of the rectangle = l u b = 24 cm u 6 cm = 144 cm2

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Area and Volume

Area of 2 triangles at two ends = 2× 1 u b uh = 6 cm u 8 cm = 48 cm2
2

? Area of the figure = 144 cm2 + 48 cm2 = 192 cm2.

b) Area of the rectangle = l u b = 20 cm u 7 cm = 140 cm2

Radius of each semi-circle = 7 cm = 3.5 cm
2
Area of two semi-circles 1 22
=2 × 2 π r2 7 u 3.5 u 3.5 cm2 = 38.5 cm2


? Area of the figure = 140 cm2 + 38.5 cm2 = 178.5 cm2.

Example 3: Find the area of the shaded regions in the following figures.

a) D 20 cm C b) S R c) D C
E H G
12 cm 10 cm T
16 cm F
8 cm B
14 cm E 10 cm
A B A 25 cm
15 cm P Q
Solution:

a) Area of the trapezium ABCD = 1 h (l1 + l2)
2

= 1 u 12 cm (15 cm + 20 cm)
2

= 6 cm u 35 cm = 210 cm2

Also, the area of 'AEB = 1 ubuh = 1 u 15 cm u 12 cm = 90 cm2
2 2

? Area of the shaded region = Area of trapezium ABCD – Area of 'AEB

= 210 cm2 – 90 cm2 = 120 cm2

b) Area of the parallelogram PQRS = b u h

= 14 cm u 10 cm = 140 cm2
1
Also, Area of 'PST = 2 ubuh
? Area of the shaded region
= 1 u 14 cm u 10 cm = 70 cm2
2
= Area of parallelogram ABCD – Area of 'PST

= 140 cm2 – 70 cm2 = 70 cm2

c) Area of rectangle ABCD =lub
Also, area of rectangle EFGH = 25 cm u 16 cm = 400 cm2
? Area of the shaded region
=lub = 80 cm2
= 10 cm u 8 cm

= Area of rectangle ABCD – Area of rectangle EFGH

= 400 cm2 – 80 cm2 = 320 cm2

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Area and Volume

Example 4: The perimeter of a squared floor of a room is 48 ft.

a)Find its length b) Find its area

c) Find the cost of carpeting the floor at Rs 44 per sq. ft.

Solution:

a) The perimeter of the squared floor = 48 ft

or, 4l = 48 ft
48
or, l = 4 ft = 12 m
? Length of the floor is 12 ft.

b) Area of the floor = l2 = (12 ft)2 = 144 sq. ft.

c) The cost of carpeting in the floor = Area of floor u Rate of cost

= 144 u Rs 44 = Rs 6,336

Example 5: A rectangular garden is twice as long as its breadth and its perimeter
is 66 m.
a) Find its length and breadth b) Find its area
c) Find the cost of growing grass in the garden at Rs 40.50 per sq. metre.

Solution:
a) Let the breadth of the garden (b) = x m

? The length of the garden (l) = 2x m

The perimeter of the garden = 66 m

or, 2 (l + b) = 66 m

or, 2 (2x + x) = 66 m

or, 6x = 66 m
or, x 66
= 6 m = 11 m
? Breadth of the garden
= x = 11 m

Length of the garden = 2x = 2 u 11 m = 22 m

b) Area of the garden = l u b = 22 m u 11 m = 242 m2

c) The cost of growing grass in the garden = Area u Rate of cost
= 424 u Rs 40.50 = Rs 9,801

EXERCISE 20.1
General Section – Classwork

1. Let’s say and write the answers as quickly as possible.
a) Base of a triangle is a cm and height is b cm, area is .......................................
b) Base of a parallelogram is x cm and height is y cm, area is ............................
c) Diagonals of a rhombus are a cm and b cm, area is ........................................
d) Diagonals of a kite are x cm and y cm, area is ...............................................
e) Length of two parallel sides of a trapezium are a cm and b cm, and height is
h cm , area is ................................................
f) Diagonal of a quadrilateral is x cm, two perpendiculars from the opposite vertex
to the diagonal are a cm and b cm, its area is .......................................

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Area and Volume

Creative Section - A

2. Find the area of the following figures:

A P c) I 10.5 cm J
a)
b)

8 cm 12 cm
7 cm

B D C Q 6 cmR T K
10 cm D
d)
A e) f) g) C
6 cm
R U
BT
3 2 cm
8 cm

B 14.5 cm C E 9 cm S A J W
O

h) A i) M H j) J V P k) A 5 cm D

12cm 4cm 10cm

C

B 4 2 cm A 16 cm TU 9 cm MB E15 cm C
D n) K
l) W 8cm T m) A P o)
8cm
4.5cm Q 6.5cm 8cm S 15cm
10.5cm
C
Z 7.5cm I 6cm E
B
7cm R
6cm
X Y A r) Q T
p) P q)

Q 24cm 8.5cm S B D 4.5cm10cmND
R 11 cm C 6.5cm
UM
A

3. Find the area of the following figures: P

a) A b) c) A D
Q E
10 cm 10 cm 18 cm T 18 cm

B 12 cm D B 25 cm C
8 cm 10 cm
R 15 cm S
C

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Area and Volume

d) 5cm D C 5cm e) S R f) H G

E 15 cm F 14 cm 7 cm

A 30 cm B P 20 cm Q E 12 cm F

Creative Section -B

4. Find the area of shaded regions in the following figures:

a) b) c) d)
16 cm

1100ccmm 15 cm 17 cm
55 ccmm
15 cm

12 cm
9 cm

16 cm 18 cm 8 cm 14 cm
e) f) g) 9 cm h) 11 cm

6 cm 3cm4cm 5 cm 8 cm
6 cm
24 cm 10 cm 15 cm l) 4 cm 4 cm
i) 12 cm 16 cm k)

6 cm

16 cm
4 cm
8 cm
18 cm
j)

5cm
10 cm 4cm 4cm
12 cm
7 cm 5 cm
5 cm 5 cm
5cm

4 cm 4 cm 15 cm 14 cm 4 cm 12 cm

5. a) The perimeter of a square room is 40 m.

(i) Find its length. (ii) Find the area of its floor.

(iii) Find the cost of carpeting the floor at Rs 70 per sq. metre.

b) The perimeter of a square ground is 100 m. Find the cost of plastering the ground
at Rs 80 per sq. metre.

6. a) A rectangular park is 40 m long and 28 m broad.

(i) Find its area.

(ii) Find the cost of paving tiles all over it at Rs 90 per sq. metre.

b) A rectangular field is 32 m long and its perimeter is 114 m.

(i) Find the breadth of the field. (ii) Find the area of the field.

(iii) Find the cost of plastering the field at Rs 75 per sq. metre.

c) A rectangular room is 12 m broad and its perimeter is 54 m. Find the cost of
carpeting its floor at Rs 50 per sq. metre.

d) The length of the floor of room is two times its breadth and the perimeter of
the floor is 48 m. Find the cost of paving marbles on the floor at Rs 125 per
sq. metre.

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Area and Volume

e) The length of a rectangular ground is three times its breadth. If the perimeter of
the ground is 96 m, find the cost of growing grass on it at Rs 65 per sq. metre.

7. a) A rectangular park of length 60 m and breadth 50 m encloses a volleyball court of
length 18 m and breadth 10 m. Find the area of the park excluding the volleyball
court. Also, find the cost of paving stone in the park excluding the court at the
rate of Rs 110 per square meter.

b) Mr. Thapa built a circular pond of diameter 28 ft. for fish farming in his
rectangular field of length 70 ft. and breadth 40 ft. Find the area of the field
excluding the pond. Also, estimate the cost of planting vegetable at Rs 11 per
square feet.

It’s your time - Project work
8. a) Take a rectangular sheet of photocopy paper. Measure its length and breadth.

Draw a circle by using a compass with an arc of 3.5 cm radius in the middle
of the paper. Lay the circle out by the help of scissors. Find the area of paper
remained in the sheet.
b) Let’s take a measuring tape and measure the outer length and breadth of white
board of your class. Also measure the length and breadth of white part of the
board and calculate the area of its wooden frame.

20.4 Solids and their nets

Cube, cuboid, cylinder, sphere, cone, pyramid, triangular prism, etc. are a few examples
of solids. Solids have three measurable parts that we call the dimensions. So, solids are
also called 3D shapes.
Solids have faces (sides), edges and vertices (corners). The exception vertex
is the sphere which has no edges or vertices.
edge
face

The net of a solid is what it looks like if it is opened
out flat. A net can be folded up to make a solid.

Cube Net of cube

Prisms Cuboid
(Rectangular prism)
A prism is a polyhedron for which the top and
bottom faces are congruent polygons and these
congruent faces are called the base of the prism.
All the other faces of a prism are rectangles which
are called the lateral faces. But, a cylinder has a
curved lateral face. Such prisms are also called the Cube
right prisms. If the lateral faces are not rectangle (Square prism)
the prism is called non-right
prism.

A prism is described by the shape Triangular prism Cylinder
of its base. For example, a cube (Circular prism) Non-right prism
has square base, so it is a square
prism. A cylinder has circular
base; so, it is a circular prism.

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Area and Volume

(i) Cube and its net

A cube is a square prism. It has six square faces. When
we fold up the net of a square prism, a cube is formed.

Net of cube Cube

(ii) Cuboid and its net Net of cuboid Cuboid

A cuboid is a rectangular prism. It has six
rectangular faces. When the net of a rectangular
prism is folded up a cuboid is formed.

(iii) Triangular prism and its net Net of triangular Triangular prism
prism
A triangular prism has two bases which
are triangles. A triangular prism has three
rectangular lateral faces.

(iv) Cylinder and its net

A cylinder is also like a prism. It has two circular bases and

a curved lateral face.

Net of cylinder Cylinder

Pyramids

A pyramid is a polyhedron for which the base is a polygon and all

lateral faces are triangles. A pyramid is described by the shape of its

base. For example, a triangular pyramid has a triangular base, square

pyramid has a square base. Square pyramid

(i) Square pyramid and its net

A square pyramid has a square base. Its four lateral

faces are the isosceles triangles. Net of a square A square pyramid
pyramid

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Area and Volume

(ii) Triangular pyramid and its net

A triangular pyramid has a triangular base. It has

three congruent triangular faces. The tetrahedron is

a triangular pyramid having congruent equilateral A triangular
pyramid
triangles for base and for each of its faces. Net of a
triangular pyramid

(iii) Pentagonal pyramid and its net A pentagonal
pyramid
A pentagonal pyramid has a pentagon
base and five congruent triangular faces.

Net of a
pentagonal pyramid

(iv) Hexagonal pyramid and its net

A hexagonal pyramid has a hexagon base

and six congruent triangular faces.

Net of a hexagonal A hexagonal
pyramid pyramid

(v) Cone and its net

A cone is also like a pyramid. But it has a
circular base and a curved face.

Net of a cone A cone

EXERCISE 20.2 c)

General Section -Classwork

1. Name the following solids:
a) b)

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Area and Volume
d) e) f)
g) h) i)

2. Let’s write the names of the base and the lateral faces of the following solids:
a) E
b)

DC D
CF

A E
B
AB

3. Let’s study the nets of solids carefully. Then name the solids formed by each of these
nets.

a) b) c)

d) e) f)

g) h) i)

4. Trace the nets of the following solids: c) cone d) cylinder
a) Cube/hexahedron b) cuboid
e) tetrahedron f) triangular prism g) square based pyramid

It’s your time - Project work
5. Make the groups of your friends. Take the hard paper and cut and make the nets of

cube, cuboid, cone, cylinder, etc. and paste with glue. Then submit the solids you or
your group prepared to your subject teacher.

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Area and Volume

20.5 Area of solids

For any three dimensional figure (3D solids), we can calculate the three types of surface
area. They are area of the base, lateral surface area, and total surface area.

The top and buttom faces of a solid which are parallel to each other are the bases. The
remaining faces of the solid are the lateral surfaces. So, the combination of bases and the
lateral surfaces form the total surface of the solid.

(i) Area of cube

A cube is bounded by six regular plane square faces. In the given D C
H
figure, ABCD, EFGH, EFBA, HGCD, EHDA, and FGCB are the six l
l B
faces of the cube. Here, ABCD and EFGH are two bases and the A
remaining faces are the lateral surfaces of the cube. G

Let, l be the length of each square face of the cube. Then, the area E l
of each face = l2 F

Now, the area of the 4 lateral faces = 4l2

And, the area of 6 faces = 6l2

Thus, the total surface area of a cube = 6l2 and lateral surface area of a cube = 4l2.

(ii) Area of cuboid DC

A cuboid has 6 rectangular faces. In the given figure, The faces A Bh
ABCD and EFGH are the bases. The faces EFBA and HGCD are
two opposite lateral surfaces. Also, the faces EHDA and FGCB H G
b
are the remaining two opposite lateral surfaces. F

Here, the area of 2 bases = area of (ABCD + EFGH) = 2lb E l

The area of 2 opposite lateral surfaces = Area of (EFBA + HGCD) = 2lh

The area of 2 other opposite lateral surfaces = Area of (EHDA + FGCB) = 2bh

Thus, the total surface area of the cuboid = 2lb + 2lh + 2bh

= 2 (lb + lh + bh)

(iii) Area of cylinder

Take a rectangular sheet of paper and bend it as shown in the figure. In this way, you
will get a curved surface figure with two circular faces. It is called a cylinder. The two
circular faces of a cylinder are its bases and its lateral surface is the curved surface.

Area of circular base
= Sr

Area = l × b b Area = l × b b = h Area of curved surface
l = 2Sr ×h = 2Srh

l = 2Sr Area of circular base
=Sr

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Area and Volume

From the above illustrations, it is clear that
The curved surface area of a cylinder = 2Srh
The total surface area of a cylinder = curved surface area + 2 × area of base

= 2Srh + 2Sr2
= 2Sr (r + h)

Here, r is the radius of the circular base and h is the height of the cylinder.
Thus, curved surface area of a cylinder = 2Srh

Total surface area of a cylinder = 2Sr (r + h)

(iv) Area of triangular prism

In a triangular prism, it has two triangular bases and three rectangular lateral surfaces.

The adjoining prism is a triangular prism. c

(i) Its lateral surface area = Area of 3 rectangular surfaces a

= l.a + l.b + l. c b
= l (a + b + c) l

= Perimeter of triangular base u length = P.l

(ii) Total surface area of a triangular prism

= Lateral surface area + 2 u Area of triangular base
1
= P.l + 2 u 2 b u h

= P.l + b u h

Thus, lateral surface area of a triangular prism = P.l h l
b
Total surface area of a triangular prism = P.l + b u h

Where, P = Perimeter of triangular base
l = Length of the prism
b = Base of the triangular face
h = Height of the triangular face

Worked-out examples

Example 1: Find the total surface area of the following solids:

a) b) c) d) 8 cm 6 cm
4 cm
4 cm
5 cm
10 cm

7 cm
4 cm 10 cm 3 cm r = 3.5 cm

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Area and Volume

Solution:
a) Here, the length of the side of the cube (l) = 4 cm
? Area of the cube = 6l2 = 6 × 42 = 96 cm2.
b) Here, the length of the cuboid (l) = 10 cm

the breadth of the cuboid (b) = 3 cm
the height of the cuboid (h) = 5 cm

Now, area of the cuboid = 2 (lb + lh + bh)

= 2 (10 × 3 + 10 × 5 + 3 × 5)

= 190 cm2

c) Here, radius of the circular base of the cylinder (r) = 3.5 cm

Height of the cylinder (h) = 10 cm

? Total surface area of the cylinder = 2Sr (r + h)
22
= 2 u 7 u 3.5 (3.5 + 10) cm2

= 22 u 13.5 cm2 = 297 cm2

d) By using Pythagoras theorem to find side BC in right angled 'ABC. A
BC = AB2 + AC2 = 82 + 62 = 64 + 36 = 100 = 10 cm
8 cm
Now, total surface area of the triangular prism B 6 cm
= P.l + b.h
C
= (8 + 6 + 10) cm u 7 cm + 8 cm u 6 cm A' C' 7 cm
= 168 cm2 + 48 cm2

= 216 cm2 B'

EXERCISE 20.3
General Section – Classwork

1. Let’s say and write the answers as quickly as possible.

a) The parallel faces of a solid are called ..............................

b) If a be the length of each face of a cube, the area of the bases = ..............................

c) If l be the length of each face of a cube, the lateral surface area = ..............................

d) If l, b, and h be the length, breadth and height of a cuboid,

(i) its area of base = ..............................

(ii) its lateral surface area = ..............................

(iii) its total surface area = ..............................

e) If r be the radius of the base and h the height of a cylinder,

(i) its area of the bases = ..............................

(ii) its curved surface area = ..............................

(iii) its total surface area = ..............................

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Area and Volume

f) If a, b, and c be the length of triangular sides of a prism of length l,

(i) its lateral surface area = ..............................

(ii) its total surface area = ..............................

Creative Section

2. Find the total surface area of the following solids:

a) b) c)
3 cm 4 cm
15 cm

3 cm 12 cm 6 cm 2 cm 3.5 cm

3 cm

3. Find the area of the base, curved surface area and total surface area of the following

cylinders: b) c) r = 21 cm
35 cm
a)

20 cm 40 cm

r = 14 cm

r = 7 cm

4. Find the lateral surface area and total surface area of the following triangular prism.

a) b) c) 12 cm
8 cm
5 8 cm
18 cm 9 cm
cm 15 cm 14 cm

4 cm

6 cm 6 cm

5. a) A cubical block 10 cm long is placed on the table. How much area does it cover
on the surface of the table?

b) Find the total surface area and lateral surface area of a chalk box whose height,
length and breadth are 25 cm, 20 cm and 15 cm respectively.

6. The radius of the circular base of a thermos is 10.5 cm and it is 18 cm high.

Find (i) its curved surface area (ii) total surface area.

7. What is the surface area of a triangular prism where the base area is 25 m2, the base
perimeter is 24 m, and the length is 12 m.

8. The sides of the triangular base of a prism are 10 cm, 8 cm and 6 cm respectively and
it is 15 cm long. Find the total surface area of the prism.

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Area and Volume

It’s your time - Project work

9. a) Let’s measure the length, breadth and thickness (or height) of the following
objects and estimate the total surface area of the objects.

(i) Mathematics book (ii) Instrument box

b) Let’s take a rectangular sheet of a chart paper and measure it’s length and breadth
to find its surface area. Now roll it to form a cylinder and measure its diameter
and radius. Then, find the curved surface area. Is the curved surface area is same
as the area of rectangle?

20.6 Volume of solids
The space occupied by any object is called its volume. In the case of a solid,
obviously, its base covers the area of surface and its height occupies the region
above the surface. So, its area of the base times height gives its volume.

Thus, volume of a solid = Area of its base × its height

h h

b b
l l
The base of the cuboid occupies
The base of a cuboid is covering its height times the space of water
the surface area of water which is inside the water which is the volume
equal to the area of the base of the of a cuboid.
cuboid.

(i) Volume of cube

The area of the base of a cube is l2, where l is the length of the sides of l
the base. The height of the cube is also equal to the length of the base. l

? Volume of a cube = area of the base × height = l2 × l = l3 l
If the cube is of unit (say 1 cm) length, breadth and height, then its

volume is 1 cm3. So, cm3, m3, etc. are the unit of the measurement of volume.

(ii) Volume of cuboid h
b
Let, l, b and h be the length, breadth and height of a cuboid
respectively. Then, area of the base of the cuboid = l × b l
? Volume of the cuboid = area of base × height

=l×b×h

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(iii) Volume of cylinder Area and Volume
h
Let r be the radius of the base of a cylinder with height h.
Then, area of its circular base = Sr2 r
? Volume of the cylinder = area of the base × height
length
= Sr2 × h
= Sr2h

(iv) Volume of triangular prism

The base of a triangular prism is a triangle. So, the area of its base
means the area of the triangle.

? Volume of triangular prism = Area of triangle × height (or length)

Worked-out examples

Example 1: Find the volume of the following solids: r = 7 cm d) 8 cm 6 cm
a) b) c)
6 cm 6 cm
4 cm 6 cm 12 cm
15 cm 15 cm

6 cm

Solution:

a) The length of the sides of the cube (l) = 6 cm

? Volume of the cube = l3 = (6 cm)3 = 216 cm3.

b) The length of the cuboid (l) = 15 cm

breadth of the cuboid (b) = 4 cm

height of the cuboid (h) = 6 cm

? Volume of the cuboid = l × b × h

= 15 cm × 4 cm × 6 cm = 360 cm3

c) Here, radius of the circular base of the cylinder (r) = 7 cm

Height of the cylinder (h) = 15 cm

Now, volume of the cylinder = Sr2h = 22 u 7 u 7 u 15 cm3
7
= 2310 cm3

d) Here, the triangular base is a right angled triangle.
1
? Area of the triangular base = 2 u 8 cm u 6 cm = 24 cm2

Now, volume of triangular prism = Area of triangular base u height

= 24 cm2 u 12 cm = 288 cm3

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Area and Volume

Example 2: The base area of a cubical box is 2.25 m2. Find the volume of the box.

Solution:

Here, area of the base of the cubical box = 2.25 m2

or, l2 = 2.25 m2

or, l = 2.25 m2 = 1.5 m

Now, the volume of the box = l3 = (1.5)3 = 3.375 m3.

Example 3: A rectangular water tank is 3 m long, 2.5 m broad, and 2 m high. How
much water does it hold?

Solution:

Here, length of the tank (l) = 3 m

breadth of the tank (b) = 2.5 m

height of the tank (h) = 2 m

Now, volume of the tank = l × b × h = 3 m × 2.5 m × 2 m = 15 m3

The capacity of the tank is 15 m3. So, it holds 15 m3 of water.

Example 4: A carton is 50 cm long, 20 cm broad, and 30 cm high. How many
cubical boxes each of 10 cm long can be put inside the carton?

Solution:

Here, length of the carton (l) = 50 cm

breadth of the carton (b) = 20 cm

height of the carton (h) = 30 cm

Now, volume of the carton = l × b × h

= 50 cm × 20 cm × 30 cm = 30000 cm3.

Also, the volume of each cubical box = (10 cm)3 = 1000 cm3

Again, the number of cubical boxes = Volume of carton
Volume of each box

= 30000 cm3 = 30
1000 cm3

Hence, 30 cubical boxes can be put inside the carton.

Example 5: The length of a box is twice the breadth and thrice the height. If the
volume of the box is 288 cm3, find the length breadth and height of the
box.

Solution:

Let the length of the box (l) = x cm
Then,
and the breadth of the box (b) = x cm
the height of the box (h) = 2x cm
3
the volume of the box (V) = 288 cm3

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Area and Volume

Now, volume of the box (V) =x× x × x
288 cm3 2 3
or, = x3
or, 6
1728 = x3

or, x3 = 1728 cm3

or, x = 1728 cm3 = 12 cm

? The length of the box (l) = 12 cm

the breadth of the box (b) = x = 12 = 6 cm
2 2
x 12
the height of the box (h) = 3 = 3 = 4 cm

Example 6: Find the volume of the given solid. 4 cm 4 cm

Solution:

Here, volume of the whole solid is the sum of the volume

of 3 smaller solids. 4 cm
5 cm
Volume of the lowermost one cuboid = l × b × h 10 cm

= 12 cm × 5 cm × 4 cm = 240 cm3.

Also, length of each uppermost cuboid = 4 cm 12 cm
breadth of each uppermost cuboid = 5 cm

height of each uppermost cuboid = (10 – 4) cm = 6 cm

? Volume of 2 uppermost cuboids = 2 (4 cm × 5 cm × 6 cm)

= 2 × 120 cm3 = 240 cm3

Now, volume of the whole cuboid = 240 cm3 + 240 cm3 = 480 cm3

EXERCISE 20.4

General Section - Classwork

Let’s say and write the answers as quickly as possible.
1. a) If the area of the base of a cube is a2 cm2, its volume is ...................................

b) If the length of a cube is 2 cm, its volume is ...................................
c) If the area of the base of a cuboid is (x × y) cm2 and height is z cm, its volume

is ...................................
d) If the area of the base of a cuboid of height 3 cm is 10 cm2, its volume

is ...................................
e) If the length, breadth and height of a box is 5 cm, 4 cm and 3 cm respectively,

its volume is ...................................
f) If the area of the base of a cylinder of height 10 cm is 154 cm2, its volume

is ...................................
g) If the area of the triangular base of a prism of length 10 cm is 6 cm2, its

volume is ...................................

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Area and Volume

2. a) If the volume of a cube is 8 cm3, its length is ................................... and area
of the base is ...................................

b) If the volume of a cuboid is 60 cm3 and height is 3 cm, its area of the base
is ...................................

c) If the volume of a cuboid is 60 cm3 and its area of the base is 20 cm2, its
height is ...................................

d) If the volume of a cylinder is 1540 cm3 and height is 10 cm, its area of the
base is ...................................

e) If the volume of a cylinder is 1540 cm3 and area of the base is 154 cm2, its
height is ...................................

f) If the volume of a triangular prism is 60 cm3 and length is 10 cm its area of
the base is ...................................

g) If the volume of a triangular prism is 60 cm3 and area of the base is 6 cm2, its
length is ...................................

Creative Section - A

3. Find the volume of the following solids:

a) b) c)

4 cm 8 cm

4 cm 10 cm 5 cm 7 cm

d) 4 cm e) 6 cm f) 15.5 cm
d = 28 cm
r = 6.3 cm
r = 8.4 cm

14 cm 40 cm
4 cm
20 cm

g) h) i)

10 cm 15 cm 9 cm 14 cm 6 cm

8 cm 3 cm 8 cm

4. a) Find the volume of the cubes of the given lengths.

(i) 5 cm (ii) 4.5 cm (iii) 8 cm (iv) 2.5 m

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Area and Volume

b) Find the volume of the cuboids of the following measurements.

Length breadth height
(i) 5 cm 4 cm 3 cm
(ii) 1.5 m 20 cm 50 cm
(iii) 42 cm 30 cm 2.5 m

(iv) 1.2 m 1m 80 cm

c) Find the volume of the cylinders of the following measurements.

(i) radius = 7 cm, height = 10 cm

(ii) radius = 3.5 cm, height = 14 cm

(iii) diameter = 21 cm, height = 20 cm

(iv) diameter = 2.8 m, height = 3 m

5. a) The area of the base of a cubical box is 100 cm2, find the volume of the box.

b) If the volume of a cube is 3375 cm3, find its height and area of the base.

c) A cubical water tank is 1.2 m high. How much water does it hold?

6. a) The measurement of a rectangular vessel is 40 cm × 30 cm × 50 cm.
How many litres of oil can be put inside it to fill it completely?
(1000 cm3 = 1 litre)

b) A rectangular water tank is 1.5 m × 1.4 m × 2 m. How many litres of water
does it hold when it is completely filled?

7. a) A cuboid is 12 cm broad and 10 cm high. If its volume is 1800 cm3, find its
length.

b) A rectangular box is 1 m long and 75 cm high. If the volume of the box is
375000 cm3, find the width of the box.

c) The area of the base of a rectangular tank is 2.4 m2. If the capacity of the tank
is 3.6 m3, find the height of the tank.

d) A rectangular vessel can hold 40 litres of milk when it is full. If the area of
the base of the vessel is 2000 cm2, find the height of the vessel.

Creative Section - B

8. a) A carton is 60 cm long, 40 cm broad, and 50 cm high. How many cubical
boxes each of 20 cm long can be put inside the carton?

b) A cubical box is 75 cm high. How many rectangular packets each of
25 cm × 15 cm × 5 cm are required to fill the box?

c) A rectangular box is 5 cm long and 6 cm wide. How many cubical packets
each of 1 cm3 can be placed on the floor of its base? If 10 layers of such
packets are placed on the floor, how many packets are there?

d) A wall is 15 m long, 0.2 m wide and 2 m high. How many bricks of size
15 cm × 8 cm × 5 cm each are required to construct the wall?

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Area and Volume

9. Find the volume of the following solids.

a) 12 cm b) 4 cm
5 cm
7 cm

2cm 10 cm 3 cm
3 cm
8 cm 15 cm

c) 3 cm d)
3 cm

3 cm 5 cm 10 cm

6 cm 7 cm 5 cm
5 cm

e) 6 cm f) 6 cm 6 cm
3 cm
8 cm
7 cm
3 cm 3 cm 2 cm3 cm 12 cm 6 cm
12 cm
3cm
10 cm

3 cm 4 cm 4 cm

10. a) The ratio of length, breadth and height of a cuboid is 3:2:1. If the volume of
the cuboid is 48 cm3, find the base area of the cuboid.

b) The length, breadth and height of a rectangular room are in the ratio 4:3:2. If
the room contains 192 m3 of air, find the area of floor of the room.

c) The length of a room is three times the height and breadth is twice of its
height. If the volume of the room is 384 m3, find its length, breadth, and
height.

d) The length of a rectangular carton is twice the breadth and thrice the height.
If the volume of the carton is 4,500 cm3, find its length, breadth, and height.

It’s your time- Project work!
11. a) Let’s measure the length, breadth and thickness of your mathematics book

and find its volume.
b) Let’s measure the length, breadth and height of a cuboidal soap and calculate

the volume of such 10 soaps.
c) Let’s measure the length, breadth and height of cubical or cuboidal objects

found in your home or in your classroom. Find the volume of each object.

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Unit Bearing and Scale Drawing

21

21.1 Bearing – Looking back
Classroom - Exercise

Let’s say and write the answers as quickly as possible.

1. a) How many directions are there in compass bearing ? What are they ?

.................................................................................................................................

.................................................................................................................................

b) Which is the base line direction to find the bearing of any place ?

.................................................................................................................................

2. a) Tell and write the directions represented by

(i) NE ......................... (ii) SE ................................

(iii) SW ..................... (iv) NW ............................

b) Tell and write the bearing of these direction

(i) East ......................... (ii) South .......................... (iii) West .............................

In the adjoining compass NOS represents North- North (N)
South and EOW represents East-West directions.
The angle between N and E is 90q. The direction North West (NW) North East (NE)
NE lies exactly in between N and E. So, the angle
between N and NE is 45q. Let’s take O as the point West (W) East (E)

South West (SW) South (S) South East (SE)

of reference and ON (the North line) as base line, then, the direction of NE from O in

clockwise direction is 045q. It is called bearing of NE from O.

Similarly, the bearing of E from O is 090q, the bearing of SW from, NW N
O is 225q and so on. W 045° NE
SW E
Thus, the direction in which an object (or place) is sighted may 255° SE
be specified by giving angle in degrees that the direction makes
with north line in clockwise is called the bearing of the object S
(or place).

Of course, bearing of a place from another place with north line as the base tells the
distance between these two places in terms of angles written in 3 digits. So, it is also
called three-digit bearing (or three-figure bearing).

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Bearing and Scale Drawing

Worked-out examples

Example 1: Write down the bearing of the places, A, B, C, and D from O.

a) N A b) N N c) d) 35°
O 162° O 140° ND
70° B C O
O

Solution:
a) Bearing of A from O is 070q
b) Bearing of B from O is 162q
c) Bearing of C from O is (360q – 140q) = 220q
d) Bearing of D from O is (360q – 35q) = 325q.

Example 2: In the given diagram, if the bearing of A from B is N N1
065°, find the bearing of B from A. A
060°
Solution: B

Here, bearing of A from B is 060q.

Here, BN // AN1. So, ‘NBA and ‘N1AB are co-interior angles.
? ‘N1AB = 180q – 60q [Sum of co-interior angles is 180q]

= 120q

? Bearing of B from A = 360q – 120q = 240q.

Example 3: An aeroplane was flying in the bearing N
of 085° from Kathmandu. After flying
over a certain distance, it changed its 085° P1
direction and flew in the bearing of K
240°. By what angle did the aeroplane
change its direction? 240°
P2 c)
Solution:

Here, bearing of P1 from K = 085q

Bearing of P2 from K = 240q

? The direction changed by the plane from P1 to P2

= 240q – 085q = 155q

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Bearing and Scale Drawing

Example 4: If the bearing from Birganj to Dharan is 084°, find the bearing from
Dharan to Birgunj.

Solution: N N1
Let B represents Birganj and D represents Dharan.

Here, bearing from B to D = ‘NBD = 084°

Now, ‘NBD + obtuse ‘N1DB = 180° [NB // N1D and the sum of 084°
co-interior angles]

or, 084° + obt. ‘N1DB = 180° D

? Obt. ‘N1DB = 180° – 084° = 096° B

Again, bearing from D to B = 360° – obt.‘N1DB

= 360° – 096°

= 264°

So, bearing from Dharan to Birganj is 264°.

EXERCISE 21.1

General Section – Classwork

1. Let’s tell and write the bearing of the following directions. N
NW NE

a) NNE ............... b) NOE ............... c) NSE ............... W O E

d) NOS ............... e) NSW ............... f) NOW ............... SW S SE

g) NNW ...............

2. Let’s tell and write the bearing of A from O.

a) N b) N c) N
A
40° A 150°
O O 60°

A

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Bearing and Scale Drawing
3. Study the map of Nepal given below and answer the following questions.

Baitadi
NepalgDuhnajngadhhi
JomsomJumla

Pokhara Dharan
Kathmandu Biratnagar

Birgunj

a) If Kathmandu is the point of reference, write the compass direction of:
(i) Dharan .................................... (ii) Birgunj .....................................
(iii) Dhangadhi .............................. (iv) Jomsom ..................................

b) If Nepalgunj is the point of reference, write the compass direction of :

(i) Jumla .................................... (ii) Pokhara .....................................

(iii) Biratnagar ............................ (iv) Baitadi ......................................

c) If Dharan is the point of reference, write the compass direction of:

(i) Birgunj ............................ (ii) Pokhara ............................

(iii) Biratnagar ............................ (iv) Kathmandu ............................
Creative Section - A
4. Write the bearings of the following compass directions.

E.g., Bearing of NE = 090
Bearing of NNW = 270 + 45 = 315

a) N and NE (NNE) b) N and SE (NSE)
c) N and S (NS) d) N and SW (NSW)
e) N and W (NW) f) N and NW (NNW)

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Bearing and Scale Drawing
5. Write down the bearings of the aeroplanes from A shown below.

a) b) c) N d) N
N N A 160°
125°
60° 90° A
A A

e) f) g) h)
N N N N

150° A 130° A 90° 42°
A
A

6. In the adjoining figure, calculate the bearings of A, B, C, D N
and E from O. EA

35° 40°

O 60° B

W 10° 30°20° 80°
D C S

7. The bearings of A from B are given in the figures. Find the bearings of B from A.

a) N N1 b) N1 c) N d)
N N
115° B 150° N1
A 078° A B N1
040° B A A

B

e) N f) N g) N1 h) N1 N
N1 N1
A
NA

B B A
200° A 270°

B 285° B 330°

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Bearing and Scale Drawing

Creative Section -B
8. a) If the bearing of a place A from the place B is 055q, find the bearing of B from A.

b) If the bearing of a place P from the place Q is 170q, find the bearing of Q from P.

c) If the bearing of Kathmandu from Pokhara is 095q, what is the bearing of Pokhara
from Kathmandu?

d) If the bearing of your school from your house is 082q, what is the bearing of your
house from the school?

e) An aeroplane was flying in the bearing of 075q from Kathmandu. After flying over
a certain distance it changed its direction and flew in the bearing of 310q. By what
angle did the aeroplane change its direction?

It’s your time - Project work!
9. a) Let’s estimate the bearing of the following places from your home.

(i) your school (ii) nearby health post or hospital (iii) nearby police station

b) Let’s estimate the bearing of the following places from your school.

(i) your home (ii) nearby health post or hospital (iii) nearby police station

21.2 Scale drawing – Review

It is not possible to draw the actual distance between two places in drawing. In such
cases, we take a convenient scale and reduce the distance between the places in the
drawing. It is known as scale drawing. The scale which is taken to reduce (or enlarge)
the actual size is scale factor. Scale factor is the ratio of the size of the drawing to the
actual size of the object. For example,

(i) A distance of 5 km (ii) A length of 80 m

Scale: 1 cm to 1 km (or 1:100000) Scale 1 cm to 20 m (or 1:2000)

Scale drawing Scale drawing

0 1 2 3 4 5 km 0 20 40 60 80 m

Worked-out examples

Example 1: What is the actual distance between two places which is represented

by 5.4 cm on a map which is drawn to the scale 1:1000000?

Solution: 1000000
100
Here, the scale 1:1000000 means, 1 cm represents 1000000 cm = × 1000 km

= 10 km

Now, 1 cm represents 10 km

? 5.4 cm represents 5.4 u 10 km = 54 km
So, the distance between two places is 54 km.

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Bearing and Scale Drawing

Example 2: An aeroplane flies 500 km in the bearing of 040° and 600 km in the
bearing of 110°. Find the distance between the two places and the
bearing of the starting place from the last place. (Scale 1 cm = 100 km)

Solution:
In the figure, O is the starting place and B is the last place.
Here, 100 km = 1 cm
? 500 km = 5 cm
Also, 600 km = 6 cm

N1

N2

110°
N

A
6 cm

5 cm B

040°

O

By measurement, OB = 9 cm
Now, the distance between two places= 9 u 100 km

= 900 km
Again, by measurement, ‘N2BO = 360° - 101° = 259°
So, the bearing of the starting place from the last place is 259°.

EXERCISE 21.2

Genral Section – Classwork

1. Let’s say and write the answers as quickly as possible.
a) When the scale is 1 : 100, the actual length of 5 cm is ..........................................
b) When the scale is 1 : 200, the actual length of 3 cm is ..........................................
c) When the scale is 1 : 500, the actual length of 2 cm is ..........................................

2. a) When the scale is 1 : 100, the map / drawing length of 4 m is .............................
b) When the scale is 1 : 200, the map / drawing length of 8 m is ..............................
c) When the scale is 1 : 500, the map / drawing length of 10 m is ............................

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Bearing and Scale Drawing

Creative Section

3. Copy and complete the table.

Length of map/drawing Scale Actual length
a) 2 cm 1:5000 .......................... m

b) 3 cm 1:10000 .......................... m

c) 4.2 cm 1:100000 ..........................km
d) 5.5 cm 1:2000000
.......................... km

4. Copy the complete the table.

Actual length Scale Length on map/drawing
a) 24 m 1:200 .......................... cm
.......................... cm
b) 650 m 1:1000 .......................... cm
.......................... mm
c) 40 km 1:500000

d) 80 km 1:1000000

5. a) What is the actual length which is represented by 4.5 cm on a scale drawing
with scale of 1 cm to 5 m?

b) What is the actual distance between two places which is represented by 3.2 cm
on a scale drawing with scale of 1:1000000?

c) The scale drawing distance between Kathmandu and Biratnagar is 4.7 cm. If the
scale factor is 1:10000000, find the actual distance between these two places.

6. a) An aeroplane flies 60 km from A to B in the bearing of 070q and then 90 km from
B to C in the bearing of 105q. Represent this information by drawing in a scale
of 1 cm = 10 km and answer the following questions.

(i) How far is C from A? (ii) What is the bearing of C from A?

(iii) What is the bearing of A from C?

b) A bus is at P, a place 250 km east from Q. Traveling a distance of 400 km to due
west of P, it reaches at R. Express this information by scale drawing and find

(i) the distance from Q to R (ii) the bearing of Q from R.

It’s your time - Project work!

7. a) Let’s choose an appropriate scale and draw a road map of your school from your
home.

b) Let’s sketch the layout of ground floor of a house with an appropriate scale to
the following structures.

sitting room - 12 m by 10 m, a bed-room - 10 m by 8 m,

a kitchen - 8 m by 6 m, a dinning room - 8 m by 6 m,

a washroom - 4 m by 3 m, guest room - 10 m by 8 m.

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Unit Statistics

22

22.1 Review

Statistics is a branch of mathematics in which data of facts and information are collected,
sorted, displayed, and analysed. Statistics are used to make decision and prediction
about the future plans and policies. Statistical information helps to understand the
economic problems and formulation of economic policies.

The word ‘statistics’ comes from the word ‘state’, largely because it was the job of the
state to keep records and make decisions based on census result.

22.2 Collection of data

The marks obtained by 24 students of class 8 in Mathematics out of 20 full marks are
given below.

15, 10, 17, 14, 19, 12, 15, 10, 16, 20
14, 18, 15, 10, 17, 16, 18, 17, 20, 13

Such numerical figures are called data.

Data should be presented in a proper order so that it is easier to get the necessary
information for which they are collected. When the data are arranged either in ascending
or in descending order, they are said to be in proper order. The properly arranged data
are called arrayed data; otherwise, they are said to be raw data.

22.3 Frequency table

The data given below are the hourly wages (in Rs) of 20 workers of a factory:

75, 90, 65, 80, 70, 65, 75, 75, 80, 70, o Raw data
75, 80, 90, 80, 70, 70, 75, 75, 70, 75

65, 65, 70, 70, 70, 70, 70, 75, 75, 75, o Arrayed data
75, 75, 75, 75, 80, 80, 80, 80, 90, 90

Here, Rs 65 is repeated 2 times. So, its frequency is 2.
Rs 70 is repeated 5 times. So, its frequency is 5.
Rs 75 is repeated 7 times. So, its frequency is 7.
Rs 80 is repeated 4 times. So, its frequency is 4.
Rs 90 is repeated 2 times. So, its frequency is 2.

In this way, a frequency is the number of times a value occurs. Data and their frequencies
can be presented in a table called frequency table.

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Statistics
Now, let’s present the above wages of 20 workers in a frequency table.

Wages (in Rs) Tally marks Frequency
65 || 2
70 |||| 5
75 |||| || 7
80 ||| 4
90 || 2
20
Total

Tallying is a system of showing frequencies using diagonal lines grouped in fives. Each
time five is reached, a horizontal line is drawn through the tally marks to make a group
of five. The next line starts a new group. For example,

1 o| 6 o |||| |
2 o || 7 o |||| ||
3 o ||| 8 o |||| |||
4 o |||| 9 o |||| ||||
5 o |||| 10 o |||| ||||

22.4 Grouped and continuous data

Let following are the marks obtained by 20 students in a Mathematics exam of full
marks 50.

27, 38, 25, 18, 9, 24, 48, 15, 27, 35

23, 45, 32, 16, 26, 39, 20, 33, 40, 37

The above mentioned data are called individual or discrete data. Another way of
organizing data is to present them in a grouped form. For grouping the given data, we
should first see the smallest value and the largest value, then we have to divide the data
into an appropriate class–interval. The numbers of values falling within each class–
interval give the frequency. For example,

Marks Tally marks Frequency
0 – 10 | 1
10 – 20 ||| 3
20 – 30 |||| || 7
30 – 40 |||| | 6
40 – 50 ||| 3
Total 20

In the above series, 9 is the smallest value and 48 is the largest value. So, the data are
grouped into the interval of 0 – 10, 10 – 20, etc., so that the smallest and the largest
values should fall in the lowest and the highest class–interval respectively.

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Let’s consider a class–interval 10 – 20.
Here, 10 is called the lower limit and 20 is the upper limit of the class–interval.
The difference between two limit is called the length or height of each class–interval.
For example, in 10 – 20 the length of the class–interval is 10.
Again, let’s take class–intervals, 0 – 10, 10 – 20, 20 – 30, …
Here, the upper limit of a pervious class–interval has repeated as the lower limit of the
consecutive next class–interval. Such an arrangement of data is known as grouped and
continuous data.

22.5 Cumulative frequency table

The word ‘cumulative’ is related to the word ‘accumulated’, which means to ‘pile up’.
The table given below shows the marks obtained by 20 students in a mathematics test
and the corresponding cumulative frequency of each class–interval

Marks Frequency Cumulative o 2 students obtained marks less than 10.
(f) frequency (c.f) o 7 students obtained mark less than 20.
0 – 10 o 15 students obtained marks less than 30.
10 – 20 2 2 o 19 students obtained marks less then 40
20 – 30 5 5+2=7 o 20 students obtained marks less than 50.
30 – 40 8 8 + 7 = 15
40 – 50 4 15 + 4 = 19
1 19 + 1 = 20
Total
20

Thus, cumulative frequency corresponding to a class–interval is the sum of all frequencies
up to and including that class–interval.

22.6 Graphical representation of data

We have already discussed to present data in frequency distribution tables. Alternatively,
we can also present data graphically. Different types of diagrams are used for this
purpose. Here, we shall discuss two types of diagrams: line graph and pie chart.

(i) Line graphs

Data can also be represented by plotting the corresponding frequencies in the graph
paper. The line so obtained by joining the points is called the line graph.

While constructing a line graph, the frequencies of the items are plotted along y–axis.
The line graph given below represents the daily wags of the workers of a company.

Wages in Rs. 40 50 60 70 80 90 100

No. of workers 15 8 22 18 10 8 5

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Y

25

20

No. of workers 15

10

5

0 40 50 60 70 80 90 100 X
wages (in Rs)

EXERCISE 22.1

General Section – Classwork

Lets say and write the answers as quickly as possible.

1.a) A branch of mathematics which deals with the collection, presentation, analysis
and interpretation of data is known as ............................................

b) The properly arranged data are called ............................................

c) The number of times that a particular observation occurs in the data
is the ...........................................

d) In a class interval 40 - 50, the lower limit is ....................................... and the
upper limit is ....................................

2. The adjoining line graph shows 120
the average rain fall in mm during
6 months of the year 2020.

(i) In which month was the 100
minimum average rainfall? 80
On which month was it
maximum?

................................................ Rainfall (mm) 60
40
(ii) How much was the average
rainfall recorded in August?

................................................

(iii) Write a paragraph about the 20
common trend of rainfall
during these months in your 0 Apr May June July Aug Sep
exercise book. Months

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Creative Section

3. a) From the marks given below obtained by 20 students in Mathematics, construct
a frequency distribution table with tally marks.

15, 18, 12, 16, 18, 10, 15, 16, 15, 12

10, 12, 15, 12, 16, 18, 12, 15, 12, 16
b) The marks obtained by 40 students in mathematics in SEE examination are

given below. Group the data into the class intervals of length 10 and construct a
cumulative frequency distribution table.
42, 68, 80, 45, 92, 36, 8, 17, 49, 30
5, 26, 98, 74, 53, 65, 72, 28, 55, 46
86, 70, 62, 27, 16, 44, 85, 59, 51, 73
66, 78, 38, 81, 97, 77, 69, 45, 33, 67

c) The marks obtained by 50 students of class 8 in Mathematics are shown in the
table given below. Construct a cumulative frequency table to represent the data.

Marks 10 20 30 40 50

No. of students 4 9 20 15 2

d) The hourly wages of 40 workers in a factory are shown in the table given below.
Show their wages in a cumulative frequency table.

Wages (in Rs.) 50 60 70 80 90

No. of workers 7 14 11 5 3

4. a) Class 8 students conducted a survey and recorded the number of children in
primary level in their school in the following table. Construct a line graph to
show the numbers in different classes.

Classes I II III IV V
No. of children 25 20 40 50 45

b) The table given below shows the velocity of a bus at different interval of time.
Draw a line graph to show the velocity–time graph.

Time (in second) 5 10 15 20 25 30
Velocity (in m/s) 5 15 20 30 10 25

It’s your time - Project work!

5. a) Let’s collect and write the marks obtained by your friends in the recently
conducted mathematics exam. Group the data into the class interval of length
10 and show them in a cumulative frequency distribution table.

b) Let’s collect and write the number of students from class 1 to 8 in your school.
Show the data in a line graph.

(ii) Pie chart

A pie chart is a circular statistical graphic which is divided into sectors and the angles
of the sectors represent the frequency.
Constructing pie–charts

Follow these steps to draw a pie chart.

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– Add all the frequencies and write each frequency as a fraction of the total frequency.

– Change each fraction into a number of degrees multiplying by 360q. (There are 360q

in a circle) corresponding frequency
Total frequency
i.e. number of degrees of each fraction = × 360q

– Tabulate the angles in ascending or descending order.

– Draw a circle of convenient size. Then draw a radius as a starting point.

– Use a protractor to construct the angles at the centre corresponding to each sector.

Worked-out examples

Example 1: The table given below shows the number of students in classes 1 to 5 of
a school. Draw a pie chart to represent the numbers.

Class I II III IV V

No. of students 40 50 30 45 15

Solution:

Here, total number of students = 40 + 50 + 30 + 45 + 15 = 180

Number of degrees of the sector of class I = 40 × 360° = 80° Class II
180 100°
Class I
Number of degrees of the sector of class II = 50 × 360° = 100° Class III 80°
180

Number of degrees of the sector of class III = 30 × 360° = 60° 60° 30° Class V
180 90°

Number of degrees of the sector of class IV = 45 × 360° = 90° Class IV
180

Number of degrees of the sector of class V = 15 × 360° = 30°
180

Example 2: The table given below shows the monthly income of a family from
different sources. Represent the data in a pie chart.

Source House rent Business Agriculture Salary Others

Income (in Rs) 1500 1700 1200 2500 300

Solution:

Here, total monthly income = Rs 7200

Number of degrees of the sector of house rent = 1500 × 360° = 75° Salary
Number of degrees of the sector of business 7200 125°
Business
Number of degrees of the sector of agriculture = 1700 × 360° = 85° 85° Othe1r5°
7200
Agriculture
House rent 60°

75°

Number of degrees of the sector of salary = 1200 × 360° = 60°
7200
2500
Number of degrees of the sector of others = 7200 × 360° = 125° = 300 × 360° = 15°
7200

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Now, tabulating the angles in descending order: Statistics

Source Salary Business House rent Agriculture Others
60q 15q
Angle 125q 85q 75q

Example 3: The monthly budget of a family is shown in Education Food
the given pie chart. If the total expenditure is 95° 120°
Rs 10800, calculate the expenditure on each item
and show in a table. Rent Miscellaneous
75° 30°
Solution:
Here, total expenditure = Rs 10800 Fuel
40°

Expenditure on food = Rs 10800 × 120 Whole circle represents the total
360 expenditure.
i.e. 360q represents Rs 10800
= Rs 3600
10800
Expenditure on education = Rs 10800 × 95 1q represents Rs 360
360
120q represents Rs 10800 × 120
= Rs 2850 360

Expenditure on rent = Rs 10800 × 75 = Rs 2250
360

Expenditure on fuel = Rs 10800 × 40 = Rs 1200
360

Expenditure on miscellaneous = Rs 10800 × 30 = Rs 900
360

Now, the table given below shows the actual expenditure.

Items Food Education Rent Fuel Miscellaneous
Expenditure (Rs) 3600 2850 2250 1200 900

EXERCISE 22.2 Farming
General Section – Classwork 180°

1. The given pie-chart shows the income of a family from three Business Services
different sources. Answer the following questions. 60°

120°

a) How many degrees represent the total income of the family? ..............................

b) From which source does the family have maximum income? ..............................

c) From which source does the family have minimum income? ..............................

d) If the annual income of the family is Rs 4,50,000, how much is the income from
farming? ..............................

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Creative Section

2. a) The table given below shows the number of students of a school from class 4 to 8.
Represent the data in a pie chart.

Class IV V VI VII VIII

No. of students 50 45 40 35 10

b) The monthly budget of a family is given below.

Food – Rs 2100 Clothing – Rs 3300

Miscellaneous – Rs 2250 Saving – Rs 3150

Represent the above data in a pie chart.

c) Mr. Limbu’s spends Rs 10,800 in a month. His monthly expenditure on different

headings are given below:

Food – Rs 2,400 Education – Rs 3,600

Rent – Rs 4,200 Miscellaneous – Rs 600

Show the above data in a pie chart.

d) Represent the following data in a pie chart. Monthly expenditure of a family.

Item Education House rent Food Others

Amount (Rs) 8000 4000 10000 14000

e) World consumption of energy is given below:

Natural gas – 20 % Nuclear – 5 %

Coal – 25 % Hydroelectric – 10 %

Oil – 40 %

Represent the above data in a pie chart.

3. a) Mrs. Rai’s monthly expenditure is Rs 10,800. The Food Rent
diagram on the right is a pie chart showing her 150° 70°
expenditure on different headings. Work out how
much was spent under each heading. 30° Miscellaneous

60° 50°
Study Transportation

b) The diagram on the right is a pie chart showing the expenses Wages Fuel
of a small manufacturing firm. The total expenses were 150° 40°
Rs 1,44,000 in a month. Calculate the expenditure on
each heading. Raw Rent
materials 50°

120°

c) The pie chart given alongside shows the votes secured by YX
three candidates A, B and C in an election. If A secured 120°
5,760 votes,
i) how many votes did C secure? Z
140°
ii) who secured the least number of votes? How many votes
did he secure?

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d) The given pie chart shows the composition of different Cotton
materials in a type of cloth in percent. 90° Nylon
54°
i) Calculate the percentage of each material found in the
cloth. 144° 72°
Polyester Others
ii) Calculate the weight of each material contained by a
bundle of 50 kg of cloth.

It’s your time - Project work!

4. a) Make a group of your friend. Take the number of students from classes 1 to 5

separately and present the data you obtained in pie-chart.

b) Collect the number of students of your class who secured A+, A, B+, B, etc.
grades in Mathematics exam and show the data in pie-chart.

22.7 Measures of central tendency

The measure of central tendency gives a single central value that represents the
characteristics of entire data. A single central value is the best representative of the
given data towards which the values of all other data are approaching.

Average of the given data is the measure of central tendency. There are three types of
averages which are commonly used as the measure of central tendency. They are: mean,
median, and mode.

22.8 Arithmetic mean

Arithmetic mean is the most common type of average. It is the number obtained by
dividing the sum of all the items by the number of items.

i.e. mean = sum of all the items
the number of items

(i) Mean of non–repeated data

If x represents all the items and n be the number of items, then mean (x) = ∑x
n

(ii) Computation of combined mean

We can compute a single mean from the means of different sets of data. Such mean is

called a combined mean. and x1 be its mean. Also,
its mean. If the combined
Lneutm, nb1erbeofthiteemnusminbethr eofseitceomnsd in the first set of data nm2 ebaenthbee
x , then, set of data and x2 be

x= n1 x1 + n2 x2
n1 + n2

Worked-out examples

Example 1: Calculate the average of the following marks obtained by 10 students
of a class in mathematics.
25, 18, 35, 24, 15, 20, 33, 28, 22, 30

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Solution:

Here, 6x = 25 + 18 + 35 + 24 + 15 + 20 + 33 + 28 + 22 + 30 = 250

n = 10 ∑x 250
n 10
Now, mean (x) = = = 25

Example 2: The average weight of 90 students from class I to V of a school is 20 kg

and the average weight of 60 students from class VI to X is 35 kg. Find

average weight of the students of the school.

Solution:
Here, n1 = 90 and x1= 20 kg , n2 = 60 and x2 = 35 kg

? Combined mean x = n1 x1 + n2 x2
n1 + n2

90 × 20 + 60 × 35 = 1800 + 2100 = 26 kg
= 90 + 60 150

Hence, the average weight of the students of the school is 26 kg.

Example 3: If the average of the following wages received by 5 workers is Rs 35,

find the value of p.
30, 36, p, 40, 44.

Solution:
Here, 6x = 30 + 36 + p + 40 + 44 = 150 + p

n =5 ∑x
n
Now, average =

or, 35 = 150 + p
5

or, 150 + p = 175
or, p = 25

So, the required value of p is Rs 25.

(ii) Mean of individual repeated data (Mean of a frequency distribution)

In the case of repeated data, follow the steps given below to calculate the mean.

– Draw a table with 3 columns

– Write down the items (x) in ascending or descending order in the first column and
the corresponding frequencies in the second column.

– Find the product of each item and its frequency (fx) and write in the third column.

– Find the total of f column and fx column.

– Divide the sum of fx by the sum of f (total number of items), the quotient is the
required mean.

Example 4: From the table given below, calculate the mean mark.

Age (in years) 6 7 8 9 10 11

No. of students 3 5 4 9 7 2

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Solution:
Calculation of average age:

Marks (x) No. of students (f) fx
6 3 18
7 5 35
8 4 32
9 9 81
10 7 70
11 2 22
6fx =258
Total N = 30

Now, mean mark (x) = ∑fx = 258 = 8.6 years
N 30

So, the required average age is 8.6 years.

Example 5: If the mean of the data given below be 17, find the value of m.

x5 10 15 20 25 30
f2 5 10 m 4 2

Solution: f fx
x 2 10
5 5 50
10 10 150
15 m 20m
20 4 100
25 2 60
30 N = 23 + m 6fx = 370 + 20m

Total

Now, mean (x) = ∑fx
N

or, 17 = 370 + 20 m
23 + m

or, 391 + 17m = 370 + 20 m

or, 3m = 21

or, m = 7

So, the required value of m is 7.

(iii) Mean of grouped and continuous data

In the case of grouped and continuous data, we should find the mid–values (m) of
each class interval and it is written in the second column. The mid–value of each class

interval is obtained as: Mid–value = lower limit + upper limit
2

Then, each mid–value is multiplied by the corresponding frequency and the product
fm is written in the fourth column. The process of calculation of mean is similar to the
above mentioned process.

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Example 6: Calculate the mean from the table given below.

Solution: Marks 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50
No. of students 3 8 12 7 2

Calculation of mean

Marks (x) mid–value (m) No. of students (f) fm

0 – 10 5 3 15
10 – 20 15 8 120
20 – 30 25 12 300
30 – 40 35 7 245
40 – 50 45 2 90

Total N = 32 6fm = 770

Now, mean marks (x) = ∑ fx = 770 = 24.06
N 32

EXERCISE 22.3
General Section – Classwork

1. Let’s say and write the answers as quickly as possible.

a) Average of 4 and 6 is ........................ b) Average of 2, 5 and 9 is .......................

c) Average of 4, 6, 8, 10 is ........................ d) Average of 3 and x is 5, x = ..................

e) Average of p and 4 is 3, p = ...............................

2. a) 6fx = 50 , n = 5, x = .............. b) 6fx = 60, n = 10, x = ..............

c) 6fx = 80, n = 8, x = .............. d) 6fx = 150, n = 25, x = ..............

e) 6fx = 40, x = 5, n = ............... f) 6fx = 70, x = 14, n = ..............

Creative Section A

3. a) The ages of Ram, Hari, Shyam, Krishna, and Gopal are 12, 18, 13, 16, and 6 years
respectively. Find their average age.

b) Find the mean value of the following:
5, 11, 14, 10, 8, 6

c) Find the mean from the data given below.
57, 74, 83, 76, 60

4. a) If mean (x) = 9, 6x = 80 + p and N = 10, find the value of p.
b) In an individual series, if 6x = 60 + a, N = a – 4 and mean (x) = 5, find the
value of a.
c) If 6x = 400 – m, N = 18 + m and x = 10, find the value of m.

5. a) The average age of 5 students is 9 years. Out of them the ages of 4 students are
5, 7, 8 and 15 years. What is the age of the remaining student?

b) If 7 is the mean of 3, 6, a, 9, and 10, find the value of a.

c) Find x if the mean of 2, 3, 4, 6, x, and 8 is 5.

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6. a) The average rainfall in Kathmandu valley in the first 6 months of the year 2077
was 75 mm and the average rain fall of the last 6 months was 15 mm. Find the
average rainfall in Kathmandu in the whole year.

b) The average weight of 30 girls in class 8 is 42 kg and that of 20 boys is 45 kg.
Find the average weight of the students of the class.

c) The average height of x number of girls and 15 boys is 123 cm. If the average
height of boys is 125 cm and that of girls is 120 cm, find the number of girls.

7. a) If 6fx = 40 + a, N = 4 + a and x = 5, find the value of a.

b) If the mean of a series having 6fx = 100 – k and N = k – 4 is 15, find the value
of k.

Creative Section - B

8. a) Find the mean from the given table

Marks obtained 15 25 35 45 55
No. of students 7 8 12 7 6

b) The ages of the students of a school are given below. Find the average age.

Age (in years) 5 8 10 12 14 16

No. of students 20 16 24 18 25 15

c) Compute the arithmetic mean from the following frequency distribution table.

Height (in cm) 58 60 62 64 66 68

No. of plants 12 14 20 13 8 5

9. a) Find the mean of the following frequency distribution.

Class interval 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60

Frequency 75 6 12 8 2

b) Find mean.

Wages (Rs.) 40 – 50 50 – 60 60 – 70 70 – 80 80 – 90 90 – 100

No. of workers 3 6 12 7 8 4

c) Find the mean from the following data.

Marks obtained 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80

No. of students 4 5 2 4 3 2

d) Compute the mean from the table given below.

Age (years) 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70

No. of people 2 5 7 6 3 2

e) The table below gives the daily earnings of 110 workers in a textile mill.

Daily earning 200 – 300 300 – 400 400 – 500 500 – 600 600 – 700

No. of workers 40 20 15 25 10

Find the average weekly earning.

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f) The ages of workers in a factory are as follows

Age in years 18 – 24 24 – 30 30 – 36 36 – 42 42 – 48 48 – 54

No. of workers 6 8 12 8 4 2

Calculate the average age of the groups.
10. a) Find the mean by constructing a frequency table of class interval of 10 from the

data given below.

7, 47, 36, 39, 31, 19, 41, 49, 9, 51, 29, 22,

59, 17, 49, 21, 24, 12, 31, 8, 36, 18, 32, 16, 23
b) Construct a frequency table of class interval of 10 from the given data and find

the mean.
23, 5, 17, 28, 39, 52, 16, 22, 69, 75

41, 33, 9, 49, 34, 59, 72, 46, 65, 58

60, 48, 64, 32, 50, 73, 57, 51, 63, 36

It’s your time - Project work!

11. a) Let’s ask your parents and write the ages of your family members. Then, calculate
the average age of your family members.

b) Let’s collect the marks obtained by your any 10 friends in the recently conducted
mathematics exam and calculate the mean mark.

c) Let’s collect and write the number of students in each class from class 1 to 8
in your school. Then find the average number of students in each class in your
school.

22.9 Median

Look at the following series. 5, 9, 13, 17 21, 25, 29
3 items Middle item 3 items

In the above series, the numbers are arranged in ascending order. Here, the fourth item
17 has three items before it and three items after it. So, 17 is the middle item in the
series. 17 is called the median of the series.

Thus, median is the value of the middle–most observation, when the data are arranged
in ascending or descending order of magnitude.

(i) Median of ungroupped data

To find the median of an ungroupped data, arrange them in ascending or descending
order. Let the total number of observation be n.

– If n is odd, the median is the value of the n+1 th
2 observation.

– If n is even, the median is the average of the n th n +1 th
2 and 2 observation.

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Worked-out examples

Example 1: The weights in kg of 7 students are given below. Find the median
weight.

35, 45, 43, 30, 52, 40, 37
Solution:
Arranging the weights in ascending order, we have,

30, 35, 37, 40, 43, 45, 52

Here, n = 7 n+1 th
Now, the position of median = 2 item

= 7+1 th
2 item = 4th item
i.e., 4th item is the median.
? Median = 40 kg.

Example 2: The marks obtained by 6 students are given below. Calculate the
median mark.
15, 25, 10, 30, 20, 35

Solution:
Arranging the marks in ascending order, we have,

10, 15, 20, 25, 30, 35

Here, n = 6 th
item
Now, the position of median = n+1
2

= 6+1 th
2 item = 3.5th item

3.5th item is the average of 3rd and 4th items.

? Median = 20 + 25= 24 = 22.5
2 2

(ii) Median of Discrete series

To compute the median of a discrete series of frequency distribution, we should display

the data in ascending or descending order in a cumulative frequency table. Then, the

median is obtained by using the formula:
th
Median = value of N+1 item
2

Example 3: Compute the median from the table given below.

x4 8 12 16 20 24
f3 54762

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Solution:
Cumulative frequency table

xf c.f.

4 3 3
6 5 8
12 4 12
16 7 19
20 6 25
24 2 27
Total N = 27

Now, position of median = N+1 th 27 + 1 th = 14th item
2 2
item = item

In c.f. column, the c.f. just greater than 14 is 19 and its corresponding values is 16.
? Median = 16.

22.10 Quartiles

Quartiles are the values that divide the data arranged in ascending or descending order
into four equal parts. A distribution is divided into four equal parts by three quartiles.

– athbeovfierswt hoirchlo7w5er%qoufatrhtieleit(eQm1)s is the point below which 25 % of the items lie and
lie.

– The second quartile (Q2) is the point below which 50 % of the items lie and above
which 50 % of the items lie. Of course, the second quartile is the median.

– The third or upper quartile (Q3) is the point below which 75 % of the items lie and
above which 25 % of the items lie.

If N be the number of items in ascending (or descending) order of a distribution, then in

the case of discrete data = N+1 th
the position of the first quartile (Q1) 4
item

the position of the second quartile (Q2) = 2(N + 1) th N+1 th
4 2
= item

the position of the third quartile (Q3) = 3(N + 1) th
4
item

Similarly, in the case of grouped data, the positions of quartiles are obtained in the

following ways. th

The first quartile (Q1) = value of N item
4

The second quartile (Q2) = 2(4N)thitem = value of N th
The third quartile (Q3) = value of 3(4N)thitem 2
item

After finding the positions, the process of computing the quartiles is as similar as the

process of computing median.

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Example 4: Find the first quartile (Q1) and the third quartile (Q3) from the data
given below.

7, 17, 10, 20, 13, 28, 24

Solution:

Arranging the data in ascending order,

7, 10, 13, 17, 20, 24, 28

Here, n = 7 n+1 th = 7+1 th
The position of the first quartile (Q1) = 4 4
The value of 2nd item is 10. item item = 2nd item

? The first quartile (Q1) = 10. n+1 th
4
Again, the position of the third quartile (Q3) = 3 item = 6th item
The value of 6th item is 24.

? The third quartile (Q3) = 24.

Example 5: The marks obtained by 10 students of class 8 in Mathematics are given

below. Compute Q1 and Q3.
18, 14, 16, 10, 15, 12, 8, 5, 11, 20

Solution:

Arranging the marks in ascending order,

5, 8, 10, 11, 12, 14, 15 16, 18, 20

The position of the Q1 = n + 1 th = 10 + 1 th = 2.75th item
Here, the 2nd item is 14 and 34rd 4
item item

item is 16.

? Q1 = 14 + (16 – 14) × 75% = 14 + 2 × 75 = 15.5
100 3 × 2.75th
aonf dQ93 t=h it3emnis+4118. th

Again, the position item = item = 8.25th item
Here, 8th item is 16

? Q3 = 16 + (18 – 16) × 25% = 16 + 2 × 25 = 16.5
100
Example 6: Compute the first and the third quartiles from the table given below.

Marks 30 40 50 60 70 80

Solution: No. of students 46 10 12 5 2

Cumulative frequency distribution table

Marks (x) No. of students (f) c.f.
30 4 4
40 6 10
50 10 20
60 12 32
70 5 37
80 2 39

Total N = 39

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 297 Vedanta Excel in Mathematics - Book 8

Statistics

Now, the position of the first quartile (Q1) = N+1 th 39 + 1 th
4 4
item = item = 10th item

In c.f. column, the corresponding value of the c.f. 10 is 40.

? The first quartile (Q1) = 40 +4its1cotrhirteesmpo=nd3i0ntgh

Again, the position of the third quartile (Q3) =3 N item is 60.
In c.f. column, the c.f. just greater than 30 is 32 and value

? The third quartile (Q3) = 60

22.11 Mode

The mode of a set of data is the value with the highest frequency. A distribution that has
two modes is called bimodal. The mode of a set of data is denoted by Mo.

(i) Mode of discrete data

In the case of discrete data, mode can be found just by inspection, i.e. just by taking an
item with highest frequency.

Example 7: Find the mode for the following distribution.
25, 18, 20, 18, 22, 18, 20, 18, 20

Solution:
Arranging the data in ascending order.

18, 18, 18, 18, 20, 20, 20, 22, 25
Here, 18 has the highest frequency.
? Mode = 18.

22.12 Range

The difference between the largest and the smallest score is called range.
? Range = Largest score – Smallest score

Example 8: The marks obtained by 10 students of class 8 in Mathematics are given
below. Find the range.

78 36 27 95 43
15 69 84 72 51
Here, the highest marks = 95
The lowest marks = 15

? Range = highest score – lowest score

= 95 – 15 = 80

Vedanta Excel in Mathematics - Book 8 298 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur