Prime Mathematics 5

6. (a) 16l 245ml (b) 461l 825ml (c) 1l 750ml (d) 1l 400ml
7. (a) 75ml 250ml (b) 20l 400ml (c) 2l 315ml (d) 124 ml

1. (a) 9340 gm Exercise: 5.3 (d) 65 day
(b) 12450 gm (c) 85 hg (e) 20089 mg

(f) 352048 cg (g) 584 dg (h) 2025 gm

2. (a) 50.25 gm (b) 24.7 cg (c) 48.9gm (d) 16.65 kg (e) 4.84024 kg

(f) 62.56 kg (g) 75.26575 kg (h) 155044 mg

3. (a) 37 kg 1 gm 60 mg (b) 44 kg 225 gm (c) 41 kg 500 mg

(d) 41 kg 201 gm 190 mg (e) 57 kg 221 gm 370 mg

(f) 22 kg 186 gm 595 mg

4. (a) 45kg 225 gm 200 mg (b) 14 kg 429 gm 530 mg

(c) 467 kg 99 gm 380 mg (d) 7 kg 799 gm 820 mg

Exercise: 5.4
1. (a) 42 kg 72 gm 42 mg (b) 19264 kg 400 gm (c) 102kg 560 gm (d) 276 kg 640 gm
(e) 42 gm 76 cg 2 mg (f) 602 gm 400 mg (g) 78 hg (h) 109 dag 45 dg 35 mg

2. (a) 1 kg 94 gm (b) 730 gm (c) 4 gm 4dg 3cg (d) 3 dag 2 gm 2 dg
(e) 1 kg 392 gm (f) 1 kg 625 gm (g) 1 kg 585 mg (h) 59 dg 8 cg 4 mg

3. (a) 52 kg 200 gm (b) 52 kg 100 gm (c) 3 kg 240 gm (d) 2 kg 250 gm
(e) 195 mg 870 mg (f) 1 dg 2 cg 4 mg

1. (a) 16 hrs 23 mins 20 sec Exercise: 5.5
(b) 55 hrs 31 min 25 sec (c) 29 yrs 2 months 5 days

(d) 22 yrs 2 months 10 days (e) 9 yrs 24 weeks 2 day (f) 19 yrs 19 weeks 4 days

2. (a) 3 hrs 49 mins 30 secs (b) 3 hrs 51 mins 48 secs (c) 4 yrs 8 mons 23 days

(d) 3 yrs 10 months 2 days (e) 7 weeks 5 days 18 hours

3. (a) 23 hrs 47 mins 30 secs (b) 104 hrs 8 mins 24 secs (c) 136 days 21 hrs

(d) 97m 23 day 18 hrs (e) 186 yrs 2 months (f) 26 yrs 26 weeks 6 days

4. (a) 1 hrs 45 mins 45 secs (b) 1 hrs 8 mins 27 secs (c) 1 day 7 hrs 54 mins

(d) 2 yrs 4 months 6 days (e) 3 yrs 19 week 1 day

5. (a) 4 hrs 21 mins 18 secs (b) 8 yrs 9 month (c) 1 hrs 40 mins

(d) 2055 - 11 - 3 B.S. (e) 32 hrs 40 mins (f) 16 hrs 33 mins 12 secs
(g) 3 hrs 24 mins

1. (a) Rs. 94 and 45p Exercise: 5.6 (e) Rs. 148 and 50p
(b) Rs. 134 and 25p (c) Rs. 434 and 40p (f) Rs. 445.82
(d) Rs. 371 and 80p (e) Rs. 225.80
2. (a) Rs. 273 and 55p (c) Rs. 232 and 70p (f) Rs. 974.75
(b) Rs. 332 and 40p (d) Rs. 467.75 (e) Rs. 2421.72
(c) Rs. 1294 and 32p (f) Rs. 17243.28
3. (a) Rs. 186 and 90p (d) Rs. 381.70 (c) Rs. 9 and 95p
(b) Rs. 1574 and 64p (b) Rs. 11 and 40p

4. (a) Rs. 55 and 62p (b) Rs. 14 and 75p (c) Rs. 123 and 75p
(d) Rs. 368 and 85p (e) Rs. 55 and 45p

5. (a) Rs. 1679 and 85p
(d) Rs. 92 and 50p

96 Prime Mathematics Book - 5

Un6it Mensuration

Estimated periods − 12

Objectives
At the end of this unit, the students will be able to:
• find the perimeter of plane figure.
• find the perimeter of regular figure like equilateral triangle square and rectangle using
formula.
• find the area of squares and rectangles using formula.
• find the volume of cubes and cuboids using formula.
• solve the simple word problems related to perimeter , area and volume.

Teaching Materials
Models of squares, rectangles, cubes and cuboids

Activities
It is better to:
• discuss about the length of surrounding or boundaries of figure in real field and find
their perimeters
• discuss about the perimeter of regular figures relating with formula
• discuss about area and volume relating with formula.
• use geoboards and conduct an activity about making shapes and models.
• make playcards and design a game for matching them.

Prime Mathematics Book - 5 97

Perimeter

How long should a wire be to fence a D C
rectangular garden ABCD of length 50m
and breadth 30m?

Total length of wire needed 30m

= AB + BC + CD + DA

= 50m + 30m + 50m + 30m A 50m B
= 160m

Here, the total length around the garden, AB + BC + CD + DA is its
perimeter.

The total length of the boundary of a plane figure is called its perimeter.
[Periphery = boundary in Latin]

A AD

BC BC
For ∆ABC. For a quadrilateral ABCD
Perimeter = AB + BC + CA Perimeter = AB + BC + CD + DA

A

BE

CD
For the petagon ABCDE
Perimeter = AB + BC + CD + DE + EA

Perimeter of regular plane figures.
A In an equilateral triangle ABC.

let, AB = BC = AC = a, then

B C Perimeter = AB + BC + AC = a + a + a = 3a.

98 Prime Mathematics Book - 5

D C In a square ABCD,

let, AB = BC = CD = DA = a then

Perimeter = AB + BC + CD + DA

AB = a + a + a + a = 4a.

Dl C In a rectangle ABCD.

Let, length AB = DC = l and breadth BC = AD = b, then
b b Perimeter = AB + BC + CD + DA

Al B =l+b+l+b
= 2l + 2b = 2(l + b)

Example 1: Find the perimeter of the triangle ABC.

Solution: In the given triangle ABC, A
AB = 5cm, BC = 6cm and AC = 7cm 7cm
∴ Perimeter = AB + BC + AC 5cm
B 6cm
= 5cm + 6cm + 7cm C

= 18cm

Thus, the perimeter of the triangle ABC is 18cm.

Example 2: Find the perimeter of the given quadrilateral ABCD.
Solution: Here, in the given quadrilateral ABCD

AB = 6cm, BC = 4cm, CD = 5cm and AD = 3cm D 5cm C
∴ Perimeter = AB + BC + CD + DA 3cm 4cm
B
= 6cm + 4cm + 5cm + 3cm A 6cm

= 18cm

∴ The perimeter of the given quadrilateral is 18cm.

Example 3: Find the perimeter of an equilateral triangle of side 4cm.
Solution: Length of a side of an equilateral triangle (a) = 4cm

∴ Perimeter of the equilateral triangle = 3a
= 3 × 4cm
= 12cm

Therefore, the perimeter of the equilateral triangle is 12cm.

Prime Mathematics Book - 5 99

Example 4: Find the perimeter of a rectangle having length 6cm and
breadth 4.5cm.

Solution: Here,
Length of the rectangle (l) = 6cm

Breadth of the rectangle (b) = 4.5cm
Now, we have,
Perimeter of rectangle = 2(l + b)

= 2(6cm + 4.5cm)
= 2 × 10.5cm
= 21cm
∴ The perimeter of the rectangle is 21cm.

Example 5: Find the perimeter of the square of side 5cm.
Solution: Here, length of side of square (l) = 5cm

Now, perimeter of the square = 4l
= 4 × 5cm
= 20cm

∴ The perimeter of the square is 20cm.

Exercise 6.1

1. Find the perimeter of the following figures:

(a) A (b) P 3cm S
7cm 4cm 5cm
3.5cm
6.5cmQ 5cm R
B 5.5cm C

(c) P 5cm U (d) A 4cm E
4cm 4cm
B 6cm
QT 8cm D
9cm
3cmR 2cm S 3cm C

100 Prime Mathematics Book - 5

2. Find the perimeter of the following figures:

(a) K (b) T S

L 9.5cm M U 6cm V
(c) D
C (d) P U
A 6cm
5cm QT
B
R 4.5cm S

3. Find the perimeter of the rectangles having:
(a) length = 6cm, breadth = 4cm
(b) length = 10cm, breadth = 6cm
(c) length = 22cm, breadth = 16cm
(d) length = 18.5cm, breadth = 8.5cm
(e) length = 60m, breadth = 28m
(f) l = 112m, b = 64m
(g) l = 100m , b = 45m
(h) l = 62.5m, b = 44m

4. Find the perimeter of the square having sides:

(a) 4cm (b) 6.5cm (c) 26m

(d) 12m (e) 151 cm (f) 181 m
(g) 40m 2 4

(h) 112.5cm

5. (a) If the length of the rectangular field is 25m and breadth is 20m,
find its perimeter.

(b) If the length and breadth of a rectangle are 8.2cm and 5.5cm
respectively, find the perimeter of the rectangle.

(c) The length of a square stadium is 75m. Find its perimeter.
(d) Find the perimeter of the square whose length of a side is

17.25cm.

Prime Mathematics Book - 5 101

6. (a) If the perimeter of the triangle ABC is A 6cm C
18.5cm, find the value of a. 5cm a

B
A 3x D
(b) If the perimeter of the quadrilateral
ABCD is 56cm, find the length of the
sides of the quadrilateral. 2x 5x
B C

4x

(c) If the perimeter of the given rectangle A D

is 36cm, find the breadth (x) of the x C
rectangle. B

10cm

(d) Perimeter of the square PQRS is 68cm. PQ
Find its length of side.

SR
7. (a) Find the length of the fence around a rectangular garden 120m

long and 90m broad.

(b) If the length of a rectangle is twice its breadth, and the perimeter
is 60cm, find the length and breadth of the rectangle.

Area

Ropani, anas, bigahas, katthas, square meter, square miles, etc are the units
of area.

Consider a square of length 1cm and Total area of Nepal is
breadth 1cm. Its extension or 1,47,181 sq. km.
area is taken as 1cm × 1cm =

1cm 1cm2 or 1 square cm or 1 sq.cm.
1cm Similarly area of a square of

length 1m and breadth 1m is =
1m × 1m = 1m2 or 1 square m or 1 sq.m.

In the given figure, length of the rectangle 4cm
is 5cm and its breadth is 4cm which can be
divided into 20 squares of side 1cm. Thus area
of the rectangle is 20cm2. Multiplying length
and breadth, 5cm × 4cm = 20cm2.

∴ Area of rectangle = length × breadth 5cm
=l×b

102 Prime Mathematics Book - 5

In a square, length (l) = breadth (b)
∴ Area of square = l × l = l2

Note: To find the area, units of length and breadth should be same.

4. Addditional information: In Terai Region
1 Bigha 20 Kattha
In Hilly Region 1 Kattha 20 Dhur
1 Ropani 16 Aana

1 Aana 4 Paisa
1 Paisa 4 Dam

Example 1: Find the area of the rectangle given 4cm 3cm
in the figure alongside. 80cm

Solution: Here,
Length of rectangle (l) = 4cm
Breadth of the rectangle (b) = 3cm
We have,
Area of rectangle = l × b
or, A = 4cm × 3cm
or, A = 12cm2.

∴ Area = 12cm2.

Example 2: A side of a square is 9cm. Find its area.
Solution: Here,

Length of side of square (l) = 9cm.
We have,
Area of square = l2

= (9cm)2 = 81cm2.
∴ Area of the square = 81cm2.

Example 3: Find the area of the rectangular mat
of length 2.4m and breadth 80cm.

Solution: Here, 2.4m
LBNeroenwag,dtwhth eoofh ftahtvheee ,remcatat n(gbu) l=ar8 m0camt (=l)1 8=00 02.m4m= 0.8m

Area of rectangle = l × b

= 2.4m × 0.8m

= 1.92m2.

Therefore, the area of the mat is 1.92m2.

Prime Mathematics Book - 5 103

Example 4: Area of a rectangular room is 108m2. If its breadth is 9m, find
its length.

Solution: Here,

Area of rectangular room (A) = 108m2.

Breadth of the room (b) = 9m

Length of the room (l) = ?

We have,

l×b=A
A
or, l = b

or, l = 108m2
9m

or, l = 12m

Therefore, length of the room is 12m.

Example 5: If the area of a square room is 64m2, find its length.
Solution: Here,

Area of the square room (A) = 64m2
Length of the room (l) = ?
We have,

l2 = A
or, l2 = 64m2
or, l = 64m2
or, l = (8m)2
∴ l = 8m
Therefore, length of the square room is 8m.

Example 6: If the perimeter of square is 24cm, find its area.

Solution: Here,
Perimeter of square (p) = 24cm
Area of the square (A) = ?
If length of the square is l, we know

4l = P
or, 4l = 24cm

104 Prime Mathematics Book - 5

or, l = 24 6

cm
∴ l = 64cm

Now,

Area of the square (A) = l2 = (6cm)2

= 36cm2

Therefore, the area of the square is 36cm2.

Exercise 6.2 4cm
1. Find the area of the following rectangles:

(a) (b)
1cm

3cm 8cm
(c) (d)

12cm 60cm

16.5cm 1.2m

2. Find the area of the following squares: (d)
(a) (b) (c)

2cm 11cm 3.2cm 0.8cm

3. Find the area of the rectangles of the following measures:

(a) Length = 10cm and breadth = 7cm

(b) Length = 16.4cm and breadth = 4.2cm
(c) Length = 912 cm and breadth = 8cm
(d) Length = 2.4m and breadth = 75cm

4. Find the area of the squares of the following sides:

(a) Length = 6cm (b) length = 24.5cm
(c) Length = 18m (d) Length = 721 m

Prime Mathematics Book - 5 105

5. (a) A rectangular plot of land is 27m long and 18m broad. Find the
area of the plot.

(b) A mattress is 3.4m long and 75cm broad. Find the area of the
mattress in square metre.

(c) A rectangular field is 9 metres broad. If the perimeter of the field
is 63m, find its area.

(d) Area of a rectangular towel is 315cm2. If its length is 22.5cm, find
its breadth.

6. (a) Find the area of a square handkerchief of side 18cm.
(b) Find the area of a square plot of land of side 31.5m

(c) Perimeter of a square is 56.4cm, find its area.
(d) If the area of a square is 121cm2, find its length.

Volume 1cm
We know that every matter occupy space. The space 1cm 1cm
occupied by an object is called its volume. A cube of side
1cm is supposed to have a volume of 1cm × 1cm × 1cm
= 1cm3.

1cm A cuboid 6cm × 2cm × 1cm contains
12 cubes of volume 1cm3. So its
6cm 2cm volume (v) = 12cm3.

We observe that 12cm × 2cm × 1cm =12cm3.

Similarly, a cuboid 4cm × 2cm × 3cm contains 3cm
24 cubes of volume 1cm3. So, its volume is V
= 24cm3. Also we observed that

4cm × 2cm × 3cm = 24cm3

∴ Volume (V) = l × b × h 4cm 2cm
Volume of cuboid = length × breadth × height

or, V = l × b × h

A cube has its length, breadth and height equal
∴ Volume of a cube = length × breadth × height

= length × length × length
= l × l × l = l3

106 Prime Mathematics Book - 5

Note:

- Units of length, breadth and height should be same.
- Volume is measured in cm3, m3 etc
- Volume of liquids is measured in litres

∴ 1 litre = 1000cm3.

Conversion of units in volume:

Cubic m to cubic cm

1m3 = (1m)3

= (100cm)3 [ 1m = 100cm]

= 1000000cm3

Cubic cm to cubic metre

1cm3 = (1cm)3 3 1cm = 1 m
100
= 1 m
100

= 10001000m3

Cubic cm to litre. Cubic metre to litre.

1000cm3 = 1 litre 1m3 = (1m)3
∴ 1 cm3 = 1 litre
= (100cm)3
1000
Litre to cubic cm. = 1000000cm3
1 litre = 1000cm3
= 1000000 litre
Litre to m3. 1000

= 1000 litre

1 litre = 1000cm3

= 1000 (1cm)3

= 1000 × 13
100 m

= 1000 m3 = 1 m3
1000000 1000

Prime Mathematics Book - 5 107

Example 1: Find the volume of the solid given 4cm
alongside.

Solution: The given solid is a cuboid

where, Length (l) = 8cm

Breadth (b) = 6cm 6cm

Height (h) = 4cm 8cm

Now, we have

Volume = l × b × h

or, V = 8cm × 6cm × 4cm = 192cm3.

Therefore, the volume of the solid is 192cm3.
Example 2: Find the volume of a cube of side 7cm.
Solution: Here,
Length of side of a cube (l) = 7cm

∴ volume of the cube (V) = l3

= (7cm)3

= 343cm3

Therefore, the volume of the cube is 343cm3.

Example 3: Volume of a cuboidal box is 280cm3. If its breadth and height
are respectively 7cm and 5cm, find its length.

Solution: Here,
Volume of cuboidal box (V) = 280cm3

breadth (b) = 7cm

height (h) = 5cm

length (l) = ?

We have

l×b×h=V

or, l × 7cm × 5cm = 260cm3

or, l = 280cm3
7cm × 5cm

∴ l = 8cm
Therefore, the length of the box is 8cm.

Example 4: Length, breadth and height of a cuboidal box are 1.2m, 56cm
and 40cm respectively. Find the volume of the box in cu.m.

Solution: Here,
Length of the cuboid (l) = 1.2m

108 Prime Mathematics Book - 5

breadth (b) = 56cm = 56 m = 0.56m
100

and height (h)= 40cm = 14000m = 0.4m
Now, we have

Volume of a cuboid (V) = l × b × h

= 1.2m × 0.56m × 0.4m

= 0.2688m3.

Therefore, the volume of the cuboid = 0.2688 cu.m.

Exercise 6.3

1. Find the volume of the following cuboids:

(a) (b)
3cm

5cm 4cm 4cm 6.2cm
(c) 10.5cm

(d)

4cm 2.2cm

6cm 3cm 4.5cm 6.5cm

2. Find the volume of the following cubes of given sides:

(a) (b)

5cm

8cm
(c) (d)

1.8cm 12.5cm

Prime Mathematics Book - 5 109

3. Find the volume of the cuboids having following measuers:
(a) Length (l) = 7cm, breadth (b) = 5cm, height (h) = 4cm
(b) Length (l) = 12cm, breadth (b) = 4.5cm, height (h) = 6cm
(c) Length (l) = 25cm, breadth (b) = 14cm, height (h) = 10cm
(d) Length (l) = 18.5cm, breadth (b) = 10.2cm, height (h) = 5.5cm

4. Find the volume of the cubes of sides:

(a) 4cm (b) 9cm (c) 6.4cm

(d) 13.5cm (e) 0.7m (f) 2.8m
5. From the given volume and sides of the cuboids, find the unknown
sides:

(a) Volume (V) = 90cm3, length (l) = 6cm, breadth (b) = 5cm, height (h) = ?

(b) Volume (V) = 96cm3, length (l) = ?, breadth (b) = 6cm, height (h) = 4cm

(c) Volume (V) = 8cm3, length (l) = 5cm, breadth (b) = ?, height (h) = 4cm

(d) Volume (V) = 455m3, length (l) = ?, breadth (b) = 5.2m, height (h) = 7m

6. Find the volume of the following objects:
(a) (b)

4.5cm 1.5cm
(d)
4.5cm 4.5cm
(c)
7cm 24cm
2.4m

15cm 0.5m 0.9m

7. (a) An aquarium is 1.5m long, 90cm broad and 72cm high. How many
liters of water is required to fill the aquarium?

(b) A cubical metal tank is 2.5m long. How many litres of water can it
hold when it is full?

(c) Total surface area of a cube is 54cm2. Find it volume.
(d) The volume of a cube is 1728cm3. Find its total surface area.

110 Prime Mathematics Book - 5

Unit Revision Test

1. Define: (b) volume
(a) perimeter

2. Find the perimeter of the following:

(a) A (b) 5.1cm D 4.2cm

3.8cm 5.8cm EC
4.5cm
B 4.9cm C A1.4cm B3.8cm

3. Find the perimeter of:
(a) Rectangle having length 6.4cm and bredth 4.2cm
(b) Squares of side 3.8 cm

4. Find the area of:
(a) Square of side 21 cm
(b) Rectangle having length 6.2cm and breadth 4.3cm.

5. Find the volume of
(a) Cuboid having length (l) = 13cm and breadth (b) = 8.5cm and
height 4 cm
(b) Find the volume of a cube of side 5.4cm.

6. (a) A rectangular field is 18 m long. If the perimeter of the field is
60m, find the area of the field.

(b) A cuboidal metal tank is 125 cm long, 80cm broad and 60 cm high.
Find the capacity of the tank in liters.

Prime Mathematics Book - 5 111

Answers:

1. (a) 16cm (b) 17 cm Exercise: 6.1 (d) 33.5cm
2. (a) 28.5cm (c) 21cm
3. (a) 20 cm
(b) 24 cm (c) 22cm (d) 27 cm
(f) 352m
4. (a) 16cm (b) 32 cm (c) 76cm (d) 54cm (e) 176m
(g) 290m (h) 213m
(f) 73m
5. (a) 90m (b) 26cm (c) 104m (d) 48m (e) 62cm
6. (a) 7.5cm (g) 160m (h) 450cm

(c) 8cm (b) 27.4 cm (c) 300m (d) 69m
7. (a) 420m
(b) AB = 8 cm, BC = 16 cm , CD = 20 cm, AD = 12 cm
(d) 17cm

(b) 20 cm, 10 cm

1. (a) 3cm2 (b) 32 cm2 Exercise: 6.2 (d) 7200cm2
2. (a) 4cm2 (c) 198 cm2 (d) 0.64cm2
3. (a) 70cm2 (d) 1.8m2
4. (a) 36cm2 (b) 121cm2 (c) 10.24cm2 (d) 56.25m2
5. (a) 486m2 (d) 14cm
6. (a) 324cm2 (b) 68.88cm2 (c) 76cm2 (d) 11cm

(b) 600.25cm2 (c) 324m2

(b) 2.55m2 (c) 202.5m2

(b) 992.25m2 (c) 198.81cm2

1. (a) 60cm3 Exercise: 6.3 (d) 64.35cm3
(b) 260.4cm3 (c) 72cm3

2. (a) 125cm2 (b) 512cm3 (c) 5.832cm3 (d) 1953.125cm3

3. (a) 140cm3 (b) 324cm3 (c) 3500cm3 (d) 1037.85cm3

4. (a) 64cm3 (b) 729cm3 (c) 262.144cm3 (d) 2460.375cm3
(e) 0.343m3 (f) 21.952m3

5. (a) 3cm (b) 4cm (c) 0.4cm (d) 12.5m

6. (a) 91.125cm3 (b) 3.375cm3 (c) 2520cm3 (d) 1.08m3

7. (a) 972 liters (b) 15625 liters (c) 27cm3 (d) 864cm2

112 Prime Mathematics Book - 5

Un7it Fractions, Decimals and
Percentage

Estimated periods − 23

Objectives

At the end of this unit, the students will be able to:
• perform the fundamental operations on fractions and decimals.
• change decimal, fraction and percentage to each other .
• solve simple problems involved with fractions decimal and percentage.

Teaching Materials
Models of fraction, square papers, grids.

Activities
It is better to:
• discuss about and improper fractions, like and unlike fractions before teaching
fundamental operation of fraction
• discuss about the process of changing fractions, decimals and percentages to each other
• associate fundamental operations on fractions , decimals and percentage to solve
simple words problems.
• design a game using playcards and let the students match them in group.

Prime Mathematics Book - 5 113

Fractions

Like and unlike fractions

Consider the fractions 61 , 2 and 56 .
6

= 1 = 2 = 5
6 6 6

Here, total number of division of whole circle which is 6 parts is same in each
ptharetsfr16acatrioenal51s,o
and unit same in each.
Consider 2 3
5 and 5

123
555

Here, also total number of division of the whole which is 5 parts is same in each
1
and unit parts 5 are also same in each. Such fractions are like fractions.

Thus, the fraction having same denominators are called like fractions.

Again consider fraction 1 , 1 and 2
2 3 5

= 1 = 1 = 2
2 3 5

Here, in each figure, the wholes are same but the total number of division are
different and unit parts are not equal. Such fractions are unlike fractions.

Thus, the fractions having different denominators are called unlike fractions.

Example 1: Say whether the following fractions are like or unlike.
72 , 4 6 21 , 5 3
(i) 7 and 7 (ii) 6 and 8

Solution: (i) In the fractions 2 , 4 and 6 , denominators are same,
7 7 7
hence these fractions are like.
1 5 3
(ii) In the fractions 2 , 6 and 8 , denominators are all different,

so the fractions are unlike fractions.

114 Prime Mathematics Book - 5

Converting unlike fractions into like fractions:
1 1
Consider two fractions 2 and 3

1 1 The fractions are unlike.
2 3
= = • Unlike fractions have

different denominators.
• Unit parts are unequal.

Dividing each figure into six equal parts, we get all the unit parts equal and

thus the new fractions 3 and 2 are like fractions.
6 6

13
26

12
33

Process
(i) Find the L.C.M of the denominators.
(ii) Multiply numerator (N) and denominator (D) by certain number to make

D equal to L.C.M.

Example 2: Convert 1 and 1 into like fractions.
2 3

Solution: L.C.M of 2 and 3 = 2 × 3 = 6

∴ 1 = 1 × 3 = 3 (Multiply N and D by 3 to make D = 6)
2 2 × 3 6

1 = 1 × 2 = 2 (Multiply N and D by 2 to make D = 6)
3 3 × 2 6

∴ 3 and 2 are like fractions.
6 6

Prime Mathematics Book - 5 115

Example 3: Convert the fractions 1 , 2 and 3 into like fractions.
2 3 4

Solution: The fractions 1 , 2 and 3 are unlike
2 3 4

Here, L.C.M of denominators = 2 × 3 × 2 = 12 2 2, 3, 4
1, 3, 2

Now, 1 = 1 × 6 = 6
2 2 × 6 12

2 = 2 × 4 = 8
3 3 × 4 12

3 = 3 × 3 = 9
4 4 × 3 12

∴ 162, 8 and 9 are like fractions.
12 12

Comparison of fractions:

A. When the fractions are like fractions, the fraction with bigger
numerator is greater.
1 3
Example 4: Compare 5 and 5 .

Solutions: In the fractions 1 and 3 , denominators are same and 1 < 3
5 5

∴ 1 < 3
5 5

B. When the fractions are unlike, convert the fractions into like fractions
(make same denominators). Now the fraction with bigger numerator
is greater.
1 2
Example 5: Compare 3 and 5

Solution: 1 2
3 5
Fraction and are unlike where,

L.C.M of denominators = 3 × 5 = 15

∴ 1 = 1 × 5 = 5
3 3 × 5 15

and 2 = 2 × 3 = 6
5 5 × 3 15

Since 6 > 5

116 Prime Mathematics Book - 5

∴ 6 > 5 i.e 2 > 1
15 15 5 3

Addition and subtraction of fractions:
A. Addition and subtraction of like fractions.

Process:
- Write single (common) denominator
- Add or subtract the numerators.

Example 6: Add 3 and 1 . + =
5 5

Solution: 3 + 1 = 3 + 1 = 4 3 + 1 = 4
5 5 5 5 5 5 5
1
We can show the addition in number line as - 3

3 + 1
5 5

0 1 2 3 4 1
5 5 5 5

= 4
5
1 2
Example 7: Subtract 3 from 3 . =

Solution: 2 - 1 = 2 - 1 = 1 2
3 3 3 3 3
1
1 2 - 3
3 3

01 21
33

= 1
3

B. Addition and subtraction of unlike fractions:
Process:
- Convert the unlike fractions into like fractions.
- Perform the process as in addition / subtraction of like fractions.

Example 8: Add 2 and 1 .
3 2

Solution: Here, 2 and 1 are unlike fractions.
3 2

L.C.M of 3 and 2 = 3 × 2 = 6

Prime Mathematics Book - 5 117

∴ 2 = 2 × 2 = 4
3 3 × 2 6

And 1 = 1 × 3 = 3
2 2 × 3 6

Now, 2 + 1
3 2

= 4 + 3 = 4 + 3 = 7 = 1 1
6 6 6 6 6

Example 9: Simplify: 3 - 1
4 6

Solution: Here, 34 and 1 are unlike fractions.
6

Where, 4 = 2 × 2
6=2×3

∴ L.C.M of 4 and 6 = 2 × 2 × 3 = 12

∴ 3 = 3 × 3 = 9
and 4 4 × 3 12
Now,
1 = 1 × 2 = 2
6 6 × 2 12

3 - 1 = 9 - 2 = 9-2 = 7
4 6 12 12 12 12

C. Addition and subtraction of fractions involved with mixed fractions:
- Convert mixed fractions into improper fractions.
- If fractions are unlike, convert the unlike fractions into like fractions.

- Perform the process of addition / subtraction of like fractions.

Example 10: Add 1 and 1 2 .
2 5

Solution: 1 + 1 52 = 1 + 7 1 25 = 5 × 1 + 2 = 7
2 2 5 5 5

L.C.M of 2 and 5 is 2 × 5 = 10

∴ 1 = 1 × 5 = 5
2 2 × 5 10

7 = 7 × 2 = 14
5 5 × 2 10

118 Prime Mathematics Book - 5

Now, 1 + 7 = 5 + 14
2 5 10 10

= 5 + 14
10

= 19 10) 19 (1 = 1190
10 -10

= 1190 9

Alternately, separating whole numbers and fraction:

Solution: 1 + 1 2 L.C.M of 2 and 5 = 2 × 5 = 10
2 5

= 1 + 1 + 2 ∴ 1 = 1×5 = 5
2 5 2 2×5 10

= 1 + 1 + 2 2 = 2×2 = 4
2 5 5 5×2 10

=1+ 1 + 2
2 5

=1+ 5 + 4
10 10

=1+ 5+4
10

= 1 + 9
10

= 1190

Example 11: Simplify: 2 1 - 5
4 6

Solution: 2 41 - 5
6

= 9 - 5
4 6

L.C.M of 4 and 6 = 2 4, 6
2, 3

=2×2×3
= 12

Prime Mathematics Book - 5 119

∴ 9 = 9×3 = 27
4 4×3 12

and 5 = 5×2 = 10
6 6×2 12

= 9 - 5 = 27 - 10
4 6 12 12

= 27 - 10 = 17 = 1 5
12 12 12

Exercise 7.1

1. Say whether the following fractions are like or unlike:
1 1 51 , 2 3
(a) 3 and 4 (b) 5 and 5

(c) 112, 5 and 3 (d) 12 , 1 and 1
12 12 4 8

(e) 3 and 4 (f) 1 1 and 2 3 (g) 1 2 and 2 4
10 10 4 5 3 3

2. Convert the following unlike fractions into like fractions:
2 1 1 5 2 and 3110
(a) 3 and 4 (b) 3 and 6 (g) 2 5

(c) 3 and 1 (d) 5 and 3 (h) 1 2 and 7
4 6 8 4 7 14

(e) 1 , 5 and 3 (f) 3 , 5 and 4
3 9 4 4 8 6

3. Add:

(a) 1 and 2 (b) 10 and 2 (g) 5 and 7
4 4 13 13 24 24

(c) 18 and 6 (d) 3 and 2 (h) 5 and 9
25 25 8 8 14 14

(e) 2 and 4 (f) 18 and 2
7 7 27 27

4. Subtract:

(a) 2 from 7 (b) 5 from 11 (c) 4 from 11
12 12 18 18 15 15

120 Prime Mathematics Book - 5

(d) 11 from 5 (e) 5 from 8 (f) 5 from 3
21 21 11 11 14 14

5. Perform the following additions:

(a) 1 + 2 (b) 3 + 5 (c) 4 + 5
4 5 4 6 6 8

(d) 4 + 5 (e) 3 + 2 (f) 1 + 2 + 3
9 18 7 3 2 3 4

(g) 5 + 3 + 1 (h) 3 + 1 + 5 (i) 3 1 from 2 3
10 8 4 4 9 12 5 5

6. Perform the following subtractions:

(a) 5 - 3 (b) 5 - 3
4 4 6 12

(c) 5 - 3 (d) 17 - 4
12 16 28 21

(e) 13 - 3 (f) 5 - 2
15 6 14 7

(g) 5 1 - 3 1 (h) 7 1 - 2 3
2 6 4 4

7. Perform the following task:

(a) 1+ 3 (b) 2- 1
4 3

(c) 6- 2 (d) 5+ 5
3 7

(e) 2 + 1 1 (f) 4 - 2 1
2 3

(g) 4 1 - 2 3 (h) 8 1 - 4 1
4 8 2 4

(i) 1 3 + 2 7 + 3 4
5 10 15

Prime Mathematics Book - 5 121

8. Simplify:

(a) 3 + 1 (b) 7 - 5
4 4 9 9
1 1 2 1
(c) 2 + 5 (d) 3 - 5

(e) 1 + 1 - 1 (f) 2 + 1 - 5
2 3 4 3 4 6
9 2 1 2 5 8
(g) 10 - 5 - 10 (h) 3 - 6 - 9

9. Simplify:

(a) 2 3 - 1 1 (b) 4 1 - 2 3
4 2 4 4
1 3 4 1 1 1
(c) 1 5 + 2 5 - 1 5 (d) 1 4 + 3 4 - 2 4

(e) 4 3 - 2 1 - 1 1 (f) 3 1 + 2 3 - 1 1
4 2 4 2 4 2
3 2 2 5
(g) 7 - 4 5 - 1 5 (h) 12 - 2 4 + 3 8

10. Solve the following word problems:

(a) The weight of text books in a bag is 2 1 kg and weight of copies is
1 2
2 kg. Find the total weight in the bag.

(b) A rope is 15m long. If 4 3 m is cut from it, what is length of the rope
5
is left?

(c) What must be added to 4 3 to make 6170 ?
5
1 1
(d) A shopkeeper has 40 2 kg of fruits. If he sells 25 2 kg, how much

fruit is left unsold?

(e) Suresh bought 3 43thkegyoaf tferu4it12s kfgro omf farusihto, phoawnd m5u34ckhg froufitfsr uiist sleffrto?m
another shop. If

(f) What should be subtracted from 3 4 to get 1 23 ?
5
5 53 3
(g) Raunak purchased kg of sugar and Rajat purchased 5 4 kg of

sugar. Who purchased more sugar and by how much?

122 Prime Mathematics Book - 5

Multiplication of fractions
SLientc'se cmonuslitdipelri cmatuilotnipilsictahteiorne p3 e×a t31ed addition of same number,

3× 3 = 3 + 3 + 3 Diagrammatically,
4 4 4 4
3
= 3 + 3 + 3 3× = 4 + +
4

= 9 = 2 1 3 + 3 + 3
4 4 4 4 4

= = 2 1
4

It shows that 3 × 3 = 9 = 2 1
4 4 4

3× 3 is same as 3 ×3
4 4

If is considered to be a whole, then 3
can be shown diagrammatically as.

Now, 3 of 3 =
4

=

3 = 2 + 1 1 + 1 + 1
4 4 4

= 2 1
4

3× 3 3 ×3
4 4
3
= Three times 4 = Three fourth of 3

A whole number is a fraction with denominator 1

∴3× 3 = 3 × 3 = 9 = 2 1
4 1 4 4 4

Prime Mathematics Book - 5 123

Again let's consider the multiplication 1 × 1 which is 1 of 1
3 4 3 4

1 of 1
3 4

i.e 1 of
4

=

= 1
12
1 1
Also 4 × 3

= 1 of 1 = 1 of =
4 3 4

1 1 1 1 1 1 × 1 = 1
3 4 4 3 12 4 3 12
1 × 1
Thus, × = × = = 4 × 3

To multiply two fractions, form a new fraction by multiplying corresponding
numerators and denominators.

Worked out Examples

Example 1: Multiply 1 by 4.
Solution: 1 3
= 3 × 4

= 1 × 4
3 1
=43 = 1 13

Example 2: Find the product 7 × 10.
Solution: 7 15
= 15 × 10

= 7 × 102 [ Cancellation by 5]
15 3
7 × 2 14 2
= 3 = 3 = 4 3

124 Prime Mathematics Book - 5

Example 3: Multiply 1 2 × 214 × 65.
3
2 214 5
Solution: 1 3 × × 6

=35 × 49 × 65

= 5 ×39 × 5 8)25(3
3 ×4 × 62 - 24

= 25 = 3 1 1
8 8

Example 4: There are 32 students in a class. If 5 of them are boys, how
many boys are there? 8

Solution: 5 of 32
8
= 5 × 32
8

4
5 × 32
= 8 = 5 × 4 = 20

Therefore, 20 students are boys.

Exercise 7.2

1. Multiply: 1 2 1 2 231 1 2 1
3 3 3 3 5 3 2
(a) 4 by (b) by 18 (c) by (d) by (e) 2 by 3

2. Perform the following multiplications:
5
(a) 2 × 1 (b) 3 × 16 (c) 1 × 5 (d) 2 × 9
3 3 3

(e) 1 × 3 (f) 2 × 3 (g) 1 × 3 (h) 1 × 4
4 4 5 8 6 4 12 7

(i) 134 × 145 (j) 10 × 254 (k) 667 × 5 (l) 235 × 1319
21 16

3. Find the value of: 5
3 6
(a) 8 times 4 (b) 42 times

(c) 28 times 1 kg (d) 35 times 2
2 5

Prime Mathematics Book - 5 125

(e) 1 of 16 (f) 3 of 4
2 4 6

(g) 4 of Rs. 25 (h) 2 of 54 students
5 3

4. Solve the following word problems:

(a) There are 48 students in a class. If 11 of them are girls, find the
number of girls in the class. 16

(b) Bal Mukund can do 1 of a work in a day. How much work can he
do in 2 days? 4

(c) AFi57nmd oontfh tahe nipsuem1r1i2bm eoerft ieasr y3oe5fa. arW. sWhqauhtaa ritse ptoahfrets niodufe ma b14yeeraf e?r eits. 3 months ?
(d)
(e)

Division of fraction
Li.eet's21 c÷on4sider division of 12 by 4.
Wmeucdhiviisdoen12e (phaarltf.) Winetoo b4 teaqinuaitl ptoarbtse a81nd find how

so, 1 ÷ 4 = 1
2 8

This shows that 1 ÷ 4 = 1 × 1 = 1
2 2 4 8

Here, 41 is called reciprocal of 4. 1
8

A4g÷ai12n mleet'as ncso,nhsoidwerm dainviysi12on( h4a ÷lf )21a, re there in 4.
Thus, dividing each whole into two halves, we

get altogether 8 such halves.

126 Prime Mathematics Book - 5

∴ 4 ÷ 1 = 8 1 1 1
2 2 2

HTheirses 2h oisw tshteh aretc4ip÷ro21ca=l 4of× 212.= 8 1 1 1
2 2

1 1 1
2 2

1 1 1
2 2

Note: Reciprocal of a fraction is the fraction formed by exchanging its numeritor
14. 1
and denominator. Reciprocal of 4 is Reciprocal of a is a.

0 (zero) has no reciprocal.

Thus, division of a fraction by other fraction is the product of the first
fraction(dividend) and reciprocal of the second fraction(division).
Note: A whole number is a fraction having denominator 1.

Example 1: Divide 1 by 4. Example 2: Find the value of 3 ÷ 51.
Solution: 3 Solution:
1 1
3 ÷ 4 3 ÷ 5

= 1 × 1 = 3 × 5
3 4 1

= 1 = 15 = 15
12 1

Example 3: Complete the division 4 ÷ 38.
Solution: 9
4 8
9 ÷ 3

= 1943× 31
82
1 1
= 3 × 2

= 1 × 1 = 1
3 × 2 6

Prime Mathematics Book - 5 127

Example 4: Find the value of 214 ÷ 112.
214 ÷ 121
Solution:

= 9 ÷ 3
4 2

= 3492× 21
31
3 121
= 2 =

Example 5: The product of two fractions is 6. If one of them is 1 131, find
the other fraction.

Solution: Here, product is 6
One of the fractions is 1131

∴ Other fraction is 6 ÷ 1131

= 6 ÷ 14
11

= 36 × 11 7
14

= 33 = 4 5
7 7

Therefore, the other fraction is 4 5 .
7

Exercise 7.3

1. Show the following divisions in suitable figure and write the result:

(a) 2 ÷ 2 (b) 1 ÷ 3 (c) 2 ÷ 1 (d) 4 ÷ 1
5 4 3 5

2. Find the reciprocals of the following:

(a) 5 (b) 1 (c) 7 (d) 145
3 11

3. Perform the following divisions:

(a) 36 ÷ 24 (b) 128 ÷ 48 (c) 20 ÷ 4 (d) 24 ÷ 6
5 13 5 7

128 Prime Mathematics Book - 5

(e) 214 ÷ 3 (f) 5 ÷4 (g) 1212 ÷ 15 (h) 5165 ÷ 27
13

(i) 8 ÷ 4 (j) 12 ÷ 6 (k) 135 ÷ 151 (l) 871 ÷ 3 4
9 5 13 5 5

4. Solve the following word problems:

(a) The product of two fractions is 1114 . If one of the fraction is 5 ,
find the other fraction. 7

(b) The product of two fractions is 18 . If one of the fraction is 1 4 ,
find the other fraction. 25 5

(c) The length of a rope is 11 1 m. If it is cut into 9 equal parts, find
the length of each piece. 4

(d) The cost of 5156 m of clothes is Rs. 340, find the cost a metre of
the cloth.

Decimals

Introductory review:

Let a unit length, 1 be divided into 10

equal parts, one part is 1/10 (read as one 110=0.1
120=0.2
tenth) and is also written as 0.1 (read as
130=0.3
zero point one) sum of two such parts 140=0.4

1 + 1 = 2 (two tenths)
10 10 10

or, 0.1 + 0.1 = 0.2 (zero point two) 150=0.5

Sum of three such part 160=0.6

1 + 1 + 1 = 3 (three tenths ) 170=0.7
10 10 10 10 180=0.8
190=0.9
or, 0.1 + 0.1 + 0.1 = 0.3 (zero point three) 1100=1

10 such parts = 10 = 1.
10

If 1 is divided into 100 equal parts, one part is 1 (read as one hundredth)
100
and is also written as 0.01 (read as zero point zero one). Two such parts
is 1020 (two hundredths) = 0.02 (zero point zero two). Similarly 42 such

Prime Mathematics Book - 5 129

si0psu.a0cdr0htisv1pi =d(azer1etd40rs20oi=n p(t14oo602i70n1 80ht0 u0z=n0ed0rer.o6eq z7due8tarhlaospn )o adonrrets so)0,,.o o43nn25. e s(zupecarhro tp piasorti1ns0 t1i0 sf0 o1u(03or05n 0tewotorh)0o s.u0ism3a5inladartnlhdys )6if7o 81r
The fractions having powers of 10 as denominators are called decimal
fractions.

A decimal fraction has two parts, whole number part and decimal part
separated by decimal point.

Whole Decimal
number part

part

123.4567

Decimal point
Place value in decimal numbers

Decimal point
123.4567

Hundreds Ten thousandths
Tens Thousandths
Ones Hundredths
Tenths
Given number 123.4567

thoTuesnandths
Thousandths
Hundredths

Tenths

Ones

Tens
Hundreds
PlaNceumtbyepre

Decimal 1 2 3 4567
fraction 10 100 1000 10000

Decimal 100 20 3 0.4 0.05 0.006 0.0007
number

Thus, the number 123.4567 in words is one hundred twenty three point four
f1i0v6e00 si+x 1s0e07v0e0n., and its expanded form is 1 × 100 + 2 × 10 + 3 × 1 + 140 5
+ 100 +

Note: The number after decimal point is less than 1, so the number after
decimal is read one by one.

130 Prime Mathematics Book - 5

Converting decimal fraction into decimal
- Write the number in the numerator .
- A whole number has decimal point at the end. Shift the position of

decimal point to the left as many digits as there are number of zeros
after 1 in denominator .

e.g. 52 = 5.2
10

52 = 0.52
100

52 = 0.052
1000

Converting fractions into decimals:

- Convert the denominator into power of 10 by multiplying numerator
and the denominator by suitable number.
[if denominator is powers of 2 or 5]

- Write the number in the numerator.

- Shift the decimal point from the end to the left after as many digits as
there are zeros after 1 in denominator.

e.g. 1 = 1 × 5 = 5 = 0.5 Direct division method:
2 2 × 5 10
1 = 2) 1.0 (0.5
1 1 × 2 2 2 1.0
5 = 5 × 2 = 10 = 0.2
×

1 = 1×5 = 5 = 0.05 - Here, 1 = 1.0
20 20 × 5 100 - 1 is not divisible by 2 put zero

2 = 2×4 = 8 = 0.08 and decimal in the quotient.
25 25 × 4 100 - Now divide 10 by 2.

Converting decimal into decimal fraction:

- Write the decimal number

- Put 1 as denominator and put as many zeros after decimal after 1 as
there are digits after decimal in the number.

- Remove decimal from numerator and denominator.

e.g. 0.25 = 0.25 = 25
1.00 100

0.234 = 0.234 = 234
1.000 1000

Prime Mathematics Book - 5 131

Comparing decimal numbers:

- Compare whole number part ( bigger the whole number bigger the
decimal number )

- If whole numbers are same, compare the tenths digit.

- If tenths digits are also same, compare hundredths digit.

- If hundredths digits are also same compare thousandths digits and so on.

e.g. to compare 12.345 and 12.3298

Here, whole number parts are same

Tenths digits are also same.

Hundredths digit 4 > 2

∴ 12.345 > 12.3298 [> stands for greater than]

Example 1: Write the decimal number 134.567 in words.
Solution: 234.507
In words: Two hundred thirty four point five zero seven.

Example 2: Show the given numbers in place value table
(i) 5031.72
(ii) 975.3105

Solution :
TthhooTuuessnaannddtthhss
Hundredths

Tenths

Ones

Tens

Hundreds
Thousands

(i) 5 0 3 1 7 2
(ii) 9 7 5 3 1 0 5

Example 3: Write 3125.579 in expanded form.

Solution: 3125.579 1 1 10100.
10 100
= 3 × 1000 + 1 × 100 + 2 × 10 + 5 × 1+ 5 × + 7 × + 9 ×

= 3 thousands + 1 hundred +2 tens +5 ones +5 tenths + 7

hundredths + 9 thousandths.

Example 4: Compare the following decimal numbers.
(i) 683.25 and 89.985
(ii) 342.3642 and 342.368

132 Prime Mathematics Book - 5

Solution: (i) To compare 683.25 and 89.985
Comparing whole number parts
683 > 89
So, 683.25 > 89.985

(ii) To compare 342.3642 and 342.368
Here, whole number 342 = 342

Tenths digits 3 = 3
Hundredths digits 6 = 6

Thousandths digits 8 > 4
So, 342.368 > 342.3642.

Example 5: Convert the following fractions into decimal numbers.
1 (ii) 552
(i) 4

Solution: (i) 1 By direct division:
4 4) 1.00 (0.25

=41 ×× 2255 -8
20
=12050
-20
×

= 0.25 ∴ 1 = 0.25
4

(iI) 5 2 By direct division:
5 2 27
5 5 = 5
2×2
=5+ 5×2 5) 27.00 (5.4

=5+ 4 - 25
10 20

= 5 + 0.4 -20

= 5.4 ×
2
∴ 5 5 = 5.4

Example 6: Convert the following decimal numbers into fractions.
(i) 7.8 (ii) 12.048
1 8 7180.
Solution: (i) 7.8 = 7 wholes and 8 tenths = 7 + 8 × 10 = 7 + 10 =

(ii) 12.048 = 12 wholes and 48 thousandths

= 12 48
1000

Prime Mathematics Book - 5 133

Exercise: 7.4

1. Write the following decimal numbers in words:

(a) 0.75 (b) 1.234 (c) 1765.0508 (d) 97.0005

(e) 246.085 (f) 5.975 (g) 0.111 (h) 111.111

2. Write the following decimal number in place value table:
(a) 24.68 (b) 7.521 (c) 0.405 (d) 8273.654
(e) 504.691 (f) 1000.001 (g) 40.506 (h) 8.689

3. Write the following decimal numbers in expanded forms:
(a) 0.123 (b) 9.876 (c) 24.24 (d) 543.217
(e) 2345.08 (f) 12.5792 (g) 1.2345 (h) 543.2

4. Compare the following decimal numbers using sign >,=,<:

(a) 34.52 and 34.6 (b) 246.9 and 248.998

(c) 23.675 and 23.68 (d) 246.8215 and 246.822

(e) 220.005 and 220.01 (f) 5791.555 and 7581.559

(g) 7.531 and 6.987 (h) 207.3057 and 207.30509

5. Convert the following decimal numbers into fractions:

(a) 0.5 (b) 2.4 (c) 0.225 (d) 4.25

(e) 16.75 (f) 20.05 (g) 5.0435 (h) 13.025

6. Convert the following decimal numbers into fractions:

(a) 0.5 (b) 4.5 (c) 8.2 (d) 5.6

(e) 0.25 (f) 1.25 (g) 4.45 (h) 12.64

7. Convert the following fractions into decimal numbers:

(a) 12 (b) 7 (c) 25 (d) 4
10 10 100 100

(e) 8 (f) 1246 (g) 246 (h) 1
1000 1000 1000 2

(i) 1 (j) 2 (k) 3 (l) 64
4 5 20 25

(m) 1 (n) 12 3 (o) 15116 (p) 2
8 8 125

134 Prime Mathematics Book - 5

Fundamental operations in decimals
Addition and subtraction of decimal numbers:
Addition and subtraction of decimal numbers is similar to addition and subtraction
of whole numbers. Successively, from right to left the digits of corresponding
place value are to be added / subtracted along with carry over / borrow.

Example 1: Find the sum of 5.6 + 2.3 Grouping whole number
Solution: Arranging vertically: and decimal parts

5.6 5.6 + 2.3
+2.3 = 5 + 2 + 0.6 + 0.3
= 7 + 0.9
7.9 = 7.9
∴ 5.6 + 2.3 = 7.9

Example 2: Find the sum of 23.345 + 432.54 + 9.8712
Solution: Arranging vertically:
Or, 23.3450
23.345
432.5400
432.54
+9.8712
+ 9.8712
465.7562
465.7562
∴ The sum is 465.7562.
∴ The sum is 465.7562.

For our convinience we can add zero to the decimal part
at the end. Zero / zeros at the end of decimal part are not
significant.

Example 3: Subtract 27.42 form 54.39.
Solution: 54.39 - 27.42

=54.39

- 27.42

26.97
∴ 54.39 - 27.42 = 26.97

Example 4: Perform the subtraction 236.6 -156.72
Solution: Arranging vertically

1 12 15 15 10

2 13 16 1. 6 10
–156.72

79.88
∴ 236.6 – 156.72 = 79.88

Prime Mathematics Book - 5 135

Example 5: Simplify: 46.342 – 12.56 + 9.75

Solution: 46.342-12.56+9.75 1 1.000 5 10
46.342
= 46.342+9.75-12.56 +9.750 56.092
56.092
= 56.092-12.56 -12.560

= 43.532 43.532

Example 6: A rope is 52.65m long. Two pieces, one 13.5 m and other
14.75m are cut out from it. What is the length of the rope
left?
1 4 12
Solution: The length of the rope left 13.50 52.65

= 52.65m – 13.5m – 14.75m +14.75 -28.25

= 52.65m – 28.25m 28.25 24.40

= 24.40m

∴ The length of the rope left is 24.40m.

Exercise: 7.5

1. Find the sum:

(a) 3.19 (b) 47.301 (c) 72.60 (d) 74.070
9.53 0.590
+10.37 +86.235
+124.75 +9.475

(e) 97.752 + 24.52 (f) 12.85 + 16.63
(g) 1.314 + 2.756 + 0.864 (h) 214.7 + 15.86 + 8.275

2. Perform the following subtractions:

(a) 8.7 (b) 13.6 (c) 8.72 (d) 512.20
-5.4 -9.8 -4.58 -78.95

(e) 6.6 – 1.4 (f) 29.24 – 18.5
(g) 150 - 78.25 (h) 92 – 0.87

3. Add the following: b. 0.84 and 5.076
a. 65.4 and 32.6 d. 100 and 85.78
c. 0.846 and 0.296 f. 172.37,289.49 and 7.805
e. 72.6,89.52 and 784.472 h. 723.28 kg, 874.54 kg and 568.27 kg.
g. 12, 127.7 and 2345.89

136 Prime Mathematics Book - 5

4. Subtract: b. 19.464 from 49.005
a. 8.024 from 14.46 d. 94.715 from 164.238
c. 5.64 from 10 f. 0.001 from 100
e. 0.963 from 963 h. 0.275 from 1
g. 0.421 from 124

5. Simplify: b. 732.118 – 154.974
a. 15.684 + 12.04 d. 1.324 + 0.126 – 0.83
c. 0.06 – 3.47 + 5.2 f. 20.01 – 19.001 + 0.784
e. 25 – 13.708 + 0.2 h. 36.167 – 9.2 + 20.72
g. 100 - 91.347 + 10.143

6. Solve the following problems:
a. What is the sum of 9.99m and 0.01m ?
b. The cost of an exercise copy is Rs 25.75 and the cost of a graph
copy is Rs 13.20. Find the total cost of an exercise copy and a
graph copy.
c. Lengths of sides of a triangle are 9.50 cm, 12.70 cm and 7.20 cm.
Find the perimeter of the triangle.
d. Naresh has to pay Rs 745.50 for a pair of shoes. If he gave two Rs
500 notes to the shopkeeper, how much should he get back?
e. What should be added to 35.435 liters to make it equal to 50
liters?
f. Subtract 26.36 from the sum of 21.38 and 31.62.
g. Perimeter of a triangle is 27.9cm. If two sides are 9.2cm and 8.7
cm, find the length of the third side.
h. An item costs Rs 314.75. If Rs 45.25 is discounted and Rs 30.40 tax
is allowed, what is the actual cost of the item?

Multiplication of decimal numbers:

A. Multiplying decimal numbers by 10, 100, 1000 etc.
Let's consider 0.006 × 10, 0.006 × 100 and 0.006 × 1000
0.006 × 10 = 10600 60
Here, × 10 = 1000 = 0.060 = 0.06

0.006 × 100 = 6 × 100 = 600 = 0.600 = 0.6
1000 1000

0.006 × 1000 = 6 × 1000 = 6000 = 6.000 = 6
1000 1000

Prime Mathematics Book - 5 137

We observed that
0.006 × 10 = 0.006 = 0.06 (decimal point shifts a place to right)
0.006 × 100 = 0.006 = 0.6 (decimal point shifts two places to right)
0.006 × 1000 = 0.006 = 6.0 (decimal point shifts, three places to right)

Thus, when a decimal number is
multiplied by 10 , 100, 1000 etc, the

decimal point shifts towards right
by as many places as there are zeros

after 1 in the multiplier.

Thus, 0.006 × 10000 = 0.00600 = 60.0 = 60
Note: At the end of decimal part we can add any number of zeros.

B. Multiplication of decimal number by whole numbers.
Let's consider the product 3.2 × 3

Here, 3.2 × 3 = 10 × 3.2 × 3 32 × 3 96 Any number multiplied
10 10 10 by 10 and divide by 10
= = = 9.6 make no difference.

Now, let's consider the product 3.24 × 4.

Here, 3.24 × 4 = 100 × 3.24 × 4 = 324 × 4 = 1296 = 12.96
100 100 100

Lets consider the product 3.214 × 12.

Here, 3.124 × 12 = 1000 × 3.214 × 12 = 3214 × 12 = 38568 = 38.568
1000 1000 1000

Thus, we observe that while multiplying a decimal number by a whole
number
- Irrespective of position of decimal point, multiplication is as

usual.
- Position of decimal point in the product remain same as in the

multiplicand.
e.g. to multiply 4.5 × 9 = 40.5

138 Prime Mathematics Book - 5

4.5 Multiply 4.5 by 9.
×9 There is a decimal after one
digit in the multiplicand so
40.5 locate the decimal point in
the product accordingly.

C. Multiplying a decimal number by other decimal number:

Let's consider the multiplication of 0.4 and 0.3

Here, 0.4 × 0.3 = 140 × 3 = 12 = 0.12
10 100

Again consider 0.12 × 0.3

Here, 0.12 × 0.3 = 11020 × 3 = 36 = 0.036
10 1000

Process:
• Convert each multiplicand into the

decimal fraction.
• Multiply the fractions so formed.
• Convert the product which is also a

decimal fraction into decimal number.

Direct method: From the above process we can conclude a direct process as
• Multiply the multiplicands directly as with whole numbers.
• Locate the decimal point before as much digits from right to left as the

total number of digits after decimal in the multiplicands.

0.12 × 0.3 Total number of digits after
= 0 . 12 ← Multiply as usual decimal in the multiplicands
× 0.30 = 3 so put the decimal before
3 digits from right to left.
036
+0000
0.036

Prime Mathematics Book - 5 139

Example 1: Find the product of the following:

(i) 1.23 × 10 (ii) 1.475 × 100

Solution: (i) 1.23 × 10 Direct Method:
1.23 × 10 = 12.3
= 123 × 10
100 There is one zero after 1 in 10 so
decimal point shifts after 1 digit.
=1120300

= 12.30 = 12.3

(ii) 1.475 × 100 Direct Method:
1.475 × 100
= 1475 × 100 = 147.5
1000
There are two zero after 1 in 100
= 147500 so decimal point shifts after 2
1000 digit toward right.

= 147.500
= 147.5

Example 2 : Carry out the following multiplications:

(i) 4.12 × 8 (ii) 15.025 × 12

Solution: (i) 4.12 × 8 Direct method
4.12 × 8
= 412 × 8 = 4.12
100
×8
= 3296 32.96
100
∴ 4.12 × 8 = 32.96
= 32.96

(ii) 15.025 × 12 Direct method 15025
15.025 × 12 × 12
= 15025 × 12 = 180.300
1000 = 180.3 180300
∴ 15.025 × 12 = 180.3
= 180300
1000

= 180.300
= 180.3
∴ 15.025 × 12 = 180.3

140 Prime Mathematics Book - 5

Example 3: Carry out the following multiplications:

(i) 0.8 × 0.7 (II) 0.84 × 0.6

Solution: (i) 0.8 × 0.7 Direct method
0.8
= 8 × 7
10 10 ×0.7
56
= 56
100 000
0.56
= 0.56 ∴ 0.8 × 0.7 = 0.56
∴ 0.8 × 0.7 = 0.56

(ii) 0.84 × 0.6 Direct method Total number

= 84 × 6 0.84 × 0.6 of digits after
100 10 0.84 decimal = 3.
× 0.6
= 504 504 So, we put the
1000 decimal point
0000 before 3 digits.
= 0.504 0.504
∴ 0.84 × 0.6 = 0.504 ∴ 0.84 × 0.6 = 0.504

Example 4: If a pencil costs Rs. 5.75, find the cost of 6 such pencils.
Solution: Cost of 1 pencil = Rs. 5.75

∴ cost of 6 pencils = Rs. 5.75 × 6
= Rs. 34.50

∴ The cost of 6 pencils is Rs. 34.50

1. Find the product: Exercise: 7.6 (c) 2.452 × 10
(a) 0.247 × 10 (f) 0.004 × 100
(d) 12.005 × 10 (b) 0.085 × 10 (i) 0.0025 × 1000
(g) 5.731 × 100 (e) 0.438 × 100 (l) 63.2108 × 1000
(j) 0.347 × 1000 (h) 34.0506 × 100
(k) 7.6543 × 1000 (c) 0.963 × 5
(f) 8.421 × 26
2. Perform the following multiplications:
Prime Mathematics Book - 5 141
(a) 0.8 × 7 (b) 0.96 × 6

(d) 1.6 × 8 (e) 7.439 × 12

(g) 0.0008 × 75 (h) 32.192 × 14 (i) 23 × 3.094
(j) 35.076 × 43 (k) 124.37 × 16 (l) 72.468 × 85

3. Find the product:

(a) 1.4 × 0.5 (b) 0.9 × 0.7 (c) 1.5 × 3.84

(d) 4.2 × 5.732 (e) 1.23 × 2.34 (f) 7.004 × 0.34

(g) 0.015 × 0.66 (h) 123.4 × 0.23 (i) 3 × 1.25 × 5.67

(j) 100 × 3.245 × 1.5 (k) 10 × 1.5 × 0.006 (l) 1.24 × 3.57 × 100

4. Solve the following problems:
a. The cost of a pen is Rs. 40.25. What is the cost of 15 such pens ?
b. A metre has 100 cm. How many centimeters are there in 4.034 mertres?
c. If 1kg = 1000 gm, how many grams are there in 10.75 kilograms ?
d. The cost of an exercise book is Rs. 23.5. Find cost of a dozen of
exercise books.
e. Find the area of rectangle having length 6.2 cm and breath 4.3 cm.
f. If length, breadth and height at of a cuboid are 10.5 cm, 6.8 cm
and 4.5 cm respectively, find the volume to the cuboid.
g. Circumference of a circle is πd. If value of π is 3.14 and diameter
of the circle is d = 15.2 cm, find the length of the circumference
of the circle.
h. A dollar is equivalent to Rs.72.45. Find the value of 9.5 dollars in
Nepali Rs.

Rounding off decimal numbers

1 year is 365 days 6 If I divide Rs 100 among 32
hours. But we say 1 students each will get 100÷32
= Rs 3.125 i.e Rs. 3.12 paisa
year is 365 days. and 5/100th of 1 paisa but we
My son's age is 12 years 4
months and 14 days. But I say say each will get Rs 3
Rubina had Rs. 50 with her. She
he is 12 years old. gave 50 paisa to a beggar. She has
actually Rs 49 and 50 paisa but she

says I have Rs 50.

142 Prime Mathematics Book - 5

Why are they all lyings ?
They are not lying, they are rounding off the numbers. To make a quantity
easier to understand and remember, we express them to the nearest
convenient figure. The process is called rounding off numbers or approximation
of numbers.
Quantities in whole numbers are rounded off to the nearest tens, hundreds,
thousands, etc.

160 168 170

100 168 200
If we round off Rs. 168 to its nearest tens, we say Rs. 168 is nearly equal to
Rs. 170 ( as 68 is near to 70 and far from 60 ). Mathematically we also write
Rs. 168 Rs170.

If we have to round off Rs. 168 to its nearest hundreds, we say Rs. 168 is
nearly equal to Rs. 200. Mathematically we write Rs. 168 Rs. 200.
Decimal numbers are rounded off to the nearest tenths (one decimal place),
nearest hundredths (two decimal places), nearest thousandths (3 decimal
places) etc.
To round off decimals, you find the place to which the number is to be
rounded off.
- Rewrite the digits to the left of the required place
- Drop/omit all the digits to the right of the required place.
- If the first digits (left most digit) dropped is 5 or more, increase the

digit of the required place by 1 and if it is less than 5 just drop it and
digits after it.
e.g. to round off the decimal 21.6736 to its nearest required decimal
places.
Solution:
To round off 21.6736 to its nearest hundredths.
In the given decimal 21.6736 digit in the hundredths place is 7 and digits to
drop are 36. The first digit to drop i.e. 3 is less than 5.

∴ 21.6736 21.67
Again to round off 21.6736 to its nearest thousandths .The digit in the
thousandths place is 3 and the digit / digits to drop is 6 > 5

∴ 21.6736 21.674

Prime Mathematics Book - 5 143

To round off 21.6736 to its nearest ones, the digit in the ones place is 1 and
digits to omit are 6736 where the first digit 6 > 5

∴ 21.6736 22.

Example 1: Round off 47.46 to its

1. nearest ones 2. nearest tens.

Solution: 1. To round off 47.46 to its nearest ones, ones digit is 7 and
digits to drop are 4 and 6 where the first digit 4 < 5
∴ 47.46 47

2. To round off 47.46 to its nearest tens, the tens digit is 4
and digits to drop are 7,4 and 6 where the first digit 7 > 5.
∴ 47.46 50.

Example 2: Round off the decimal 25.527 to its

Solution: 1. nearest tenths 2. nearest hundredths
1. To round off 25.527 to its nearest tenths, the digit in the

tenths place is 5 and digits to drop are 2 and 7 where first
digit 2 < 5

∴ 25.527 25.5
2. To round off 25.527 to its nearest hundredths, the digits in

the hundredths places is 2 and the digit to drop is 7 > 5

∴ 25.527 25.53

Example 3: Round of 7.1357 to

1. 1 decimal place 2. 2 decimal places

3. 3 decimal places

Solution: 1. To round off 7.1357 to 1 decimal place the digits to omit
are 3,5 and 7 where first digit is 3 < 5

∴ 7.1357 7.1 (rounding off to 1 decimal place)

2. To round off 7.1357 to 2 decimal places the digits to drop
are 5 and 7 where first digit 5 = 5

∴ 7.1357 7.14 ( rounding off to 2 decimal places)

3. To round off 7.1357 to 3 decimal places, the digit to drop
is 7 > 5.

∴ 7.1357 7.136 ( rounding off to 3 decimal places)

144 Prime Mathematics Book - 5

Exercise: 7.7

1. Round off the following to the nearest ones:

(a) 56.81 (b) 162.35

(c) 238.125 (d) 519.98

2. Round off the following to the nearest tens:

(a) 745.452 (b) 1234.612

(c) 489.34 (d) 9.9054

3. Round off the following to the nearest hundreds:

(a) 125.7 (b) 490.52

(c) 4567.91 (d) 12391.857

4. Round off the following to the nearest tenths:

(a) 1.257 (b) 25.8702

(c) 0.3029 (d) 245.0675

5. Round off the following to the nearest hundredths:

(a) 42.363 (b) 72.57802

(c) 105.317 (d) 0.154

6. Round the following decimals to the nearest thousandths:

(a) 0.12342 (b) 2.780875

(c) 59.98715 (d) 215.00097

7. Round off the following decimals:
(a) 8.1762 ( to 1 decimal places )
(b) 21.4281 ( to 2 decimal places)
(c) 12.3456 ( to 3 decimal places )
(d) 125.293539 ( to 4 decimal places)

8. Round off 146.34863 to (b) Nearest tens
(a) Nearest ones (d) The nearest tenths
(c) Nearest hundreds (f) The nearest thoundths
(e) The nearest hundredths
(g) 3 decimal places Prime Mathematics Book - 5 145