Prime Mathematics 3

F. Convert into metres and centimetres:

345 cm = 300 cm + 45 cm 848 cm = ...........
= 3m + 45 cm = ...........
= 3m 45cm = ...........

708 cm = 1234 cm = ...........
= = ...........
= = ...........

879 cm = 720 cm = ...........
= = ...........
= = ...........

1342 cm = 1020 cm =
= =
= =

G. Convert kilometers into metres:

(a) 4 km =4 × 1000m = 4000m
(b) 5km
(c) 8km = ........ × ........... = ...........
(d) 10km = ........ × ........... = ...........
(e) 15km = ........ × ........... = ...........
(f) 24km = ........ × ........... = ...........
(g) 47km = ........ × ........... = ...........
(h) 64km = ........ × ........... = ...........
= ........ × ........... = ...........

146 Prime Mathematics Book − 3

H. Change the following into kilometres:

(a) 2000 m = 2000 ÷ 1000 km = 2 km.
(b) 3000 m = ÷ km.
(c) 8000 m = ÷ km = ........... km.
(d) 12000 m = ÷ km.
(e) 19000 m = ÷ km = ........... km.
(f) 21000 m = ÷ km.
(g) 80000 m = ÷ km = ........... km.
(h) 93000 m = ÷ km.
km = ...........

km = ...........

km = ...........

km = ...........

I. Convert the following into metres:

(a) 7km 850m = 7 × 1000m + 850m = 7000m + 850m = 7850m

(b) 2km 340m = ==

(c) 1km 234m = ==

(d) 8km 15m = ==

(e) 9km 5m = ==

(f) 12km 75m = ==

(g) 14km 8m = ==
(h) 16km 234m = ==

Prime Mathematics Book − 3 147

J.Convert into kilometres and metres as shown in the first one:
(a) 7524m = 7000m + 524m = 7km + 524m = 7km 524cm
(b) 1234m = .......... + ......... = ....... + ......... = ................
(c) 3579m = .......... + ......... = ....... + ......... = ................
(d) 9876m = .......... + ......... = ....... + ......... = ................
(e) 10980m = .......... + ......... = ....... + ......... = ................
(f) 2474m = .......... + ......... = ....... + ......... = ................
(g) 15379m = .......... + ......... = ....... + ......... = ................
(h) 6005m = .......... + ......... = ....... + ......... = ................

K. Perform the following sum:

m cm m cm m cm
4 25 21 57 55 48
+7 39 + 42 34 + 44 42
= = =

= = =

km m km m km m
15 150 7 456 413 705
+ 12 350 +113 514 +247 225
= = =

= = =

148 Prime Mathematics Book − 3

m cm m cm m cm
120 85 43 92 205 82
− 49 28 − 23 75 − 66 48
= = =

= = =

km m km m km m
9 349 28 788 535 125
− 4 155 − 17 579 − 210 25
= = =

= = =

L. Add: (d) 2km 456m and 123km 344m.
(e) 48m 58cm and 69m 32cm.
(a) 48m 58cm and 69m 33cm. (f) 246km 474m and 642km 232m.
(b) 50m 72cm and 9m 13cm.
(c) 22m 55cm and 44m 30cm.

M. Subtract: (d) 126km 265m from 235km 420m.
(e) 78km 358m from 146km 711m.
(a) 871m 20cm from 900m 48cm. (f) 211m 25cm from 345m 32cm.
(b) 126m 46cm from 236m 98cm.
(c) 50m 61cm from 122m 93cm.

N. Word problems on addition of lengths:

(a) If two pieces of clothes 7m 40cm and 5m 25cm long are joined,
what will be the total length?

(b) A bench is 2m 55cm long and second bench is 3m 30cm long. What
is the total length when they are connected together?

(c) Sangita walks 2km 520m on foot and travels 8km 230m by bus. What
is distance she travel altogether?

(d) The place B is 25km 350m east of the place A and the place C is 41km
450m east of the place B. What is the distance between A and C?

Prime Mathematics Book − 3 149

Word problems on subtraction of lengths:

(a) Kristina’s school is 7km 420m from her home. She walks 1km 100m
on foot and then she takes school bus to reach her school. How far
does she travel by bus?

(b) Anima bought 7m 88cm of ribbon. She gave 2m 45cm to Binu. What
length of ribbon is left with her?

(c) Ashok’s height is 1m 65cm and Binay’s height is 1m 42cm. By how
much is Ashok taller than Binay?

(d) Distance between two places is 79 km 840m. If 40 km 220m road
between the two places is pitched and rest is just gravelled. What
length of the road that is not pitched?

Unit Revision Test

A . Fill in the blanks.

a) 10 mm = cm b) cm = 1 m
c) 1 km = m d) mm = 2cm

Convert: b) 12m into cm.
a) 3cm into mm. d) 5m 60cm into cm
c) 800 cm into m f) 5km 720m into m
e) 378cm into m and cm

Perform the following task :

m cm km m m cm km m
5 35 4 120 84 44 7 734
+6 47 − 49 38 456
+16 450 −2

Solve the following problems:

a) Two pipes 15m 240cm and 16m 360cm long are joined. Find the
total length of the pipe.

b) A shopkeeper had 59m 67cm of cotton cloth. He sold 34m 28cm of
the cloth. How much cloth is left?

150 Prime Mathematics Book − 3

Unit Estimated periods − 9

10

AREA AND VOLUME

Objectives

• At the end of this unit, the students will be able to :
• find the area of a figure by counting the number of squares covered by it
• find the capacity of pots using standard vessels (25ml, 50ml, 100ml, 200ml, 500ml, 1 litre)
• convert litres into millitres and millilitres into litres
• convert millilitres into litres and millilitres
• find the volume of solids (cubes/cuboids) by counting the cubes and using formula.

Teaching Materials
• Graph paper, Square paper, Geo-board, vessels of standard capacities, blocks, dice

(cubes) etc.

Activities
It is better to:

• explain about the area using square papers, graphs and geo-board.
• demonstrate the vessels of standard capacities and give the concept of litres and

milliliters.
• pile up the cubes and give the concept of volumes of cubes or cuboids.

Area

Study and learn

The given figure is a square with all sides 1cm. It covers 1cm
a surface of 1 square cm i.e the area of the square is 1 1cm
square cm = 1 sq.cm or 1cm2.

Similarly, area of square of all sides 1m is 1 1m
square m or 1 sq.m = 1m2

If units of length and breadth are not mentioned, 1m
unit of area is written as square unit.

1cm 1cm 1cm 1cm 1cm
1cm
1cm2 1cm 1cm 1cm 1cm
1cm 1cm 1cm 1cm 1cm 1cm
2cm2 1cm
1cm 4cm2 6cm2
3cm2

Area of square with length 1cm and breadth 1cm is 1 square cm. Area of two
such squares is 2 square cm. Area of three such squares is 3 square cm.

Diagonal divides the square in two equal parts. So,
21pasrqt .icsm12.
area of the shaded sq.cm or 0.5cm2. and
unshaded part also
1cm
Two such parts make 1cm
1 1
2 sq.cm + 2 sq.cm = 1 sq.cm.

The given figure has two squares and two half 3cm2
squares. Its area is 3 sq.cm.

152 Prime Mathematics Book − 3

Total number of square rooms = 12 1cm 1cm 1cm 1cm
Area of each square room = 1cm2 1cm
Total area of all square rooms = 12 × 1cm2
3cm 1cm
= 12cm2

Shortly, 1cm
We calculate the total area as 4cm × 3cm = 12cm2 4cm

Exercise - 10.1

A. Count the square each 1cm2 and write the area
of the figures.

(a) (b)

Area = cm2 Area = cm2
(c) (d)

Area = cm2 Area = cm2

Prime Mathematics Book − 3 153

Find the area of the following figures where
length of each square is 1 cm.

(a) (b)

Area = sq.cm Area = sq.cm
(c) (d)

Area = sq.cm

Area = sq.cm
(e)

Area = sq.cm
.
Fill in the blanks in the following.

(a) The area of a square of length 1m is .

(b) The area of a square field of length 1km is .

(c) 4 half sq.cm make sq.cm.

(d) The surface covered by an object or a figure is called

154 Prime Mathematics Book − 3

Capacity

Study and learn. 1000 litres 1 litre 500 ml

When a cubical vessel 1cm long 1cm wide and 1cm high is filled with a
liquid, the liquid is said to occupy a space of 1 cubic centimetre or 1 cu.
cm or 1cm3 or 1 millilitres or 1ml.

Similarly, when a cubical vessel 1cm 1m
1m in length is filled with liquid, 1cm 1cm
the liquid is said to occupy a
space of 1 cubic metre or 1cu.m 1m 1m
or 1m3.

The volume of liquid that a vessal can hold is called its capacity.
Standard units of measuring the capacity of liquid are litres and milliliters.
1000 millilitres = 1 litre

Exercise - 10.2

A. Convert litres into millilitres.

a) 2l = 2 × 1000 ml = 2000 ml.

b) 7l = 7 × ml = ml.

c) 14l = 7 × ml = ml.

d) 16l = 7 × ml = ml.

e) 25l = 7 × ml = ml.

f) 33l = 7 × ml = ml.

Prime Mathematics Book − 3 155

B. Change milliliters into litres 3l
(a) 3000ml = 3000 ÷ 1000 l =

(b) 5000ml = ÷ =l

(d) 15000ml = ÷ =l

(e) 20000ml = ÷ =l

(f) 36000ml = ÷ =l

(h) 84000ml = ÷ =l

C. Convert the following into millilitres.

(a) 4l 235ml = 4 × 1000 ml + 235 ml = 4000 ml + 235 ml

= 4235 ml

(b) 8l 428ml = × ml + ml = ml + ml
= ml

(c) 9l 150ml = × ml + ml = ml + ml
= ml

d) 11l 570ml = × ml + ml = ml + ml
= ml

(e) 52l 25ml = × ml + ml = ml + ml
= ml

(f) 75l 5ml = × ml + ml = ml + ml
= ml

156 Prime Mathematics Book − 3

Study and learn.
To convert ml into l and ml
9 7 5 3 ml

Separate three digits from right to left.
}

}

9000ml + 753ml Remaining will be thousand millilitre.
Leave as it is

Convert thousand ml to litres by dividing by 1000.
= 9l + 753ml
= 9l 753ml

A.Convert the following in litres and millilitres.

(a) 6334ml = 6000 ml + 334 ml = 6 l + 334 ml = 6 l 334 ml.

(b) 9126ml = ml + ml = l + ml = l ml.

(c) 1357ml = ml + ml = l + ml = l ml.

(d) 7008ml = ml + ml = l + ml = l ml.

(e) 12007ml = ml + ml = l + ml = l ml.

(f) 44074ml = ml + ml = l + ml = l ml.

B. Complete the following: ml (d) l ml
(a) l ml (b) l ml (c) l 275 437 200
385 + 63 560
74 373 225 705 25
+19 78 + 6 75 +375 ml
456
93 451 543

= 93l 451ml = = =

(e) l ml (f) l ml (g) l (h) l ml
315 250 9 379 478 87 200
315 3 +78 250
+365 +12 +43

= ===

Prime Mathematics Book − 3 157

C. Complete the following:

(a) l ml (b) l ml (c) l ml (d) l ml
987 94 89
86 445 144 379 26 477 74 19
19 245 88 295 12

= ===
(e) l
ml (f) l ml (g) l ml (h) l ml
124 348 500
94 109 456 789 432 234 8 480
234 567 123 94 3

= ===

D. Add as given in the example:
To add 50l 230ml and 25l 370ml.

l ml Arrange vertically in the column of litres
and millitres.
50 230
+ 25 370

75 600
= 75l 600ml

(a) 160 litres 265ml and 60l 130ml. (d) 79 litres 174ml and 45l 243ml.
(b) 144 litres 880ml and 28l 100ml. (e) 129 litres 465ml and 69l 233ml.
(c) 115 litres 150ml and 711l 325ml. (f) 5 litres 255ml and 7l 325ml.

E. Subtract:

(a) 13l 292ml from 46l 487ml (d) 144l 357ml from 274l 474ml
(b) 26l 625ml from 52l 925ml (e) 94l 149ml from 129l 392ml
(c) 74l 345ml from 125l 450ml (f) 8l 340ml from 407l 450ml

F. Solve the following word problems:

A. (a) A cow gives 5 L 500 ml milk and a buffalo gives 4 L 200 ml milk. Find the
(b) total milk.
A shopkeeper sold 16l 235ml of oil on Sunday and 25l 265ml on
Monday. Find the total amount of oil he sold in two days.

158 Prime Mathematics Book − 3

(c) A bucket contains 12l 345ml of water and second bucket contains
23l 435ml. Find the total quantity of water contained in two
buckets.

(d) A bottle contains 1 litre 235 ml of juice and
second bottle contain 2l 725ml. The juice of
B. (a) two bottles are poured into a single bottle.
(b) Find the total quantity of juice in the bigger
(c) bottle.
(d) A cow gave 10l 450ml of milk in a day. The
owner of the cow sold 6l 125ml of milk. How much milk was left?
The fuel tank of a motor bike contained 12 litres 500 millititre of
petrol. After a journey, 3 litres 240 milllilitres of petrol was left in
the tank. How much petrol was used?
A jug contained 2l 725ml of juice in the morning.
In the evening 1l 225ml of juice was left. How
much juice was consumed?
A shopkeeper has 50l 800ml of kerosene. He sold
35l 350ml of kerosene. How much kerosene is left unsold?

Volume of solids:
Study and learn

Space occupied by an object is called volume.

Let’s take a solid cube of side 1cm. Its volume
is considered to be 1 cubic centimetre. Which in
1cm

short is written as 1cu.cm or 1cm3. 1cm 1cm

Two such cubes have volume 2cm3 and three such
cubes have volume 3 cm3.

Let’s arrange 12 cubes of volume 1cu.cm each, as in the figure.

Number of cubes along length = 3

Number of cubes along breadth = 2

Number of cubes along height = 2
We get
3 × 2 × 2 = 12 cubes

Prime Mathematics Book − 3 159

Again let’s arrange 24 such cubes as in the figure.
Here,
Number of cubes along length = 4
Number of cubes along breadth = 2
Number of cubes along height = 3
We get,
4 × 2 × 3 = 24 cubes.
Therefore, counting the units along length, breadth and height and
multiplying them, we get the volume.

∴ Volume = l × b × h

12 cubes of side 1cm are arranged as shown the figure, forming a cuboid.
Here,

Length of the cuboid (l) = 3cm 2cm
Breadth of the cuboid (b) = 2cm 3cm 2cm
Height of the cuboied (h) = 2cm

∴ Volume of the cuboid =l×b×h
∴ Volume = 12cu.cm = 3cm × 2cm × 2cm
= 12cm3 = 12cu.cm

Exercise - 10.3

A. Count the cubes and find the total volume:

(a) (b)

Volume = cu.units Volume = cu.units
(c) (d)

Volume = cu.units Volume = cu.units

160 Prime Mathematics Book − 3

(e) (f)

Volume = cu.units Volume = cu.units
(g) (h)

Volume = cu.units Volume = cu.units

B. Count the units along length, breadth, and height
of the following cuboids and find the volume by using
formula (V) = l × b × h, where unit = cm.

(a) (b)

l = cm, b = cm l = cm, b = cm
h = cm h = cm
∴ Volume = l × b × h ∴ Volume = l × b × h

= ×× == × ×
= cm3. = cm3.

(c) (d)

l = cm, b = cm l = cm, b = cm
h = cm h = cm
∴ Volume = l × b × h ∴ Volume = l × b × h

= ×× = ××
= cm3. = cm3.

Prime Mathematics Book − 3 161

(e) (f)

l = cm l = cm
b = cm b = cm
h = cm h = cm
∴ Volume = l × b × h ∴ Volume = l × b × h

= ×× = ××
= cm3. = cm3.

C. Measure the length, breadth and height of the
following and also find the volume:

(a) l = cm
b= cm
h= cm

∴ Volume =

=

=

(b) Hear,

l = cm
b = cm
h = cm
∴ Volume =

=
=

162 Prime Mathematics Book − 3

(c) Hear,

l = cm
b = cm
h = cm

∴ Volume =
=
=

(d) Hear,

l = cm
b = cm
h = cm

∴ Volume =

=

=

D. Find the volume of the following cubes or cuboids:
(a) length (l) = 4cm, breadth (b) = 2cm, height (h) = 1cm
(b) length (l) = 3cm, breadth (b) = 3cm, height (h) = 3cm
(c) length (l) = 4cm, breadth (b) = 2cm, height (h) = 2cm
(d) length (l) = 6cm, breadth (b) = 4cm, height (h) = 2cm

Unit Revision Test

A.Count the squares each of 1cm2 and write the area.
(a) (b)

Area = cm2 Area = cm2

Prime Mathematics Book − 3 163

B. Find the area of the following figures.(Each square
is 1 sq. unit)

(a) (b)

Area = Area =

C. Convert:

(a) 3l 244ml into millilitres.
(b) 1246 millilitres into litres and millilitres.

D. Perform the following task:

(a) Add: 12l 444ml and 9l 345ml
(b) Subtract: 69l 357ml from 100l 417ml

E. Count the units along length, breadth and height
of the following and find the volume (1 unit = 1cm)

(a) (b)
l = cm
l = cm b = cm
b = cm h = cm
h = cm ∴ Volume =
∴ Volume =
=
= =
=

F. Perform the following task:

(a) Find the volume of the cuboid where length (l) = 6cm, breath (b) =
4cm and height (h) = 3cm.

(b) Find the volume of the cube where l = 6cm.

164 Prime Mathematics Book − 3

Unit Estimated periods − 7

11

WEIGHT

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

• measuretheweightsofvariousobjectsusingdifferentstandardweightupto1kgand
physical balance.

• convert grams to kg and kg to grams.
• add and subtract weights in kg and gm.

Teaching Materials
• standard weights 50 gm to 1 kg, physical balance.

Activities
It is better to

• show the standard weights up to 1 kg before starting the lesson.
• demonstrate and let the students practise weighing objects.

Weight

50kg 500gm 500gm

Study and Learn:
These are the standard weights issued by the bureau of measurements.

50gm 100gm 200gm 250gm

500gm 1kg 2kg 5kg

10kg 20kg 50kg

• Weights of lighter objects are measured in grams and weights of
heavier objects are measured in kilograms.

• Kilogram is written as ‘kg’ and gram is written as ‘g’ or ‘gm’ in
short.
∴ 1000 gm = 1 kg × 1>000

• To convert kilograms into grams we Kg gm
multiply by 1000 and to convert grams
into kilogram we divide by 1000. >
÷ 1000

166 Prime Mathematics Book − 3

Example 1: Convert 5 kilograms into grams
Solution: We know, 1 kg = 1000 gm

∴ 5 kg = 5 × 1000 gm = 5000 gm
Example 2: Convert 6000 grams into kilogram.

Solution: We know 1000 gm = 1 kg
∴ 6000 gm = 6000 ÷ 1000 kg

= 6 kg

Exercise - 11.1

A. Practical work
1. Provide different standard weights and balance, let the students weigh

the given objects.
(apple, orange, book/books, tiffin box, banana, etc)
2. Let the students estimate the weights of the given object (apple,

orange, banana, book, tiffin box etc.)
Note: Prize can be given to the nearest estimate

B. Convert the following into grams:
(a) 2 kg = 2 × 1000 gm = 2000 gm

(b) 5 kg = ....... × ............. = ................
(c) 8 kg = ....... × ............. = ................
(d) 20 kg = ....... × ............. = ................
(e) 24 kg = ....... × ............. = ................
(f) 50 kg = ....... × ............. = ................

C. Convert the following into kilogram:

(a) 3000gm = 3000 ÷ 1000 kg = 3 kg

(b) 7000gm = ............ ÷ ............. = ..............

(c) 10000gm = ............ ÷ ............. = ..............

(d) 14000gm = ............ ÷ ............. = ..............

(e) 26000gm = ............ ÷ ............. = ..............

(f) 32000gm = ............ ÷ ............. = ..............

Prime Mathematics Book − 3 167

D. Convert kilograms and grams into grams:

(a) 2 kg 250 gm = 2 kg + 250 gm
= 2 × 1000 g + 250 gm
= 2000 gm + 250 gm

= 2250 gm

(b) 4 kg 400 gm (c) 9 kg 125 gm
(d) 16 kg 240 gm (e) 20 kg 200 gm
(f) 42 kg 750 gm (g) 50 kg 550 gm
(h) 80 kg 100 gm (i) 12 kg 990 gm

E. Convert grams into kilograms and grams:

(a) 1234 gm = 1000gm + 234gm
= 1 kg + 234 gm
= 1 kg 234 gm

(b) 2345 gm (c) 6210gm
(d) 8035gm (e) 12345gm
(f) 22200gm (h) 30205gm
(i) 3450gm (j) 5555gm

168 Prime Mathematics Book − 3

F. Perform the following task:

kg gm kg gm kg gm
4 235 4 370
2 240 + 6 455 + 8 430
= =
+ 5 350 = =
= 7 590
= 7kg 590gm

kg gm kg gm kg gm
15 345 23 629 4 150
+ 8 90 + 45 21 + 50 500
= = =
= = =

kg gm kg gm kg gm
234 385 500 789 750 245
+ 456 185 + 276 + 25 15
= = 8 =
= = =

G. Add the following: (d) 23 kg 8 gm and 7 kg 880 gm.
(e) 49 kg 750 gm and 7 kg 150 gm.
(a) 4 kg 240 gm and 7 kg 360 gm. (f) 123 kg 605 gm and 654 kg 155 gm.
(b) 1 kg 700 gm and 9 kg 280 gm.
(c) 9 kg 12 gm and 15 kg 88 gm.

H. Perform the following task:

kg gm kg gm kg gm
18 340 8 456 25 570
− 5 120 − 5 167 − 13 180

= 13 220 = =
= 13kg 220gm = =

Prime Mathematics Book − 3 169

kg gm kg gm kg gm
27 855 143 408 432 725
− 14 257 − 97 283 − 234 530

= = =
= = =

kg gm kg gm kg gm
246 700 642 841 765 200
− 98 206 − 352 261 − 175 165

= = =
= = =

I. Subtract:

(a) 64 Kg 174 gm from 95 kg 393 gm. (d) 235 Kg 105 gm from 500 kg 300 gm.
(b) 45 Kg 345 gm from 662 kg 535 gm. (e) 94 Kg 148 gm from 138 kg 493 gm.
(c) 405 Kg 729 gm from 800 kg 800 gm. (f) 77 Kg 345 gm from 144 kg 370 gm.

J. Word Problems:

kg gm

a) Rashmi is carrying 3 kg 340 gm of books and 1 kg

450 gm of fruits in her bag. Find the total weight

she is carrying in her bag.

She carrying .

kg gm

b) Mother bought 12kg 600gm of rice and 9kg 175gm

of pulses. What is the total weight of rice and

pulses?

The total weight is .

170 Prime Mathematics Book − 3

c) Asha’s bag weighs 5 kg 345 gm and Amar’s bag kg gm

weighs 4 kg 645gm. What is the total weight of the

two bags?

The total weight of two bags is .

d) A porter with sack weighs 120kg 675gm. kg gm
If the weight of the sack is 50kg 185gm, gm
find the weight of the porter.
The weight of the porter is
.

e) A fruit seller had 38kg 600gm of apples. If 13 kg 235gm kg

of apples were rotten, find the weight of the good

apples.

The weight of the good apples is .

f) A sack contains 32kg 500gm of potatoes. kg gm
If 12 kg 600gm of potatoes are removed kg gm
how much potatoes are left in the sack?
of potatoes is left in the

sack.

g) Priya's weight including her bag is 35kg 255gm. If Priya’s

weight is 31kg 150gm, find the weight of her bag.

Weight of Priya’s bag is .

Prime Mathematics Book − 3 171

Unit Revision Test

A. Convert 7kg into grams. Convert 25000gm into kilogram.
7kg = ....... × ........ = ............
25000gm= ......... ÷ ........ = .......

B. Convert 3kg 450gm into grams: Convert 23456gm into kilogram and grams.

3kg 450gm = . .. ....... kg + . . . ....... gm 23456gm = . .. ....... kg + . . . ....... gm
= ... ....... gm + . . ........gm = ... ....... gm + . . ........gm

= .......... gm = .......... gm

C. Perform the following task:

a) Kg gm b) Kg gm c) Add 17kg 280gm and 25kg
47 560 234 456 320gm.
375 − 50 344
− 15 d) Subtract 72kg 400gm from
100kg 750gm.

D. Perform the following:

A porter is carrying 43kg 450gm. If 17kg 450gm is added, kg gm

how much weight is he carrying in total?

Now he is carrying .

E. Perform the following:

kg gm

Jitendra’s weight was 67kg 789gm.With hard exercise

in a week, 5kg 679gm of weight was reduced. What is

his weight now?

Now his weight is .

172 Prime Mathematics Book − 3

Unit Estimated periods − 20

12

FRACTION AND DECIMAL

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

• gain the concept of numerator and denominator of fractions
• compare the fraction with same denominators
• gain the concept of tenths, hundredths and their corresponding decimals
• add and subtract the fractions with same denominators.

Teaching Materials
Paper, models of fractions, figures representing fraction.

Activities
It is better to:

• demonstrate figures, paper models of fractions before starting the lesson.
• showthefiguresofdifferentfractionsandletthestudentswritethefractionwith

their numerator and denominator.
• askthestudentstowritefractionswithsamedenominatorandwritethemin

ascending and descending order after discussion.
• discusswithstudentsandaskthemtowritethefractionswithsamedenominatorand

then add and subtract.
• showthefigureswith10and100equalpartsandgivetheconceptoftenthsand

hundredths and their decimals.

Fraction

Study and learn.

A piece or part of an object is called a fraction.

A circular paper is divided into four equal

parts. One part is shaded. So, one out of four

is shaded. The shaded part is one fourth of
tahsre14e.
the whole and is written And remain-
ing part is 3 whole or fourth of the
whole. 4 of

A rectangular piece of paper is divided

into six equal parts. Two parts out of

six are removed. The removed parts

are two out of six or two sixth and is
2
written as 6 .

A rectangular is divided into 2 equql

parts . Each part is called half of the

whole rectangle. One half of the rect-

angle is shaded One half is written as 1 .
2

1 circle is divided into 3 equal parts.
Tclheeesahcahdepdarpt airstciasll13edoof ntehethwirhdoloef
cir-
the
1
whole circle. One- third is written as 3

174 Prime Mathematics Book − 3

The non-shaded part is two third of the whole circle.

Two third parts is written as 2
3

1 circle is divided into 8 equal parts.

3 parts are shaded or taken. The fraction of shaded part= 3
8

The number of taken part 3 3 is numerator
8 8 is denominator
Total number of part

In a fraction 1 , 3 and 2 , where 1, 3 and 2 are numerators (the
4 4 6
number written above is the numerator) which shows the parts

taken in consideration and the total number of parts where 4 and

6 are denominators. Unless we compare with the whole part, ev-

ery object or every part is whole (single)

It is a whole/single or 1 It is a whole/single or 1

Prime Mathematics Book − 3 175

Exercise - 12.1

A.Write the fraction for the shaded part.

1
2

B. Shade the parts to represent the given fraction:

1 31 2 2
3 84 5 3

C.Write the following fractions in words:

2 = Two fifth 4 =
5 7
7 1
10 = 4 =

1 = 2 =
2 3
5 1
7 = 8 =

D.Write the following in fraction:

One half = 1 five eight =
2

one tenth = seven twelfth =

three fourth = two fifth =

176 Prime Mathematics Book − 3

E. Draw suitable diagram and shade to represent the
following fractions.

(a) 2 (b) 3 (c) 7 (d) 3
3 8 10 4

F. Complete the following:

(a) Fraction is 21

Numerator = 1
Denominator = 2

(b) Fraction of shaded part of whole =

Numerator =

Denominator =

(c) Fraction of shaded part of whole =

Numerator =
Denominator =

(d) Fraction of shaded part of whole =

Numerator =
Denominator =

(e) Fraction of shaded part of whole =

Numerator =
Denominator =

G. Numerator (N) and denominator (D) are given.
Write the fraction:

(a) N = 5, D = 7, Fraction = 5 (d) N = 2, D = 9, Fraction =
7

(b) N = 3, D = 4, Fraction = (e) D = 6, N = 1, Fraction =

(c) N = 1, D = 7, Fraction = (f) D = 3, N = 10 Fraction =

Prime Mathematics Book − 3 177

Like and unlike fractions:

Learn the following:

• IIrnne pterhaeecs heg nifvrteaencd tfibiogynut rdheeesn,f otrmhaceitn siaohtnaosdre 41ids a4pn.o drSto34i,o .n14s aarned 3
4

are like fractions.

• If the denominators of two or more fractions

are same, they are called like fractions.

Example: 14 and 3
4

• If the denominators of two or more fractions

are not same, they are called unlike fractions.

Example: 41 and 1
6

• Like fractions are fractions of same or identical objects.

1 21 3
4 44 4

Different objects Identical objects

• In like fractions, size of unit fractions are same.

equal

In like fractions, fractions with greater numerators are greater
than the other.
4 3
= 8 = 8

3 and 4 are like fractions with denominator 8.
8 8
Here, 4 > 3
4 3
So, 8 > 8

178 Prime Mathematics Book − 3

Exercise - 12.2

A.Tick ( ) the like fractions and cross ( × ) the unlike
fractions.

(a) 1 and 2 (b) 1 and 3
3 3 4 5

(c) 3 and 4 (d) 131 and 3
5 5 11

(e) 4 and 5 (f) 112and 11
7 8 12
B. Write the fractions for the shaded parts and compare.

(a) (b)

3 < 4
5 5

(c) (d)

(e) (f)

C.Compare the following (Use >or< or = between them)

(a) 4 5 (b) 2 2
6 6 4 5

(c) 4 3 (d) 1 7
5 5 8 8

(e) 2 2 (f) 170 6
7 7 10

Prime Mathematics Book − 3 179

D. Arrange the following in increasing (ascending)

order (one is done for you).

(a) 170, 130, 5 (b) 4 7 and 5
10 8 8 8
,

Here, 3 < 5 < 7. So, the fractions Here, ..........................
are in ascending order ∴ , and

∴ 3 , 5 and 7
10 10 10

(c) 182, 1112and 1 (d) 165, 125, 175and 4
12 15
Here, .......................... Here, ..........................

∴ , and ∴ , , and

E. Arrange the following fractions in decreasing
(descending) order:

161, 1101, 171, 2 145, 185, 155, 9
11 15

Here, 10 > 7 > 6 > 2

∴ The fractions are in
descending order

1110, 171, 161, 2
11

2 , 1 , 5 , 4 2 , 1 , 5 , 4
7 7 7 7 9 9 9 9

180 Prime Mathematics Book − 3

Addition of fractions having same
denominators

Study and learn

A circle is divided into 8 equal parts.

Fraction of shaded part with red is 3 .
8
Fraction of shaded part with blue is 28 .
5
Total shaded part is 8 of the whole.

So, 3 + 28 = 3 + 2 = 5
8 8 8

Again a rectangle is divided into 6 equal parts.

Fraction of shaded part with green is 16 .
Fraction of shaded part with red is 26 .
Fraction of total shaded part is 36 of the whole.
1 + 26 1 + 2 3
Let’s see 6 = 6 = 6

For adding fractions with same denominators, add the numerators and keep
the denominator common.

Example 1: Add : 1 + 53
5
1 + 35
Solution : 5

= 1 + 3
5
4
= 5

Prime Mathematics Book − 3 181

Example 2: Add : 3 + 27 + 17
7
3 + 27 + 71
Solution : 7

= 3 + 2 + 1 3 21
7 7 77
6
= 7 6
7
Note:
Addition of fraction is same as adding shaded parts in the same figure.

Exercise - 12.3

A. Add the following fractions:

(a) 13 + 13 (b) 51 + 25 (c) 83 + 28

(d) 173 + 133 + 123 (e) 49 + 91 + 92 (f) 165 + 145 + 135

B. Add the following fractions. Show the addition in
the figure by shading.

(a) 152 + 132 = + =

(b) 2 + 18 = +=
8

(c) 130 + 150 = +=

(d) 1 + 82 + 38 = ++ =
8

182 Prime Mathematics Book − 3

Subtraction of fraction having same denominators:

Study and learn

• Fraction of total shaded part = 58 = 5 – 2
• Fraction of removed shaded part = 28 8 8
5 − 2
• Fraction of remaining shaded part = 8

= 3
8
5 2 5 − 2 3
= 8 – 8 = 8 = 8

• For subtraction of like fraction, the numerator of the smaller fraction
is subtracted from the numerator of the greater fraction, keeping the
denominator common.

Exercise - 12.4

A. Perform the following task :

(a) 23 − 31 (b) 45 − 53 (c) 57 − 27
(d) 29 − 19 (f) 1183− 178
(g) 48 − 28 (e) 170− 130 (i) 187− 177

(h) 8 − 3
9 9

B. Subtract:

(a) 2 from 5 (b) 3 from 5
6 6 8 8

(c) 185from 1135 (d) 1103from 12
13

(e) 159from 199 (f) 137from 7
17

Prime Mathematics Book − 3 183

C. Show the following subtraction in figure:

(a) 3 − 1 = 2
4 4 4

(b) 4 − 1 =
5 5

(c) 4 − 3 =
6 6

(d) 5 − 3 =
8 8

(e) 170− 4 =
10

(f) 192− 5 =
12

184 Prime Mathematics Book − 3

Decimals

Study and learn

Tenth
• A rectangle is divided into ten equal parts and one
Wipsah1r1it0c ihs (isrseh aaaldds eoad sw. orTnihtete entnet nahtseh 0)fr.1a c(trieoandf oars tzheeros hpaodinetd opnaer)t
• A rectangle is divided into ten equal parts and two
ppWSiaamhrritticsliahsar ilr1sy2e 0aw sl s(heoraec dwaaednrdi ta.wtseTr itnhtw eeaons1 3 t00teh. n 2e=t (hf0rr)e.a3acdt ioasn zfeorrot phoeinsth atdweod)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

• In a number line, if space between 0 and 1 is divided into ten equal
1
parts, each part is taken as 0.1 which is in fraction equal to 10 . Thus
fractions and their decimal equivalents are given as

Decimal 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Fraction 1 2 3 4 5 6 7 8 9 10
10 10 10 10 10 10 10 10 10 10

• A decimal fraction is a fraction whose denominator is 10 or power of

10 (10, 100, 1000 etc.)

Exercise - 12.5

A. Write down the fractions for the shaded parts of the
following figures and express them in decimal forms.

Fraction: 7 Fraction:
10 Decimal:
Decimal: 0.7
Prime Mathematics Book − 3 185
Read as : Zero decimal Seven Read as :
or Zero Point Seven or

Read as : Fraction: Read as : Fraction:
or Decimal: or Decimal:

Read as : Fraction: Read as : Fraction:
or Decimal: or Decimal:

B. Write the decimal numbers.
a) Zero point one five .
b) Zero decimal zero three .
c) Zero point one two six .
d) Zero point one zero two three .
e) Zero decimal zero five two .

C. Write the following fractions in decimals.

(a) 2 = (b) 5 =
10 = 10 =
= =
(c) 8 (d) 9
10 10

(e) 3 (f) 6
10 10

186 Prime Mathematics Book − 3

D.Write the fractions equivalent to the following decimals:

(a) 0.2 (b) 0.5 (c) 0.7 (d) 0.8

(e) 0.4 (f) 0.6 (g) 0.9 (h) 0.1

E. Write the missing decimal numbers in the given
number lines:

(a) 0.5 0.6 0.9 1

0 0.1

(b) 0.6 8 1
35 10 1
0 10 10

(c) 0.2 0.4 0.7

0 0.2 0.4 0.7

(d)
0

F. Write the fraction and decimal of the following:

(a) two tenth = 2 = 0.2 (d) one tenth = =
10

(b) seven tenth = = (e) eight tenth = =

(c) five tenth = = (f) six tenth = =

G. Shade the given figure corresponding to the given
decimal numbers.

(a) 0.3 (b) 0.6

(c) 0.2

(d) 0.4

Prime Mathematics Book − 3 187

Study and learn the following:

Hundredth
• The given figure is divided into 100 equal parts out

of which 7 are shaded. The shaded part in fraction
= 1700. In decimal system 1700 can be written as 0.07
(read as zero decimal zero seven/zero point zero
seven/point zero seven).

• In the given figure, there are 100 equal parts. Out

of 100 parts 25 are shaded. Shaded part in fraction
12050. 25
= In decimal system 100 can be written as 0.25.

(read as zero decimal two five/ Zero point two

five/point two five).

• The hundredth fraction has 100 as its denominator and in decimal, two

digits after decimal.

1 = 0.01, 2 = 0.02, 1300 = 0.03,
100 100

1040 = 0.04, 10 = 0.10, 11 = 0.11,
100 100

12 = 0.12, 11030 = 0.13 and so on.
100

Exercise - 12.6

A. Write the fractions of the shaded and unshaded parts
and write them in decimal system:

Shaded part:

Fraction = Decimal =

Unshaded part:

Fraction = Decimal =

Shaded part:

Fraction = Decimal =

Unshaded part:

Fraction = Decimal =

188 Prime Mathematics Book − 3

Shaded part: Decimal =
Fraction = Decimal =
Unshaded part:
Fraction =

Shaded part: Decimal =
Fraction = Decimal =
Unshaded part:
Fraction =

B.Write down the following fractions as decimals:

(a) 31 = (b) 19 = (c) 24 =
100 100 100
37 44 52
(d) 100 = (e) 100 = (f) 100 =

(g) 63 = (h) 80 = (i) 98 =
100 100 100

C. Convert the following into fractions:

(a) 0.06 (b) 0.09 (c) 0.25 (d) 0.52
(e) 0.77 (f) 0.87 (g) 0.99 (h) 0.60

Mixed fraction

Learn the following:

In the given figure, there are two equal rectangles,

each divided into ten equal parts. In the first all the I
parts are shaded which is one whole and is written as

1. In the second figure out of ten, 3 parts are shaded II
1fr13a0ctwiohnich130 ca.nO bnee wwrhitotleena ansd 1t.h3r iene dteecnimthails.
which is in
written as

Prime Mathematics Book − 3 189

In the given figure, there are three equal

rectangles each divided into ten equal parts. Two

are shaded whole and third is shaded 6 parts out of 10 parts. We can write
6
the shaded parts in fraction as 2 10 which can be written as 2.6 in decimal.

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

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0.1 0.2 0.3 2

1 3 = 1130 = 1.3
10
In the above number line, the space between 0 and 1 and 1 and 2 are

divided into 10 equal parts. The indicated parts is 1130 = 1.3 or 1 + 0.3 = 1.3

In the given figure, two equal squares
are divided into 100 equal parts, first is
shaded whole and second is shaded 15
out of 100. The shaded parts = 111050

= 1.15

Exercise - 12.7

A. Write the fraction and decimal represented by the
shaded parts in the given figures.

Fraction =
Decimal =

Fraction =
Decimal =

190 Prime Mathematics Book − 3

Fraction = 0 1 2
Decimal = 0 12 3

Fraction =
Decimal =

Fraction =
Decimal =

Fraction =
Decimal =

B.Compare the following:

(a) 1.6 < 2.1 (i) Compare the
(b) 3.5 whole number
(c) 5.1 4.0 part.
(d) 25.4 (ii) Intfhu temhesba ewmrhepo,alrecto smapreare
(e) 59.1 > 1.5 tenth digits.
(f) 7.49 (iii) If tenth digits
(g) 3.24 21.5 are also same,
(h) 9.35 49.32 compare
(i) 6.18 7.5 hundredth digits.
3.3
9.36

6.2

Prime Mathematics Book − 3 191

C.Complete the following:

1.5 3.7 3.3 0.3 To add: Otennetsh
+2.2 + 4.2 + 4.4 + 0.4
7.12 1.4 + 2.3 = 1.4
0.5 5.3 3.4 +0.34 2.3
+ 0.3 + 0.2 + 8.5 − − AdsrAetdrrdcaiagdinmii ggtt.haeetl nsvl tei3(hnpr. teoa7i.icnnadtl )lt yhl ieseon i ntoh naets
12.34 13.53 34.42 83.41
+2.23 +0.44 +23.57 +14.27

D.Complete the following subtraction:

5.3 8.7 9.4 To subtract 2.34 from 7.56.
− 2.2 − 5.4 − 5.0 Arrange vertically with
4.86 hundredth, tenth,
6.7 6.15 − 1.16 decimal, ones. Tens digits
− 3.7 − 2.12 etc. in straight line.

7.56 Subtract with
− 2.34 the usual
process.
5.22

16.34 0.75 27.67
− 0.21 − 0.53 − 5.15

192 Prime Mathematics Book − 3

E. Show the following in number line and write the result.

0.7 + 0.2 = 0.9

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0.1 0.2 0.3 0.4 0.5 0.6

0.8 − 0.3 = 0.5

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0.1 0.2 0.3 0.4 0.5 0.6

1.2 + 0.3 =

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0.1 0.2 0.3 0.4 0.5 0.6

1.3 – 0.8 =

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0.1 0.2 0.3 0.4 0.5 0.6

0.6 + 0.5 =

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0.1 0.2 0.3 0.4 0.5 0.6

0.9 – 0.4 =

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0.1 0.2 0.3 0.4 0.5 0.6

1.4 + 0.5 =

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0.1 0.2 0.3 0.4 0.5 0.6

1.6 – 0.7 =

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0.1 0.2 0.3 0.4 0.5 0.6
Prime Mathematics Book − 3 193

Unit Revision Test

Write five hundredth in fraction.

Write how you read the
fraction 130.
Write the fraction of the shaded part and write the
numerator and denominator.

Fraction = , Numerator =

Denominator =

Compare the fractions:

(a) 3 1 (b) 5 9
4 4 10 10

Add : 2 2
3 7 7 10
(a) 7 and (b) 10 and

Complete the given subtraction:

7 – 4 = =
12 12

Convert into decimals:

(a) 3 = (b) 12 =
10 100

Write in fraction:

(a) 0.03 (b) 0.5

Complete the following:

4.2 0.73 3.25 7.64
+ 1.2 − 0.22 + 4.53 − 3.42

194 Prime Mathematics Book − 3

Unit Estimated periods − 7

13

UNITARY METHOD AND BILL

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

• findthecostorvalueofsetofsameobjectsonthebasisofcostorvalueofunitobject.
• findthecostofunitobjectonthebasisofthecostorvalueofsetofsameobjects.
• get the informations from bills.

Teaching Materials
• Price lists of objects, different types of bills.

Activities
It is better to:

• discuss about price lists, labelled prices of objects.
• discussthemethodtofindthecostorvalueofsetofsameobjectswhencostorvalueof

unit (1) object is known.
• discussthemethodtofindthecostorvalueofunit(1)objectwhencostorvalueofsets

of same objects is known.
• collectdifferenttypesofbillsanddiscussabouttheinformationgivenbythem.