Shubharambha Math Book 5 for CTP 2077

Exercise 4.4

1. Subtract the following b. Rs. 175.62 from Rs. 280.32
a. Rs. 924.60 from 1200 d. Rs. 231.50 from 9600

c. Rs. 436.51 from 1423

2. Sunaina bought a cake for Rs. 60.50 and she gave the shopkeeper a 100

rupee note. How much money did she get back?

3. Ramsagar visited a market. His mother gave Price List

him Rs. 2000. He bought Radio, a CFL bulb Radio = Rs. 450.60

and a led bulb. How much did he save? Watch = Rs. 957.20

4. Dorje bought a kg of Chhurpi for Rs. 485.50 he Charger = 250.50
sold it for Rs. 712.20. How much did he earn? Cfl bulb = Rs. 375.75
Led bulb = Rs. 567.50
5. Pema filled the full tank of her car, whose

capacity is 60 l. After travelling 3 hours, she again filled the fuel tank

with 17.75 l. How much fuel was left in the fuel tank after travelling

three hours?

Maths Zone - Grade 5 101

Multiplication and Division of Money

Let's discuss few example, of multiplication and division of money.

Example 1

How much does Puspa have to pay for 5 books. If the cost of a book
is Rs. 80.75.

Solution: Here, Cost of a book = Rs. 80.75

Cost of 5 books = Rs. 80.75 × 5
\ Cost of 5 books is Rs. 403.75.

2 Same number Be careful !
3 of digits after
decimal There are two digits after decimal. So we
80.75 must put decimal (dots) after two digits
×5 from the right in the product.

403.75 Multiply as ordinary process and put
decimal according to the question.


Example 2

Arjun divided Rs. 120.75 equally among the 7 girls. How much does
each girl get?

Solution: Here, Total amount to be divided = Rs. 120.75

One girl gets = Rs. 120.75 ÷ 7

17.25 First, divide 120 by 7. Put decimal
7 120.75 before bringing down 7.

–7 After dividing the number
50 in front of decimal put the
–49 decimal in the quotient.

17
–14

35
–35

0

∴ Each girls get Rs. 17.25



102 Maths Zone - Grade 5

Exercise 4.5

1. Multiply b. Rs. 15.25 by 12
a. Rs. 7.45 by 3 d. Rs. 245.35 by 20
f. Rs. 25030 × 2.50
c. Rs. 27.30 × 15

e. Rs. 57.50 by 25

2. Divide b. Rs. 121 and 66 P by 11
a. Rs. 84 and 28 P by 7 d. Rs. 178.50 by 7

c. Rs. 525 and 50 P by 10

3. The cost of a Packet of eraser is Rs. 65.75. Find the cost of a dozen
Packet.

4. The cost of a marker is Rs. 85.25. Find the cost of a score marker.

5. Samir bought a jacket for Rs. 1296.25 Find the cost of such 15 jackets.

6. Rs. 2557.84 is divided among 16 people. How much does each get?

7. The cost of a gross pencils is Rs. 1260. What is the cost of a pencil. [1
gross = 12 dozen]

8. Shyam sold 3 mobile's cover at Rs. 465 each, and sold 2 charger at Rs.
235 each. He gave this money equally to his 4 children. How much
did each child get?

Maths Zone - Grade 5 103

Maths Fun

Teachers are requested to show or display the coins or notes to the students
in the class to find out if anyone knows the names of currency from another
country. Discuss the color, size and value of the coins/notes.
Students are requested to collect the coins/rupees of SAARC countries
and also find out how much Rs. 100 will be equal to in the currency of the
chosen country.

This is the note of ………………………………

This is the coin of ………………………..
104 Maths Zone - Grade 5

Practice Zone

Group 'A'
A. Circle the correct answer of the following questions.

1. Which of the following is leap year?

a. 2001 b. 2005 c. 2004

2. 5:30PM is written as 24 - hour clock system ............

a. 16:30 hours b. 17:30 hours c. 15:30 hours

3. 25 minutes past 9 in the morning is written as

a. 9:25 Pm b. 9:25Am c. 25:9 Am

4. If 7 month is subtracted form 2067/10/15, the correct date is ....

a. 2060/10/25 b. 2067/3/25 c. 2067/10/18

5. Rs. 73.6 50P is added to Rs. 250 and 75 paisa. What is the sum?

a. Rs. 986 and 50P b. Rs. 986 and 75P c. Rs. 987 and 25P

6. 775 paisa is equal to

a. Rs. 77 and 5P b. Rs. 7 and 75 paisa c. Rs. 775

7. 10 years is equal to ............

a. 1 century b. 1 leap year c. 1 decade

8. Add 8 months 20 days to 6th Shrawan 2072. What is the new date?

a. 26th chaitra 2072 b. 20th falgun 2072

c. 26th Baisakh 2073

Maths Zone - Grade 5 105

Group 'B'
Solve the following questions.

1. How many years, months and weeks are there in 811 days?

2. A man borrowed a loan on 15th Kartik 2068 and paid back on 10th
Mangsir 2069. How long did he use the money?

3. How old are you? Write your age in year, month and days.

4. A pipe can fill a tank in 2 hour 35 minutes. How long does it take to
fill 6 similar tanks?

5. Rani completes the homework of 6 subjects in 8 hours 32 minutes and
48 seconds. How long will it take to complete the homework of one
subject? (If each subject takes equal time)

6. A tour package to Pokhara for 30 children is Rs. 3457.50. How much

does each child has to pay for it? Price List
7. Find the total price of following items. 1. Rice = Rs. 65.75/kg
2. Sugar = Rs. 87.45/kg
a. 3kg rice and 500g cheese 3. Egg = Rs. 96.25/dozen
b. 6kg potato, 2 litre oil and 1 dozen 4. Cheese = Rs. 446.35/kg
5. Potato = Rs. 46.25/kg
egg 6. Dairy Milk = 55/piece
c. 5kg sugar and 3 piece dairy milk 7. Oil = Rs. 135/litre
d. 2kg cheese, 4kg potato and 3

dozen egg.

8. A fun park charge Rs. 25.50P for children, Rs. 50.50P for adult. How

much should a family with 3 children and a couple pay to enter the
park?

106 Maths Zone - Grade 5

Answers of Unit 4

Exercise 4.1

1. a. 1850 seconds b. 20117 seconds

2. a. 3264 hours b. 546 hours

3. a. 2 hours 7 minutes 34 seconds

b. 1 hour 30 minute 26 second

4. a. 1220 days b. 293 months

c. 735 days d. 886 hours

5. a. 23H 4M 10S b. 16W 3D 8H c. 18Y 3M 7D
21Y 4M 14D
d. 12H 2M 26S e. 25W 3D 8H f. 8Y 1M 29D
3Y 11M 5D
6. a. 7H 28M 49S b. 11W 0D 20H c.

d. 15H 42M 49S e. 3W 5D 19H f.

Exercise 4.2

1. a. 60H 20M 50S b. 21H 30D 12S c. 16H 30M 50S
109M 22D 15H
d. 80M 24D 16H e. 25m 13d 6h f. 85Y 5M 10D
3M 2D 3H
g. 47Y 3M 18D h. 22Y 5M 6D i. 1Y 9M 2D
2Y 4M 8D, R - 2 days
2. a. 2H 18N 4S b. 3H 7M 42S c. 3H 52M 42S
64 minutes
d. 3M 21D 8H e. 2Y 3M 5D f.

3. a. 1Y 10M 26D b. 2Y 6M 8D c.

d. 0H 52M 28S e. 3H 20M 11S f.

4. 5 hours 15 minutes 5. 16 hours 6.

7. 2 hours 5 minutes 28 seconds

Exercise 4.3 b. 10363 paisa c. 9027 paisa
e. 43426 paisa f. 16375 paisa
1. a. 4637 paisa b. Rs. 27 65 paisa c. Rs. 342 30 paisa
d. 23545 paisa e. Rs. 44 26 paisa f. Rs. 234 26 paisa
2. a. Rs. 234 b. Rs. 89 c. Rs. 698.76
d. Rs. 573 3 paisa
3. a. Rs. 46.72

Maths Zone - Grade 5 107

d. Rs. 126.1 e. 176.45 f. 214.182

4. Rs. 99.85 c. Rs. 986.49
4. Rs. 226.70
5. Rs. 2852.52 ≈ withdraw Rs. 2853
c. Rs. 409.5
6. Rs. 1441.46 f. Rs. 62575
c. Rs. 52 paisa
Exercise 4.4 b. Rs.104.70 5. Rs. 15555
3. Rs. 606.15 8. Rs. 466.25
1. a. Rs. 275. 40
d. Rs. 9368.5 b. Rs. 183
2. Rs. 39.5 e. Rs. 1437.5
5. 42.25l b. Rs. 11 6 paisa

Exercise 4.5 4. Rs. 1023
7. Rs. 8.75
1. a. Rs. 22.35
d. Rs. 4907
2. a. Rs. 12 4 paisa
d. Rs. 25.5
3. Rs. 789
6. Rs. 159.86

108 Maths Zone - Grade 5

UNIT

5 MEASUREMENT

Specific Objective Prescribed by CDC

Distance

‰‰ T o solve the simple and daily life word problems related to the multiplication
and division of the units of distance.

‰‰ To estimate the length, breadth and height of various objects and the distance
between homes, school and other places.

Capacity

‰‰ To solve the simple and daily life word problems related to the multiplication
and division of liters and milliliters.

Weight

‰‰ To multiply and divide the kilogram and gram by a number.
‰‰ T o solve daily life word problems related to the multiplication and division

of kilogram and gram.
‰‰ T o estimate the weight of different objects.
‰‰ T o find the relation between kilogram and quintal.

Maths Zone - Grade 5 109

Lesson

1 Length and Distance

We use scale, measuring tape to measure the length.

Millimeter (mm),

centimeter (cm), and

metre (m) are the units of

measurements of length.

We use kilometer (km)

to measure the distance

between places.

1 cm is divided into 10 parts. Each part is called 1 mm.

1 cm = 10 mm

1 meter is divided into 100 parts. Each part is called 1 cm.

1 m = 100 cm

1 km is divided to 1000 parts.

Each part is called 1 meter

1 km = 1000 m

÷ 10 ÷ 100 ÷ 1000

mm cm cm mm km

× 10 × 100 × 1000



110 Maths Zone - Grade 5

Multiplication of Length

Example 1

a. Multiply 15 m 25 cm and 8 mm by 4.
b. Multiply 27 km 125 m and 50 mm by 6

Solution:

a. m cm mm b. km m cm
27 125 50
15 25 8 ×6
750 300
60 100 ×4 162 +3 –300
32 10 mm = 1 cm 753m 0 cm

+3 –30 30 mm = 3 cm 162km
60 103 2 1 m = 100 cm
+ 1 –100

61m 3 cm 2 mm

Exercise 5.1 m cm b. km m mm
200 75 9 225 70
1. Multiply. ×8
a. km ×5
7

c. km m cm d. m cm m

12 500 25 20 160 7

×6 ×7

e. km m cm f km m cm
12 130 75
35 700 60
× 12
× 15

Maths Zone - Grade 5 111

2. A rope is 2m 50 cm long. How long will it be if the rope measures five
times it length?

3. A road was completed in 7 phases. In each phase 19 km 600m was
constructed. How long was the road?

4. Pokhara Kathmandu highway is 220 km 600 m long. East west high
way is 5 times the Pohara Kathmandu highway. Find total length of
East-west highway.

5. A man walk 45 km 455 m 35 cm in a day. How long can he walk in 12
days?

Division of Length

Divide : (a) 75 cm 70 cm by 5 (b) 19 km 31 m 15 cm by 5

Solution:

a. m cm b. km m cm
15 14 23
3 606 15
5 75 70
–5 5 19 31 3 km =3000 m
25 3000+31=3031 m
–25 –15
0 70
–5 3 3031
20
– 20 –30
0
31
∴ Quotient = 5 m 14 cm
–30

1 115 1 m = 100 cm
-10 100+15=115 cm

0 15
–15

0

∴ Quotient= 3 km 606 m and 23 cm.

112 Maths Zone - Grade 5

Exercise 5.2

1. Divide :
a. 31 km 505 m by 5
b. 737 km 45m 14 cm by 18
c. 75 km 866 m 56 cm by 8
d. 35 km 22m 5 cm by 7
e. 46 km 6m 47 cm by 9
f. 5 km 600m 50 cm by 8

2. 55 km 600 m 80 cm road is divided into four equal segments for black
topped. Find the length of each equal segment.

3. Divide a 40 m 84 cm ribbon into 4 equal parts. Find the length of each
ribbon segment.

4. A wall of a tower is 53 m 28 cm tall. If it is made by 9 equal beam of
metal, what is the length of each beam?

5. How far is your house from your mama's house? Ask your parents
and estimate it.

6. Do you have measuring tape? If not, find it. Measure the length and
breadth of following.

1. Length of your study room
2. Length of your garden
3. Length & breadth of your bed.

Maths Zone - Grade 5 113

Lesson

2 Capacity

Liters and Milliliters

Look at the given objects and learn the capacity that they contains. After
learning this, also find out the capacity of your water bottle.

60 ml 500 ml 1.5 l

1 l 20 l 500 l

The amount of liquid that an object holds into it is called capacity of the
object.
Some standard vessels are used to measure the liquid.

We use these vessels to measure the liquid.

‰‰ Ask your grandfather/
grandmother and How did they
measure liquid on their time?

Pathi Mana

114 Maths Zone - Grade 5

The amount of liquid that a vessel can hold into it is called its capacity. The
commonly used units of capacity are liters and milliliters. Here are some
vessels which are used to measure capacity of liquids.

1 l 1l 2l
2 2000 ml

100 ml 200 ml 500 ml 1000 ml

1000 ml = 1 l 2000 ml = 2 l

Let's discuss few examples.

Example 1

How many ml are there in a Jug. of capacity 3 1 l
=3l 500 ml [ 1 l = 500 ml] 2
1
3 2 l 2 1000 ml is equal to 1 litre.

= (3 × 1000) ml + 500 ml 500 ml is called half litre.

= (3000 + 500) ml

= 3500 ml Multiply by 1000 to change into ml.
Divide by 1000 to change to l.

Example 2

A bottle of cold drinks contain 2l 250 ml. How much will there be
in 5 such bottles.

Solution: 1250 ml = 1000 ml + 250 ml
5 bottles contain 5 times the = 1 l + 250 ml

So, add '1' litre on the litre column.

Maths Zone - Grade 5 115

capacity of 1 bottle cold drinks. So,

2l 250 ml
×5
10 1250
+1 –1000
11 l 250 ml

Example 3

(a) Divide 40 l 500 ml by 4

(b) 6 Jugs of equal capacity are filled with 13 l 602 ml of water
contained in a bucket. What is the capacity of a Jug?

Solution:

(a) (b)

l ml l ml
10 125
4 40 500 2 267
–40
6 13 602
0 500
–4 –12
10
–8 1 602
20
–20 –1 + 1000
0
0 1602
∴ Quotient = 10l 125 ml
–1200

400

–360

42

–42

0

∴ Capacity of a jug 2 l 267 ml

116 Maths Zone - Grade 5

Exercise 5.3 l ml c. l ml
50 540 46 230
1. Multiply the following
a. l ml b. ×4 ×6
15 250
×5 l ml f. l ml
13 670 25 435
d. l ml e
14 300 ×9 × 20
×7

2. Divide the following b. l ml
5 24 204
a. l ml
5 12 250

c. l ml d. l ml
7 169 750 8 40 644

e. l ml f. l ml
9 95 580 10 35 440

3. A bottle can hold 1.5 l of water. How many bottles are required to
empty a vessel of capacity 25l 500 ml?

4. The capacity of a water tanker is 8000 l . How many water tankers are
required to fill the Pokhari of 4,30,00000 litres.

5. A bike fuel tank holds 13 l 350 ml petrol. How much petrol can hold
by such 6 motorbikes?

6. There are 8 family members in Saurav's house. Each member drinks
3l 350 ml water per day. How much water do they need per day?

Maths Zone - Grade 5 117

Lesson

3 Weight

Kilogram and Grams

What is your weight?
Is your book heavier than pen?
How can you measure your weight?
Let's take the weight of students in class.

50 g 500 g 1 kg 35 kg 50 kg

We use gram to measure the wight of lighter object like book, pen eraser,
pencil. etc. We use kilogram to measure the weight of heavier object like,
rice, sugar, your weight, etc.

The symbols use for gram is 'g' and kilogram is 'kg'.

1000 grams = 1 kilogram
1000 g = 1 kg

118 Maths Zone - Grade 5

Different objects or things have different weights. Milligram (mg), gram (g)
and kilogram (kg) are common units of measurement of weights in metric
system.

25 gm 1.5 kg 5 kg 500 kg 5000 kg

Let's discuss few examples.

Example 1

How many grams are there in 2 1 kg ? 1000 g is equal to 1 kg
2 is equal to half kg
1
Solution: 2 2 kg = 2 kg 500 g 500g

= (2 × 1000)g + 500 g

= (2000 + 500)g = 2500 g

Example 2 1750 g = 1000 g + 750 g
= 1 kg + 750 g
Multiply 25 kg 350 g by 5 So, Add 1 kg on the 'kg'
Solution: column.

kg g × 1000
25 350
×5 g kg
125 kg 1750 g
+1 –1000 ÷ 1000
126 kg 750 g
1kg = 1000 grams
1 gram = 1000 milligrams

Maths Zone - Grade 5 119

3. Divide: 32 kg 450 g by 5.

Solution:

kg g 2 kg = 2000 gm
6 490 (2000 + 450) g = 2450 gm
5 32 450
–30
2 450
–2 + 2000
0
2450
–2000

450
–450

00

∴ Quotient = 6 kg 490 gram.

Exercise 5.4

1. Multiply the following kg g c. kg g
46 250 60 650
a. kg g b. ×7
50 150 ×8
×6

d. kg g e. kg g f. kg g

25 350 37 247 58 500

× 9 × 4 × 20

2. Divide the following b. kg gms
a. kg g 6 27 318

5 6 250 d. kg g
c. kg gms 7 44 72

7 107 450

120 Maths Zone - Grade 5

e. kg g f. kg g
5 17 625 8 17 760

3. The weight of your math book is 350 gram. What will be the weight of
such 15 books?

4. In a bag there is 50 kg sugar. How many packets of 250 grams can be
made from it?

5. A bag contains 43 kg and 500 g of rice. If it is divided equally and put
into 6 packets, how much rice is their in each packet?

Relation between Ton, Quintal and kg

Let's have a look!
1000 grams = 1 kg
100 kg = 1 quintal
10 quintal (1000 kg) = 1 Ton.

Example Divide kg by 100 to
change into quintal.
a) Convert into quintal
÷ 10
3000 kg
QT
b) Convert into kg ÷ 1000 ÷ 100
9 quintal × 10

Solution: g kg Multiply quintal by 100
to change into kg.
a) 3000 kg

= 3000 kg ÷ 1000 × 100
100
= 30 quintal

b) Convert 9 quintal into kg

= 9 quintal

= (9 × 100) kg

= 900 kg

Maths Zone - Grade 5 121

Capacity of Bridge 13T

We can see this type of signal near the bridge.
What does it mean?
It means only 13 Ton is the capacity of this bridge.
or
"Trucks over 13 Tons are not allowed."

Exercise 5.5

1. Convert following into kg

a. 15 quintal b. 12 quintal c. 12.5 quintal
f. 5 quintal 80 kg
d. 40 quintal e. 16.8 quintal
c. 14000 kg
2. Convert into Quintal f. 1235 kg

a. 2000 kg b. 12000 kg c. 2007 kg
f. 140 quintal
d. 3040 kg e. 650 kg
c. 25 tons
3. Convert into Ton. b. 3278 kg f. 7.6 tons.
a. 5250 kg

d. 14.5 quintal e. 20.5 quintal

4. Convert into Quintal

a. 180 tons b. 14 tons

d. 2.8 tons e. 5.4 tons

122 Maths Zone - Grade 5

Maths Fun

• Teachers are requested to measure the
weight and height of all the students
in the class.

• Give the prices to the students under
the following conditions:

a. Having equal height but not
weight in the class.

b. Highest and lowest weight of
the class.

c. Group of students; sum of whole
weight is equal to their maths
teachers weight.

Teachers are requested to measure the
capacity of student's water bottle using
measuring cylinder and let them find the more or less capacity.

Students are requested to estimate 1 liter
the distance of their house from the
school. And teachers are requested Maths Zone - Grade 5 123
to given away the prices for the
farthest students.

Practice Zone

Group A

A. Circle the correct answer of the following questions.

1. 500 mg is equal to

a. 21 kg b. 1 g c. 1 mg
2 2

2. How many centimeters are there in 8 meters?

a. 80 cm b. 800 cm c. 8000 cm
3. How many quintals are there in 9000 kg?

a. 90 b. 9000 c. 9

4. 1 metric ton is equivalent to

a. 1000 kg b. 10000 kg c. 100000 kg

5. How many millimetres are there in 12 liters?
a. 1200 ml b. 12000 ml c. 120 ml

6. 1 kilo litre is equal to

a. 100 l b. 1000 l c. 10000 l

B. Fill in the blanks.

1. A quintal contains ....................... kg

2. A metric ton contains .................................. quintal

124 Maths Zone - Grade 5

Group B

A. Solve the following questions.

1. Find the total weight of 2 packets of milk, 4 packets of tea and 3 packets
of butter. If the weight of a packet milk, a packet tea and a packet
butter are, 450 g, 125 g and 1/2 kg respectively.

2. A container has capacity of 650 l 250 ml. If the capacity of a pot is one
fifth of the container, what is the capacity of the pot?

3. A book weight is 450 gram.

a. What will be the weight of such 12 books?

b. How many books will make 9 kg?

4. Multiply: ml b. kg g
250
a. l × 12 45 230
3

×9

5. Divide :
a. 107 kg 450 gram by 7
b. 12 kg 250 gram by 5
6. A bag contains 43 kg 500 g sugar. If it is divided equally and put into

6 packets, how much sugar is there in each packet?
7. A bottle can hold 1.5 l of water. How many bottles are required to

empty a vessel of capacity 25 l 500 ml?


Maths Zone - Grade 5 125

Answers of Unit 5

Exercise 5.1

1. a. 36km 3m 75cm b. 73km 805m 60cm c. 75km 1m 50cm

d. 142m 120cm 49mm e. 535km 509m 0cm f. 145km 569m 0cm
4. 1103km
2. 12m 50cm 3. 137km 200m

5. 545km 464m 20cm

Exercise 5.2

1. a. 6km 301m b. 40km 946m 95cm, R - 4cm c. 9km 483m 32cm

d. 5km 3m 15cm e. 5km 111m 83cm f. 0km 700m 6cm, R - 2cm

2. 13km 900m 20m 3. 10m 21cm 4. 5m 92cm

5. 489 km 300m 6. 3km 625m 7 & 8 (do yourself)

Exercise 5.3

1. a. 76l 250ml b. 202l 160ml c. 277l 380ml d. 100l 100ml

e. 123l 30ml f. 508l 700ml c. 24l 250ml
f. 3l 544ml
2. a. 2l 450ml b. 4 l 840ml - R : 4 ml

d. 5l 80ml R: 4ml e. 10l 620ml

3. 17 bottles 4. 5375 water tankers

5. 80l 100ml 6. 26l 800ml

Exercise 5.4 323kg 750g c. 485 kg 200g
148kg 988g f. 1170kg 0g
1. a. 300kg 900g b. 4kg 553g c. 15kg 350g
d. 228kg 150g e. 3kg 525g f. 2kg 220g
2. a. 1kg 250 g b. 200 packets 5. 7kg 250g
d. 6kg 296g e.
3. 5kg 250 paisa 4.

Exercise 5.5

Relation Between ton, quintal and kg

1. a. 1500kg b. 1200kg c. 1250kg d. 4000kg
d. 30.4 quintal
e. 1680kg f. 580kg d. 1.45 ton
d. 28 quintal
2. a. 20 quintal b. 120 quintal c. 140 quintal

e. 6.5 quintal f. 12.35 quintal

3. a. 2.25 ton b. 3.278 ton c. 2.007 ton

e. 2.05 ton f. 14 ton

4. a. 1800 quintal b. 140 quintal c. 250 quintal

e. 54 quintal f. 76 quintal

126 Maths Zone - Grade 5

UNIT

6 MENSURATION

Specific Objective Prescribed by CDC

Perimeter

‰‰ To calculate the perimeter of the rectangular objects by using formula.

Area

‰‰ To calculate the area of the rectangular objects by using formula and solve
the daily life word problems related to them.

Volume

‰‰ To calculate the volume of cuboids by using formula.

Maths Zone - Grade 5 127

Lesson

1 Perimeter

Perimeter of a Triangle

We know that 'Peri' means around and meter means measurements. So,

perimeter means around measurement. P

Class Discussion 4 cm 7 cm

Perimeter of a triangle is the sum of all three sides

PQ + QR + PR = Perimeter of DPQR

\ Perimeter of DPQR = PQ + QR + PR Q

= (4 + 6 + 7)= 17 cm 6 cm R

Perimeter of Quadrilateral

Perimeter of Quadrilateral is the sum of all 4 sides. C 3 cm
5 cm
\ Perimeter of Quadrilateral ABCD 3 cm

= AB + BC + CD + DA D B

= (5 + 3 + 3 + 4) cm
2 cm
= 15 cm 4 cm

4 cm

5 cm
\ Perimeter of plane shape is the sum of it's sides.

Exercise 6.1 A

1. Find the perimeter of these figures.

a. b. c.

P X X
20 cm
6 cm 3 cm 6 cm 4.5 cm 12 cm

Q 5 cm R YZ 4 cm B 16 cm C
6 cm
d. f.
e.
A 4 cm D P 3 cm Q
W 7 cm X R

5 cm T 4 cm S5 cm

C Y 6 cm Z 4 cm
B 10 cm

128 Maths Zone - Grade 5

2. Find the perimeter of these shapes.

a. b. c.

P 3 cm
6 cm 3 cm
2 cm
2 cm

5 cm
15 cm 4 cm2 cm
5 cm
8 cm 3 cm2 cm

1 cm 2 cm 4 cm

3 cm 1 cm
Q 7 cm R 2 cm

3 cm

d. e. f.

7 cm 8 cm 3 cm 4 cm

3 cm 3 cm 4 cm
5 cm
10 cm 8 cm 2 cm
5 cm
6 cm 2 cm
3 cm

Perimeter of Rectangle

ABCD is a rectangle. Rectangle has 4 sides and opposite D l C
sides are equal. b

AB = CD = l and AD = BC = b b B

\ Perimeter of Rectangle = Sum of all sides

= AB + BC + CD + AD A l

= l + b + l + b = 2l + 2b = 2 (l + b)

∴ Perimeter of rectangle = 2 (l + b)


Perimeter of Square

ABCD is a square. Square has 4 equal sides. Dl C
l
So, AB = BC = CD = DA = l l
\ Perimeter of square = sum of all sides. B

= AB + BC + CD + DA 129

Al

Maths Zone - Grade 5

= l + l + l + l
= 4l

∴ Perimeter of square = 4l
or Perimeter of square = (4 × length of sides)

Example 1

Find the perimeter of rectangle PQRS. S R
Q
Solution: P 25 cm

Length ( l ) = 25 cm 10 cm

Breadth ( b ) = 10 cm

Perimeter ( P ) = ?

We have,

Perimeter (P) = 2 (l + b)

= 2 (25 + 10)

l = 2 × 35 = 70 cm

\ Perimeter of rectangle (P) = 70 cm.

Example 2 DC
AB
Find the perimeter of square ABCD.
Solution:
Length ( l ) = 9 cm
Perimeter ( P ) = ?
We have,
Perimeter (P) = 4l = 4 × 9= 36 cm

\ Perimeter of square (P) = 36 cm.

130 Maths Zone - Grade 5

Exercise 6.2

1. Find the perimeter of these figures.

a. b. C c. Y

S RD Z

8 cm
10 cm

17 cm

P 12 cm Q A 5 cm B W 22 cm X

d. O e. B f. G

P A H

M 16 cm N D 15 cm C E 13 cm F

2. Find the perimeter of square having following sides.

a. 6 cm b. 9 cm c. 8 cm d. 19 cm

e. 23 cm f. 17 cm

3. Find the perimeter of rectangle having following dimension.

a. length = 25 cm breadth = 18 cm

b. length = 16 cm breadth= 14 cm

c. length = 30 cm breadth = 20 cm

d. length = 40 cm breadth = 35 cm

e. length = 17.5 cm breadth = 10.5 cm

4. Find the perimeter of rectangular ground having length 50m and
breadth 45 m.

5. The perimeter of a rectangle is 40 cm, if length is 12 cm, find it's
breadth.

6. The perimeter of a square is 100 cm. Find its sides.

Maths Zone - Grade 5 131

Lesson

2 Area

Area of Rectangle

Class Discussion

The unit of Area is square of unit length. 1 cm 4 cm
1 cm
\ To find the area of rectangle, we need to This is a square of 1
find the number of square in it. unit area

How many unit square boxes are there in 5 cm
adjoining rectangle?

Obviously, here are 20 sq. boxes inside
the rectangle.

\ Area of Rectangle = 20 sq units.

In fact, the area 20 cm2 is the product of
unit area along the length and breadth.

Hence Area of rectangle = length ×
breadth.


\ Area of rectangle = l × b

Area is defines as the amount of
space inside the boundary of flat
(2- dimensional) object such as triangle,
circle etc.

132 Maths Zone - Grade 5

Area of Land in Nepali System

We use our own system to measure the area of land. The system are
different in Terai and Hilly region.

Terai Region Hilly Region

1 bigha = 20 kattha 1 ropani = 16 aana
1 kattha = 20 dhur 1 aana = 4 paisa
1 paisa = 4 dam
Bigha, Kattha, Dhur Ropani, Aana, Paisa, Dam

1 Bigha = 13.31 Ropani

We use 1 Bigha = 13 Ropani in General Calculation.

Activity

1. Ask your parent about your land. How much land does your
parent have?

2. Suppose Ram Lakhan has 2 bigha land at Janakpur. How much
land will it be at Pokhara? [Use 1 bigha = 13 ropani]

3. Pema Tamang has 52 Ropani land at Sindhuli. How much land
at Kohalpur will be equivalent to it?

Example 1

Find the area of rectangle with length 20 cm, and breadth 15 cm.

Solution:

Length ( l ) = 20 cm

Breadth ( b ) = 15 cm

Area (A) = ?

We have,

Area of rectangle (A) = l × b

= 20 cm × 15 cm

= 300 cm2

\ Area of rectangle (A) = 300 cm2

Maths Zone - Grade 5 133

Example 2

What is the breadth of a rectangle whose area is 75 cm2 and length is
15 cm.

Solution:

Length ( l ) = 15 cm

Area of rectangle ( A ) = 75 cm2

We have,

Area of rectangle (A) = l × b

or, l × b = A

i.e. l × b = 75 cm2

or, 15 cm × b = 75 cm2

5 cm
75 cm2
or, b = 15 cm = 5 cm

\ Breadth (b) = 5cm

Area of Square

A square is also a rectangle having equal length D l C
and breadth. So, l l

Area of square = l × l = l2

\ A = l2

\ Area of square = (side)2 Al B

Example 3

Find the area of square having length 10 cm.
Solution:
Length ( l ) = 10 cm
Area of square ( A ) = ?
We have,

134 Maths Zone - Grade 5

Area of square (A) = l2
= 102
= 100 cm2.

\ Area (A) = 100 cm2

Example 4

Find the length of square having area 36 cm2.
Solution:
Area ( A ) = 36 cm2

Length ( l ) = ?

We have,
Area of square (A) = 36 cm2
or, l2 = 36 cm2
or, l2 = (6 cm)2
\ l = 6 cm.

\ Length (l) = 6 cm

Example 5

Find the Area of shaded part. D C

Solution: Outer Rectangle

Length ( l ) = 20 cm

Breadth ( b ) = 12 cm 8 cm
12 cm
We have, Area of outer rectangle ABCD

= l × b A 16 cm B
= 20 cm × 12 cm = 240 cm2 20 cm
Inner Rectangle

Length (l) = 16 cm

Breadth (b) = 8 cm

Area of inner rectangle = 16 cm × 8 cm = 128 cm2

\ Area of shaded part = Area of outer rectangle – Area of inner rectangle

= 240 cm2 – 128 cm2 = 112 cm2.

\ Area of shaded part = 112 cm2

Maths Zone - Grade 5 135

Example 6

Find the area of given shape. 3 cm 3 cm

Solution:

Area of 1st rectangle = 9 cm × 3 cm 9 cm 3 cm
3 cm
= 27 cm2 3 cm
9 cm
Area of 2nd square = 3 × 3 = 9 cm2 3 cm+

Total area of given shape = (27 cm2 + 9 cm2) 6 cm
⇒ (27 cm2 + 9 cm2)
= 36 cm2
= 36 cm2



Exercise 6.3

1. Find the area of following rectangles.

a. b. c.

S RD CZ

5 cm Y
12 cm X
9 cm
P 10 cm Q A 7 cm B W 18 cm

2. Find the area of following square.

a. b. G c. C

P OH D

M 13 cm N E 15 cm F A 17 cm B

136 Maths Zone - Grade 5

3. Find the Area of the rectangles having the following measurements:

a. length = 20 cm breadth = 18 cm

b. length = 17 cm breadth = 13 cm

c. length = 24 cm breadth = 20 cm

d. length = 22 cm breadth = 12 cm

4. Find the area of squares having the following measurements.

a. length = 14 cm b. length = 19 cm

c. length = 21 cm d. length = 2.5 cm

5. If the area of a rectangle is 120 cm2 and its breadth is 8 cm. Find its
length.

6. What is the length of a rectangle whose area and breadth are 1365 cm2
and 35 cm respectively?

7. Find the side of a square having area 1225 cm2
8. What is the length of side of a square having the area 625 cm2.

9. Find the Area of shaded parts of the given shapes.

a. b.

2 cm
5 cm

4 cm 6 cm
8 cm 10 cm

c. d.

2 cm 6 cm
12 cm
3 cm
15 cm

Maths Zone - Grade 5 137
5 cm 8 cm
5 cm
3 cm

8 cm

e. f.

12 cm4 cm 6 cm
4 cm

4 cm
2 cm

6 cm b. 10 cm

10. Find the Area of following 3 cm
a. 2 cm

4 cm

2 cm

8 cm
4 cm

10 cm

2 cm

11. Find the Area of rectangular plot with length 40 haat and breadth 20
haat. (Haat is local unit used in hilly region)


138 Maths Zone - Grade 5

Lesson

3 Volume

Class Discussion

The measure of the space occupied by a solid object is called its volume.

Let's take a cube with 1 cm length, l cm breadth and 1
1 cm height. The amount of space occupied by this 11
cube is called 1 cubic centimetre or 1 cm3.

We can find the volume of a solid object by counting the no.of unit cubes
contained in it.

1 cm3 2 cm3 3 cm3 4 cm3

Volume of Cuboid

3 cm

2 cm
4 cm
In bottom layer there are 4 × 2 = 8 cubes and there are 3 such layers. So
total number of unit cubes are 8 × 3 = 24. The volume of the cuboid is 24
cm3. We can calculate the volume by multiplying 4 cm, 2 cm and 3 cm.
Volume = 4 cm × 2 cm × 3 cm = 24 cm3

∴ Volume of cuboid = length × breadth × height
V=l×b×h

Maths Zone - Grade 5 139

Volume of Cube

Length, breadth and height of cube are equal.

∴ Volume of cube = length × length × length l
ll
=l×l×l

= l3

\ Volume of Cube (V) = (l)3

Example 1

Find the volume of cube having a side 5 cm.

Solution:

Length ( l ) = 5 cm

Volume of Cube (V) = ?

We have,

Volume of cube (V) = l3

= 53 = 5 × 5 × 5

= 125 cm3

\ Volume of Cube (V) = 125 cm3

Example 2

Find volume of cuboid having length 10 cm, breadth 8 cm and height

6 cm.

Solution:

Length ( l ) = 10 cm

Breadth (b) = 8 cm

Height (h) = 6 cm

Volume (V) = ?

We have,

Volume of cuboid (V) = l × b × h

= (10 × 8 × 6 ) cm3

= 480 cm3

\ Volume of cuboid (V) = 480 cm3

140 Maths Zone - Grade 5

Example 3

The volume of a cuboid box is 1080 cm3. If length and breadth are 15
cm and 12 cm respectively, find the height.

Solution:

Volume of cuboid ( V ) = 1080 cm3
Length ( l ) = 15 cm
Breadth (b) = 12 cm
Height (h) = ?
We have, Volume (V) = 1080 cm3
or, l × b × h = 1080 cm3
or, 15 × 12 × h = 1080

or, 180 × h = 1080

or, h= 1080 cm3 = 6cm
180 cm2

\ Height (h) = 6cm.



Exercise 6.4

1. Find the volume of following cuboids or cubes.
a. b.

5 cm

5 cm

6 cm 4 cm 6 cm

12 cm

Maths Zone - Grade 5 141

c. d.

10 cm
12 cm

10 cm 10 cm 12 cm 12 cm

2. Find the volume of cuboid having following dimenstion:

a. length = 7 cm breadth = 4 cm height = 6 cm

b. length = 12 cm breadth = 8 cm height = 6 cm

c. length = 10.5 cm breadth = 5 cm height = 3 cm

3. Find the volume of the cubes

a. length = 8 cm b. length = 12 cm

c. length = 15 cm d. length = 6 cm

4. The volume of a cuboid is 24 cm3, if length and breadth are 4 cm and
3 cm respectively, find its height.

5. A match box has length 5 cm, breadth 3 cm and height 2 cm. Find its
volume.

6. If the volume of the cube is 125 cm3. Find its length.

142 Maths Zone - Grade 5

Maths Fun

• Take a measuring tape, find the length of the light wire required to
place in the boarder of the white/black board of your class.

• If 1 meter of this wire is Rs.50, find the cost of placing the light wire
in that board.

• In a school function you want to decorate the same board by the color
paper of size 15cm x 15cm. How many piece of papers are required to
cover it.


Discuss in class room, estimate the approximate number of pieces of paper.
If the cost of a paper is Rs. 30, calculate the cost of covering the board.
Note to the teacher: Give the award to those students who can estimate
more correctly.

Maths Zone - Grade 5 143

Practice Zone

Group 'A'
A. Circle the correct answer of the following questions.

1. Perimeter of rectangle is obtained by using the formula

a. l × b b. 4l c. 2(l + b)

2. The perimeter of square having 11cm length is

a. 121 cm b. 44cm c. 22cm

3. Area of square with length 'l' is

a. l × b b. 2l c. l2

4. The perimeter of a square is 24cm. What is its length of each side?

a. 6cm b. 8cm c. 12cm

5. The volume of cuboid is

a. l × b × h b. l × b c. 2(l +b)

6. The volume of cube with length 5cm is

a. 25cm3 b. 125cm3 c. 20cm3

7. Area of square is 25cm2. What is it's perimeter?

a. 25cm b. 20cm c. 100cm

8. Area of base of a cuboid is 80cm2 and height is 5cm. What is its volume?

a. 85cm3 b. 16cm3 c. 400cm3

9. The volume of a cubical block is 216cm3. What is its thickness

a. 6cm b. 8cm c. 7cm

144 Maths Zone - Grade 5

Group 'B'
Solve the following questions.

1. Find the volume of a cuboid having length 15cm, breadth 7.2cm and
height 5.5cm.

2. The length of a rectangle is 2xcm and it's breadth is xcm. If perimeter
of the rectangle is 24cm, find its length and breadth.

3. Find the area of shaded part of rectangle. 2cm

4. The perimeter of a square is 36cm. 8cm
i. Find its length 5cm

ii. Find its Area.

5. The volume of a block is 24cm3. If the Area of base is 12cm2 , find it's
height.

6. Find the Area of the following shapes. 2 cm
a. b.

3 cm

2 cm

3 cm 9 cm 2 cm
5 cm
2 cm

2 cm 3 cm
8 cm

7. A rectangular plot of land is 52m long and 32.5m wide. What is its
Area?

8. A square and a rectangle have equal perimeter. If the square has
length 12cm and the rectangle has length 16cm, find he breadth of the
rectangle.

9. How many cubes of length 2 cm can be fitted into a box of length 6 cm,
breadth 4 cm and height 2 cm?

10. How many packets of chalk of length 8 cm breadth 4 cm and height 5
cm can be fitted in a box having the length 24 cm, breadth 15 cm and
height 12 cm?

Maths Zone - Grade 5 145

Answers of Unit 6

Exercise 6.1

1. a. 14cm b. 14cm c. 48cm d. 25cm

e. 22.5cm f. 16cm

2. a. 30cm b. 26cm c. 21cm d. 29cm

e. 26cm f. 24cm

Exercise 6.2

1. a. 40cm b. 30cm c. 78cm d. 64cm e. 60cm f. 52cm
2. a. 24cm b. 36cm c. 32cm d. 76cm e. 92cm f. 68cm
3. a. 86cm b. 60cm c. 100cm d. 150cm e. 56 cm
4. 190m 5. 8cm 6. 25cm

Exercise 6.3

1. a. 50cm2 b. 84cm2 c. 162cm2
2. a. 189cm2 b. 225cm2 c. 289cm2
3. a. 360cm2 b. 221cm2 c. 480cm2 d. 264cm2
4. a. 196cm2 b. 361cm2 c. 441cm2 d. 6.25cm2
5. 15cm 6. 39cm 7. 35cm 8. 25cm
9. a. 32cm2 b. 64cm2 c. 86cm2 d. 87cm2 e. 56cm2
f. 28cm2
10. a. 20cm2 b. 46cm2
11. 800 square haat

Exercise 6.4 b. 360cm3 c. 1000 cm3­­ d. 1728cm3
b. 576cm3 c. 157.5 cm3 d. 216 cm3
1. a. 120cm3 b. 1728cm3 c. 3375cm3
2. a. 168cm3 5. 30cm2 6. 5cm
3. a. 512cm3
4. 2cm

146 Maths Zone - Grade 5

UNIT

7 FRACTION, DECIMAL
AND PERCENTAGE

Specific Objective Prescribed by CDC

Fraction

‰‰ To convert mixed numbers (fractions) in to improper fraction and vice-versa.
‰‰ T o add, subtract and multiply (simple) two mixed numbers.
‰‰ To solve the simple daily life word problems related to addition and

subtraction of the fraction.

Decimal

‰‰ T o convert decimal numbers in to fractions and vice-versa.( up to three
decimal places)

‰‰ To add and subtract the decimal numbers (up to three decimal places)
‰‰ T o solve daily life word problems related to addition and subtraction of

decimal numbers.
‰‰ To round off the decimal numbers in the given position.

Percentage

‰‰ To convert the fractions into percentage and vice-versa.
‰‰ To solve daily life word problems related to percentage.

Maths Zone -Grade 5 147

1Lesson Fraction

Class Discussion

Let's have discussion in different fractions.

Proper, Improper and Mixed Fractions

Numerator < Denominator Proper Fraction
11
=4 =2

4 Numerator = Denominator Improper
=4 4 Numerator > Denominator Fraction

3

Whole number and Mixed Fraction
Proper fraction

2
15

Like and Unlike Fractions

4, 3, 1, 5 Having same denominators Like Fraction
7777 Having different denominators Unlike Fractions
3, 4, 2, 1
9586

148 Maths Zone -Grade 5

Equivalent Fractions

3 3 ×2 6 4 ÷2 2
5 5 10 10 5
= = =

×2 ÷2

3 = 3×2 = 6 4 = 4÷2 = 2
5 5×2 10 10 10 ÷ 2 5

Multiplying both Numerator and Dividing both Numerator and
Denominator by the same number Denominator by the common number.

Reducing a fraction into their lowest term

12 = 2×2×3 = 2 40 = 40 = 4
18 2×3×3 3 50 50 5
18 18 ÷ 2 9
2 4 =2+ 2×2 =2 2 18 18 9 3 3 20 = 20 ÷ 2 = 10
6 2×3 3 24 2412 4
= =

4

Cancelled the common Cancelled by common Divided by HCF

prime factors number or zero

Conversion of Mixed and Improper fractions.

Mixed to Improper Fraction Improper to Mixed Fraction

15 2 1
7 7
3 2 × 5 + 3 13 = 7 15 =2
5 5 5 –14
2 = =
1
3 12 ×7 + 3 87
12 7 = 7 = 7 18 15 + 3 15 3 3 3
5 = 5 = 5 + 5 =3+ 5 =3 5

Conversion of Unlike into like fractions.

2 and 1 3 and 1 5 and 3
5 4 8 2 8 10

2×4 and 1×5 3×1 and 1×4 5×5 and 3×4
5×4 4×5 8×1 2×4 8×5 10 × 4

8 and 5 3 and 4 25 and 12
20 20 8 8 40 40

Maths Zone -Grade 5 149

Exercise 7.1

1. Classify the following fractions as Proper Fractions, Improper

Fraction and Mixed Fraction.

1 , 7 , 4 , 2 3 , 1 1 , 8
4 7 6 7 2 3


Proper Fraction

Improper Fraction

Mix Fraction

2. Classify the Like and Unlike Fractions

3 , 4 , 2 , 4 , 4 , 5
7 5 7 8 8 7


Like Fractions Unlike Fractions

3. Write four equivalent fractions of the following fractions.

a. 2 b. 4 c. 5 d. 24 [Division Process]
3 5 7 32

4. Reduce the fraction into their lowest term (Use your best method)

a. 12 b. 10 c. 27 d. 60
16 15 30 80

5. Convert the mixed fractions into Improper Fractions.

a. 1 2 b. 1 3 c. 5 1 d. 7 5
3 7 6 8

6. Convert the Improper Fractions into Mixed Fractions.

a. 25 b. 32 c. 43 d. 87
7 9 6 8

7. Convert the Unlike Fractions into Like Fractions.

a. 3 and 1 b. 3 and 5 c. 7 and 5 d. 3 and 5
4 5 7 6 12 6 8 10

150 Maths Zone -Grade 5