Vedanta Excel in Mathematics Book 5 Final (2078) NEW

4. Let's say and write the missing decimal numbers quickly. Decimal

a) 0.3 + = 0.7 b) 0.4 + = 0.9 c) + 0.2 = 0.8

d) 0.02 + = 0.05 e) 0.05 + = 0.08 f) 0.7 – = 0.3

g) 0.9 – = 0.4 h) 0.06 – = 0.03 i) 0.08 – = 0.02

5. It's your time! Let's add your decimal numbers to get the given sums.

a) + = 0.6 b) + = 0.06

c) + = 0.9 d) + = 0.009

6. Let's complete these triangle patterns where each number is the sum of
the two above it. For example, 0.9 + 0.3 = 1.2

a) 0.9 0.3 0.7 0.8 b) 1.3 1.2 0.6 1.8
1.2

Section B
7. Let's add these decimal numbers.

a) 0.5 + 0.3 b) 0.05 + 0.03 c) 0.005 + 0.003
f) 2.9 + 1.6
d) 1.4 + 0.8 e) 1.36 + 0.75 i) 10.369 + 15.85
l) 20.42 + 7.963
g) 4.7 + 2.26 h) 3.002 + 1.01

j) 25.07 + 12.009 k) 18.5 + 17.005

8. Let's subtract these decimal numbers.

a) 0.8 – 0.5 b) 0.8 – 0.05 c) 0.8 – 0.005

d) 1.7 – 0.9 e) 1.7 – 0.09 f) 1.7 – 0.009
g) 2.45 – 1.6 h) 3.28 – 1.5 i) 12.32 – 6.4
j) 9.003 – 2.07 k) 15.008 – 8.09 l) 27.35 – 13.575

9. Let's simplify.

a) 0.2 + 0.5 – 0.1 b) 0.9 – 0.3 – 0.4 c) 0.08 – 0.05 + 0.06

d) 1.5 – 0.4 – 0.6 e) 2.005 + 0.05 – 0.5 f) 3.6 – 1.08 – 0.9

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Decimal

10. a) Mother cut a chocolate bar into 10 equal pieces. She gave 4 pieces to her
daughter and 3 pieces to her son. Express these pieces in decimals and ind
the sum of the decimal numbers.

b) A teacher divided a paper strip into 100 equal parts. He asked you to shade
5 parts with red and 12 parts with green colours. Express these shaded
parts in decimals and ind the sum of the decimal numbers.

11. a) The cost of a book is Rs 280.70 and the cost of a pen is Rs 45.50. Find the
total cost of these two items.

b) Dakshes bought a packet of cookies for Rs 120.75 and an ice-cream for
Rs 65.50.

(i) Find the total cost of these two items.

(ii) If he gave Rs 200 to the shopkeeper, what changes did the shopkeeper
return to him ?

c) Mrs. Khadka went to the grocery store and spent Rs 650.85 on fruits and
vegetables and now she has Rs 349.15 left. How much money did she have
to begin with ?

12. a) 1 US dollar is equal to Rs 112.45 and 1 Australian dollar is equal to Rs 82.68.
By how much is the US dollar more expensive than the Australian dollar ?

b) Bijaya Yadav downloaded two apps which were 720.63 kb total. If one app
was 356.31 kb, how big was the other app ?

c) The price tag of a T-Shirt was Rs 835.25 but you paid only Rs 775.75 after
getting some discount. How much discount did you get ?

It's your time - Project work !

13. a) Let's cut a few number of rectangular paper strips each of 10 cm long. Divide
each strip into 10 equal parts. Now, shade the parts with different colours to
show the following operations.

(i) 0.3 + 0.4 (ii) 0.2 + 0.7 (iii) 0.8 + 0.6 (iv) 0.9 + 0.7

b) Let's play the card matching game of sums or difference of decimal numbers.

Each of 2 students in a group should prepare at least 10 rectangular lash
cards of equal size from a chart paper. Each student of the group should write
sums of addition and subtraction of decimal numbers in 5 cards, and the
answer of each sum in other 5 cards as shown below.

0.2+0.6 0.03+0.04 0.5+0.07 0.6–0.4 0.09–0.05 0.3+0.7 0.05+0.02 0.2+0.05 0.8–0.3 0.07– 0.04

0.8 0.07 0.57 0.2 0.04 1.0 0.07 0.25 0.5 0.03

Each student shuf le their cards well and exchange to each other. Now, the

student who match the sums and the answers cards irst, is the winner !

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Decimal

5.7 Multiplication of decimal numbers by whole numbers

Let's study the following illustrations and learn about the multiplication of
decimal numbers by whole numbers.

2 times 0.3 = 2 × 0.3 = 0.3 + 0.3 = 0.6

3 times 0.4 = 3 × 0.4 = 0.4 + 0.4 + 0.4 = 1.2

Similarly, 1 0.2

one digit after 1.2 × 9 → means 9 times 1 whole block and 9 times 2 small
decimal blocks
1 . 2 decimal point

×9 = 9 × 1 + 9 × 2 decimal blocks = 9 + 18 decimal blocks

10 . 8 one digit after = (9 + 1) whole blocks + 8 decimal blocks = 10 + 0.8 = 10.8

decimal point in
the product

one digit after 1.4 × 8 → means 8 times 1 whole block and 8 times
4 decimal blocks
1 . 4 decimal point
= 8 × 1 + 8 × 4 decimal blocks
×8 = 8 + 32 decimal blocks
= (8 + 3) whole blocks + 2 decimal blocks
11 . 2 one digit after decimal = 11 whole blocks + 2 decimal blocks = 11 + 0.2 = 11.2

point in the product

Again, let's multiply 2.35 by 7. I understood!
2.35 has two decimal places.
two digits after So, the product 16.45 also has
two decimal places!!
2 . 35 decimal point

×7

16.45 two digits after decimal

point in the product

5.8 Multiplication of decimal numbers by 10, 100 and 1000

Let's study the following illustrations and learn about the tricky way of
multiplication of decimal numbers by 10, 100, and 1000.
10 × 0.2 o

10 × 0.2 11
= 2.0
10 has one zero. So, we shift decimal
point one digit to the right.

Similarly, 10 × 0.05 = 0.5, 10 × 0.48 = 4.8, and so on.

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Decimal

Again, let's multiply a) 100 × 0.003 b) 1000 × 0.072

100 has two zeros. So, we shift decimal

a) 100 × 0.003 = 0.3 point two digits to the right.

b) 1000 × 0.072 = 72.0 1000 has three zeros. So, we shift
decimal point three digits to the right.

In this way, when a decimal number is multiplied by 10, 100 or 1000, we should
shift the decimal point as many number of digits to the right as there are zeros
in 10, 100 or 1000.

5.9 Multiplication of decimal numbers by decimal numbers

Let's study the following illustrations and investigate the rule of multiplication

of decimal numbers by decimal numbers. 0.06

a) Let's multiply 0.3 by 0.2.

0.2 × 0.3 means 0.2 of 0.3

And, it is 6- hundredths = 0.06

0.3 There are two digits after decimal 0.2 = 2 and 0.3 = 3
×0.2 point altogether in 0.3 and 0.2 10 10
0 . 06
So, the product 0.06 has two digits 0.2 × 0.3 = 2 × 3 = 6 = 0.06
after decimal point. 10 10 100

0.12

b) Let's multiply 0.4 by 0.3

0.3 × 0.4 means 0.3 of 0.4

And, it is 12- hundredths = 0.12

0.4 There are two digits after decimal 0.3 = 3 and 0.4 = 4
×0.3 point altogether in 0.4 and 0.3 10 10
0 . 12
So, the product 0.12 has two digits 0.3 × 0.4 = 3 × 4
after decimal point. 10 10

= 12 = 0.12
100

c) Let's multiply 12.25 by 1.5

12 . 25 12.25 and 1.5 have three digits 1225 15
×1.5 altogether after the decimal point. 100 10

12.25 = and 1.5 =

6125 1225 × 15 = 18375
100 10 1000
12250
18.375 The product has three digits after the = 18.375
decimal point.

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Decimal

EXERCISE 5.3

Section A - Class work
1. Each cube represents 0.1. How many times the decimal cubes and how

many altogether ?

a) b)
4 × 0.2 = 0.8 ×=

c) d)

×= ×=

2. Each cube represents 0.01. How many times the decimal cubes and how
many altogether ?

a) b)
4 × 0.02 = ×=

c) d)
×= ×=

Let's say and write the products as quickly as possible.

3. a) 2 × 0.2 = b) 2 × 0.02 = c) 2 × 0.002 =

d) 3 × 0.3 = e) 3 × 0.03 = f) 3 × 0.003 =

g) 4 × 0.3 = h) 4 × 0.03 = i) 4 × 0.003 =

4. a) 10 × 0.3 = b) 10 × 0.03 = c) 10 × 0.003 =

d) 10 × 0.25 = e) 10 × 0.025 = f) 100× 0.25 =

g) 100 × 0.025 = h) 1000 × 0.004 = i) 1000 × 0.072 =

5. a) 0.2 × 0.1 = b) 0.2 × 0.2 = c) 0.2 × 0.3 =

d) 0.3 × 0.3 = e) 0.2 × 0.01 = f) 0.3 × 0.02 =

g) 0.4 × 0.3 = h) 0.5 × 0.2 = c) 0.6 × 0.4 =

103Approved by Curriculum Development Centre, Sanothimi, Bhaktapur vedanta Excel in Mathematics - Book 5

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Section B
Let's multiply and ϐind the products.

6. a) 2 × 0.3 b) 3 × 0.2 c) 2 × 0.4 d) 3 × 0.03
h) 5 × 0.05
e) 4 × 0.002 f) 3 × 0.4 g) 4 × 0.6 l) 9 × 14.07
d) 100 × 0.6
i) 3 × 1.2 j) 4 × 2.6 k) 6 × 8.15 h) 1000 × 0.3
7. a) 10 × 0.24 b) 10 × 0.052 c) 10 × 0.009 l) 1000 × 0.296
d) 0.3 × 0.6
e) 100 × 0.8 f) 100 × 0.07 g) 100 × 0.005 h) 9.8 × 0.8

i) 1000 × 0.5 j) 1000 × 0.04 k) 1000 × 0.002
8. a) 0.2 × 0.4 b) 0.3 × 0.2 c) 0.4 × 0.3

e) 1.3 × 0.2 f) 2.6 × 0.4 g) 7.2 × 0.5

i) 10.4 × 1.2 j) 14.7 × 1.5 k) 18.24 × 1.6 l) 25.17 × 2.4

9. Let's convert tenths, hundredths, and thousandths into fractions. Then
multiply and ϐind the products.

0.4 × 10 = 4 × 10 =4 0.3 × 0.05 = 3 × 5 = 15 = 0.015
10 10 100 1000

a) 0.3 × 10 b) 0.03 × 100 c) 0.003 × 1000 d) 0.5 × 10

e) 0.05 × 100 f) 0.005 × 1000 g) 1.6 × 10 h) 2.35 × 100

i) 0.4 × 0.2 j) 0.5 × 0.7 k) 0.2 × 0.07 l) 0.6 × 0.09

10. a) A bottle of cold drink holds 0.5 litre of cold drink. How much cold drink do
3 such bottles hold?

b) A bakery used 0.36 kg of lour to make a cake. How much lour did the bakery
use to make 5 cakes?

c) A 9 years old child needs to drink 1.6 litres of water per day. How much
water does a child drink in 10 days?

d) 1 teaspoon of salt is about 2.4 grams. What is the estimated amount of salt
in 6 teaspoons?

e) A packet of milk contains 0.5 litres of milk. A family consumes 3 packets of
milk everyday. How much milk does the family consume in a week?

f) A cetamol tablet costs Rs 1.15. Find the cost of 100 tables.

11. a) The rectangular surface of a table is 0.6 m long and 0.4 m broad. Find its
area in square metres. (Area of rectangle is length × breadth)

vedanta Excel in Mathematics - Book 5 104 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

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b) Find the perimeters of the given plane shapes. [perimeter of equilateral

triangle = 3 × length of a side, perimeter of a square = 4 × length of a side, perimeter
of a rectangle = 2(l + b)]

(i) 2.5 cm (ii) 3.6 cm (iii) 2.3 cm
3.6 cm
2.5 cm 3.6 cm 3.6 cm

2.5 cm 4.2 cm

It's your time - Project work!

12. a) Let's cut a few number of rectangular paper strips each of 10 cm long. Divide
each strip into 10 equal parts. Now, shade the parts with different colours
to show the following multiplications of decimal numbers. Then ind the
products.

(i) 2 × 0.4 (ii) 3 × 0.5 (iii) 4 × 0.6 (iv) 5 × 0.7

b) Let's draw a few number of squares each of 10 cm long and 10 cm broad in a
chart paper. Divide each square vertically and horizontally into 10/10 equal
parts as shown in the following diagram. Now, shade the parts with different
colours to show the following multiplications. Then, ind the products.

(i) 0.3 × 0.2 (ii) 0.2 × 0.4 (iii) 0.3 × 0.5 (iv) 0.4 × 0.6

0.5 × 0.4 = 0.20

5.10 Division of decimal numbers by whole numbers

Let's study the given illustrations and learn about the division of decimal
numbers by whole numbers.

a) Divide 0.2 into 2 equal parts. 0.2 ÷ 2 0.1 0.1
= 0.1 0.2
Each part is 0.1

2 0.2 0. o 2 0.2 0.1
–2
0

105Approved by Curriculum Development Centre, Sanothimi, Bhaktapur vedanta Excel in Mathematics - Book 5

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b) Divide 0.2 into 4 equal parts. 0.2 ÷ 4 0.05
Each of 4 parts are 0.5 = 0.20 ÷ 4

= 0.05 0.2

4 0.2 0. o 4 0.2 0.0 o 4 0.20 0.05 c) 0.08 ÷ 5
– 20
0

Example 1 : Divide a) 0.6 ÷ 4 b) 2.7 ÷ 3

Solution

a) 4 0.6 0.15 b) 3 2.7 0.9 c) 5 0.08 0.016
–4 – 2.7 –5
20 0 30
– 20 – 30
0 0

5.11 Division of decimal numbers by 10, 100, and 1000

Let's study the following illustrations and learn about the tricky way of
division of decimal numbers by 10, 100 and 1000.

a) 2 ÷ 10 2 ÷ 10 b) 0.2 ÷ 10 0.2 ÷ 10
= 2.0 ÷ 10 = 0.02
= 0.2 = 2 × 1 = 2 × 1
10 10 10

= 2 = 0.2 = 2 = 0.02
10 100

c) 2 ÷ 100 2 ÷ 100 d) 0.2 ÷ 100 0.2 ÷ 100

= 02.0 ÷ 100 1 = 00.2 ÷ 100 = 2 × 1
100 = 0.002 10 100
= 0.02 = 2 ×

= 2 = 0.02 = 2 = 0.002
100 1000

Example 2: Divide a) 1.5 ÷ 10 b) 32.5÷ 100 c) 12.6 ÷ 1000
Solution

a) 1.5 ÷ 10 = 0.15 10 has one zero. So, decimal point is
shifted one digit to the left.

b) 32.5 ÷ 100 = 0.325 100 has two zeros. So, decimal point
is shifted two digits to the left.

vedanta Excel in Mathematics - Book 5 106 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

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c) 12.6 ÷ 1000 1000 has three zeros. So, decimal
= 012.6 ÷ 1000 = 0.0126 point is shifted three digits to the left.

In this way, when a decimal number is divided by 10, 100, or 1000, we should
shift the decimal point as many number of digits to the left as there are zeros
in 10, 100, or 1000.

5.12 Division of whole numbers and decimal numbers by decimal
numbers

Let's study the given illustrations and learn about the division of whole
numbers or decimal numbers by decimal numbers.

a) 2 ÷ 0.2 = 2 ÷ 2 = 2 × 10 10
10 2 20 20
2 ÷0.2 = 0.2 = 2
= 10
= 10
102
b) 3 ÷ 0.5 = 3 ÷ 5 = 3 × 5 6
10
30 30
= 3×2 3 ÷0.5 = 0.5 = 5

=6 =6

c) 4 ÷ 0.02 = 4 ÷ 2 = 24 × 100 400 =24020
100 2 0.02
4 ÷0.02 =

= 2 × 100 = 200

= 200

d) 0.4 ÷ 0.2 = 4 ÷ 2 = 24 × 10 0.4 ÷0.2 = 0.4 = 42
10 10 10 2 0.2 2

=2 =2

e) 0.9 ÷ 0.03 = 9 ÷ 3 = 39 × 100 0.9 ÷0.03 = 0.90 =3930
10 100 10 3 0.03

= 3 × 10 = 30
= 30

f) 1.2 ÷ 0.4 = 12 ÷ 4 = 312 × 10 1.2 ÷0.4 = 1.2 = 123
10 10 10 4 0.4 4

=3

=3

107Approved by Curriculum Development Centre, Sanothimi, Bhaktapur vedanta Excel in Mathematics - Book 5

Decimal

Example 3 : Divide a) 15 ÷ 0.5 b) 2.4 ÷ 1.2

Solution 150
0.5
a) 15 ÷ 0.5 = Divisor 0.5 has one digit after the decimal point.
So, write one zero at the end of the dividend 15
150 3 and remove the decimal point from the divisor.
5
=

= 30

b) 2.4 ÷ 1.2 = 2.4 Dividend 2.4 and divisor 1.2 have the equal
1.2 number of digits after the decimal points. So,
remove decimal point from the dividend and
= 24 2 divisor.
12

=2

EXERCISE 5.4
Section A - Class work
1. Let's say and write the quotients as quickly as possible.

a) b) c)

0.3 0.4 0.6

0.3 ÷ 3 = 0.4 ÷ 2 = 0.6 ÷ 2 =

d) e) f)

0.3 0.4 0.5

0.3 ÷ 2 = 0.4 ÷ 5 = 0.5 ÷ 2 =

2. Let's say and write the answers as quickly as possible.

a) How many 0.1s are there in 1 ? 0.1

So, 1 ÷ 0.1 = 1
b) How many 0.1s are there in 2 ? 0.1

So, 2 ÷ 0.1 = 2
c) How many 0.2s are there in 1 ?
0.2

So, 1 ÷ 0.2 = 1

vedanta Excel in Mathematics - Book 5 108 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

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d) How many 0.5s are there in 1 ? 0.5

So, 1 ÷ 0.5 = 1
e) How many 0.4s are there in 2 ? 0.4

So, 2 ÷ 0.4 = 2
f) How many 0.5s are there in 2 ?
0.5

So, 2 ÷ 0.5 = 2

Let's say and write the quotients as quickly as possible.

3. a) 0.4 ÷ 2 = b) 0.04 ÷ 2 = c) 0.004 ÷ 2 =

d) 0.30 ÷ 3 = e) 0.48 ÷ 4 = f) 0.70 ÷ 2 =
4. a) 3 ÷ 10 = b) 0.3 ÷ 10 = c) 0.03 ÷ 10 =

d) 5 ÷ 100 = e) 0.5 ÷ 100 = f) 2 ÷ 1000 =

g) 20 ÷ 1000 = h) 200 ÷ 1000 = i) 18.9 ÷ 10 =
c) 3.0 ÷ 0.2 =
5. a) 2.0 ÷ 0.4 = 20 ÷ 4 = 5 b) 2.0 ÷ 0.5 =

d) 3.0 ÷ 0.3 = e) 4.0 ÷ 0.8 = f) 6.0 ÷ 0.6 =

6. a) 0.2 ÷ 0.2 = b) 0.02 ÷ 0.02 = c) 0.002 ÷ 0.002 =

d) 0.6 ÷ 0.3 = e) 0.06 ÷ 0.03 = f) 0.006 ÷ 0.003 =

g) 1.2 ÷ 0.4 = h) 1.5 ÷ 0.3 = i) 2.4 ÷ 0.6 =
Section B

Let's divide and ϐind the quotients as shown in the examples.

7. 0.8 ÷ 2 = 8 ÷2 = 48 × 12= 4 = 0.4
10 10 10

a) 0.2 ÷ 2 b) 0.4 ÷ 2 c) 0.3 ÷ 3 d) 0.9 ÷ 3 e) 1.2 ÷ 3

f) 1.5 ÷ 5 g) 1.6 ÷ 4 h) 2.4 ÷ 8 i) 3.6 ÷ 4 j) 4.5 ÷ 9

8. 0.8 ÷ 2 → 2 0.8 0.4 0.32 ÷ 4 → 4 0.32 0.08 e) 0.18 ÷ 6
–8 –0 j) 2.5 ÷ 5
0 32 o) 9.8 ÷ 7
–32
0

a) 0.6 ÷ 2 b) 0.8 ÷ 4 c) 0.12 ÷ 3 d) 0.15 ÷ 5
f) 0.24 ÷ 2 g) 0.42 ÷3 h) 0.48 ÷ 4 i) 1.8 ÷ 3
k) 4.8 ÷ 6 l) 3.6 ÷ 3 m) 5.2 ÷ 4 n) 7.5 ÷ 5

109Approved by Curriculum Development Centre, Sanothimi, Bhaktapur vedanta Excel in Mathematics - Book 5

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9. 6 ÷ 0.2 = 6.0 ÷ 0.2 = 360 = 30 c) 5 ÷ 0.5 d) 4 ÷ 0.5
2 i) 3.5 ÷0.7
a) 4 ÷ 0.2 b) 6 ÷ 0.3 e) 8 ÷ 0.4
j) 5.6 ÷ 0.8
f) 1.2 ÷ 0.6 g) 1.4 ÷0.7 h) 2.4 ÷ 0.4

10. Let's divide and ϐind the quotients. d) 49.65 ÷ 15
a) 25.6 ÷ 8 b) 32.4 ÷ 6 c) 15.6 ÷ 12 h) 4.32 ÷ 1.8
e) 16.8 ÷ 1.2 f) 33.6 ÷ 1.6 g) 1.56 ÷1.3

11. a) The cost of 5 pencils is Rs 37.50. Find the cost of 1 pencil.
b)
If the voice-call charge for GSM prepaid is Rs 0.60 per minute, how many
c) minutes can you talk for Rs 4.20 ?

d) The length of a rope is 30 m. If it is cut into 12 equal pieces, ind the length
of each piece.

How many pieces of ribbon each of 6.5 cm long can be cut from a 65 cm
long ribbon ?

12. a) A bread is cut into 10 equal parts. How many children can share all parts
of the bread if each child takes 0.2 parts ?

b) A bread is cut into 10 equal parts. What decimal parts of the bread would
each of 5 children equally get ?

c) 2 pizzas are cut into 20 equal pieces. How many students can share all
slices if each student takes 0.5 parts ?

d) 2 pizzas are cut into 20 equal pieces. What decimal parts of the pizzas
would each of 4 children equally get ?

It's your time - Project work

13. a) Let's cut a few number of rectangular paper strips each of 10 cm long
from a chart paper. Divide each strip into 10 equal parts. Then shade the
parts to show the quotients of the following divisions.

(i) 0.3 ÷ 3 (ii) 0.4 ÷ 2 (iii) 0.6 ÷ 3 (iv) 0.8 ÷ 4

b) Let's do the same activities. Now, shade the parts with different colours to
show the quotients of the following divisions.

(i) 1 ÷ 0.2 (ii) 1 ÷ 0.5 (iii) 2 ÷ 0.4 (iv) 3 ÷ 0.5

5.13 Rounding off decimal numbers

Let's divide Rs 10 between 3 friends equally. Each friend will share

Rs 10 ÷ 3 = Rs 3.33, … We can write it as Rs 3.30 to the nearest tenths or Rs 3.00
to the nearest whole number. It is called rounding off a decimal number.

vedanta Excel in Mathematics - Book 5 110 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

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Thus, rounding off a decimal number means to express the decimal number to
the nearest tenths, hundredths, thousandths or to the nearest whole number.
It makes easier to understand the number of quantities.

Now, let's learn the following rules to round off decimal numbers.

Rule 1: If the digit to be rounded off is less than 5, its place is considered as
zero and the digit at the higher place remains unchanged.

Example 1: Round off 25.24 to one decimal place and to the nearest
whole number.
Solution
25.24 25.2 In 25.24, 4 is rounded off to zero.

25 In 25.2, 2 is rounded off to zero.

Remember that we usually use the symbol while rounding off a decimal
number. The symbol tells 'approximately equal'.

Rule 2: If the digit to be rounded off is 5, or greater than 5, it is considered
zero and 1 is added to the digit at the higher place.

Example 2 : Round off 36.748 to two and one decimal places, and to the
nearest whole number.

Solution

36.748 36.75 8 is rounded off to zero and 1 is added to 4.

36.8 5 is rounded off to zero and 1 is added to 7.

37 8 is rounded off to zero and 1 is added to 6.

5.14 Use of decimals

We use decimal numbers in different types of measurements such as the
measurement of money, length, weight, capacity, and so on. We specially use
decimal numbers in the conversion of units of measurements into its higher
or lower units.

Example 3: Convert a) 75 p into rupees (Rs) b) Rs 0.50 into paisa.

Solution It's easy!
a) 75 p
= Rs 75 = Rs 0.75 1 paisa is one-hundredths of Re 1.
100
So, 75 p is 75 of Re 1!!
100

b) Rs 0.50 = 0.50 × 100 p Re 1 = 100 p. So, Rs 0.50 = 0.50 × 100 p!!
= 50 p 111 vedanta Excel in Mathematics - Book 5

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Decimal

Example 4 : Convert a) 6 mm into cm b) 0.48 m into cm.

Solution 6 I have remembered!
a) 6 mm 10
= cm = 0.6 cm 1 mm is one-tenth of 1 cm.

So, 6 mm is 6-tenths of 1cm = 6 cm !!
10

b) 0.48 m = 0.48 × 100 cm I have also remembered!
= 48 cm 1 m = 100 cm. So, 0.48 m = 0.48 × 100 cm!!

Example 5 : Convert a) 465 g into kg b) 0.75 l into ml

Solution

a) 465 g = 465 kg 1 Very simple! 450
1000 1000 1000
1 g = kg, so, 465 g = kg!!

= 0.465 kg

b) 0.75 l = 0.75 × 1000 ml I got!
= 750 ml 1 l = 1000 ml. So, 0.75 l = 0.75 l × 1000 ml!!

EXERCISE 5.5
Section A - Class work
1. Let's round off these decimal numbers to the nearest whole numbers.

a) 3.2 b) 4.7 c) 9.4 d) 15.5

e) 17.3 f) 28.8 g) 56.1 h) 95.6

2. Let's round off these decimal numbers to one decimal place.

a) 5.36 b) 6.43 c) 8.25 d) 10.52

e) 14.74 f) 18.18 g) 45.69 h) 84.81

3. Let's round off these decimal numbers to two decimal places.

a) 2.163 b) 1.476 c) 3.218 d) 7.542

e) 4.827 f) 6.955 g) 8.381 h) 9.654

vedanta Excel in Mathematics - Book 5 112 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

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4. Let's say and write the correct values in the blank spaces.

a) Rs 0.01 = p b) 0.1 cm = mm c) 0.01 m = cm

d) 0.001 km = m e) 0.001 kg = g f) 0.001 l = ml

Section B
5. Let's round off these decimal numbers to one decimal place, then to the

nearest whole numbers.

7.46 7.5 and 7.5 8

a) 1.32 b) 4.69 c) 5.83 d) 8.44 e) 10.47
6. Let's round off these decimal numbers to two and one decimal places, then

to the nearest whole numbers.

3.245 3.25, 3.25 3.3 and 3.3 3

a) 2.168 b) 5.475 c) 6.714 d) 9.439 e) 12.647
7. Let's convert the units of measurement to the higher or lower units as

indicated .

a) 65 p (in Rs) b) 50 p (in Rs) c) Rs 0.25 (in paisa) d) 7 mm (in cm)

e) 0.5 cm (in mm) f) 54 cm (in m) g) 0.75 m (in cm) h) 250 m (in km)
i) 0.685 km (in m) j) 415 g (in kg) k) 0.268 kg (in g) l) 750 ml (in l)
m) 0.375 l (in ml) n) 0.5 m (in cm) o) 500 m (in km)

It's your time - Project work !

8. Let's visit to the available website such as www.google.com and ind out the
exchange rates of the following currencies with Nepali currency.

a) 1 U.S. dollar ($) b) 1 Indian rupee ( ) c) 1 U.K. Pound ( )

d) 1 Australian dollar ($) e) 1 Chinese Yuan (Y) f) Malaysian Ringgit (RM)

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Unit Percent

6

6.1 Percent - How many out of 100 ? - Looking back

Classwork - Exercise

1. There are 100 square rooms in each of the following square grids. Let's
say and write the fractions and percents of the different coloured rooms.

a) b) =
=

==

=
=

2. Let's say and write how many percent.

a) 45 out of 100 students are boys. are boys.

b) 55 out of 100 students are girls. are girls.

c) 10 out of 100 eggs are broken. eggs are broken.

d) 95 marks out of 100 marks. marks.

In this way, percent means 'per hundred' or 'out of hundred'. 40 students

out of 100 students got A+ grade in an exam means 40 percent or 40% students
got A+ grade. We use the symbol '%' for the word 'percent'.

6.2 Conversion of percent into fraction 40
100
We have already discussed that 40% means 40 out of 100. So, 40% = .

Similarly, 25% = 12050, 90% = 90 and so on. Now, let's investigate and say the
100

rule of conversion of percent into fraction from the following examples:

Example: Convert a) 5% b) 40% c) 75% into fraction.

Solution 753
100 4
a) 5% = 5 b) 40% = 240 c) 75% =
1002 1005
1 2 3
= 20 = 5 = 4

Thus, to convert a percent into a fraction, we should divide the given percent

by 100. Then, the fraction is reduced to its lowest terms wherever necessary.

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6.3 Conversion of fraction into percent
Study the following examples and investigate the rule of conversion of a

fraction into a percent.

Example 2: Convert a) 1 b) 4 c) 3 into percents.
4 5 10
Solution
Another process

a) 1 = 1 × 100 or, 1 × 25 % 1 = 1 × 25 = 25 = 25%
4 4 100 4 4 4 25 100
100

= 25% 4 = 4 × 20 = 80 = 80%
5 5 20 100
4 4 2
b) 5 = 5 × % = 4 × 20%
100

= 80% 3 = 3 × 10 = 30 = 30%
10 10 10 100
3 3
c) 10 = 10 × 100 % = 3 × 10%

= 30%

So, to convert a fraction into percent, we should multiply the given fraction by
100 %. Then, it is simpli ied to get the percent in the simpler form.

6.4 Conversion of percent into decimal and decimal into percent.

The process of converting a percent into decimal and a decimal into percent is
same as the conversion between percent and fraction. Let's learn the processes
from the following examples.

Example 3: Convert a) 4% b) 51% into decimals.

Solution I know the rule!
I should divide the given percent by 100!!
a) 4% = 4 = 0.04
100 I also learned the rule!
I should divide 51% by 100 and remove % symbol.
b) 51% = 51 = 0.51
100

Example 4 : Convert a) 0.07 b) 0.99 into percents.
Solution

a) 0.07 = 0.07 × 100% = 7% We got the value!

b) 0.99 = 0.99 × 100% = 99% We should multiply the given decimal by 100%!!

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EXERCISE 6.1
Section A - Class work

1. Let's say and write the fractions, decimals, and percents of the different
coloured square rooms of the square grid.

a) How many square rooms are there in the square
grid ?

b) Fraction, decimal, and percent of blue rooms are
, and

c) Fraction, decimal, and percent of pink rooms are
, and

d) Fraction, decimal, and percent of green rooms are ,
and

2. Let's say and write the fractions and decimals quickly.

a) 7% = b) 9% =

c) 13% = d) 47% =

e) 63% = f) 99% =

3. Let's say and write the percents quickly.

a) 5 = b) 24 = c) 53 =
100 100 100

d) 77 = e) 81 = f) 98 =
100 100 100

4. Let's say and write the percents as quickly as possible.

a) 0.02 × 100% = b) 0.08 × 100% =

c) 0.25 × 100% = d) 0.36 × 100% =

e) 0.66 × 100% = f) 0.89 × 100% =

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Section B
5. Let's answer the following questions.

a) What is a percent ? De ine with examples.
b) Out of 100 students, 65 students are girls. How many percent are girls ?
c) Out of 100 mangoes, 10 are rotten. How many percent are fresh mangoes?
d) Write a rule to convert a percent into a fraction or decimal.
e) Write a rule to convert a fraction or decimal into a percent.
6. Let's convert these percents into fractions of their lowest terms.

a) 2% b) 4% c) 5% d) 10% e) 15% f) 20%
k) 60% l) 90%
g) 25% h) 30% i) 40% j) 50%

7. Let's convert these percents into decimals.

a) 3% b) 8% c) 18% d) 29% e) 35% f) 47%

g) 52% h) 66% i) 70% j) 81% k) 90% l) 99%

8. Let's convert these fractions and decimals into percents.

a) 1 b) 1 c) 1 d) 3 e) 4 f) 7
2 4 5 4 5 10

g) 3 h) 4 i) 9 j) 0.01 k) 0.09 l) 0.12
20 25 50

m) 0.27 n) 0.33 o) 0.56 p) 0.75 q) 0.8 r) 0.9

9. a) 2 of the number of students of a class are girls.
5
b)
c) (i) Find the percentage of girls. (ii) Find the percentage of boys.

Solution

(i) The percentage of girls = 2 × 2 % = 2 × 20% = 40%
5
100

(ii) The percentage of boys = 100% – 40% = 60%

3 of the number of students of a school are boys.
4
(i) How many percentage of the students are boys ?

(ii) How many percentage of the students are girls ?

Deejina saves 4 of her pocket money everyday.
5
(i) What percentage of her pocket money does she save everyday ?

(ii) What percentage of her pocket money does she spend everyday ?

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10. a) Out of 40 full marks, Shreyasha obtained 30 marks in maths.

Express her marks in percent.

Solution 30
40
30 marks out of 40 full marks =

Now, percent of her marks = 30 × 25 % = 3 × 25% = 75%
40
100

b) Out of 50 full marks, Bishal obtained 45 marks in science. How many

percentage did he obtain in science ?

c) There are 45 students in a class. Among them, 27 are girls.
(i) Find the percentage of girls (ii) Find the percentage of boys.

11. Let's read a monthly progress report Monthly Progress Report
of Pooja Gurung and solve the given
problems. Subject Full Marks
English Marks Obtained
a) Express her marks in each subject in
percent. 30 24

b) In which subject did she have better Nepali 20 16
25 25
performance ? Maths 25 20

c) Find the total of full marks and the total
of her obtained marks. Then, express in Science

percent.

It's your time - Project work !

12. a) How many students are there in your class ?
b) How many percent of them are girls ?
c) How many percent of them are boys ?
d) How many teachers are there in your school ?
e) How many percent of them are male teacher ?
f) How many percent of them are female teacher ?

13. a) Draw a rectangle in a chart paper and divide it into 10 equal parts. Colour
10% parts with red, 20% parts with blue and 30% parts with green colours.

b) Draw another rectangle in the same chart paper and divide it into 4 equal
parts. Shade 50% parts with green colour and the rest with red colour.
What percent parts did you colour with red ?

c) Draw another rectangle in the same chart paper and divide it into 5 equal

parts. Shade 40% parts with red and the rest with blue colour. What

percent parts did you shade in blue ?

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6.5 Finding the value of the given percent of a quantity

Let's study the following examples and learn the process of inding the value
of the given percent of a quantity.

Example 1 : Find a) 5% of Rs 80 b) 45% of 200 people We know!
5
Solution 5% = 100 !!

a) 5% of Rs 80 = 5% × Rs 80 =20201=0501405×0 4 Rs 4 45% = 45 !!
b) 45% of 200 people = 45% × 45 × 100
Rs 80 =
× 200 = 2 = 90 people

In this way, to ind the value of the given percentage of a quantity (or a number),
we should multiply the quantity by the given percent. Then we should convert
the percent into fraction and simplify to get the required value.

EXERCISE 6.2

Section A - Class work
1. Let's say and write the values as quickly as possible.

a) 10% of Rs 90 = 10 × Rs 90 = 1 × Rs 9 = Rs 9
100

b) 10% of Rs 40 = 10 × Rs 40 = × =
100 =
=
c) 10% of Rs 150 = 10 × Rs 150 = × =
100 =

d) 20% of 30 students = 20 × 30 = × students
100 people
km
e) 50% of 80 people = 50 × 80 = ×
100

f) 60% of 60 km = 60 × 60 = ×
Section B 100

2. a) There are 480 students in a school and 55% of them are girls.

(i) Find the number of girls. (ii) Find the number of boys

Solution 1505011×

(i) Number of girls = 55% of 480 = Rs 24 = 11 × 24 = 264

480

2
(ii) Number of boys = 480 – 264 = 216

So, there are 264 girls and 216 boys in the school.

b) There are 30 students in a class. 60% of them are boys. Find the following:
(i) the number of boys (ii) the number of girls.

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c) Out of 80 full marks Rekha Bhatta got 90% marks in science. How many
marks did she get in science ?

d) Mr. Yadav is a Mathematics Teacher in a school. His monthly salary is
Rs 30,000. He spends 40% of his salary in a month.
(i) How many rupees does he spend in a month ?
(ii) How many rupees does he save in a month ?

e) There are 25,000 people living in a village. If 48% of them are male, ind
the male and female population of the village.

3. a) In the occasion of a festival, a supermarket gives 10% discount on all of its

items. If the listed price of a pair of shoes is Rs 1,500,

(i) ind the amount of discount.

(ii) ind the cost of the shoes after discount.

b) The listed price of a school bag is Rs 1,200 and the shopkeeper gives 20%
discount. Find the cost of the bag after discount.

4. a) Mother earns Rs 24,000 in a month. She spends 10% of the earning on
your school fee and 45% to run the family.
(i) How much money does she spend on your school fee ?
(ii) How much money does she spend to run the family ?
(iii) How much money does she spend altogether in a month ?
(iv) How much money does she save in a month ?

b) There are 700 students in a school. 20% of them are in secondary l e v e l ,
30% are in Lower Secondary level and the rest are in Primary level.
(i) How many students are there in secondary level ?
(ii) How many students are there in Lower Secondary level ?
(iii) How many students are there altogether in these two levels ?
(iv) How many students are there in Primary level ?

5. Let's read Sunayana's terminal progress Terminal Progress Report
report and solve the given problems.
Subject Full Marks
a) Find her marks in each subject. Marks Obtained
English
b) Find the total of full marks and the total Nepali 75 80%

of her marks, then express her marks in 80 75%

percent.

Maths 80 90%

Science 65 60%

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Unit Buying and Selling

7

7.1 Cost price (C. P.) and selling price (S. P.) - Looking back

Classwork - Exercise

1. Let's say and write the answer of these questions quickly.

A stationer buys a book for Rs 250 and she/he sells it for Rs 300.

a) What is the buying price of the book?
b) What is cost price (C. P.) of the book?
c) What is the selling price (S. P.) of the book?

Thus, when we buy something, we should pay money for it. This money is
buying or purchasing price. It is also called the cost price (C. P.).

Similarly, when we sell something, we take money from the buyer. This money
is called selling price (S. P.).

7.2 Profit and loss - Looking back

2. Let's say and write the answer of these questions quickly.
A fruit seller buys some fruits for Rs 120 per kg and sells for Rs 130 per kg.
a) What is the cost price (C. P.) of 1 kg of fruits?
b) What is the selling price (S. P.) o f 1 kg of fruits?
c) Which one is greater, C. P. or S. P.?
d) Is there pro it or loss to the fruit seller?
e) How much is the pro it or loss?
Thus, when we gain money by selling something, it is called pro it.

When selling price (S. P.) is higher than cost price (C. P.), we make pro it.

Pro it = Selling price – Cost price = S. P. – C. P.

On the other hand, when we lose money by selling something, it is called loss.
When selling price (S. P.) is less than cost price (C. P.), we make loss.

Loss = Cost price – Selling price = C. P. – S. P.

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7.3 Profit percent and loss percent

Let's discuss about the answers of these questions.

a) You buy a pen for Rs 100 and you sell it at a pro it of Rs 10. What is your
pro it in percent?

b) Sita buys a book for Rs 100 and she sells it at a loss of Rs 15. What is her
loss in percent?

Thus, buying a pen for Rs 100 and selling it at a pro it of Rs 10 means
pro it is 10 %.
Also, buying a book for Rs 100 and Selling it at a loss of Re 15 means loss
is 15 %.
In this way, we make pro it and loss always on cost price (C. P.). When we make
a pro it from the C. P. of Rs 100, it is called pro it percent. Similarly, when we
make a loss from the C.P of Rs 100, it is called loss percent.

Remember! We do not calculate pro it percent or loss percent from
the selling price (S. P.). We always calculate it from cost price (C. P.).

Now, let's study the following examples and investigate the rules to ind pro it
or loss percents.
a) When C. P. is Rs 200 and pro it is Rs 10, ind pro it percent.

Here, C. P. is Rs 200, and pro it = Rs 10

C. P. is Re 1, and pro it = Rs 10 By unitary method
200

C. P. is Rs 100, and pro it = Rs 10 5 u 100 pro it u 100
200 C. P.

1

= Rs 5

Rs 5 is the pro it in the C. P. of Rs 100. So, pro it percent is 5%.

b) When C. P. is Rs 400 and loss is Rs 40, ind loss percent.

Here, C. P. is Rs 400, and loss = Rs 40

C. P. is Re 1, and loss = Rs 40 By unitary method
400

C. P. is Rs 100, and pro it = Rs 4400010u 100 loss u 100
C. P.
1

= Rs 10

Rs 10 is the loss in the C. P. of Rs 100. So, loss percent is 10%.

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Pro it percent = pro it u 100% and loss percent = loss u 100%
C. P. C. P.

Again, let's study the following examples and learn more about pro it, loss,
pro it percent, and loss percent.

Example 1: Mrs. Rana bought a school bag for Rs 720 and sold it for
Rs 900.

a) Find her proϔit or loss b) Find her proϔit or loss percent

Solution:

C.P. of the bag = Rs 720

S.P. of the bag = Rs 900

a) Pro it = S. P. – C. P.

= Rs 900 – Rs 720 = Rs 180 20

b) Now, pro it percent = pro it u 100% = 180 u 100 = 20%
C.P. 900
1
Hence, her pro it is Rs 180 and pro it percent is 20%.

Example 2: A shopkeeper bought a mobile for Rs 2,500 and sold it for
Rs 2,300.

a) Find his proϔit or loss

b) Find his proϔit or loss percent

Solution:

C.P. of the mobile = Rs 2,500

S.P. of the mobile = Rs 2,300

a) Loss = C. P. – S. P.

= Rs 2,500 – Rs 2,300 = Rs 200

b) Now, loss percent = loss u 100% = 200 u 4
C.P. 2500
100

1

= 2 u 4%

= 8%

Hence, his loss is Rs 200 and loss percent is 8%.

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EXERCISE 7.1
Section A - Classwork

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

a) A stationer buys a packet of colour pencils for Rs 120 and sells it for Rs 150.

(i) What is the cost price (C. P.) of pencils?
(ii) What is the selling price (S. P.) of pencils?
(iii) How much is the pro it or loss to the stationer?
b) Bishwant bought a calculator for Rs 900 and he sold it to Pratik for Rs 800.
(i) What was the cost price (C. P.) of the calculator?
(ii) What was the selling price (S. P.) of the calculator?
(iii) How much was the pro it or loss to Bishwant?
2. Let's say and write the pro it or loss in each of the following cases.

a) C. P. = Rs 40, S. P. = Rs 45, then profit = Rs 5

b) C. P. = Rs 60, S. P. = Rs 55, then =

c) C. P. = Rs 100, S. P. = Rs 120, then =

d) C. P. = Rs 250, S. P. = Rs 225, then =

3. Let's say and write the pro it or loss percent quickly.

a) C. P. = Rs 100, pro it is Rs 5, then pro it percent =

b) C. P. = Rs 100, loss is Rs 12, then loss percent =

c) C. P. = Rs 100, pro it is Rs 15, then pro it percent =

d) C. P. = Rs 100, S. P. = Rs 90, then loss percent =

e) C. P. = Rs 100, S. P. = Rs 125, then pro it percent =

Section B

4. Let's answer the following questions.

a) Do you make pro it or loss, if the cost price (C. P. ) of an item is higher than
its selling price (S. P.)?

b) Do you make pro it or loss, if the selling price (S. P.) of an item is higher

than its cost price (C. P.)?

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c) What do you mean by pro it percent and loss percent?

d) Write the rules (formulae) to ind pro it and pro it percent.

e) Write the rules (formulae) to ind loss and loss percent.
5. Let's ind pro it or loss in each of the following cases.

a) C. P. = Rs 180, S. P. = Rs 200 b) C. P. = Rs 200, S. P. = Rs 180

c) C. P. = Rs 320, S. P. = Rs 450 d) C. P. = Rs 990, S. P. = Rs 840

e) C. P. = Rs 1,860, S. P. = Rs 2,150 f) C. P. = Rs 5,470, S. P. = Rs 5,290
6. Let's ind pro it or loss percents.

a) C. P. = Rs 200, S. P. = Rs 220 b) C. P. = Rs 250, S. P. = Rs 200

c) C. P. = Rs 500, S. P. = Rs 560 d) C. P. = Rs 750, S. P. = Rs 840

e) C. P. = Rs 1000, S. P. = Rs 950 f) C. P. = Rs 2500, S. P. = Rs 3,000

7. a) A shopkeeper bought a school bag for Rs 500 and sold it at a pro it of Rs 80.
Find his/her pro it percent.

b) Sahayata bought a video game for Rs 1,200 and sold it to Dakshes at a loss
of Rs 60. Find her loss percent.

8. a) A fruit seller bought some mangoes for Rs 800 and sold them for Rs 880.

(i) Find his pro it or loss (ii) Find his pro it or loss percent

b) Shivani Tharu bought a watch for Rs 1,250 and sold it for Rs 1,200.

(i) Find her pro it or loss (ii) Find her pro it or loss percent

c) Raju Maharjan bought a mobile for Rs 4,000 and sold it to Teriya Magar for
Rs 4,800.

(i) Find his pro it or loss (ii) Find his pro it or loss percent

9. a) The cost price of a fan is Rs 1,700. If it is sold at a pro it of 10%, ind the
pro it amount.

Solution: 10
100
Pro it amount = 10% of C.P. = u 1700 = 10 u Rs 17 = Rs 170

b) The cost of a T-shirt is Rs 800. If it is sold at a pro it of 20%, ind the pro it
amount.

c) Mr. Anamol Kandel purchased a few packets of crayons for Rs 1500 and
sold them at a loss of 5%. Find his loss amount.

"

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Unit Unitary Method and Simple Interest

8

8.1 Unitary method - Looking back

Classwork - Exercise

1. Let's read the price tag of a pen. Then, say and write the price of more
number of pens.

a) What is the price of 1 pen? Rs 20

b) The price of 2 pens = 2 u = ?

c) The price of 3 pens = 3 u = ?

d) The price of 4 pens = =

?

e) The price of 5 pens = =

?

Thus, if we know the value of unit (1) number of quantity, we can ind the

value of more number of quantity by multiplication.

2. Let's read the price tag of the given number of different things. Then, say
and write the price of unit number of thing.

a) What is the price of 2 sweets? Rs 10
What is the price of 1 sweet? 10 y 2 =

b) What is the price of 3 pencils? y = Rs 27
What is the price of 1 pencil?

c) What is the price of 4 pens?

What is the price of 1 pen? y = Rs 100

Thus, if we know the value of more number of quantity, we can ind the value
of unit (1) number of quantity by division.

In this way, a method of inding more values by multiplication and unit values
by division is known as unitary method.

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8.2 Unit value

1 pencil, 1 laptop, 1 kg of rice, 1 litre oil, and so on are unit number of quantities.
The value of a unit number of quantity is called unit value.

If the cost of 2 pen is Rs 50, the cost of 1 pen = Rs 50 y 2 = 50 = Rs 25.
2

Here, 1 pen is the unit number of quantity and Rs 25 is unit value.

Thus, unit value = Value of more number of quantities
Number of quantities

Similarly, if the cost of 1 pen is Rs 25, the cost of 4 pens = 4 u Rs 25 = Rs 100.

Thus, more value = Number of quantities × unit value

8.3 Rate of cost

a) If the cost of 3 kg of rice = Rs 300,

the rate of cost of rice = Rs 300 = Rs 100 per kg
3

b) If the cost of 5 l of milk = Rs 400,

the rate of cost of milk = Rs 400 = Rs 80 per litre
5

Thus, the unit cost is also called the rate of cost.

Now, let's study the following examples and learn to solve the given problems
by using unitary method.

Example 1: If the rate of cost of pencils is Rs 120 per dozen, ϔind the cost
of 6 dozens of pencils.

Solution

Rate of cost of pencils = Rs 120 per dozen

The cost of 6 dozens of pencils = 6 × Rs 120

= Rs 720

Hence, the cost of 6 dozen of pencils is Rs 720.

Example 2: The cost of 4 kg of apples is Rs 600. Find the rate of cost of
apples.

Solution

The cost of 4 kg of apples = Rs 600

The cost of 1 kg of apples = Rs 600 = Rs 150
4

Hence, the rate of cost of apples is Rs 150 per kg.

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Example 3: The cost of 8 exercise books is Rs 400.

a) Find the cost of 1 exercise book.

b) Fins the cost of 5 exercise books.

Solution

a) The cost of 8 exercise books = Rs 400

The cost of 1 exercise book = Rs 400
8

= Rs 50

b) The cost of 1 exercise book = Rs 50

The cost of 5 exercise books = 5 × Rs 50

= Rs 250

Hence, the cost of 1 exercise book is Rs 50 and 5 exercise books is Rs 250.

Example 4: Which is the best buy, 5 pens for Rs 90 or 6 pens Rs 120?
Solution

The cost of 5 pens = Rs 90

The rate of cost of 5 pens = Rs 90
5

= Rs 18 per pen

Also, the cost of 6 pens = Rs 120

The rate of cost of 6 pens = Rs 120
6

= Rs 20 per pen

The rate of cost of 5 pens is cheaper than the rate of cost of 6 pens.

Hence, the best buy is 5 pens for Rs 90.

Example 5: A bus can travel 100 km with 10 litres of diesel. Find the
distance travelled by the bus with 25 litres of diesel.

Solution

The bus can travel 100 km with 10 litres of diesel.

So, the mileage of the bus = 100 = 10 km per litre
10

Now, the distance travelled with 1 litre of diesel = 10 km

the distance travelled with 25 litres of diesel = 25 × 10 km = 250 km

Hence, the distance travelled by the bus with 25 l of diesel is 250 km.

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EXERCISE 8.1

Section A - Classwork
1. Let's say and write the cost of the given number of items.

Items Unit cost Numbers Cost
(Rate of cost)

a) Erasers Rs 5 5

b) Pencils Rs 10 6

c) Exercise books Rs 40 10

d) Boxes Rs 100 3

2. Let's say and write the unit cost (rate of cost) quickly.

Items Number of Cost Unit cost
quantities per kg

a) Vegetables 2 kg Rs 50

b) Wheat lour 3 kg Rs 150 per kg
c) Milk 4 l Rs 320 per l
d) Mustard oil 5 l Rs 1000 per l

3. The price list of some items are given below. Let's complete the table.

Items Cost of 1 piece Cost of 2 pieces Cost of 3 pieces

a) Candies Rs 10

b) Biscuits Rs 40

c) Ice-creams Rs 150

d) Chocolates Rs 40

Section B vedanta Excel in Mathematics - Book 5
4. Let's answer these questions.

a) De ine unit quantity and unit value.
b) Write a rule to ind unit value.
c) What do you mean by rate of cost?
d) Write a rule to ind rate of cost.

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Unitary Method and Simple Interest

5. a) If the rate of cost of eraser is Rs 5 per piece, ind the cost of 6 erasers.

b) If the cost of 4 gel pens is Rs 60, ind the rate of cost of the gel pen.

c) The rate of cost of lady inger is Rs 60 per kg. Find the cost of 5 kg of
lady ingers.

d) The cost of 10 kg of sugar is Rs 800. Find the rate of cost of the sugar.
6. a) The cost of a pen is Rs 30, ind the cost of 4 pens.

b) The cost of 5 exercise books is Rs 200, ind the cost of 1 exercise book.

c) Rs 80 is needed to exchange 1 Australian dollar. How many rupees is
needed to exchange 20 Australian dollars?

d) Rs 1,100 is need to exchange 10 US dollars. Find the exchange rate of
US dollar with Nepali rupees.

7. a) The cost of 1 dozen of exercise books is Rs 600.
(i) Find the rate of cost of the exercise books.
(ii) Find the cost of 4 exercise books.

b) The cost of 8 kg of grapes is Rs 960.

(i) Find the rate of cost of the grapes.

(ii) Find the cost of 10 kg of grapes.

c) Mother purchased 3 kg of vegetables for Rs 120.

(i) At what rate of cost did she purchase the vegetables?

(ii) If she had purchased 5 kg of vegetables, how much money would she
have to pay?

d) A bus can travel 150 km with 15 l of diesel.

(i) Find the mileage of the bus.
(ii) How many kilometres does it travel with 30 l of diesel?

e) There are 180 minutes in 3 hours.

(i) How many minutes are there in 1 hour?

(ii) How many minutes are there in 5 hours?
8. a) The cost of 4 kg of fruits is Rs 360. Find the cost of 7 kg of fruits.

b) The cost of 4 kg of fruits is Rs 360. How much fruits can be bought for
Rs 270?

c) The cost of 6 l of petrol is Rs 648. Find the cost of 10 l of petrol.

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d) The cost of 6 l of petrol is Rs 648. How much petrol can be bought for
Rs 540?

e) A labourer gets Rs 3500 for 7 days work. How much money does he get for
6 days work?

f) A labourer gets Rs 3500 for 7 days work. How many days should he work
to get Rs 2,000?

g) A motorbike can travel 90 km with 3 l of petrol. How many kilometres does
it travel with 15 l of petrol?

h) A motorbike can travel 90 km with 3 l of petrol. How many litres of petrol
does it need to travel 300 km?

9. Let's ind which one is the best buy.

a) 6 pencils for Rs 48 or 5 pencils for Rs 50.

b) 4 kg of potatoes for Rs 120 or 6 kg of potatoes for Rs 150.

c) 9 sweets for Rs 135 or 10 sweets for Rs 160.

d) 7 kg of grapes for Rs 700 or 8 kg of grapes for Rs 720.

10. a) The cost of 1 kg of wheat ϔlour is Rs 24.
2
(i) Find the cost of 1 kg of ϔlour.

(ii) Find the cost of 1 kg of ϔlour.
3
Solution
1
(i) The cost of 2 kg of lour = Rs 24
lour = Rs 24 ÷
The cost of 1 kg of 1 = Rs 24 × 2 = Rs 48
2 1

So, the cost of 1 kg of lour = Rs 48

(ii) The cost of 1 kg of lour = 1 × Rs 48 = Rs 16
3 3
1
b) The cost of 2 kg of sugar is Rs 40.

(i) Find the cost of 1 kg of sugar. (ii) Find the cost of 1 kg of sugar.
4
1
c) The cost of 3 kg of rice is Rs 30.

(i) Find the cost of 1 kg of rice. (iii) Find the cost of 1 kg of rice.
2
1
d) 4 part of the distance between two places is 15 km.

(i) Find the whole distance between the places.

(ii) Find 1 part of the distance between two places.
3

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It's your time - Project work!

11. Let's collect the information about the rate of cost of the following items in
your local markets.

Items Milk Rice Sugar Cooking oil
Rate of cost per l per kg
per kg per l

a) Estimate the quantity of each item consumed by your family in 1 month
and in 1 year.
b)
c) Estimate the expenditure on each item in 1 month and in 1 year.

12. a) How much is the total expenditure on these items in 1 month and in
1 year?

How much is your monthly school fee? How much do your parents pay
for your school fee in one year?

b) Let's visit to the available website such as www.google.com and ind
today's exchange rates of foreign currencies. Then, calculate how much
Nepali currency is needed to exchange the following foreign currencies?

(i) Indian rupees 100 (ii) Saudi Arabian riyal 100

(iii) Qatari riyal 100 (iv) U. S. Dollar 100

8.4 Simple Interest - Introduction

Mrs. Sharma deposited Rs 10,000 in a bank. After 1 year, when she withdrew
her money, she got Rs 11,000. Here, the deposited sum of money is called
the principal. The additional sum of Rs 1,000 paid by the bank is called the
interest.

Similarly, if you borrow money from a bank, you need to pay interest for the
use of money for the certain duration of time.

Thus, the original sum of money which is deposited or borrowed is the
principal. The additional sum of money which is paid for the use of these
money is the interest.

8.5 Rate of interest

Interest is usually given as the percentage of the principal in 1 year. It is called
rate of interest.

a) 5% per year interest means principal is Rs 100 and interest is Rs 5 in year.

b) 10% per year interest means principal is Rs 100 and interest is Rs 10 in
1 year.

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Similarly,

a) When the principal is Rs 100 and interest is Rs 7 in 1 year, the rate of
interest is 7% per year.

b) When the principal is Rs 100 and interest is Rs 12 in 1 year, the rate of
interest is 12% per year.

Now, let's study the following examples and learn to calculate simple interest
of the given sums of money.

Example 1: If the rate of interest is 5% per year, ϔind the interest of the
principal of Rs 1,500 in 1 year.

Solution

At 5% per year interest,

When principal is Rs 100, interest in 1 year = Rs 5

When principal is Re 1, interest in 1 year = Rs 5
100

When principal is Rs 1,500, interest in 1 year = Rs 5 × 1500
100

= Rs 5 × 15 = Rs 75

Hence, the required interest is Rs 75.

Example 2: Find the interest of a sum of Rs 100 at the rate of 10% per
year in 3 years.

Solution

At 10% per year interest,

When principal is Rs 100, interest in 1 year = Rs 10

When principal is Rs 100, interest in 3 years = 3 × Rs 10 = Rs 30

Hence, the required interest is Rs 30.

Example 3: The principal is Rs 1,000 and the rate of interest is 12% per
year.

a) Find the interest in 1 year. b) Find the interest in 5 years.

Solution

At 12% per year interest,

a) When principal is Rs 100, interest in 1 year = Rs 12

When principal is Re 1, interest in 1 year = Rs 12
100

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When principal is Rs 1000, interest in 1 year = Rs 12 × 1000
100

= Rs 12 × 10 = Rs 120

b) Interest in 1 year = Rs 120

Interest in 5 years = 5 × Rs 120 = Rs 600

Hence, the required interest in 1 year is Rs 120 and in 5 years is Rs 600.

Example 4: Mr. Dhurmus deposited Rs 2,000 in a bank at the rate of 7%
per year interest.

a) How much interest did he get in 1 year?

b) How much interest did he get in 4 years?

Solution:

At 7% per year interest,

a) When principal is Rs 100, interest in 1 year = Rs 7

When principal is Re 1, interest in 1 year = Rs 7
100

When principal is Rs 2,000, interest in 1 year = Rs 7 × 2000
100

= Rs 7 × 20 = Rs 140

b) Interest in 1 year = Rs 140

Interest in 4 years = 4 × Rs 140 = Rs 560

Hence, he got Rs 140 interest in 1 year and Rs 560 in 4 years.

EXERCISE 8.2
Section A - Classwork
1. Let's say and write the rate of interest quickly.

a) Principal = Rs 100, Interest in 1 year = Rs 4, rate of interest =
b) Principal = Rs 100, Interest in 1 year = Rs 6, rate of interest =
c) Principal = Rs 100, Interest in 1 year = Rs 9, rate of interest =
d) Principal = Rs 100, Interest in 1 year = Rs 14, rate of interest =

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2. Let's say and write the interest in 1 year quickly.

a) Principal = Rs 100, Rate of interest = 6%, interest in 1 year =

b) Principal = Rs 100, Rate of interest = 8%, interest in 1 year =

c) Principal = Rs 100, Rate of interest = 10%, interest in 1 year =

d) Principal = Rs 100, Rate of interest = 13%, interest in 1 year =

Section B

3. Let's answer the following questions.
a) Mrs. Suntali deposited a sum of Rs 5,000 in a bank. After 2 years, she
withdrew Rs 6,200 from the bank.
(i) How much was the principal that Suntali deposited?
(ii) How much interest did Suntali get after 2 years?
b) Write the meaning of the rate of interest is 6% per year.
c) If the principal is Rs 100 and the interest in 1 year is Rs 9, what is the rate
of interest?

4. Let's calculate the interest of the given principal in 1 year.
a) Principal = Rs 200, Rate of interest = 10% per year
b) Principal = Rs 500, Rate of interest = 7% per year
c) Principal = Rs 1,000, Rate of interest = 8% per year
d) Principal = Rs 3,000, Rate of interest = 9% per year

5. Let's calculate the interest of the principal of Rs 100 for the given years.
a) Principal = Rs 100, Rate of interest = 5% per year, Time = 2 years
b) Principal = Rs 100, Rate of interest = 6% per year, Time = 3 years
c) Principal = Rs 100, Rate of interest = 11% per year, Time = 4 years
d) Principal = Rs 100, Rate of interest = 15% per year, Time = 5 years

6. a) Find the interest of a principal of Rs 1,200 at the rate of 9% per year in 1
year.

b) A farmer borrowed Rs 2,500 from a bank at the rate of interest 12% per
year. How much interest did she pay after 1 year?

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c) Manoj Limbu deposited Rs 4,000 in a bank. How much interest did he
receive in 1 year at the interest rate of 10% per year?

7. a) Find the interest of a principal of Rs 100 in 3 years at the rate of 8% per
year?

b) Rajani Thakur borrowed Rs 100 from Tashi Lama at the rate of interest
15% per year. How much interest did she pay to him after 5 years?

c) Pratik Kasaju borrowed Rs 100 from Sunayana Dahal at the rate of 14%
per year. How much interest did he pay to her after 4 years?

8. a) The principal is Rs 900 and the rate of interest is 10% per year.

(i) Find the interest in 1 year (ii) Find the interest in 2 years.

b) The principal is Rs 2,000 and the rate of interest is 5% per year.

(i) Find the interest in 1 year (ii) Find the interest in 3 years.

c) Laxmi Rai deposited a sum of Rs 5,000 in a bank at 7% per year.

(i) How much interest did she get in 1 year?

(ii) How much interest did she get in 4 years?

d) Upendra Jha borrowed Rs 7,500 from Jyoti Gurung at 13% per year.

(i) How much interest did he pay to her in 1 year?

(ii) How much interest did he pay to her in 6 years?

It's your time - Project work!
9. a) Let's copy the following table in a chart paper. If the principal is Rs 100,

calculate the interest and complete the table.
Principal is Rs 100

Time 5% Rate of interest (per year) 10%
Rs 5 6% 7% 8% 9%
1 year
2 years Rs 28
3 years
4 years Rs 50
5 years

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Unit Ratio

9

9.1 Ratio - Introduction

Let's discuss about the answer to these questions.

a) How many more red apples are there than green apples?

b) How many less green apples are there than red apples?

Here, we ind the difference between the numbers of red apples and green
apples to compare their numbers.

c) How many times are the number of red apples more than green apples?

d) How many times are the number of green apples less than red apples?

Here, 6 ÷3 = 6 =2 times the number of red apples are more than green
apples. 3

Also, 3 ÷6 = 3 = 1 times the number of green apples are less than red apples.
6 2

Thus, we compare these two numbers dividing one number by another

number. Comparison of two quantities by the process of division is know as

the ratio between two numbers.

A ratio aways compares how many times a number is more or less than
another number.

Remember!
1. A ratio is a number.
2. A ratio compares two quantities of the same unit.
3. A ratio is obtained dividing one quantity by another

quantity of the same kind.

9.2 Ways of writing a ratio
There are 3 ways of making and writing ratios.
1. Ratio of a and b is a to b, ratio of b and a is b to a.
Ratio of x and y is x to y, ratio of y and x is y to x.
Ratio of 3 and 4 is 3 to 4, ratio of 4 and 3 is 4 to 3.

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2. Ratio of a and b is a : b (read as 'a is to b').
Ratio of b and a is b : a (read as 'b is to a').

Ratio of 4 and 5 is 4 : 5 (read as '4 is to 5').

Ratio of 5 and 4 is 5 : 4 (read as '5 is to 4).

Remember, the colon (:) stands for the words 'is to' in a ratio.

3. Ratio of a and b is ab, ratio of b and a is ab.
x y
Ratio of x and y is y , ratio of y and x is x .

Ratio of 1 and 2 is 12, ratio of 2 and 1 is 21. = pq.
In this way, ratio of p and q = p to q = p : q

p tells how many time p is more or less than q and q tells how many times
q p

q is more or less than p.

9.3 Terms of a ratio
Let's take a ratio 2 to 3 = 2 : 3 = 32.
Here, 2 and 3 are called the terms of the ratio 2 : 3.

Similarly 5 and 4 are the terms of 5 : 4, x and y are the terms of x : y, and so on.

In 2 : 3 or 2 , the irst term 2 is called antecedent and the second term 3 is
3
called consequent.

Now, let's study the following examples and learn to ind the ratio of the
quantities or numbers.

Example 1: Find the ratio between 75 cm and 1 m.

Solution: 75 cm : 100 cm

1 m = 1 × 100 cm = 100 cm 75 cm
100 cm
Ratio of 75 cm to 100 cm = 75 : 100 =

= 753 = 75
1004 100

= 3 = 3 : 4 So, a ratio is a number which
4 does not have any unit.

Remember! We should express a ratio to its lowest terms wherever necessary.
Hence, 75 : 100 = 3 : 4.

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Example 2: There are 30 students in a class. Among them 18 are girls.
Find the ratio of the number of girls and boys.

Solution:

Number boys = Total number of students – Number of girls

= 30 – 18 = 12

Now, the ratio of number of girls to boys = 18 : 12 = 18 3 = 3 = 3 : 2
12 2 2

Hence, the ratio of the number of girls and boys is 3 : 2.

EXERCISE 9.1
Section A - Classwork

1. Let's say and write the ratios and also express the ratios in words.

a) Ratio of a to b = a = a : b (a is to b)
b
b
Ratio of b to a = a = b : a (b is to a)

b) Ratio of p to q = = ( )
)
Ratio of q to p = = ( )
)
c) Ratio of x to y = = ( )
)
Ratio of y to x = = (

d) Ratio of 2 to 5 = = (

Ratio of 5 to 2 = = (

2. Let's say and write the ratios of the number of things.

a) b)

Apples to oranges is Dogs to cats is
c) d)

Girls to boys is Rose lowers to marigold lowers is

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3. Let's say and write the correct answers in the blanks spaces.

a) In m : n, the irst term is and the second term is .

b) In 5 : 4 the antecedent is and consequent is .

c) In 4 : 5, the antecedent is and consequent is .

d) In 3 : 7, the irst term 3 is called

e) In 7 : 3, the second term 3 is called

Section B

4. Let's answer the following questions.
a) Write the meaning of a ratio with an example.

b) How do we get a ratio between any two numbers?

c) Can we make a ratio between 2 metres and 3 kilograms? Why?

d) What are antecedent and consequent of a given ratio? Write with an
example.

5. Let's write the ratios of two numbers under the given conditions.

a) x is two times of y o x : y = 2 : 1

y is half times of x o y : x = 1 : 2

b) p is three times of q. c) q is one-third times of p.

d) a is four times of b. e) b is one-quarter times of a.

6. Find the ratios of the following numbers or quantities in their lowest
terms (wherever necessary).

a) 15 girls to 20 boys b) 21 women to 14 men c) Rs 10 to Rs 25

d) 9 kg to 7 kg e) 18 m to 24 m f) 1 m to 50 cm

g) 60 paisa to Re 1 h) 1 l to 500 ml i) 750 m to 1 km

7. a) There are 15 girls and 10 boys in a class. Find the ratio of the number of
girls and boys.

b) A sugar-water solution contains 250 g of sugar and 750 g of water. Find the
ratio of the weights of sugar and water.

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c) The capacity of a bottle is 2 litres and the capacity of a jug is 4 litres. Find
ratio of the capacities between the bottle and the jug.

d) A table is 2 m long and 80 cm wide. Find the ratio of the length and breadth
of the table.

e) A mobile phone is 180 g and a mini laptop is 1.2 kg. Find the ratio of the
weights of the phone and the laptop.

8. a) There are 32 students in a class and 12 of them are boys.

(i) Find the ratio of the number of students and the number of boys.

(ii) Find the ratio of the number of boys and the number of girls

b) There are 27 teachers in a school and 18 of them are lady teachers. Find
the ratio between the number of gents teacher and lady teachers.

It's your time - Project work!

9. a) Let count the number of girls and boys in your class. Then, ind the ratio
between these two numbers.

b) Let's count the number of gents teachers and lady teachers in your school.
Then ind the ratio between these two numbers.

c) Let's conduct a survey inside your classroom. Ask a question to your
friends, which subject do they like the most: Maths, Science or Computer?
Then tabulate the data in the table.

Subjects Maths Science Computer

No. of students

Now, let's ind the ratios between the number of students who like -
(i) Maths : Science (ii) Computer : Maths (iii) Science : Computer
(v) Maths : Computer (vi) Computer : Science (vii) Science : Maths

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Unit Time, Money, Bill and Budget

10

10.1 Telling time - Looking back

Classwork - Exercise

1. The table given below represents a timetable. Let's say and write your times

using a. m., or p. m. Also, show the times in the given clocks by drawing hour

and minute hand. Wake up Morning study Morning meal

Daily activities Time
(00:00 a.m./p.m.)

Wake up

Morning study School time Math period Tif in time

Morning meal

School time

Maths period School time Over Dinner Evening study

Tif in time

School time over

Dinner Bed time

Evening study

Bed time

10.2 24 - hour clock system

We say and write time using a. m. (Ante Meridiem) from
12:01 in the morning to 11:59 in the morning. Then, from
12:01 afternoon to 11:59 at night, we say and write time
using p. m. (Post Meridiem).

In 24-hour clock system, we say and write time in a little
different way. Let's compare the ways of saying and writing
time in 12-hour and 24 hour clock systems.

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12-hour clock system 24 - hour clock system
12 : 00 mid night 00:00 or 24 : 00 or 0000 hours
1 : 00 a. m. 01 : 00 or 0100 hours
11 : 59 a. m. 11 : 59 or 1159 hours
12 : 00 noon 12 : 00 or 1200 hours
1 : 00 p. m. (1 + 12 = 13 : 00) 13 : 00 or 1300 hours
4 : 30 p. m. (4 : 30 + 12 = 16 : 30) 16 : 30 or 1630 hours
11 : 59 p. m. (11 : 59 + 12 = 23 : 59) 23 : 59 or 2359 hours

Thus, when we add 12 hours to p. m. time of 12 - hour clock system, we get the
time of 24 - hour clock system.

10.3 Conversion of units of time
Let's recall the following relations between different units of time. These
relations are useful in the conversion of units of time.

1 hour (h) = 60 minutes (min) 1 month = 30 days
12 months
1 minute (min) = 60 seconds (s) 1 year = 365 days
366 days
1 day = 24 hours (h) 1 year = 10 years
100 years
1 leap year =

1 week = 7 days 1 decade =

1 century =

Now, let's study the following examples and learn to convert the units of time.

Example 1: Convert 2 h 15 min into minutes.

Solution It's easier!
2h 15 min = 2 × 60 min + 15 min 1 h = 60 min
2h = 2 × 60 min = 120 min
= 120 min + 15 min

= 135 min

Example 2: Convert 105 minutes into hours and minutes.

Solution Another process

105 min = 60 min + 45 min 105 minutes = (105 ÷ 60) h
= 1 h + 45 min
= 1 h 45 min 60)105 )1 hour
minute
–60
45

? 105 minutes = 1 h 45 min

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Example 3: Convert 2 years 6 months into months.

Solution = 2 × 12 months + 6 months We got it!
2 years 6 months = 24 months + 6 months 1 year = 12 months
2 years = 2 × 12 months

= 30 months

Example 4: Convert 40 months into years and months.

Solution Another process

40 months = 3 × 12 months + 4 months 40 months = (40 ÷ 12) years
= 3 years + 4 months
= 3 years 4 months 12) 40 )3 year
month
–36
4

? 40 months = 3 years 4 months

EXERCISE 10.1
Section A - Classwork

1. Let's say and write the time in words monitoring morning, afternoon or
evening.

a) 6 : 50 a. m. It is 10 to 7 in the morning.

b) 7 : 55 a. m.

c) 12 : 15 p. m.

d) 4 : 30 p. m.

2. Let's say and write the p. m. time in 24 - hour clock system.

a) 1 : 15 p. m. = 13 : 15 1 : 15 p. m. = 1 : 15 + 12 = 13.15

b) 1 : 30 p. m. = c) 2 : 45 p. m. =

d) 5 : 10 p. m. = e) 7 : 50 p. m. =

3. Let's say and write the time in 12-hour clock system using a. m. or p. m.

a) 00 : 00 = 12 : 00 mid-night 00 : 00 = 24 – 12 = 12 : 00

b) 0.1 : 25 = c) 07 : 40 =

d) 14 : 05 = e) 16 : 50 =

f) 18 : 00 = 144 g) 22 : 45 =

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4. Let's ind how many hours?

a) between 8 : 00 a. m. and 3 : 00 p. m.

Here, 3 : 00 p. m. = 3 : 00 + 12 = 15 : 00. So, 15 : 00 – 8 : 00 = 7 hours

b) between 7 : 00 a. m. and 1 : 00 p. m.

c) between 10 : 00 a. m. and 2 : 00 p. m.

d) between 9 : 00 a. m. and 4 : 00 p. m.
5. Let's say and write the correct answers as quickly as possible.

a) 1 hour = minutes b) 1 minute = seconds

c) 1 day = hours d) 1 month = days

e) 1 year = months f) 1 year = days

g) 1 leap year = days h) 1 decade = years

Section B
6. Let's write the time in 24 - hour clock or in 12 - hour clock system.

a) 2 : 10 p. m. b) 13 : 25 c) 3 : 30 p. m. d) 15 : 05
g) 7 : 20 p. m. h) 20 : 55
e) 4 : 45 p. m. f) 17 : 50

7. Let's convert into as indicated:

a) 2 h 10 min (min) b) 1 h 30 min (min) c) 3 h (min)

d) 75 min (h and min) e) 90 min (h and min) f) 105 min (h and min)

g) 1 min 10 s (s) h) 2 min 15 s (s) i) 100 s (min and s)

8. Let's convert into as indicated:

a) 4 weeks 2 days (days) b) 45 days (weeks and days)

c) 1 years 6 months (months) d) 2 years 8 months (months)

e) 20 months (years and months) f) 42 months (years and months)

g) 3 months 10 days (days) h) 75 days (months and days)

i) 3 years (weeks) j) 2 years (days)

9. Let's ind how many hours and minutes are there?

a) between 6 : 00 a. m. and 12 : 00 noon b) between 8 : 00 a. m. and 1 : 00 p. m.

c) between 8 : 15 a. m. and 1 : 30 p. m. d) between 7 : 20 a. m. and 2 : 50 p. m.
e) between 9 : 00 a. m. and 5 : 40 p. m. f) between 10 : 10 a. m. and 8 : 45 p. m.

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10. a) A year which is divisible by 4 is a leap year. Which of these are leap years?
(i) 2014 (ii) 2016 (iii) 2019 (iv) 2020 (v) 2022 (vi) 2024

b) If 2020 is a leap year, which one is the next immediate leap year?
11. It's your time - Project work!

a) Let's make your timetable including at least 10 regular tasks such as wake
up time, morning study time, school time, ... and so on.

Compare your timetable with your friends. You can stick your timetable in
your bedroom.

b) Let's make Nepali and English calendars of your birthday month in a chart
paper. On which date and day is your birthday? Circle it.

10.4 Addition and subtraction of time - Looking back

Classwork - Exercise

1. Let's add minutes and regroup into hours and minutes.

a) 45 min + 35 min = 80 min = 60 min + 20 min = 1 h 20 min

b) 30 min + 40 min =

c) 50 min + 25 min =

d) 40 min + 50 min =
2. Let's add and ind what time is it now?

a) 8 : 40 a. m. + 35 min = 9 : 15 a. m.

b) 7 : 30 a. m. + 40 min = (40 + 35) min = 75 min
c) 9 : 45 a. m. + 45 min = 75 min = 60 min + 15 min
d) 3 : 15 p. m. + 50 min = = 1 h 15 min

(8 + 1) : 15 = 9 : 15 a. m.

3. Let's subtract as shown.

60 min

a) 1 h 15 min – 30 min = (60 + 15) min – 30 min = 75 min – 30 min = 45 min

60 min

b) 1 h 10 min – 20 min =

c) 1 h 20 min – 40 min =

d) 1 h 30 min – 50 min = 146
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Time, Money, Bill and Budget

4. Let's subtract and ind what time is it now?
a) 8 : 20 a. m. – 40 min = 7 : 80 – 40 min = 7 : 40 a. m.
b) 7 : 20 a. m. – 30 min =
c) 6 : 30 a. m. – 50 min =
d) 11 : 10 p. m. – 40 min =

Now, let's study the following examples and learn more about the addition and
subtraction of time.

Example 1: Add or subtract a) 3 h 40 min + 2 h 50 min

b) 5 h 25 min – 3 h 45 min

Solution

a) 3 h 40 min (40 + 50) min = 90 min = 1 h 30 min

+ 2 h 50 min (1 + 3 + 2) h = 6 h

5 h 90 min 3 h 40 min + 2 h 50 min = 6 h 30 min

6 h 30 min 1 h = 60 min is borrowed to 25 min.
b) 1 h=60 min So, (60 + 25) min = 85 min

5 h 25 min (85 min – 45 min = 40 min
Also, (5 – 1) h – 3 h = 1h
– 3 h 45 min

1 h 40 min

Example 2: Football match started at 3 : 50 p. m. and it was of 1 hr 30 min.

At what time was it over? I understood!
Solution 80 min = 60 min + 20 min

3 : 50 p. m. + 1 h 30 min = 1 h 20 min

= (3 + 1) : (50 + 30) = 4 : 80 (4 + 1) : 20 = 5 : 00 p. m.

= 5 : 20 p. m.

Example 3: Santosh Sherchan travelled 3 h 35 min from Kathmandu to

Mugling and 1 h 55 min from Mugling to Pokhara. How long

did he travel from Kathmandu to Pokhara?

Solution

3 h 35 min I got it!
+ 1 h 55 min 4 h 90 min = 4 h (60 min + 30 min)

4 h 90 min = (4 + 1)h 30 min
= 5 h 30 min

5 h 30 min

Thus, he travelled 5 h 30 min from Kathmandu to Pokhara.

147 vedanta Excel in Mathematics - Book 5

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Time, Money, Bill and Budget

Example 4: Roshani Karki spent 1 h 20 min to complete her maths
and science homework. If she completed maths homework in
45 min, how long did she take to complete science homework?

Solution

60 min 80 min 1 h = 60 min is borrowed to 20 min
– 45 min So, (20 + 60) min = 80 min
1 h 20 min And (80 – 45) min = 35 min
– 45 min =

35 min

Hence, she took 35 min to complete her science homework.

Example 5: A maths class starts at 12:45 p. m. and gets over at 1 : 25 p. m.
How long is the class conducted?

Solution

Here, 1 : 25 p. m. = (1 + 12) : 25 p. m. = 13 : 25 p. m. = 13 h 25 min

13 h 25 min = 12 h 85 min

– 12 h 45 min – 12 h 45 min

40 min
Hence, the class is conducted for 40 minutes.

Example 6: The school time of a school is of 6 h 30 min everyday. If the
school is over at 4 : 10 p. m., at what time does it start?

Solution

Here, 4 : 10 p. m. = (4 + 12) : 10 p. m. = 16 h 10 min

16 h 10 min = 15 h 70 min

– 6 h 30 min – 6 h 30 min

9 h 40 min
Hence, the school starts at 9 : 40 a.m.

Example 7: Mr. Bishwakarma borrowed a loan from a bank join 17

Mangsir 2076 and paid on 20 Falgun 2077 B. S. How long did

he use the money?

Solution Y MD

2077 Falgun 20 = 2077–11–20 Falgun is the 11th month
and Mangsir is the 8th

– 2076 Mangisr 17 – 2076 – 8–17 month of the year.

1–3–3
Hence, he used the money for 1 year 3 months and 3 days.

vedanta Excel in Mathematics - Book 5 148
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur