Shubharambha Math Book 5 for CTP 2077

Example 2

Find the HCF of 24 and 36 by prime factorization method.
Solution: First, we find the prime factors of 24 and 36.

2 24 and 2 36
2 12 2 18
26 39

3 3

∴ 24 = 2 × 2 × 2 × 3
36 = 2 × 2 × 3 × 3
The common prime factors = 2 × 2 × 3
∴ HCF = 12

Example 3

Find the HCF of 15 and 25 by division method.

Solution:

1 Step I :
15 25
Take the smaller number i.e 15 as
– 15 divisor and the bigger number i.e 25 as
dividend.
10 15 1
Step II:
10
The remainder 10 becomes the divisor
5 10 2 and the divisor 15 becomes the
dividend.
10

×

∴ HCF = 5

Step III:

Repeat this process till the remainder
becomes zero. The last divisor is the
HCF.

Maths Zone - Grade 5 51

Example 4

Find the HCF of 16, 18 and 24 by
(i) Possible factors method
(ii) Prime factorization method
(iii) Division method

Solution I:

Possible factors of 16 = 1, 2, 4, 8, 16 1 × 16, 2 × 8, 4 × 4

Possible factors of 18 = 1, 2, 3, 6, 9, 18 1 × 18, 2 × 9, 3 × 6

Possible factors of 24 = 1, 2, 3, 4, 6, 8 12, 24 1 × 24, 2 × 12, 3 × 8, 4 × 6

Common factors among these = 1, 2

Highest Common Factors = 2

∴ HCF = 2

Solution II:

First, we find the prime factors of 16, 18 and 24.

2 16 2 18 2 24 Interesting Fact
28 39 2 12 16 = 24
24 33 26 18 = 2 × 32
22 33 24 = 23 × 3
1 Common Prime Factors = 2
1 1 Least power of 2 = 2
∴ HCF = 2
∴ 16 = 2 × 2 × 2 × 2

18 = 2 × 3 × 3

24 = 2 × 2 × 2 × 3

∴ The common prime factors = 2

∴ HCF = 2

52 Maths Zone - Grade 5

Solution III: Step I :

16 18 First, we consider the first two
numbers and follow the same step
– 16 1, 2 and 3 of the above example of
2 16 8 division method.

16 Step II:
×
The HCF of the first two numbers
Now, 2 24 12 which is 2 becomes the divisor
–2 and the third number 24 becomes
4 dividend. This process is repeated
–4 till the remainder becomes O. HCF
is the last divisor.
∴ HCF =×2

Example 5

Find HCF of 11 and 17. 1 is HCF of prime numbers or co-
Solution : prime numbers. (relatively prime)
Possible factors of 11 = 1, 11
Possible factors of 17 = 1, 17 E.g. : 4 and 5 are co-prime numbers.
Common factors = 1
∴ HCF = 1 They do not have any common
factors other than 1.

Lowest Common Multiples (LCM)

Least (Lowest) Common Multiple (LCM) of two or more numbers is the
smallest number which can be exactly divided by each of the given number.

Let us consider two numbers 6 and 8.

Multiples of 6 = 6, 12, 18, 24, 30, 36, 42, 48, .......

Multiples of 8 = 8, 12, 24, 32, 40, 48, .......

Common Multiples = 24, 48

Lowest Common Multiples = 24

∴ LCM = 24

Maths Zone - Grade 5 53

Finding LCM
1. Common Multiples Method
2. Prime Factorization Method
3. Division Method

Example 1

Find the LCM of 4 and 6 by Common Multiples Method.

Solution:
Multiples of 4 = 4, 8, 12, 16, 20, 24, .....
Multiples of 6 = 6, 12, 18, 24, ......
Common Multiples = 12, 24
Lowest Common Multiples = 12

∴ LCM = 12

Example 2

Find the LCM of 12 and 18 by prime factorization method.

Solution: First, we find the prime factors of 12 and 18.

2 12 2 18 Remember !
26 39
33 33 LCM = Common Factors ×
1 remaining factors
∴ 12 = 2 × 2 × 3 1

18 = 2 × 3 × 3

The common prime factors = 2 × 3

The remaining prime factors = 2 × 3

∴ LCM = 2 × 3 × 2 × 3 = 36

54 Maths Zone - Grade 5

Example 3

Find the LCM of 18 and 24 by division method.

Solution: Here all prime factors

2 18, 24 • Write the given numbers in a horizontal line, separating
3 9, 12
them by commas.
3, 4 • Divide them by a suitable prime number which exactly
∴ LCM = 2 × 3 × 3 × 4
divides at least two of the given numbers.
• We put the quotient directly under the numbers in the

next row. If the number is not divided exactly, we bring
it down in the next row.
• We continue the process of step II & III until co-prime

∴ LCM = 72 numbers are left in the last row.

• We multiply all the prime numbers by which we have
divided and the co-prime numbers left in the last row.

This product is the least common multiples of the given

numbers.

Example 4



Fine the LCM of 30, 36 and 40 by

(I) Prime factorization method (II) Division method

Solution 'I':

First, we find the prime factors of 30, 36 and 40.

2 30 2 36 2 40
3 15 2 18 2 20
55 39 2 10
33 55
1
1 1

∴ 30 = 2 × 3 × 5

36 = 2 × 2 × 3 × 3

40 = 2 × 2 × 2 × 5

The common prime factors among three = 2

The common prime factors between any two = 2 × 3 × 5

The remaining factors = 2 × 3

∴ LCM = 2 × 2 × 3 × 5 × 2 × 3

∴ LCM = 360

Maths Zone - Grade 5 55

Solution II:

Interesting Fact

2 30, 36, 40 2 30, 36, 40

2 15, 18, 20 2 15, 18, 20

3 15, 9, 10 2 15, 9, 10

5 5, 3, 10 3 15, 9, 5

1, 3, 2 3 5, 3, 5

5 5, 1, 5

1, 1, 1

∴ LCM = 2 × 2 × 3 × 5 × 3 × 2 ∴ LCM = 2 × 2 × 2 × 3 × 3 × 5

∴ LCM = 360 ∴ LCM= 360

Example 5

Find the LCM of 13 and 19.
Solution: First, we find the prime factors of 13 and 19.
13 13 19 19 As both are prime numbers.

11

∴ 13 = 13 × 1 In case of prime numbers,
19 = 19 × 1 their product is LCM.
∴ LCM = 13 × 19 = 247

Quick Method of HCF and LCM

Example 1

Find the HCF and LCM of 6 and 12

For HCF (I) For LCM (II)

By Division Method By Common Multiple Method

6 12 2 M6 = 6, 12
12 M12 = 12
LCM = 12 (Bigger one)
×

∴ HCF = 6 (Smaller one)

56 Maths Zone - Grade 5

We can find HCF & LCM by Inspection Method of two numbers.

Between two numbers, if one is multiple of another, then smaller is HCF
and bigger is LCM.

Exercise 2.6

1. Find the HCF of the following numbers by Possible Factors Method.

(a) 4, 6 (b) 8, 12 (c) 9, 12

(d) 16, 24 (e) 4, 6, 8 (f) 6, 9, 12

2. Find the LCM of the following numbers by Common Multiples
Method.

(a) 6, 8 (b) 10, 15 (c) 6, 9

(d) 12, 15 (e) 2, 3, 4 (f) 5, 8, 10

3. Find the HCF and LCM by Prime Factorization Method.

(a) 18, 27 (b) 45, 75 (c) 36, 48

(d) 72, 120

4. Find the HCF and LCM by Division Mehtod.

(a) 9, 15 (b) 22, 66 (c) 27, 36

(d) 25, 45 (e) 8, 12, 32 (f) 60, 90, 120

5. Find the HCF and LCM by Inspection Method (without any process).

(a) 8, 16 (b) 20, 40 (c) 32, 16

(d) 60, 120 (e) 12, 24, 48 (f) 10, 20, 40

6. Solve the following problems.
(a) What is the greatest number that divides 16 and 24 without
leaving remainder?
(b) Find the smallest number which is exactly divisible by 12 and 18.

Maths Zone - Grade 5 57

Lesson Square, Square Root,
Cube and Cube Root
5

Class Discussion

Square Number:

The product of a number multiplied by itself is called square number.
E.g: 1, 4, 9, 16, etc

Let's investigate the idea for square numbers.

1×1=1 2×2=4 3×3=9 4 × 4 = 16 5 × 5 = 25

Diagrammatically,

Shape number 12 3 4 5
Numbers of small squares
1 1 + 3 = 4 4 + 5 = 9 9 + 7 = 16 16 + 9 = 25

1=1 1=1 First odd number
4=1+2+1 1+3=4 Sum of first two odd numbers

9=1+2+3+2+1 1+3+5=9 Sum of first three odd numbers

16 = 1 + 2 + 3 + 4 + 3 + 2 + 1 1 + 3 + 5 + 7 = 16 Sum of first four odd numbers

25 = 1 + 2 + 3 + 4 + 5 + 4 + 3 + 2 + 1 1+3+5+7+9 = 25 Sum of first five odd numbers

By next way,

1 1+2 +1 1 + 2 + 3 + 2 + 1 1+2+ 3 + 4 + 3 + 2 + 1 1+2+3+4+5+4+3+ 2 + 1

58 Maths Zone - Grade 5

Square root:

The opposite process of finding a square number is a square root number.
The symbol for square root ' '.
E.g. : The square of 4 = 4 × 4 = 16
The square root of 25 = 25 = 5 × 5 = 52 = 5

Example 1

Find the square of 6 and 20.
Solution:
Square 6 = 62 = 6 × 6 = 36
Square of 10 = 102 = 10 × 10 = 100

Example 2

Solution:
Find the square root of 25 and 49.
Square root of 25 = 25 = 5 × 5 = 52 = 5
Square root of 49 = 49 = 7 × 7 = 72 = 7

Example 3

Find the square root of 225 by prime factorization method.

Solution:

First we have to find the prime factors of 225.

∴ 225 = 3 × 3 × 5 × 5
3 225
3 75 Taking one from each two
5 25
5 ∴ 225 = 3 × 5

Square root of 225 = 15

= 225

= 3 × 3 × 5 × 5

Maths Zone - Grade 5 59

= 32 × 52
= 3 × 5
= 15

Cube and Cube root

Cube number:

The product, of a number multiplied by itself three times is called cube
numbers. E.g. : 1, 8, 27 etc.

Let's investigate the idea for Cube numbers.

1 × 1 × 1 = 1 2 × 2 × 2 = 8 3 × 3 × 3 = 27 4 × 4 × 4 = 64
Diagrammatically,

1 2 3 4
11 22 3
3 4
4

Shape number 1 2 3 4
Number of 1 8 27 64
small cubes

1 = 13 = 1 First odd number

8 = 23 =3 + 5 Sum of second and third odd number

27 = 33 = 7 + 9 + 11 Sum of 4th , 5th and 6th odd number

64 = 43 = 13 + 15 + 17 + 19 Sum of 7th , 8th, 9th , 10th odd number

125 = 53 = 21 + 23 + 25 + 27 + 29 Sum of 11th, 12th, 13th , 14th , 15th odd number

Cube root:

The opposite process of finding a cube number is a cube root. The symbol
for cube root is '3 '

E.g.: The cube of 4 = 4 × 4 × 4 = 64
The cube root of 125 = 3 125

60 Maths Zone - Grade 5

= 3 5 × 5 × 5
= 3 53
= 5

Example 1

Find the Cube of 5 and 10
Solution:
Cube of 5 = 53 = 5 × 5 × 5 = 125
Cube of 10 = 103 = 10 × 10 × 10 = 1000

Example 2

Find the cube root of 64 and 216.
Solution:
Cube root of 64 = 3 64 = 3 4 × 4 × 4 = 3 43 = 4
Cube root of 216 = 3 216 = 3 6 × 6 × 6 = 3 63 = 6

Example 3

Find the cube root of 1728 by prime factorization method.

Solution:

First, we have to find the prime factors of 1728.

Alternative Method
2 1728 1728 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
2 864 = 2 × 2 × 3 (Taking one from each three)
2 432 = 12
2 216 3 1728 = 12
2 108
2 54
3 27
39
3

Cube root of 1728

Maths Zone - Grade 5 61

= 3 1728
= 3 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
= 3 23 × 23 × 33

= 2 × 2 × 3

= 12

Exercise 2.7

1. Find the square of the following numbers.

(a) 2 (b) 5 (c) 7

(d) 11 (e) 15 (f) 25

2. Find the cube of the following numbers.

(a) 3 (b) 4 (c) 8

(d) 10 (e) 12 (f) 15

3. Find the square root of the following numbers by prime

factorization method.

(a) 25 (b) 81 (c) 225

(d) 484 (e) 576 (f) 625

4. Find the cube root of the following numbers by prime factorization

method.

(a) 64 (b) 125 (c) 216

(d) 729 (e) 1000 (f) 1728

5. (a) Find the number when multipled by itself makes the product

196.

(b) 1024 students were arranged in equal number of rows and

column. Find the number of students in each row.

6. (a) When a number multiplies continuously 3 times the product

becomes 512. Find the number.

(b) The volume of a cubical tank is 27000m3. Find the length of the

tank.

62 Maths Zone - Grade 5

Maths Fun

This is a 4 by 4 magic square. Find out the missing numbers. What does
each line add up to?

16 3 2
11 8

9 12
4 15 14 1

Read the instruction and fill the missing numbers from 1 to 25 on the
following 5 by 5 magic square.
Hints: 1 digit number less then 5 in first row.
Square numbers in second row.
Odd numbers from 10 to 15 in third row.

23 20 19
22 14
5 13 21

12 10
73

Maths Zone - Grade 5 63

Practice Zone

Group A
A. Fill in the blanks of the following questions.

1. The numbers which have more than two factors are
called……………….

2. Rewrite and put commas according to international system: 1 5 7
9 2 1 3 2 5 ....................................................

3. Write the place value of the underlined digit 2 4 3 7 9 8 3 7
………….......

4. 6 is a square root of…………………
5. Find the square root of 225……………..
6. What is the cube of 4 ? …………….
7. The cube root of 125 is ………..
8. Round off 368 to the nearest 10 gives you ……………
9. HCF of 8 and 9 is………………
10. LCM of 5 and 7 is ……………..
11. Write the possible factors of 9……………………
12. Write the first five multiples of 4…………………………………..
13. The smallest number of one digit is ………………
14. The greatest number of 7 digits is …………………..

64 Maths Zone - Grade 5

Group 'B'
A. Answer the following questions.

1. Write the number names of the number 27859628 in National
and International Number System.

a. National:………………………
b. Internatinal:……………………..
2. Find the sum and differernce of the largest and the smallest

numbers formed by the digits 2, , 7, 0 , 4 , 5 and 9.
3. List the number divisible by 2, 3, 5 or prime number in the proper

place from the following
10 33 21 13 25 35 27
14 19 22 31 39

Divisible by 2 Divisible by 3 Divisible by 5 Prime number
10 27 10 13

4. Find the HCF and LCM of 24 and 32.
5. Find the prime factors of 210
6. If 225 students are arranged in a square field, how many students

are there in the first row?
7. A number has three identical factors. If one of them is 5, find the

number.

Maths Zone - Grade 5 65

Answers of Unit 2

Exercise 2.1

1. (a) Face value : 3
Place value : 30000000 (3 Crore)
(b) Face value : 4
Place value: 40,00,00,000 (Forty Crore)
(c) Face value : 1
Place value: 10000000 (One crore)
(d) Face value : 4
Place value : 40,00,00,000 (Forty crore)
2. (a) 43,72,91,885
Forty three crore, seventy lakh, ninety one thousand, eight hundred

and eighty five.
(b) 57,98,12,320
Fifty seven crore, ninety eight lakh, twelve thousand three hundred

and twenty
(c) 37,37,42,198
Thirty seven crore, thirty seven lakh, forty two thousand one hundred

and ninety eight.
(d) 50, 20, 40, 333
Fifty crore, twenty lakh, forty thousand three hundred and thirty

three.
3. (a)

T.C C T.L L T.Th Th H T O
466533241

∴ 46, 65, 33, 241
(b)

T.C C T.L L T.Th Th H T O
372345973

∴ 37, 23, 33, 241
(c)

T.C C T.L L T.Th Th H T O
504036280

∴ 50, 40, 36, 280
(d)

66 Maths Zone - Grade 5

T.C C T.L L T.Th Th H T O

670092048

∴ 67, 00, 92, 048

4. (a) 36, 25, 23, 415 (b) 80, 73, 46, 237 (c) 63, 00, 46, 910

(d) 44, 80, 00, 302

5. (a) %%,#%,^),@#$ (b) @),$$,^&,*() (c) &%,^),$^,(*&

6. (a) rf}tL; s/f]8 ;t} L; nfv rjGg xhf/ 5 ;o ;q

(b) kt} L; s/f]8, rjfnL; nfv krkGg xhf/ tLg ;fo afx|

(c) aL; s/f]8, c7f;L nfv pgGrfnL; xhf/ rf/ ;o

7. (a) Smallest number of 8 digits : 10000000

Greatest number of 8 digits : 99999999

Sum : 109999999

(b) Greatest number = 9 7 6 5 4 3 2 1

Smallest number = 1 2 3 4 5 6 7 9

Difference =85308642

(c) Greatest number = 9 8 7 6 4 3 2 0

Smallest number = 2 0 3 4 6 7 8 9

Exercise 2.2

1. (a) 213, 517, 347

∴ Two hundred, thirteen million, Five hundred seventeen thousand

three hundred and forty seven.

(b) 425, 618, 205

∴ For hundred twenty five million, six hundred eighteen thousand two

hundred and five

(c) 800, 700,444

∴ Eight hundred million, seven hundred thousand for hundred and forty

four.

(d) 918, 763, 258

∴ Nine hundred eighteen million, seven hundred sixty three thousand two

hundred fifty and eight.

2. (a)

H.M T.M M H .Th. T.Th Th. H T O

463 2 97524

∴ 463, 297, 524

Maths Zone - Grade 5 67

(b)

H.M T.M M H .Th. T.Th Th. H T O

846 4 56973

∴ 846, 456, 973

(c)

H.M T.M M H .Th. T.Th Th. H T O

700 4 40686

∴ 700, 440, 686

3. (a) 218, 503, 682
(b) 476, 300, 290

(c) 510, 398, 805

4. (a) Nepali system : Place name : Crore

Place value : Three crore

International system : Place name : Ten millions

Place value : Thirty millions

(b) Nepali system : Place name : Ten crores

Place value : Twenty crores

International system : Place name : Hundred millions

Place value : Two hundred millions

5. (a) 100 thousands (b) 1 Million
(c) 50 lakhs (d) 500 millions

Exercise 2.3

1. (a) 20 (b) 40 (c) 210 (d) 570

2. (a) 500 (b) 700 (c) 4600 (d) 4800

3. (a) 5000 (b) 2000 (c) 27000 (d) 26000

4. (a) 587670000 (b) 587700000

(c) 59,00,00,000 (d) 588000000 (e) 5900,00,000

5. (a) 60,000

(b) CPN - UML 32,00,000

Nepali Congress 31,00,000

Maoist Centre 13,00,000

SSF 5,00,000

RJP 5,00,000

Others 10,00,000

Exercise 2. 4

A. 1. a 2. d 3. d 4. c 5. c 6. b

7. d 8. d 9. b 10. b 11. a 12. c

B. Answer show to your teacher.

68 Maths Zone - Grade 5

Exercise 2.5

1. Figure show to your teacher.

(a) 16 = 2 × 2 × 2 × 2 (b) 27 = 3 × 3× 3

(c) 40 = 2 × 2 × 2 × 5 (d) 90 = 2 × 3 × 3 × 5

(e) 210 = 2 × 3 × 5 × 7

2. (a) 42 = 2 × 3 × 7 (b) 72 = 2 × 2 × 2 × 3 × 3

(c) 96 = 2 × 2 × 2 × 2 × 2 × 3 (d) 144 = 2 × 2 × 2 × 2 × 3 × 3

(e) 215 = 2 × 2 × 2 × 3 × 3 × 3 (f) 215 = 5 × 43

(g) 315 = 3 × 3 × 5 × 7 (h) 420 = 2 × 2 × 3 × 5 × 7

Exercise 2.6

1. (a) 2 (b) 4 (c) 3 (d) 8 (e) 2 (f) 3

2. (a) 24 (b) 30 (c) 18 (d) 60 (e) 12 (f) 40

3. (a) HCF = 9, LCM = 54 (b) HCF = 15, LCM = 225

(c) HCF = 12, LCM = 144 (d) HCF = 24, LCM = 360

(e) HCF = 8, LCM = 96 (f) HCF = 5, LCM = 360

4. (a) HCF = 3, LCM = 45 (b) HCF = 32, LCM = 66

(c) HCF = 9, LCM = 108 (d) HCF = 5, LCM = 225

(e) HCF = 4, LCM = 96 (f) HCF = 30, LCM = 360

5. (a) HCF = 8, LCM = 16 (b) HCF = 20, LCM = 40

(c) HCF = 16, LCM = 32 (d) HCF = 60, LCM = 120

(e) HCF = 12, LCM = 48 (f) HCF = 10, LCM = 40

6. (a) 8 (b) 36

Exercise 2.7

1. (a) 4 (b) 25 (c) 49 (d) 121 (e) 225 (f) 625
2. (a) 27 (b) 64 (c) 512 (d) 1000 (e) 1728 (f) 3375
3. (a) 5 (b) 9 (c) 15 (d) 22 (e) 24 (f) 25
4. (a) 4 (b) 5 (c) 6 (d) 9 (e) 10 (f) 12
5. (a) 14 (b) 32
6. (a) 8 (b) 30m

Maths Zone - Grade 5 69

UNIT

3 BASIC OPERATIONS
OF MATHEMATICS

Specific Objective Prescribed by CDC

‰‰ To solve number and word problems with two operations among +, -, × and
÷ and two brackets {( )}.

70 Maths Zone - Grade 5

Lesson

1 Addition and Subtraction

Class Discussion

While adding or subtracting numbers first we have to arrange them
according to their place and perform accordingly.

Example 1

Add : 3572098 and 6765320

Solution: Carry over  Arrange the number vertically
according to their places
11 1
 Add the numbers of respective
3572098 places.

+ 6765320

10337418


Example 2
Subtract: 28763459 from 83452306 To subtract the numbers
 Arrange the numbers
Solution:
vertically according to their
17 12 1314 11 12 9 16 Renaming places.

83452306

–28763459  Subtract the numbers at

54688847 respective places.
 Rename the number if

Example 3 necessary and subtract.

According to the population census 2068(2011), the male population
of Nepal is 1,28,49,041 and female population is 1,36,45,463. Find the
total population of Nepal.

Solution:

Male population = 1,28,49,041
Female population = + 1,36,45,463
Total = 2,64,94,504

Maths Zone - Grade 5 71

Example 4

In a school, there are 874 students. If 398 students are boys. Find the
number of girls.

Solution:

Total number of students = 874

Number of boys = 398

Number of girls =?

Number of girls = Total Number of students – Number of boys

= 874 – 398

∴ Number of girls = 476.

Example 5

A man earns Rs. 6, 75, 790 in a year. He spends Rs. 2,35,630 on Food,

2,00,750 on Education and 75, 560 on other things. How much does

he save? = 2, 35, 630 1 111
= 2,00,750
Solution: = 75, 560 235630
Expense on Food = 5, 11, 940 200750
Expense on Education + 75560
Other expanses 5, 1 1, 9 4 0
∴ Total expenses
4 17

6, 7 5, 7 9 0

Total saving = Total income – Total expenses – 5, 1 1, 9 4 0

= 6,75,790 – 5,11,940 163850

= Rs. 1,63,850

Exercise 3.1

1. Add the following. c.

a. b. 83762586 4876534
6357890 +94356872 546857
+576325
+ 3542683

72 Maths Zone - Grade 5

d. e. 657205637
5304627 634276321
837654 + 8345367
67398
+ 48 b.
64308743
2. Subtract the following. –5785568
a.
538324056
–264785247

c d.
48763240 4634200

–29875465 –1857342

3. Find the value of following (simplify)

a. 65976321 – 3467572 + 437654 b. 3245607 – 246735 – 687573

c. 65327615 – 46848326 + 999 d. 76532 – 405643 + 975432

4. a. The yearly income of Sagun's family is Rs. 4, 45, 650. The total
expenditure is Rs. 2, 46, 389. Find the saving amount.

b. The population of City is 8,97,645. If the male population is 4, 57,
c. 247, find the population of female.
d. Rangila's father bought a house for Rs. 2, 49, 43, 250 and a land
5. a. for Rs. 85, 68, 645. How much did he spend altogether?
In a mobile game, Sanjana scored 5, 46, 34, 250 and Ranjan scored
3,87,56,368. What is the total score made by them?
Saurav earns Rs. 4,87,620 per year and he spends Rs. 2,60,530
on food Rs. 153480 on education of children and Rs. 60,520 on
miscellaneous. How much does he save?

b. Kaji earns Rs. 6,02,028 per year. His annual expenditure on food
and education is Rs. 1,23,016 and Rs. 2,32,512 respectively. If he
does not save money and spend remaining on miscellaneous,
how much does he spend on miscellaneous?

Maths Zone - Grade 5 73

Lesson

2 Multiplication and Division

Class Discussion

While multiplying and dividing we have to remember the following rules.
☻ 5 × 1 = 5 (The product of a number and 1 is the number itself.)
☻ 5 × 0 = 0 (The product of a number and zero is always zero.)
☻ 50 × 300 = 15000 (5 × 3 = 15. So, 50 × 300 = 15000)
☻ 7 ÷ 1 = 7 (when a number is divided by 1, the quotient is the number itself.)
☻ 0 ÷ 8 = 0 (when 0 is divided by a number, the quotient is zero.)
☻ 2400 ÷ 80 = 240 ÷ 8 = 30

Example 1

Multiply the following:

a. 34620 by 435 b. 56000 by 250

a. Solution:

34620

×435 (435 = 400 + 30 + 5)

173100 (34620 × 5)
1038600 (34620 × 30)
+13848000
(34620 × 400)

15059700 Something = (∞, Undefined)
b. Solution: 0

56000 Multiply only 56 × 25
×250 We get 1400
Put 4 zero's on 1400 we get
00000 14000000 is the product.
2800000
+11200000
14000000

74 Maths Zone - Grade 5

Example 2 b. 339670 by 278
Dividend = Divisor × Quotient
Divide the following :
\ 73025 = 127 × 575
a. 73025 by 127
Dividend = Divisor × Quotient +
a. Solution: Remainder
\ 339670 = 1221 × 278 + 232
575
127 73025

– 635
952

– 889
635

– 635
×

b. Solution:

1221
278 339670

– 278
616

– 556
607

– 556
510
278
232

Example 3

The cost of a bus is Rs. 32,69,320 . What is the cost of such 200 buses?

Solution:

The cost of a bus = Rs. 32,69,320 3269320

\ The cost of such 200 buses = Rs. 3269320 × 200 × 200

= Rs. 65,38,64,000 653864000

Maths Zone - Grade 5 75

Example 4

A company earns Rs. 6, 54, 77,700 as a profit which should be divided
equally to 450 shareholders. How much does each shareholder get?

Solution:

Total profits = Rs. 6,54,77,700

No of share holders = 450

\ The share of profit to each shareholder

= 65477700 = 145506
450 450 65477700

– 450

2047

– 1800

2477

– 2250

2277

– 2250

270

–0

2700

– 2700

×

\ Each shareholder gets = Rs. 1,45,506.

Exercise 3.2

1. Find the product of b. 70356 × 135 c. 23506 × 436
a. 46789 × 235

d. 65324 × 150 e. 34670 × 250 f. 45000 × 360

2. Find the Quotients of

a. 4673 ÷ 54 b. 3520 ÷ 25 c. 43465 ÷ 45

d. 73025 ÷ 127 e. 54360 ÷ 450 f. 23746 ÷ 135

3. The cost of a computer is Rs. 52450. What is the cost of such 225
computers?

4. Rama has 420 motorbikes in her garage. The total cost of bikes is Rs.
10,50,00,000. What is the cost of a bike?

76 Maths Zone - Grade 5

Lesson

3 Simplification

Sign Rules

Let's learn some rules of sign (+, – , ×, ÷ )

a. The numbers with same signs are always added.

(+) + (+) = (+)

(–) + ( –) = (–)

Example: (+3) + (+5) = (+8)

(–4) + (–3) = (–7)

b. The numbers with the different signs are always subtracted and
the sign of greater number.

Example: (+5) + (–3) = (+2)

(+7) + (–12) = (–5)

c. The product or the quotient of two numbers with same sign is
always positive.

Rule Example:
(+) × (+) = (+) (+ 6) × (+ 4) = ( + 24)
(–) × (–) = (+) (– 3) × (- 4) = (+ 12)

Rule Example:

(+) ÷ (+) = ( + ) (+ 10) ÷ (+ 2) = (+ 5)

(–) ÷ ( – ) = ( + ) (– 12) ÷ (– 4) = (+ 3)

d. The product or quotient of two numbers with different signs are
always negative.

Rule Example:

(+) × (–) = (–) (+3) × (–3) =(–9)

(–) × (+) = (–) (–4) × (+5) = (–20)

Maths Zone - Grade 5 77

Rule Example:
(+) ÷ (–) = (–) (+4) ÷ (–2) =(–2)
(–) ÷ (+) = (–) (–6) ÷ (+2) = (–3)

Order of Operation
Lets learn the example solved by Rima and Sima.

3×5+5 3×5+5
= 15 + 5 = 3 × 10
= 20 = 30



Rima and Sima got different answer.

Who is correct? Who is wrong?

Are both correct?

Two operations are included. Here, at the same time, more than two
operations may be included.

So, to solve these types of problems we should follow the order
of operation. We should do divide, multiplication, addition and
subtraction respectively. This is called 'DMAS' Rule.

D Stands for Division (÷)

M Stands for Multiplication (×) Ohh!
A Stands for Addition (+)
S Stands for Subtraction (–) Rima is
correct.

78 Maths Zone - Grade 5

Example 1

Simplify : 4 + 18 ÷ 3 × 5 – 2 [ First Division, 18 ÷ 3 = 6]
Solution: [Then, Multiplication, 6 × 5 = 30]
[Addition, 4 + 30 = 34]
4 + 18 ­÷ 3 × 5 – 2 [Subtraction, 34 – 2 = 32]
= 4 + 6 × 5 – 2
= 4 + 30 – 20
= 34 – 2
= 32

Example 2

Simplify : 140 ÷ 10 + 5 × 4 ÷ 2 – 6 × 2
Solution:

140 ÷ 10 + 5 × 4 ÷ 2 – 6 × 2

= 14 + 5 × 2 – 6 × 2 [ Division, 140 ÷ 10 = 14, 4 ÷ 2 = 2]

= 14 + 10 – 12 [Multiplication, 5 × 2 = 10, 6 ×2 = 12]

= 24 – 12 [Addition, 14+ 10 = 24]

= 12 [Subtraction, 24 – 12 = 12]

Example 3

The sum of 8 and 7 is subtracted from the product of 5 and 4. Find
the result.

Solution: According to question
(5 × 4) – (8 + 7)

= 20 – 15

= 5



Maths Zone - Grade 5 79

Exercise 3.3

1. Find the sum b. ( + 15) + (12) c. (+15) + (+10)
a. (+ 8) + ( + 9)

d. (– 15) + (–19) e. (–10) + (–12 ) f. (–22) + (–16)

2. Simplify b. (– 32) + (+16) c. (+18) + (–14)
a. (+ 8) + (– 12)

d. (– 15 ) – ( + 12) e. (+ 24) – (–16) f. ( + 20) – (–16)

3. Find the product of b. ( + 8) × ( + 4) c. ( + 6) × (+4)
a. ( – 6) × (+ 2)

d. ( – 7) × (– 3) e. ( – 5) × (– 4) f. (+ 12 ) × (–5)

4. Find the quotient : b. (– 48) ÷ (+6) c. (– 24) ÷ (–3)
a. ( + 36) ÷ (–4)

d. (+54) ÷ ( +3) e. (– 27) ÷ (+9) f. ( + 18) ÷ (–6)

5. Simplify : b. 10 × 20 ÷ 4 + 5 – 3
a. 3 × 60 + 12 – 48

c. 15 ÷ 5 × 3 + 7 – 15 d. 67 – 564 ÷ 12 + 10 × 4

e. 24 + 24 ÷ 3 – 3 × 10 f. 4 × 7 – 63 ÷ 9 + 20

g. 625 ÷ 5 × 6 + 25 – 25 h. 360 ÷ 24 + 24 × 24 ÷ 4 + 25

i. 144 ÷ 12 – 3 × 15 ÷ 3 + 4 j. 140 ÷ 10 + 5 × 4 ÷ 2 – 6 × 2

k. 84 ÷ 12 × 2 + 7 – 3 × 4 l. 6 + 3 × 8 – 12 – 27 ÷ 9

6. Write the mathematical sentences and simplify:
a. Subtract 20 from the sum of 16 and 31.

b. The product of 5 and 18 is added to 25, then 42 is subtracted.

c. The sum of the product of 6 and 12 and the product of 8 and 21
is subtracted from 349.

d. 5 times 6 is added to the quotient when 12 is divided by 12.

e. Divide 25 by 5 and subtract 6 from the quotient then add the
product of 5 and 7.

80 Maths Zone - Grade 5

Use of brackets in Simplification

We use 'DMAS' to simplify the mathematics sentence, sometimes we can
not simplify the problem according to the order of operation. To preserve
the theme of problem we use brackets.

Example: Multiply the difference of 5 and 3 with 7. We write. '5 – 3 × 7'
According to DMAS role. We must do multiply first but question says, first
do the difference. So the correct form of above questions is

(5 – 3) × 7 = 2 × 7 = 14

So, at first, we do the work inside the brackets.

We use three types of brackets.

() Small bracket 'Remove ( ) first' (small brackets)
Curly bracket 'Then Remove { }' (Curly bracket)
{}

[ ] Square bracket Finally remove [ ] (Square brackets)

First →( ) Inside the brackets
Second →{ } use 'DMAS'
Third →[ ]

While Simplifying the expression involving, follow the 'BODMAS' rule.

B stands for Brackets Maths Songs
O stands for Off
D stands for Division Same sign eP hf8] \g]
M stands for Multiplication different sign df 36fpg]
A stands for Addition hf]8 jf 36fp u/];L
S stands for Subtraction 7"nfs] f] Sign ptf k9fpg]

• If bar (—) is used in the expression, firstly, we should do the the work
inside the bar before using 'DMAS' rule.

Maths Zone - Grade 5 81

Example 1

Simplify : 88 + [63 ÷ 7 + {(16 – 13) + 6}]
Solution:

88 + [63 ÷ 7 + {(16 – 13) + 6}]

= 88 + [63 ÷ 7 + {3 + 6}] [ First, remove the ( ), 16 – 13 = 3 ]

= 88 + [ 63 ÷ 7 + 9 ] [then { }, {3 + 6} = 9]

= 88 + [9 + 9]

= 88 + 18 [Finally [ ]

= 106

Example 2

Simplify : 2 × [14 – {(5 × 7) ÷ (2 + 3) – 5} + 10]
Solution:

2 × [14 – {(5 × 7) ÷ (2 + 3) – 5} + 10]

= 2 × [14 – {35 ÷ 5 – 5} + 10] [Working inside the ( ) ]

= 2 × [14 – {7 – 5} + 10] [Inside { } using DMAS]

= 2 × [ 14 – 2 + 10]

= 2 × [24 – 2]

= 2 × 22

= 44

Exercise 3.4

1. Simplify: b. (93 ÷ 3 × 2) + 62 – 124
a. 20 × (17 – 6 + 3)

c. (5 – 2) × (11 – 5) d. 5 + 4 × {(8 – 2) –3} ÷ 3

e. 4 + 3 [16 ÷ 4{26 – 8 (8 – 5)}] f. 3 + [4 × {(7 – 1) +2} ÷ 16]

g. 25 – 28 + {5 + (8 – 3) × 2}

h. 10 + 3 [{4 + 3 (15 – 5) ÷ 5} – 7] ÷ 9

82 Maths Zone - Grade 5

i. 50 – [6 + 2 {25 ÷ 5 (5 – 2 + 6 ÷ 3)}]
j. 45 + [63 ÷ 7 + {(16 – 13) + 6}].
k. 60 ÷ 3 – {3 + (6 + 9 – 5 + 3}
l. 12 – [20 – {40 – 2 (14 – 10 – 6)}] 
m. 95 + 7 [49 ÷ (56 ÷ 8) – 5]
n. 7 of 15 ÷ [17 – {5 + (2 × 6 – 7)}]
2. Convert the following problems into mathematical sentences and
simplify:
a. 6 times the difference of 14 and 2 is divided by 24.
b. 4 times the difference of 20 and 9 is added to 5 and the result is

divided by 7.
c. The sum of the product of 6 and 12 and the product of 8 and 21

is subtracted from 436.
d. 6 times the difference between 19 and 4 is subtracted from 3 times

the sum of 45 and 15. Find the difference.
e. The product of 5 and 17 is subtracted from the quotient of 450

divided by 5. Then the remainder is added to 15. What is the
sum?
Expand Your Knowledge
A frog is climbing a well 17 metres deep. It
climbs 5 meters in one hour but then slides
back 2 meter as it rest for an hour. How long
will it take to climb out of the well?

Maths Zone - Grade 5 83

Maths Fun

Teachers are requested to arrange the students in the file on one side of the
ground. Arrange the questions with answer sheet and pencil in another
side for which is 50 m far from the students. According to the instruction
given by the teacher, students will run and solve the questions and come
back in their initial position, who came first with correct answer will be the
winner.

20 × (17 - 6 + 3)
20 × (17 - 6 + 3)
20 × (17 - 6 + 3)

20 × (17 - 6 + 3)
20 × (17 - 6 + 3)
20 × (17 - 6 + 3)

About 50 m

84 Maths Zone - Grade 5

Practice Zone

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

1. (–20) ÷ (-5) is equal to

a. - 4 b. 4 c. 5

2. We use ........... rule inside the bracket c. AMDS
a. DMAS b. MDAS

3. 5x(3 + 6) - 15 is equal to

a. 30 b. 6 c. 15

4. 3 times the sum of 4 and 5 is expressed as

a. 3 × 4 + 5 b. 4 + 5 × 3 c. 3x (4 + 5)

5. In simplification first we use DMAS inside the ........... bracket.

a. Square b. Curly c. Small

6. Which is the correct form to get answer 24 of 7 – 3 × 6

a. 7 – 3 × 6 b. 7 – (3 × 6) c. (7 – 3) × 6

7. What is the least number to be subtracted from 75, so that it is exactly
divisible 9?

a. 1 b. 2 c. 3

8. When a number is divided by 7, the quotient is 8 and leaves 2
remainder.,

a. 16 b. 56 c. 58

Maths Zone - Grade 5 85

Group 'B'
A. Solve the following questions.

1. Simplify 25 - [90 - 5{9 - (14 - 12)}] ÷ 5

2. Make mathematical question and simplify.

a. The difference between 15 and 9 is subtracted from 5 time the
sum of 7 and 3.

b. 5 times the difference between 11 and 7 is divided by 10.

c. The difference of 14 and 6 is multiplied by one-fourth of the sum
of 11 and 9.

3. A man earns Rs. 7,55,603 in a year. He spends Rs. 2,46,035 on food,
Rs. 2,03,760 on his children education and Rs. 87,602 on other things.
How much does he save?

4. Find the product of

a. 63201 × 205 b. 4632 × 132

5. Find the quotient of

a. 32900 ÷ 140 b. 19272 ÷ 132

6. Find the quotient of

a. 63201 and 205 b. 4632 × 132

7. Simplify : 10 + [15 - {12 - 6)} ÷ 3

8. Simplify: 5 + 2 [4 of 7 + (8 × 2 – 36 ÷ 3)]

9. Simplify: 14 ÷ {10 - 2 × (7 - 6 – 3)}

86 Maths Zone - Grade 5

Answers of Unit 3

Exercise 3.1 c. 8966074 d. 6209727

1. a. 693415 b. 178119458 c. 18887775 d. 2776858
e. 1299827325 c. 18480288 d. 646321
2. a. 273538809 b. 58523175 c. 33511895 d. 93390618
3. a. 62946403 b. 2311299
4. a. 199261 b. 440398
5. a. 13090 b. Rs. 246500

Exercise 3.2

1. a. 10995415 b. 9498060 c. 10248616 d. 9798600
e. 8667500 f. 16200000

2. a. Q 86 Remainder 29 b. Q - 140 R - 20

c. Q - 965 R - 40 d. Q - 575 R - 0

e. Q - 120 R - 360 f. Q - 175 R - 121

3. Rs. 1,18,01,250 4. Rs. 2,50,000

Exercise 3.3

1. a. +17 b. + 25 c. +4 d. - 27 e. + 40 f. + 96
f. +96
2. a. -4 b. -16 c. +4 d. -27 e. +40 f. -60
f. -3
3. a. -12 b. 32 c. 24 d. 21 e. 20 f. 41
l. 15
4. a. -9 b. - 8 c. 8 d. 18 e. -3

5. a. 144 b. 52 c. 1 d. 60 e. 2

g. 750 h. 184 i. 1 j. 12 k. 9

6. a. 27 b. 73 c. 109 d. 31 e. 34

Exercise 3.4

1. a. 280 b. 0 c. 18 d. 9 e. 10 f. 5
g. 12 h. 11 i. 42 j. 63 k. 4 l. 12
m. 109 n. 15
c. 196
2. a. 8 b. 7 d. 90 e. 20

Maths Zone - Grade 5 87

UNIT

4 TIME AND MONEY

Specific Objective Prescribed by CDC

Time

‰‰ To multiply and divide (simple) for the units of time and solve daily life word
problems related to them.

Money

‰‰ To add, subtract, multiply and divide (simple) daily life word problems
related to rupee and paisa.

88 Maths Zone - Grade 5

Lesson Time

1

Class Discussion

Second, Minutes, Hours and Days.
A day has 24 hours. 1 hour consists 60 minutes.
1 minutes consists 60 seconds.
Therefore, 60 seconds = 1 minute
60 minutes = 1 hour
24 hour = 1 day

Leap year and Non Leap Year

In one year there are 365 1 days.
4 afonudrleyaevares41
We take 365 days in a year days in each year.

It makes complete 1 day in

1 + 1 + 1 + 1 = 1 day
4 4 4 4

So in fourth year there are 366 days and this year is

called leap year.

1 day is increased in the month February. The leap
year is divisible by 4. The year 2000 is a leap year.

365 days = 1 year

366 days = 1 leap year.

Days, Week, Month, Decade, Century and Millennium

We have,

7 days = 1 week 30 days = 1 month

52 weeks = 1 year 12 months = 1 year

10 years = 1 decade 10 decades = 1 century (100 years)

10 centuries = 1 millennium (1000) years.

Maths Zone - Grade 5 89

Different English months has different number of days.
28 days = Feb.
29 days = Feb (Leap year)
30 days = April, June, September, November
31 days = Jan, March, May, July, August, October, December.
But, there are not fix number of days in Nepali month. Generally we
use 30 days for a month. i.e., 1 month = 30 days.

A.M. and P.M.

In 12 hours clocks, we use A.M. and P.M.

A.M. = Anti meridian = before noon

P.M. = Post meridian = after noon.

The time from midnight to mid-day is denoted by A.M.

The time from mid-day to midnight is denoted by P.M.



Example 1

Convert 4 month 15 days into hours.

Solution: 4 month + 15 days
(4 × 30) days + 15 days
4 month 15 days = 120 days + 15 days
= 135 days.
4 month = 4 × 30 days = 135 × 24 hours .
= 3240 hours.
= 120 days

120 days = 120 × 24 hours

= 2880 hours.

and 15 days = 15 × 24 hours.

= 360 hours

\ 4 month 15 days = (2880 + 360) hours = 3240 hours.

90 Maths Zone - Grade 5

Example 2

Convert 6745 seconds into hours, minutes and seconds.

Solution:

6745 seconds. 112 minute

= 6745 minutes 60 6745 sec
60 – 60
74
= 112 minutes 25 seconds. – 60
145
= 112 hours 25 seconds. 120
60 25 sec.

= 1 hours 52 minutes 25 seconds.

\ 6745 seconds = 1 hour 52 minutes 25 112 minute
seconds.
60 112 min 1 hour
Example 3 – 60
52 min.

Add 8 hrs 23 minutes 35 sec and 3 hrs 30 minutes 20 sec.

Solution:

Add hours to hours. Hrs Min Sec
Add minutes to minutes. 8 23 35
Add seconds to seconds.
+ 3 30 20
Example 4 11 53 55

Subtract 13 hours 37 min 28 sec from 15 hours 48 min 39 sec.
Solution:

Hrs Min Sec ‰‰ Subtract second from seconds.
15 48 39 ‰‰ Subtract minutes from minutes.
– 13 37 28
2 11 11 ‰‰ Subtract hours from hours.

Maths Zone - Grade 5 91

Example 5

Add 6 years 8 months 25 days and 8 years 7 months 22 days.
Solution:

Years Months days ‰‰ Add years, months days separately.

6 8 25 ‰‰ Subtract 30 days from 47 days and add
+ 8 7 22 it's equivalent 1 months to 15 months

14 15 47 ‰‰ Subtract 12 months from 16 months
+1 –30 add it's equivalent 1 year to 14 years.

14 16 17
+1 –12
15 4 17

Example 6

Subtract 6 year 9 month 27 days from 14 years, 8 months 10 days.
Solution:

Year Month days ‰‰ We can't subtract 27 sec from 10 sec. and
9 months from 8 months. So, we have to
14 8 10 rearrange that month and days.

–6 9 27 ‰‰ Borrow 1 month form 8 months and add it
is equivalent 30 days with 10 days.
??
‰‰ Borrow 1 year from 14 years and add it's
Year Month days equivalent 12 months with 7 months.

14 7 +30 40 ‰‰ Now subtract days from days, month from
–6 9 27 month and year from year.

? ∴ We get 7 years, 10 months and 13 days.

Year Month days

13 19 40
– 6 9 27

7 10 13

92 Maths Zone - Grade 5

Exercise 4.1

1. Convert into seconds b. 5 hours 35 minutes 17 seconds
a. 30 minutes 50 seconds

2. Convert into hours b. 11 month, 22 days 18 hours
a. 4 month 16 days

3. Convert into hour, minutes and seconds

a. 7654 seconds b. 5426 seconds

4. Convert as indicated

a. 3 years, 4 months 5 days (into days)

b. 24 years 5 months (into months)

c. 2 year 5 days (into days)
d. 3 month 22 hours (into hours)
5. Add the following:

a. H M S b. W D H c. YM D
12 48 50
9 6 15 7 9 15
+ 10 15 20
+ 6 3 17 + 10 5 22

d. H M S e. W D H f. YM D
6 46 37 8 5 20 13 8 25
+ 6 4 12 + 7 5 19
+ 5 15 49

6. Subtract the following W D H c. YM D
15 6 15 17 10 27
a. H M S b. – 4 5 19 – 9 28 28
14 47 16
– 7 18 27

d. H M S e. W D H f. YM D
12 9 25
42 15 46 22 34 18 – 8 10 20

– 26 32 57 – 18 45 23

Maths Zone - Grade 5 93

Multiplication and Division of Time

Let's Discuss few examples of multiplication and division of time.

Example 1

Multiply 3 years 5 month and 17 days by 3

Solution:

Years Months Days First multiply simply by 3
17
3 5 • 30 days = 1 month, so subtract
×3 30 days from 51 days and add
51 it's equivalent 1 month to 15
months.
9 15 – 30
21 • 12 months = 1 year, so subtract
+ 1 12 months from 16 months and
21 days add this 1 year to 9 years.
9 16

+1 – 12

10 years 4 month

We get '10 years 4 month 21 days'

Example 2

Multiply : 15 hours 30 minutes 16 seconds by 20.

Solution: 5 min
60 320 seconds
Hours Minutes Seconds
– 300
15 30 16 20 seconds

× 20 10 hours
60 605 seconds
300 600 320
– 600
+ 5 – 300 5 seconds

300 605 20

+10 – 600

310 hours 5 minutes 20 seconds

\ 310 hours 5 minutes 20 seconds.

1.. First multiply by 20 to hours minutes and seconds respectively.
2. 60 seconds = 1 minutes. So, divide 320 seconds by 60, and subtract

300 from 320 and add 300 seconds = 5 minutes to 600 minutes
3. 60 minutes = 1 hour, so divide 605 minutes by 60 and subtract 600

from 605 and add 600 minutes = 10 hour to 300 hours.
\ We get 310 hours 5 minutes and 20 seconds.

94 Maths Zone - Grade 5

Example 3

Divide 18 hours 35 minutes 45 seconds by 5.

Solution:

3 43 9

5 18 hours 35 minutes 45 seconds

– 15

3 hour 35 minutes (3 × 60 + 35) minutes
= 180 + 35 minutes
= 215 minutes = 215 minutes.

– 20

15

– 15

0 45 seconds

45

×

\ Quotient = 3 hours 43 minutes 9 seconds.

Example 4

Divide 12 years 9 months and 18 days by 8.
Solution:

Y M D

1 7 6

8 12 9 18

– 8 4 year = 48 month,
which is added with
4 9 18 month.

– 4 + 48 1 month has 30 days,
add with days.
0 57 18
Maths Zone - Grade 5 95
– 56

1 18

–1 + 30

48

– 48

×

\ Quotient = 1 year 7 months 6 days.

Exercise 4.2

1. Multiply b. HM S c. HM S
7 10 4 8 15 25
a. H M S 3
12 4 10 × ×2
×5

d. M D H e. M D H f. MD H
10 3 2 47 5 15 20 9
×8 7
×6 ×

g. Y M D h. Y M D i. YM D
8 6 16
53 2 579
× 10
×9 ×4

2. Divide HM S b. H M S
a. 18 24 32 4 12 30 48

8

c. M D H d. M D H
9 27 18 27 5 18 15 40

e. Y M D f. Y M D
9 18 27 45 6 10 6 12

3. Divide.
a. 11 years 5 months 6 days by '6'
b. 12 years 7 months 10 days by 5
c. 9 years 5 months 4 days by 4
d. 4 hours 22 minutes 20 seconds by 5
e. 16 hours 40 minutes 55 seconds by 5
f. 15 hours 30 minutes 48 seconds by 4.

96 Maths Zone - Grade 5

4. Each period of your school is 45 minutes. How many hours and
minutes are there in 7 periods?

5. A pipe can fill a tank in 2 hours 40 minutes. How long does it take to
fill the 6 similar tanks?

6. The total college hour of a college is 5 hours forty minutes which
includes 20 minutes break. If there are 5 periods of equal interval,
how many minutes are there in a period?

7. Sunil completes trekking tour in 12 hours 32 minutes and 48 seconds.
He took 6 days to complete the tour. How long did he trek each day,
if he treks for equal length time everyday?

Maths Zone - Grade 5 97

Lesson

2 Money

Addition of Money

Let's discuss few examples of conversion and ÷ 100
addition of money.

Example 1 Paisa Rupee

Convert Rs. 65 and 80 paisa into paisa. × 100
Solution:
To change Rupee into Paisa
Rs. 65 and 80 paisa Multiply by '100'
= (65 × 100) paisa + 80 paisa To change Paisa into Rupee
= (6500 + 80) paisa divide by 100
= 6580 paisa.

Example 2

The cost of a book is Rs. 350.60 and cost of a pen is Rs. 37.50. What
is the total cost.

Solution:

Total cost = Rs. 350.60 + Rs. 37.50

\ 350.60

+ 37.50
388.10

\ Total cost = Rs. 388.10, (Rs. 388 and 10 Paisa)

Example 3

Look at the price list of a 'Suman store'. How much money did
Anusha pay if she bought 1 kg rice, 1 kg sugar and 1 kg daal?

Solution:
Here, cost of a kg rice = Rs. 72.50

cost of a kg sugar = Rs. 90.25

98 Maths Zone - Grade 5

cost of a kg daal = Rs. 235.50 Price List
Rice = Rs. 72.50 / kg
\ Total cost paid by Anisha Sugar = 90.25/kg
Tea = 445/kg
= Rs. 72.50 + 90.25 + Rs. 235.50 Daal = Rs. 235.50/ kg
72.50 Noodles= Rs. 22.75/piece
90.25 Dal milk = Rs. 816.25/ l

+ 235.50
398.25



\ Total cost = Rs. 398.25.

Exercise 4.3

1. Convert into paisa.

a. Rs. 46 and 37 paisa b. Rs. 103 and 63 paisa

c. Rs. 90 and 27 paisa d. Rs. 235.45

e. Rs. 434.26 f. Rs. 163.75

2. Convert into Rs.

a. 23400 Paisa b. 2765 Paisa

c. 34230 Paisa d. 57303 Paisa

e. 4426 Paisa f. 23426 Paisa

3. Find the sum

a. Rs. 3.47 + Rs. 43.25 b. Rs. 42.25 + Rs. 46.75

c. Rs. 259.50 + Rs. 439.26 d. Rs. 80.35 + Rs. 45.75

e. Rs. 34.15 + Rs. 145.30 f. Rs. 189.83 + Rs. 24.352

4. Sanu bought a pen for Rs. 24.50, a glue stick for Rs. 32.75 and marker

for Rs. 42.60. How much money did she spend? Price List
5. Mausami visited a super market to purchase
the goods. She forgot to take money but she Jacket = Rs. 1445.27
had ATM card. How much money should Paint = Rs. 946
she withdraw from ATM to purchase Jacket, T-shirt= Rs. 642.50
Bag = Rs. 764.75
a bag and a T-shirt?
6. Samira bought a birthday cake for Rs.1275.46. Sweator = Rs. 8563.50
Birthday cap for Rs. 125.50 and candle for Rs. Shoes = Rs. 235
40.50. How much money did she spend on Skirt = Rs. 557.25

her birthday?

Maths Zone - Grade 5 99

Subtraction of Money

Let's discuss few examples of subtraction of money.

Example 1

Ramesh had Rs. 1500, he gave Rs. 250 to Hari and bought a Cap for
Rs. 437.50. Find the money left with him.

Solution: 1500 4 10
Ramesh has = Rs. 1500
Ramesh gave to Hari = Rs. 250 1500 1500.00
Cost of Cap = Rs. 437.50 –250 – 2 5 0. 0 0
Money left = ? 1250 1 2 5 0. 0 0
\ Money left with him = Rs. 812.50 – 4 37 . 5 0
1250 812.50

250

1250.00 812.50
–437.50
812.50

437.50

Example 2

Parisha has a 500 rupee note. She bought a 'Math Zone' for Rs. 235.75.
How much money did she get back if she gave, a 500 rupee note to
shopkeeper?

Solution:

Money given = Rs. 500 4 9 9 9 10
Cost of book = Rs. 235.75
Cash back = ? 500.00
\ Parisha got Rs. 264.25. –235.75

264.25

100 Maths Zone - Grade 5