Math 4

Class Assignment

1. Circle the greater number. b. 64293 and 63165
a. 4682 and 37489 d. 9736412 and 9731237
c. 825643 and 827190 f. 30861213 and 47302811
e. 5943217 and 5943782

2. Put < or > sign in the box.

a. 3742 23481 b. 52604 53216
d. 946215 94621
c. 68432 68109 f. 621596 426313

e. 372154 372124

Ascending and descending order of numbers

Ascending order means arrangement of numbers from the smallest to the largest.
348296, 463587, 582432 are in ascending order.
It is also known as increasing order.
Descending order means arrangement of numbers from the greatest to the
smallest.
432562, 351639, 232425 are in descending order.
It is also known as decreasing order

Class Assignment

Arrange the numbers in the ascending order.
a. 5432685, 378614, 456249, ......................, ......................, .......................
b. 1623394, 2384239, 576482 ......................, ......................, ......................
Arrange the numbers in the descending order.
a. 6289543, 7624932, 5319621 ......................, ......................, ......................
b. 596312, 532463, 4815625 ......................, ......................, .........................

Oasis School Mathematics Book-4 51

Successor and predecessor of a number

The number that comes just after the given number is called successor of the

given number. Add 1 to the given number
236582 comes just after 236581 to get its successor.
\ 236582 is the successor of 236581

The number that comes just before the given number is the predecessor

of the given number. Subtract 1 from the given
841263 comes just before 841264 number to get its predecessor.
\ 841263 is the predecessor of 841264.

Class Assignment

1. Write the number which comes just before.

a. 238461 b. 653490 c. 8269542
d. 5932846
2. Write the number which comes just after. c. 4362450
c. 8396200
a. 859263 b. 9268374

3. Write the successor of:

a. 34292 b. 512399

4. Write the predecessor of: 428650
a. 372164 b.

Exercise 2.5

1. Put > or <.

a. 32731 65427 b. 284154 87321
d. 5264312 636851
c. 152762 242863 f. 8536410 437216

e. 5042645 4503201

2. Arrange the given numbers in an ascending order.

a. 538241, 230514, 824652 b. 4572663, 4877724, 929496

c. 34826351, 4267124, 12845732

52 Oasis School Mathematics Book-4

3. Arrange the given numbers in an descending order.

a. 6532041, 8542641, 5055605
b. 737775, 6563601, 52563
c. 8352417, 7382617, 73826187

4. Write the successor of the following numbers.

a. 53824 b. 846249 c. 954630
d. 5842763 e. 3742654 f. 4730429

5. Write the predecessor of the following numbers.

a. 834651 b. 5804021 c. 3273520
d. 8646052 e. 38573241 f. 60673840

Consult your teacher.

Number system

Natural number
The numbers 1, 2, 3,4, 5, …etc are the natural numbers. These numbers
are used for counting. So they are also called the counting numbers.

Whole number
The set of natural numbers including 0 are the whole numbers. 0, 1, 2, 3,
4, 5, ……etc are the whole numbers.

Remember !

• The smallest natural number is 1.
• The smallest whole number is 0.
• There is no largest natural number.
• There is no largest whole number.
• Every natural number is a whole number.

Odd and even numbers

Odd numbers are unpaired numbers. Odd numbers are not exactly divisible by 2.
1, 3, 5, 7, … etc are the odd numbers.
Even numbers are the paired numbers. Even numbers are exactly divisible by 2.
2, 4, 6, 8 ….. etc are the even numbers.

Oasis School Mathematics Book-4 53

Properties of odd and even numbers

1. Even numbers give remainder zero if they are divided by 2 and odd
numbers give remainder 1 when divided by 2.

8 ÷ 2, remainder = 0
9 ÷ 2, remainder = 1

2. The predecessor and successor of every even number is odd.

4 – 1 = 3 (Odd) 4 + 1 = 5 (Odd)

3. Odd number + odd number = even number

3 + 5 = 8 (Even), 7 + 5 = 12 (Even)

Note: Note
• If the digit in the place of one is even or 0, the number is even.
• If the digit in the place of one is odd, the number is odd.

25678 is even. 53621 is odd.

8 is even 1 is odd

4. Even number + Even number = Even number

8 + 4 = 12 (Even), 6 + 12 = 18 (Even)

5. Odd number + Even number = Odd number

7 + 6 = 13 (Odd), 9 + 8 = 17 (Odd)

6. Odd number × Odd number = Odd number

7 × 5 = 35 (Odd), 3 × 5 = 15 (Odd)

7. Even number × Even number = Even number

8 × 4 = 32 (Even), 6 × 8 = 48 (Even)

8. Odd number × Even number = Even number

9 × 6 = 54 (Even), 7 × 2 = 14 (Even)

54 Oasis School Mathematics Book-4

Class Assignment

1. Without actual mathematical operation, determine whether the result
of the given sum is odd or even.

a. 253 + 374 Odd b. 5684 + 2892

c. 53163 + 23727 d. 96842 – 1

e. 4537 × 32565 f. 2380 × 46218

g. 56123 × 2420

Prime and composite numbers

Prime number
2 is divisible only by 1 and 2 itself
2 ÷ 1 = 2 2 ÷ 2 = 1
3 is divisible only by 1 and 3 itself
3 ÷ 1 = 3 3 ÷ 3 = 1
5 is divisible only by 1 and 5 itself
5 ÷ 1 = 5 5 ÷ 5 = 1
\ 2, 3, 5 are prime numbers.
The number which is divisible by only 1 and itself is called a prime number.
So, 2, 3, 7, 11, 13, etc. are prime numbers.

Composite number
6 is divisible by 1, 2, 3, and 6 itself
6 ÷ 1 =6, 6 ÷ 2 = 3, 6 ÷ 3 = 2, 6 ÷ 6 = 1
8 is divisible by 1, 2, 4, 8 itself
8 ÷ 1 = 8, 8 ÷ 2 = 4, 8 ÷ 4 = 2, 8 ÷ 8 = 1
6, and 8 are composite numbers.

The number which is exactly divisible by other numbers except 1 and the number
itself is called a composite number.

Remember !

• 1 is neither a prime number nor a composite number.
• 2 is the only even number, which is prime.

Oasis School Mathematics Book-4 55

Prime and composite numbers from 1 to 50.
1 2 3 4 5 6 7 8 9 10
11 12 13 14 15 16 17 18 19 20
21 22 23 24 25 26 27 28 29 30
31 32 33 34 35 36 37 38 39 40
41 42 43 44 45 46 47 48 49 50

 Encircle 1
 Leaving 2, encircle all the numbers which are divisible by 2.
 Leaving 3, encircle all the numbers which are divisible by 3.
 Leaving 5, encircle all the numbers which are divisible by 5.
 Leaving 7, encircle all the numbers which are divisible by 7.

The numbers in the circle are composite numbers and the remaining
numbers are prime numbers.
Prime numbers from 1 to 50 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43
and 47.

Exercise 2.6

1. a. Write all the even numbers from 1 to 10.
b. Write all the odd numbers from 1 to 10.
c. Write all the even numbers from 11 to 20.
d. Write all the odd numbers from 10 to 20.
2. a. Write all the prime numbers from 1 to 30.
b. Write all the prime numbers from 31 to 50.
c. Write all the composite numbers from 1 to 30.
3. From the given numbers, separate the odd and even numbers:

45, 48, 96, 87, 34, 112, 115, 163, 196, 286, 343, 490, 187, 2314, 2562, 3763, 9711
4. From the given numbers, separate the prime and composite numbers:
20, 21, 30, 33, 31, 43, 17, 16, 20, 25, 36, 33, 39, 47

56 Oasis School Mathematics Book-4

5. Answer the following questions:
a. Which is the smallest natural number?
b. Which is the smallest whole number?
c. Are all the natural numbers whole numbers?
6. Write whether the given numbers are odd or even.
a. Sum of two even numbers
b. Sum of two odd numbers
c. Sum of an even and an odd numbers
d. Product of two even numbers
e. Product of two odd numbers

Consult your teacher.

Rounding off whole numbers

Rounding off means replacing the given number by another convenient
number which is easy to understand but close to the original number.
Rounding off to the nearest 10

20 21 22 23 24 25 26 27 28 29 30

Here 23 lies between 20 and 30. It is nearer to 20 than 30.
\ We round off 23 to 20.
Again, 26 is nearer to 30 than 20.
\ We round off 26 to 30.
Again, 25 is halfway between 20 and 30. We round off to the higher
number.
\ 25 rounded off nearest to ten is 30.

Remember !

To round off the number to the nearest 10
• Look at the digit in the ones place.
• If the digit at the ones place is 1, 2, 3, 4, then replace the ones digit by 0.
• If the digit at the ones place is 5, 6, 7, 8 and 9, then replace the ones

digit by 0 and increase the tens digit by 1.

Oasis School Mathematics Book-4 57

Rounding off to the nearest 100

200 210 220 230 240 250 260 270 280 290 300
Take a number 242.
Number 242 lies between 200 and 300.
242 is nearer to 200 than 300.
\ We round off 242 to 200.
Again, 278 lies nearer to 300 than 200.
\ We round off 278 to 300.
Again, 250 is halfway between 200 to 300. We round off 250 to higher number.
\ 250 rounded off nearest to hundred is 300.

Remember !
To round off the number to the nearest 100
• Look at the digit in the tens place.
• If the digit at tens place is 0, 1, 2, 3, 4, replace each of the tens and ones digit
by 0 and keep the other digit same.
• If the digit at tens place is 5, 6, 7, 8, and 9, then increase the hundreds digit
by 1 and replace each of the tens and ones digit by 0.

Class Assignment

1 Round off the given numbers to the nearest ten.

20 21 22 23 24 25 26 27 28 29 30

21 24 28

26 23 29
2. Round off the given numbers to the nearest hundred.

900 910 920 930 940 950 960 970 980 990 100

910 930 990

980 920 960

58 Oasis School Mathematics Book-4

Exercise 2.7

1. Observe the given number line and round off the circled numbers to
the nearest ten.
a.

50 51 52 53 54 55 56 57 58 59 60
b.

70 71 72 73 74 75 76 77 78 79 80
c.

30 31 32 33 34 35 36 37 38 39 40

2. Round off the following numbers to the nearest ten.

a. 32 b. 41 c. 28 d. 59 e. 382

f. 467 g. 835 h. 234 i. 221 j. 936

k. 484 l. 296

3. Observe the given number line and round off the circled numbers to
the nearest hundred.

a.

200 210 220 230 240 250 260 270 280 290 300
b.

500 510 520 530 540 550 560 570 580 590 600
c.

700 710 720 730 740 750 760 770 780 790 800

4. Round off the following numbers to the nearest hundred.

a. 226 b. 387 c. 524 d. 694 e. 836

f. 792 g. 1540 h. 1872 i. 2356 j. 3484

k. 3216 l. 4350 Consult your teacher.

Oasis School Mathematics Book-4 59

Prime factorisation 18

Take a number 18 and factorise it.
18 18

1 × 18 2 ×9 3×6

The factors of 18 are 1, 2, 3, 6, 9 and 18.
Among these factors, 2 and 3 are the prime numbers.
\ 2 and 3 are the prime factors.
Hence, a prime number which is the factor of a given number is called a prime
factor.

A factorisation in which every factor is prime is called prime
factorisation.

18 24

2 ×9 2 × 12

3× 3 2 ×6
\ 18 = 2 × 3 × 3
2× 3

\ 24 = 2 × 2 × 2 × 3
Here, 18 and 24 have been completely factorised.

\ They can be expressed as the products of prime factors.
18 = 2 × 3 × 3 prime factorisation of 18

24 = 2 × 2 × 2 × 3 prime factorisation of 24

Example 1: 12 12
Write the possible factors of 12.
Solution: 3× 4

12

1 × 12 2× 6

\ Possible factors of 12 are 1, 2, 3, 4, 6, 12.

60 Oasis School Mathematics Book-4

Example 2: Alternative method Steps:
Find the prime factors of 120.
Solution: 2 120 • Divide 120 by 2.
2 60 120 = 2 × 60
120 2 30 60 is a composite number.
3 15 • Divide 60 by 2
2 × 60 60 = 2 × 30
5 30 is a composite number
2 × 2 × 30 • Divide 30 by 2.
\ 120 = 2 × 2 × 2 30 = 2 × 15
2× 2× 2 × 15 5 ×3×5 15 is a composite number.
2 × 2 × 2× 3× • Divide 15 by 3.
15 = 3 × 5
• 5 is a prime number.

\ 120 = 2 × 2 × 2 × 3 × 5

Class Assignment

Complete the given factors tree.

72

×

××

××× 192

× × ×× ×
××

× ××

× × ××

× ××××

× ×××××

Oasis School Mathematics Book-4 61

Exercise 2.8

1. Write all the possible factors of the following numbers.
a. 9 b. 12 c. 18 d. 24 e. 32 f. 36 g. 48

2. Find the prime factors of the following numbers.

a. 48 b. 54 c. 72 d. 64 e. 68 f. 76 g. 84

h. 120 i. 150 j. 850 k. 750 l. 900 m. 936 n. 560

3. Complete the given factors tree. 80

24 2

2

Factors and multiples Consult your teacher.

Look and learn 12 is the multiple of 3 and 4. 3 and
We know that, 8 = 4 × 2 4 are the factors of 12.

multiple factor factor
Again, 12 = 3 × 4

multiple factor factor

Class Assignment

Write the factors and multiples of given multiplication facts.

Multiplication fact Factors Multiple
3 × 4 = 12
5 × 3 = 15
7 × 4 = 28
8 × 4 = 32
6 × 7 = 42
7 × 8 = 56
9 × 7 = 63

62 Oasis School Mathematics Book-4

The highest common factor (H.C.F.) and the lowest common multiple (L.C.M.)

Highest Common Factor

Let’s take any two numbers. Steps:

Say 12 and 16. • Write all the possible

The possible factors of 12 are 1, 2, 3, 4, 6, and 12 factors of given numbers.

The possible factors of 16 are 1, 2, 4, 8, and 16 • Select the common factors.

The common factors of 12 and 16 are 1, 2, and 4. • Select the largest common
The largest common factor among them is 4.
factor.

i.e. Highest Common Factor = 4.

\ H.C.F. of 12 and 16 is 4.

Example 1:

Write all the possible factors of 12 and 20.
Find the highest common factor:
Possible factors of 12 are 1, 2, 3, 4, 6, 12
Possible factors of 20 are 1, 2, 4, 5, 20.
Common factors are 1, 2 and 4
Highest common factor = 4
\ H.C. F. of 12 and 20 is 4.

Process of finding H.C.F.

H.C.F. of two or more than two numbers can be obtained by prime factorisation method.
Let’s see an example.

Example 2: Steps:

Example: Find the H.C.F. of 18 and 24. • Break down each of the numbers
into its prime factors.
Solution: 2 18 2 24
• Take out the common prime
3 9 2 12 factors.

Here, 3 2 6 • Multiply all the common factors
3 which is the H.C.F. of given
18 = 2 × 3 × 3 numbers.

24 = 2 × 2 × 2 × 3 H.C.F. is the product of all
\ H.C.F. = 2 × 3 = 6 common prime factors.

Oasis School Mathematics Book-4 63

Exercise 2.9

1. Write all the possible factors of the given numbers and find the highest

common factor:

a. 9 and 12 b. 10 and 15 c. 18 and 27

d. 12 and 18 e. 12 and 16 f. 24 and 35

2. Find the H.C.F. of the following numbers by prime factorisation method:

a. 10 and 15 b. 8 and 12 c. 12 and 20 d. 18 and 24
e. 21 and 28 f. 24 and 36 g. 27 and 36 h. 40 and 50
i. 9, 12 and 18 j. 12, 18 and 24 k. 16, 20 and 40

1. a. 3 b. 5 c. 9 d. 6 e. 4 f. 1
2. a. 5 b. 4 c. 4 d. 6 e. 7 f. 12 g. 9 h. 10 i. 3 j. 6 k. 4

Lowest common multiple (L.C.M.)

Let’s take any two numbers, say 12 and 9.
Some multiples of12 are 12, 24, 36, 48, 60, 72 ……
Some multiples of 9 are 9, 18, 27, 36, 45, 54, 63, 72, …..
Common multiples of 12 and 9 are 36, 72 ..... etc.
Lowest common multiple is 36.
\ L.C.M. of 12 and 9 is 36.

Steps:
• Write some multiples of the given numbers.
• Select the common multiples.
• Select the lowest (smallest) multiple.

Process of finding L.C.M.
L.C.M. of two or more than two numbers can be obtained by prime factorisation method:
Let's see an example and get the idea to find the L.C.M.

64 Oasis School Mathematics Book-4

Example 1: Steps:
• Break down each of the numbers
Find the L.C.M. of 18 and 24.
into their prime factors.
Solution: • Take out the common prime

2 18 2 24 factors.
• Take the remaining factors which
3 9 2 12
are not common.
3 2 6 • Multiply the factors of steps (ii)

18 = 2 × 3 × 3 3 and (iii) to get L.C.M.

24 = 2 × 2 × 2 × 3 L.C.M.
=2×3×3×2×2
\ L.C.M. = 2 × 3 × 3 × 2 × 2
common factors remaining factors
= 72

Exercise 2.10

1. Write some multiples of the given numbers and find their L.C.M.

a. 2, 3 b. 4, 5 c. 6, 8 d. 3, 6

e. 5, 10 f. 3, 7 g. 6, 9 h. 8, 10

2. Find the L.C.M. of the following numbers using prime factorisation

method.

a. 4 and 5 b. 8 and 10 c. 6 and 18 d. 12 and 18

e. 18 and 24 f. 10 and 30 g. 10 and 15 h. 15 and 20

i. 5, 10 and 15 j. 8, 12 and 20 k. 6, 18 and 27

1. a. 6 b. 20 c. 24 d. 6 e. 10 f. 21 g. 18 h. 40
2. a. 20 b. 40 c. 18 d. 36 e. 72 f. 30 g. 30 h. 60 i. 30 j. 120 k. 54

Activity

L.C.M by using the number line:
Find the L.C.M of 3 and 4.

Oasis School Mathematics Book-4 65

- Jump by 3 on the top of the number line. LCM
- Jump by 4 on the bottom of the number line.
The first place, where the two different jumps meet is 12.
\ L.C.M. of 3 and 4 is 12.

Using the given number line, find the L.C.M. of:
(i) 3 and 5

(ii) 4 and 6 LCM

(iii) 2 and 7 LCM

To the teacher:
Ask students to find the L.C.M. of other numbers using the number line.

Objective Questions

Choose the correct alternatives:

1. The number two crore four lakhs four thousand and four is

204404 2044004 20404004

2. The smallest number formed by the digits 2, 3, 8, 0, 1, 5 is

123580 102358 012358

3. The largest number of 6 digits is

999999 100000 10000000

4. Which of the following statement is not true?

10 lakh is equal to 10 million is equal 1 million is equal
1 million. to 1 crore. to 1 crore.

5. Roman numeral for 66 is

LXVI LXIV LXV

66 Oasis School Mathematics Book-4

6. One more than the greatest number of five digits is equal to

The smallest The greatest The smallest
number of five number of six number of six

digits. digits. digits.

7. Which of the following statements is not true.

Product of two odd Sum of two odd Product of an odd and
numbers is an even numbers is an even even number is an even

number. number. number.

8. Rounding off the number 347 to its nearest hundred is

350 400 300

9. Which of the following statements is not true?

The smallest The smallest Every natural
natural number whole number number is a whole

is 0. is 0. number.

10. H.C.F. of the numbers 12 and 18 is

6 3 12

Worksheet

Using the given colour code colour the figures below.

• The smallest natural number (Red)

• The largest number of 6 digits (Green)

• The smallest number of 4 digits (Yellow)

• The number XC (Blue)

• Only one prime number which is even (Skyblue)

• Rounding 584 to its nearest ten (Pink)

• H.C.F. of 8 and 12 (Brown)

• L.C.M of 12 and 18 (Purple)

Oasis School Mathematics Book-4 67

36 90 1000
2
4
999999

1 580

Unit Test Full marks -20

1. Put comma, using both Nepali and international system of numeration

and write their name according to both system. 2

a. 5426013 b. 2001506

2. Arrange the following numbers in an ascending order. 1
23615, 238102, 230051

3. Arrange the following numbers in an decreasing order. 1
5400321, 5000215, 5421005

4. Round off the following number. 2
a. 54 (to its nearest ten)
b. 584 (to its nearest hundred)

5. Find the prime factors using factor tree. 4
a. 24 b. 32

6. Write all the possible factors of: 4
a. 12 b. 18

7. Find the H.C.F. of the following. 3
a. 21 and 28 b. 27 and 36 c. 9, 12 and 18

8. Find the L.C.M. of the following numbers using prime factorisation

method: 3

a. 10 and 15 b. 8 and 12 c. 6, 12 and 18

68 Oasis School Mathematics Book-4

UNIT Four Fundamental
Operations
3

12 Estimated Teaching Hours: 15
93

6

Contents Addition
• Addition of large numbers
• Simple verbal problems on addition
Subtraction
• Subtraction of large numbers
• Simple verbal problems on subtraction
Multiplication
• Multiplication of upto 5-digit numbers with upto

3-digit numbers
• Simple verbal problems on multiplication
Division
• Dividend, Divisor, Quotient and Remainder
• Division of large numbers by two and three

digits number
Simplification

Expected Learning Outcomes

Upon completion of the unit, students will be able
to develop the following competencies:

• To add the numbers up to 7-digits with and without carry over
• To solve the simple verbal problems on addition
• To subtract the number up to 7-digits with borrowing and without

borrowing
• To solve the simple verbal problems on subtraction
• To multiply the numbers up to 5-digits by single digit, two digit and

three digit number
• To solve the simple verbal problems on multiplication.
• To identify dividend, divisor, quotient and remainder and know their

relation
• To divide the numbers up to 5-digits by one digit and two digit numbers
• To solve the simple verbal problems on division
• To simplify the mathematical expression using DMAS and BODMAS rule

Materials Required: Addition table, multiplication table, division table, etc.

Oasis School Mathematics Book-4 69

Addition (Review) Remember !

Look and learn. • 10 ones = 1 ten
Number combinations of 10. • 10 tens = 1 hundred
10 = 10+0 10 = 9 + 1 10 = 8 + 2 10 = 7 + 3 • 10 hundreds = 1 thousand
10 = 6 + 4 10 = 5 + 5 10 = 4 + 6 10 = 3 + 7 • 15 ones = 1 tens + 5 ones
• 25 tens = 2 hundreds + 5 ones
Number combinations of 11.
11 = 11 + 0 11 = 10 + 1 11 = 9 + 2 11 = 8 + 3
11 = 7 + 4 11 = 6 + 5 11 = 5 + 6 11 = 4 + 7

Class Assignment

1. Write all the number combinations of 12. I learnt all this in

11 + ................. 7 + ................. 8 + ................. class III

12 + ................. 12 9 + .................

6 + ................. 5 + ................. 10 + .................

2. Write all the number combinations of 13.

9 + ................. 10 + ................. 11 + .................

8 + ................. 13 12 + .................

7 + ................. 6 + ................. 5 + .................

3. Complete as shown.
27 ones = 2 tens + 7 ones
32 ones = ........... tens + ........... ones
54 ones = ........... tens + ........... ones
65 ones = ........... tens + ........... ones
18 tens = ........... hundreds + ........... tens
26 tens = ........... hundreds + ........... tens
35 tens = ........... hundreds + ........... tens
42 tens = ........... hundreds + ........... tens

70 Oasis School Mathematics Book-4

Exercise 3.1

1. Add ones and convert into tens and ones.

a. 9 ones + 7 ones b. 6 ones + 8 ones + 9 ones

c. 8 ones + 7 ones + 6 ones d. 9 ones + 8 ones + 7 ones

e. 5 ones + 4 ones + 9 ones f. 4 ones + 6 ones + 8 ones

2. Add tens and convert into hundreds and tens.

a. 7 tens + 8 tens b. 6 tens + 7 tens

c. 5 tens + 4 tens + 8 tens d. 9 tens + 8 tens + 7 tens

e. 5 tens + 7 tens + 8 tens f. 4 tens + 9 tens + 8 tens

3. Add:

a. 3 4 7 2 b. 5 7 6 4 c. 6 3 2 7 d. 3 2 5 6 0

+ 1 2 2 4 + 1 0 3 4 + 1 6 5 2 + 1 7 3 2 4

e. 5 7 3 0 4 f. 3 2 6 3 4 g. 3 7 2 8 4 h. 7 0 1 8 2
+ 1 2 6 8 5 + 6 4 3 1 5 + 4 2 5 1 5 + 2 8 3 0 6

Consult your teacher.

Addition with carry over

Addition of large numbers is the same as the addition of smaller numbers. Start
addition from ones place and regroup into the higher place taking carry over.

Example 1: Steps:

Add: 43628 + 27197 • Add ones: 8 + 7 = 15 ones = 1 ten + 5
Solution: ones, carry over 1 ten

1 1 1 Carry overs • Add tens: 1 + 2 + 9 = 12 tens = 1
43628 hundreds + 2 tens, carry over 1
+ 2 7 1 9 7 hundred
70825
• Add hundreds: 1 + 6 + 1 = 8 hundreds

• Add thousands : 3 + 7 = 10 thousands
= 1 ten thousands, carry over 1 ten
thousand

• Add ten thousands : (4 + 2 + 1) = 7 ten
thousands

Oasis School Mathematics Book-4 71

Example 2:

Add: 46879 + 37496 + 53642 Steps:
• 9 + 6 + 2 = 17 ones = 1 tens + 7 ones,
Solution:
1221 carry over 1 tens and follow the
Carry overs same process with other place value
also
46879
Steps:
+ 3 7 4 9 6 • Write the digits at their

+ 5 3 6 4 2 correct place value.

138017

Example 3:
Add: 468352 + 76942 + 6587 + 648
Solution:

1 2 2 2 1 Carry overs
4 6 83 5 2
+ 7 6 9 4 2
+ 6 5 8 7
+ 6 4 8
55 2 5 2 9

Exercise 3.2 b. 6 3 7 5 c. 5 6 8 7
+ 1 8 9 8 + 1 7 6 5
1. Add the following:
a. 2 8 7 6 e. 3 2 6 9 f. 3 7 8 6
+ 1 4 8 9 + 3 7 5 6 + 5 6 9 7
d. 3 2 6 8
+ 5 7 9 6 b. 8 7 6 3 5 c . 6 4 5 6 8
+ 3 6 8 4 7 + 2 8 7 5 7
2. Add the following: f. 5 7 6 4 8 5
a. 5 3 8 4 6 e. 4 7 9 5 6 4
+ 3 7 6 4 7 + 2 8 4 8 6 9 + 3 7 6 8 4 7
i. 4 5 9 4 3 2 6
d. 5 6 8 7 6 4 h. 5 8 9 4 3 2 7
+ 3 7 9 8 7 6 + 3 9 3 8 7 2 6 + 1 7 3 7 6 8 7

g. 7 6 8 2 6 5 3
+ 2 7 4 8 7 6 8

72 Oasis School Mathematics Book-4

3. Add: b. 2 4 8 3 7 6 c. 7 8 6 4 5 3
a. 2 6 8 4 7 5 + 5 7 0 8 9 + 6 4 2 8 7 6
+17863 + 6 8 4 6 + 5 4 6 2 7
+ 5643 + 6 4 5 4

d. 7 4 6 4 3 6 6 e. 2 4 8 3 7 6 f. 5 6 1 2 7 8 3

+ 5 4 3 3 3 8 4 + 2 6 5 4 9 2 + 5 6 3 8 1

+ 3 4 4 5 3 6 + 3 7 1 2 3 5 + 4 3 7 0

+ 6 35 3 7 + 1 2 3 2 5 2 + 6 6

4. Add: + 9

a. 4638 + 5342 + 6543 b. 47645 + 67342 + 56846

c. 24325 + 37842 + 57324 + 6781

d. 645724 + 153812 + 23423 + 345 + 7

e. 3254617 + 372142 + 36421 + 672 + 32

5. Write the missing digits: b. 3 5 6 1
a. 4 6 8 4 3 2 1 + 3 2 3
7 8 9 4 8 6 5
+ 5 2

9 8 8 7 6 3 5

Consult your teacher.

Verbal problems in addition

Look at the given examples properly and get the idea of solving verbal
problems related to addition.

Example 1:

There are 286419 men, 295384 women and 236418 children in a town. Find the
total population of the town.

Solution: 2 1 1 1 2

Number of men = 286419

Number of women = 2 9 5 3 8 4

Number of children = 2 3 6 4 1 8

Total population of the town = 8 1 8 2 2 1

\ Total population of the town = 818221.

Oasis School Mathematics Book-4 73

Example 2:

The monthly expenses of a household is as follows. Food: Rs 4767, clothes:

Rs. 6586 and miscellaneous: Rs. 3284. What is the total monthly

expenditure?

Solution:

Monthly expenses on, 1 2 1


Food = Rs 4 7 6 7

Cloths = + Rs 6 5 8 6

Miscellaneous = + Rs 3 2 8 4

Total = Rs 1 4 6 3 7

\ Monthly expenditure of the household is Rs 14637.

Exercise 3.3

1. The population of four towns is 463725, 326584, 131236 and 126541
respectively. Find the total population of the four towns.

2. Ankit bought a motorcycle for Rs. 1, 50, 640 and he spent Rs. 43,645 for
repair. How much money did he spend altogether?

3. A company deposited Rs. 4,37,834 on Sunday, Rs 3,28,614 on Monday
and Rs. 1,32,614 on Tuesday. Find the total money deposited by the
company.

4. A publication house published 45,374 Maths books, 65,486 Science books
and 36,839 English books. Find the total number of books published by
the publication.

5. A man has a piece of land costing Rs. 34,56,372 and a house costing
Rs. 4,38,56,423 and cash Rs. 2,68,49,623 in the bank. Find his total
property.

6. A fruit seller has 3,26,426 oranges, 53,245 mangoes and 4,362 apples. Find
the total number of fruits he has.

Consult your teacher.

Subtraction

Subtraction is the operation of finding the differences between two numbers.
Subtraction of large numbers is same as the subtraction of smaller numbers.

Subtraction without borrowing
Start subtraction from ones place, subtract ones, tens, hundreds and so on.

74 Oasis School Mathematics Book-4

Example : Steps:

Subtract: • Subtract ones: 8 – 4 = 4
3578 • Subtract tens: 7 – 1 = 6
• Subtract hundreds: 5 – 3 = 2
- 2 3 1 4 • Subtract thousands: 3 – 2 = 1

1264

Exercise 3.4

1. Subtract: b. 6 3 2 1 4 c. 4 6 1 5 2
- 2 1 1 0 3 - 2 1 0 4 2
a. 5 3 6 4
- 2 0 3 2

d. 5 0 2 4 1 e. 5 6 2 3 1 f. 4 6 3 7 2
- 2 0 1 1 0 - 3 2 1 1 0 - 1 2 2 1 0

g. 3 6 8 4 2 1 h. 5 7 6 4 2 6 i. 3 8 6 8 0 4
- 2 5 3 1 1 0 - 1 2 3 4 1 5 - 2 0 5 6 0 1

2. Subtract: b. 678421 – 167110
a. 546217 – 321006 d. 477366 – 122123
c. 486439 – 322318

Consult your teacher.

Subtraction of large numbers (with borrowing)

Review: regrouping of numbers: My logic!
6 tens + 2 ones = (6-1) tens + 1 tens + 2 ones
= 5 tens + 12 ones Reduce tens by 1 and
increase ones by 10.

7 tens + 4 ones = (7-1) tens + ( 1 tens + 4 ones)

= 6 ones + 14 tens

Again, regrouping of tens and hundreds

3 hundreds + 6 tens = (3 – 1) hundreds + 1 hundreds + 6 tens

= 2 hundreds + 16 tens I have to reduce
5 hundreds + 4 tens = (5 – 1) hundreds + 4 tens hundreds by one and
= 4 hundreds + 14 tens increase tens by 10.

Oasis School Mathematics Book-4 75

Class Assignment b. 5 tens + 7 ones =
d. 3 tens + 5 ones =
Regroup as above: f. 9 tens + 2 ones =
h. 6 tens + 5 ones =
a. 4 tens + 3 ones =
b. 5 hundreds + 4 tens =
c. 6 tens + 8 ones = d. 4 hundreds + 3 tens =
f. 6 hundreds + 5 tens =
e. 8 tens + 2 ones =

g. 6 tens + 5 ones =

Regroup as above:
a. 6 hundreds + 3 tens =

c. 7 hundreds + 8 tens =

e. 8 hundreds + 6 tens =
Remember !
• 1 ten = 10 ones
• 1 hundred = 10 tens
• 1 thousand = 10 hundreds

Example :

Subtract: 635846 from 846312 Steps:
• 2 < 6. 6 cannot be subtracted from 2, regroup tens
Solution: 5 12 10 • 1 tens is borrowed from 1 tens
• 1 tens + 2 ones = 12 ones
846312 • Subtract ones, 12 – 6 = 6 ones

- 635846 Steps:
In ones,
2 1 0 4 6 6 • 0 < 4
• Regroup tens
Example : • Borrow 1 ten from tens
Subtract: 5 0 0 0 0 0 • 1 ten = 10 ones
- 3 6 8 9 4 2 • 10 ones – 2 ones = 6 ones
Solution: • Similarly, subtract tens and hundreds by

9 9 9 9 regrouping if necessary.
4 10 10 10 10 10
5 0 0 0 0 0
- 3 6 8 9 4 2
1 3 1 0 5 8

76 Oasis School Mathematics Book-4

Exercise 3.5

1. Subtractthefollowing:

a. 4 3 2 6 5 1 b. 3 5 7 4 1 2 c. 4 7 8 2 6 1
- 3 7 1 5 8 4
- 2 1 4 3 6 4 - 2 6 8 3 7 6

d. 5 2 8 5 2 0 e. 4 6 8 1 7 4 2 f. 6 3 2 1 2 6 4
- 4 1 9 6 7 4 - 2 3 1 6 5 1 7 - 1 2 5 6 1 5 9

2. Subtract: b. 5 2 8 5 7 4 c. 3 2 2 9 4 6
a. 3 6 7 8 4 2 - 1 8 6 4 8 - 7 4 3 8 8

- 1 9 3 7 6

d. 6 3 7 4 2 8 1 e. 2 6 8 4 2 1 5 f. 6 3 2 1 2 6 4
- 2 1 4 9 3 - 7 8 9 3 - 1 2 5 6 1 5 9

3. Subtract:

a. 2684372 - 1532169 b. 3219564 - 1204337

c. 526486 - 231874 d. 689462 - 493745

e. 856942 - 7869 f. 978230 - 16294

4. Subtract:

a. 5 0 0 0 0 0 b. 4 0 0 0 0 c. 3 0 0 0 0 0
- 3 7 4 6 2 7 - 1 2 7 3 8 - 1 7 4 8 2 6

d. 7 0 0 0 e. 8 0 0 0 0 0 f. 9 0 0 0 0 0
- 5 6 8 4 - 9 6 9 2 - 2 3 7 4 3 2

Consult your teacher.

Alternative method in subtraction with zero:
500000
- 2 4 8 6 9 6

5 0 0 0 0 0 - 1 = 4 9 9 9 9 9 Reduce both the
2 4 8 6 9 6 - 1 = 2 4 8 6 9 5 numbers by 1 and
Now, subtract.

499999
– 248695

251304

Oasis School Mathematics Book-4 77

Class Assignment

Reduce both the numbers by 1 and subtract:

a. 4 0 0 0 0 b. 3 0 0 0 0 0
- 2 3 6 4 1 - 2 3 1 4 5 6

c. 2 0 0 0 0 0 d. 8 0 0 0 0
- 4 2 3 6 4 - 2 7 4 6 8

Verbal problems in subtraction

Look at these problems properly and get the idea of solving the verbal
problems related to subtraction.

Example 1 :
What should be added to 457649 to make 653480?
Solution:

5 14 12 14 7 10

6 5 3 4 8 0
– 4 5 7 6 4 9
1 9 5 8 3 1

\ 195831 should be added to 457649 to make 653480.

78 Oasis School Mathematics Book-4

Exercise 3.6

1. a. Find the difference between 268413 and 164965.
b. Find the difference between 694870 and 281765.
2. a. What should be added to 4674068 to make 7264897?
b. What should be added to 3780543 to make 4309461?
c. What should be added to 54093 to make 217416?
3. a. By how much is 27924 greater than 18392?
b. By how much is 456927 greater than 82624?
4. a. By how much is 729260 is less than 1782281?
b. By how much is 265894 less than 459618?
5. a. Swapnil earned Rs. 79,850 in a year. He spent Rs. 54,645

throughout the year. How much money did he save that year?
b. Ananya had Rs. 2,75,000. She bought a scooter for Rs. 1,60,350.

How much money is left with her?
c. Umanga deposited Rs. 3,75,340 in a bank. He withdrew Rs.

1,97,585 from the bank. How much money is left in his account?
d. A man wants to buy a car worth Rs. 12,65,000. He has only

8,97,350. How much more money he needs to buy the car?
e. Swastika has Rs. 75,670. She gave Rs. 57,980 to her friends.

How much money is left with her?
6. a. The population of a town is 85,335. There are 25,642 men,

27,376 women and the remaining are children. How many
children are there in the town?
b. Ram has Rs. 95,000. He gave Rs. 36,345 to his son and Rs 25,875
to his daughter and the rest to his wife. How much money did
he give to his wife?

Consult your teacher.

Oasis School Mathematics Book-4 79

Multiplication

Review
Multiplication is the repeated addition
5 × 2 = five 2’s = 2 + 2 + 2 + 2 + 2
6 × 4 = six 4’s = 4 + 4 + 4 + 4 + 4 + 4
3 × 7 = three 7’s = 7 + 7 + 7

Class Assignment

1. Complete as above:

4 × 6 = four 6’s = + + +

5 × 6 = ………. = + + + +

3 × 8 = ….…… = + +

6 × 3 = ………. = + + + + +

4 × 7 = ………. = + + +

5 × 3 = ………. = + + + +

6 × 4 = ………. = + + + + +

2. Multiply: I know the
multiplication table, I

can do it.

7×4= 8×6 = 7×7= 4×8=
9×3= 9×6=
9×7= 9×4= 7×6= 5×3=
5×2= 4×6=
8×4= 8×5=

5×7= 5×5=

Teacher's Signature

80 Oasis School Mathematics Book-4

Class Assignment

1. Multiply: 4 × 0 = …....… 0 × 2 = …....… 0 × 6 = ....….…
7 × 0 = …..….. 8 × 0 = ..…..… 0 × 11 = …....…
3 × 0 = …..…..
0 × 10 = …..…..

2. Identify multiplicand, multiplier and the product

12 × 5 = 60 Multiplicand = …………… Multiplier = ……………
Product = …………… Multiplier = ……………
6 × 7 = 42 Multiplicand = …………… Multiplier = ……………
Product = ……………
8 × 6 = 48 Multiplicand = …………… Teacher's Signature
Product = ……………

Multiplication of a number by zero (0)
Review

Any number × 0 = 0 In a multiplication, if the
0 × any number = 0 order of numbers is changed
Multiplicand, multiplier and product the product remains the
Look at this multiplication fact same.

5 × 6 = 30 i.e. 2 × 6 = 6 × 2

multiplicand multiplier product 8 × 5 = 5 × 8, 7 × 6 = 6 × 7 and
so on.
Multiplication of a number by 10, 20, 30, 40 ….. etc.

36 × 10 = 360. Multiply 36 and 1, 36 × 1 = 36. Write one zero right to the product.

36 × 20 = 720. Multiply 36 and 2, 36 × 2 = 72. Write one zero right to the product.

Multiplication of a number by 100, 200, 300, 400 etc.
56 × 100 = 5600. Multiply 56 and 1. 56 × 1 = 56. Write two zeros right to the product.
56 × 200 = 11200. Multiply 56 and 2. 56 × 2 = 112. Write 2 zeros right to the product.

Multiplication of a number by 1000, 2000, 3000 etc.
24 × 1000 = 24000. Multiply 24 and 1, 24 × 1 = 24. Write three zeros right to the product.
24 × 2000 = 48000. Multiply 24 and 2, 24 × 2 = 48. Write 3 zeros right to the product.

Oasis School Mathematics Book-4 81

Class Assignment

1. Multiply: 32 × 30 = I learnt all this in class
65 × 20 = 92 × 40 = III
86 × 30 = 72 × 60 =
63 × 50 =

2. Multiply: 82 × 200 =
76 × 100 = 21 × 400 =
75 × 300 = 48 × 500 =
32 × 400 =

3. Multiply: 52 × 2000 =
23 × 1000 = 84 × 4000 =
76 × 3000 =

Multiplication of a 3-digit number by a one digit number

Look at this example. 22 Multiply ones, tens and
hundreds by 4.
265

×4

1060

Exercise 3.7(A)

1. Multiply the following:

a. 3 2 4 b. 4 6 8 c. 5 2 5 d. 6 3 8 e. 4 3 7
×4
×2 ×3 ×4 ×6

f. 2 8 6 g. 6 4 3 h. 5 0 2 Consult your teacher.
×7 ×4 ×6

Multiplication of a two-digit number by a two-digit number

Look at the given example and get the idea of multiplication of a two-digit
number by a two-digit number.

45 Multiply 45 with 6 ones: 45 × 6 = 270
×26 Multiply 45 with 2 tens i.e. by 20: 45 × 20 = 900
270 Add: 270 + 900 = 1170
+ 9 0 0

1170

82 Oasis School Mathematics Book-4

Exercise 3.7(B)

1. Multiply the following:

a. 5 2 b. 2 7 c. 6 5 d. 7 5
×24 ×34
× 3 6 ×32
g. 9 3 h. 8 6
e. 8 4 f. 7 6 ×17 ×13
×36 ×24

1. a. 1872 b. 864 c. 1560 d. 2550 e. 3024 f. 1824 g. 1581 h. 1118

Multiplication of large numbers

We have learnt multiplication of 3-digit numbers by 2-digit numbers. Now,
we shall study the multiplication of more than 3 digit numbers by 3-digit or
4 digit numbers.

Example : In short Multiplication by 2 ones
Multiply: 3275 Multiplication by 3 tens
Multiplication by 4 hundreds
3275 ×432
×432 6550 Sum

Solution: 98250
432 = 400 + 30 + 2 1310000
3275
× 2 Multiply 3275 1414800
6 5 5 0 by 2 ones
3275 Add all the products
× 3 0 Multiply 3275 6550
9 8 2 5 0 by 30 98250
1310000
3275 1414800
× 4 0 0 Multiply 3275
1 3 1 0 0 0 0 by 400 \ 3275 × 432 = 1414800.

Oasis School Mathematics Book-4 83

Exercise 3.8

1. Multiply the following:

a. 2 4 5 b. 3 6 4 c. 9 8 2 d. 4 8 6
×32 ×45 ×72 ×63

e. 6 2 4 f. 7 8 6 g. 6 4 5 h. 8 4 5
×48 ×52 ×63 ×26

2. Multiply the following:

a. 2 4 6 8 b. 8 6 1 2 c. 9 7 3 2 d. 2 5 8 6
×36 ×74
×35 ×46
g. 3 2 6 5 h. 4 8 7 5
e. 1 7 8 4 f. 2 1 6 5 ×263 ×123
×342 ×252
c. 7 3 5 6 4 d. 5 2 9 3 5
3. Multiply: b. 5 3 6 4 2 ×26 ×76
a. 2 3 1 6 4 ×56
×82 g. 3 0 1 6 5 h. 5 9 6 2 0
×632 ×409
e. 3 7 5 6 9 f. 4 7 6 0 4
×283 ×512 c. 8063 × 235 d. 7608 × 432
g. 97124 × 37 h. 86034 × 54
4. Find the product:
a. 3726 × 45 b. 1053 × 49
e. 15034 × 23 f. 27046 × 62
i. 54123 × 254 j. 37205 × 362

1. a. 7840 b. 16380 c. 70704 d. 30618 e. 29952 f. 40872
g. 40635 h. 21970
2. a. 86380 b. 396152 c. 350352 d. 191364 e. 610128 f. 545580
g. 858695 h. 599625
3. a. 1899448 b. 3003952 c. 1912664 d. 4023060 e. 10632027 f. 24373248
g. 19064280 h. 24384580 4. a. 167670 b. 51597 c. 1894805 d. 3286656
e. 345782 f. 1676852 g. 3593588 h. 4645836 i. 1377242 i. 13468210

84 Oasis School Mathematics Book-4

Quick method of multiplication:

46 × 103 46 × 100 = 4600 • Split 103
46 × (100 + 3) 46 × 3 = 138 100 + 3
Add: 4738 • Multiply both by 46
• Add the products
Multiply as above:
• Split 29
a. 52 × 101 b. 97 × 102 c. 47 × 105 30 – 1
• Multiply both by 45
Again, • Subtract the products

29 × 45 30 × 45 = 1350
(30 - 1) × 45 45 × 1 = – 45
Subtract = 1305

Multiply as above:

a. 38 × 56 b. 18 × 27 c. 49 × 76

Verbal problems in multiplication

Look at the given example and get the idea of solving verbal problems in
multiplication.

Example :

There are 264 apples in a box. How many apples are there in 3465 such boxes?
Solution:
Number of apples in one box = 264

Number of apples in 3465 boxes = 3465 × 264

Here, 3465


× 2 6 4

13860

207900

+693000

914760

\ There are 914760 apples in 3465 boxes.

Oasis School Mathematics Book-4 85

Exercise 3.9

1. There are 24 hours in a day. How many hours are there in 365 days?
2. A radio costs Rs. 986. Find the cost of 72 radios.
3. There are 135 students in class IV. Each student collects Rs. 86 for their

picnic. How much money will be collected for the picnic?
4. The cost of a refrigerator is Rs. 25645. What is the cost of 32 refrigerators?
5. Garima walks 3265 m in an hour. How long distance will she cover in 243 hours?
6. There are 144 pencils in a packet. How many pencils are there in 2364 such

packets?
7. A dealer bought 6437m of cloth at a rate of Rs. 245 per metre. How much

did the dealer pay for it altogether?

1. 8760 hrs 2. Rs 70992 3. Rs 11610 4. Rs 820640 5. 793395 hrs
6. 340416 pencils 7. Rs 15,77,065

Division

Review
Division is the repeated subtraction of same number
Let's be clear from this example, 15 ÷ 3

15 - 3 = 12 First subtraction Subtracting 3, 5 times
12 – 3 = 9 Second subtraction from 15 gives 0. It means
9 – 3 = 6 Third subtraction 15 ÷ 3 = 5
6 – 3 = 3 Fourth subtraction
3 - 3 = 0 Fifth subtraction
\ 15 ÷ 3 = 5

86 Oasis School Mathematics Book-4

Class Assignment

Fill in the blanks:

a. 20 – 5 = 15 b. 21 – 7 = 14 c. 24 – 6 = 18 d. 30 – 5 = 25
18 – 6 = 12 25 – 5 = 20
15 – 5 = 10 14 – 7 = 7 12 – 6 = 6 20 – 5 = 15
6–6=0 15 – 5 = 10
10 – 5 = 5 7–7=0 ∴ ÷ = 10 – 5 = 5
5–5=0
5–5=0 ∴ 21 ÷ 7 = ∴ ÷ =

∴ 20 ÷ 5 =

Division is the inverse operation of multiplication: Teacher's Signature
5 × 6 = 30 30 ÷ 6 = 5 7 × 8 = 56 56 ÷ 7 = 8
30 ÷ 5 = 6 56 ÷ 8 = 7 A multiplication
fact gives two
division facts.

Class Assignment

Complete as above: 8 × 6 = 48 …...... ÷ …...... = …......
3 × 7 = 21 …2.1..... ÷ …7...... = …3 ...... …...... ÷ …...... = …......
…2.1..... ÷ …3...... = …7 ...... 9 × 8 = 72 …...... ÷ …...... = …......
4 × 5 = 20 …...... ÷ …...... = …...... …...... ÷ …...... = …......
…...... ÷ …...... = …......
6 × 7 = 42 …...... ÷ …...... = …......

…...... ÷ …...... = …...... Teacher's Signature

Some important facts on division:

When a number is divided by 1, the quotient is the We cannot divide a
number itself: number by zero.
5 ÷ 1 = 5, 6 ÷ 1 = 6, 12 ÷ 1 = 12, 95 ÷ 1 = 95
When a number is divided by itself, the quotient is 1.
15 ÷ 15 = 1, 5 ÷ 5 = 1, 10 ÷ 10 = 1, 65 ÷ 65 = 1
When zero is divided by any number, the quotient is 0.
0 ÷ 1 = 0, 0 ÷ 5 = 0, 0 ÷ 10 = 0, 0 ÷ 20 = 0

Oasis School Mathematics Book-4 87

Class Assignment

1. Divide:

6÷1= 5÷1= 20 ÷ 1 = 0÷6= 0÷7=

0 ÷ 100 = 6÷6= 16 ÷ 16 = 25 ÷ 25 =

Dividend, Divisor, Quotient and Remainder Teacher's Signature

In the division of numbers,

The number which is divided is the “dividend”.

The number by which a number is divided is the “divisor”.

The answer obtained after division is “quotient”.

The number left after the complete process of division is “remainder”.

30 ÷ 6 = 5 Divisor Dividend Quotient
30 ÷ 6 = 5 Reminder
6)34(5
Dividend Divisor Quotient -30

4

Class Assignment Divisor Quotient

Complete the given table:
Dividend

12 ÷ 4 = 3
40 ÷ 5 = 8
56 ÷ 7 = 8
72 ÷ 8 = 9

Teacher's Signature

Relation among Dividend, Divisor, Quotient and Remainder
Let's see an example, 25 ÷ 4

Divisor 4 ) 25 ( 6 Quotient
– 24
1 Remainder

88 Oasis School Mathematics Book-4

Here, Dividend = 25
Divisor = 4
Quotient = 6
Remainder = 1
Then,

25 = 4 × 6 + 1

i.e. Dividend = Divisor × Quotient + Remainder

If the remainder = 0,

Then Dividend = Divisor × Quotient

Division of large numbers

Division of large numbers by a single digit:

Divide: 46842 by 7

Solution:

7 ) 4 6 8 4 2 ( 6691 Steps:
• Divisor 7 > 4.Take one more digit 6
-42
48 So, divide 46 by 7
-42
64 7 × 6 = 42, 7 × 7 = 49
63
49 > 46
12
-7 Write 6 on the quotient and 42 just below 46,
remainder 4, bring 8 down and repeat the
same process.

5

Division of large numbers by 2-digit number

Example: Divide 46187 by 32 Steps:
• Divisor 32 has 2 digits
Solution:
32 ) 4 6 1 8 7 ( 1443 \ Take two digits 46 from the dividend and
-32 divide 46 by 32. 46 > 32. So, 46 can be divided
141 by 32.
-128
138 32 × 1 = 32 < 46
-128
107 32 × 2 = 64 > 46
- 96
11 Write 1 at the quotient 32 × 1 = 32, so write
32 below 46 and subtract 46 – 32 = 14. Bring
1 down.

Repeat the same process.

Oasis School Mathematics Book-4 89

Checking:
Divisor × Quotient + Remainder

= 32 × 1443 + 11
= 47176 + 11
= 46187
= Dividend

Division of large numbers by 3-digit numbers

Example: Divide 36154 by 126

Solution: Steps:
• Divisor 126 has 3 digits
126 ) 3 6 1 5 4 ( 286 \ Take 3-digits 361 from the dividend.
-252 126 × 1 = 126, 126 × 2 = 252,
1095 126 × 3 = 378 > 361
- 1008 Hence, write 2 at the quotient.
874 126 × 2 = 252
- 756 Write 252 below 361
118 Subtract 361 – 252 = 109
Bring 5 down.
\ Quotient = 286 Repeat the same process.

Remainder = 118

Division by 10, 100, 200, 1000, 2000, etc.

When the dividend ending with zeros is divided by another number ending with
zeros:

Let's be clear with the help of the given example.

Example : Steps:

80 ÷ 20 = 8 ÷ 2 = 4 • Cancel the same number of zeros from both
2000 ÷ 400 = 20 ÷ 4 = 5 dividend and divisor.
7500 ÷ 1500 = 75 ÷ 15 = 5
• Divide the remaining number of dividend
by the remaining number of divisor.

90 Oasis School Mathematics Book-4

Exercise 3.10

1. Find the dividend using the relation dividend = divisor × quotient +
remainder.

a. divisor = 12, quotient = 5, remainder = 7

b. divisor = 8, quotient = 6, remainder = 2

c. divisor = 6, quotient = 12, remainder = 3

d. divisor = 15, quotient = 5, remainder = 5
e. divisor = 18, quotient = 5, remainder = 10

2. Divide the following and check your answer:

a. 37846 ÷ 7 b. 4562 ÷ 8 c. 56782 ÷ 4 d. 36142 ÷ 5 e. 12368 ÷ 9

3. Divide:

a. 5762 ÷ 13 b. 1236 ÷ 24 c. 5764 ÷ 23 d. 23764 ÷ 72
e. 34386 ÷ 47 f. 58318 ÷ 53 g. 20415 ÷ 52 h. 47632 ÷ 35

4. Divide:
a. 200 ÷ 10 b. 3500 ÷ 50 c. 21000 ÷ 3000 d. 300 ÷ 60
e. 80000 ÷ 1600 f. 15000 ÷ 300 g. 20000 ÷ 4000 h. 16000 ÷ 4000

5. Divide the following: c. 1482 ÷ 243 d. 37215 ÷ 216
a. 5742 ÷ 124 b. 6453 ÷ 532 g. 28463 ÷ 356 h. 69472 ÷ 635
e. 64375 ÷ 315 f. 56427 ÷ 405 k. 78405 ÷ 645 l. 53821 ÷ 712
i. 16459 ÷ 194 j. 87432 ÷ 532

1. a. 67 b. 50 c. 75 d. 80 e. 100
2. a. Q = 5406, R = 4 b. Q = 570, R = 2 c. Q = 14195, R = 2
d. Q = 7228, R = 2 e. Q = 1374, R = 2 3. a. Q = 443, R = 3 b. Q = 51, R = 12
c. Q = 250, R = 14 d. Q = 330, R = 40 e. Q = 731, R = 29 f. Q = 1100, R = 18
g. Q = 392, R = 31 h. Q = 1360, R = 32
4. a. 20 b. 70 c. 7 d. 5 e. 5 f. 50 g. 5 h. 4
5. a. Q = 46, R = 38 b. Q = 12, R = 70 c. Q = 6, R = 24 d. Q = 172, R = 63
e. Q = 204, R = 115 f. Q = 139, R = 132 g. Q = 79, R = 339 h. Q = 109, R = 257
i. Q = 84, R = 163 j. Q = 164, R = 184 k. Q = 121, R = 360 l. Q = 75, R = 421

Oasis School Mathematics Book-4 91

Verbal problems in division

Look at the given example and get the idea of solving verbal problems related
to division.

Example 1:
How many times can a number 24 be subtracted from 1656?

Solution:

As division is the repeated subtraction, to get the answer, we have to divide 1656
by 24.
Here, 24 ) 1 6 5 6 ( 69

-144
216
-216
0

24 can be subtracted from 1656, 69 times.

Example 2:

The cost of 124 copies is Rs. 1860. What is the cost of 1 copy?

Solution:

Cost of 124 copies = Rs. 1860

Cost of 1 copy = Rs. 1860 ÷ 124

Here, 124 ) 1 8 6 0 ( 15
-124
620
620
0

\ Cost of 1 copy = Rs. 15.

Exercise 3.11

1. a. How many times 45 be subtracted from 20835 to get 0?
b. How many times 183 be subtracted from 4392 to get 0?
2. a. The cost of 48 watches is Rs 14160. What is the cost of a watch?
b. A man earns Rs 17220 in a year. How much does he earn in a month?
c. The cost of a pen is Rs 65. How many pens can be bought with Rs. 8125?
d. In a year a person earns Rs 107675. How much does he earn in one

day if there are 365 days in a year?

92 Oasis School Mathematics Book-4

e. The cost of 125 mobile sets is Rs 123125. What is the cost of one
mobile set?

f. In a garden, there are 2072 trees in 37 rows. If each row has the same
number of trees, find the number of trees in each row.

g. A school needs 24500 pencils a year. How many boxes containing
25 pencils must the school buy?

1. a. 463 b. 24 2. a. 295 b. 1435 c. 125 d. 295 e. 985 f. 56 g. 980

Mathematical fun related to addition and subtraction

- Choose a 3 digit number with the first digit greater than the third. Say 563

- Reverse the digits 365

- Subtract: 5 6 3

- 3 6 5 It’s interesting!
198 Every time the
result is 1089.
- Reverse the digits 891

Add: 198

+ 8 9 1

1089

Try this with other numbers and examine the result.

Oral test:
1. What should be added to 45 to make 60?
2. By how much 15 is greater than 9?
3. By how much 25 is less than 45?
4. Which number is 15 more than 20?
5. Which number is 15 less than 20?
6. How many times 5 is added to itself to make 45?
7. How many times 6 should be subtracted from 54 to get 0?
8. What is the difference between 5 two’s and 2 fives?
9. If you divide 24 chocolates equally among 4 children, how many

chocolates will each child get?

Oasis School Mathematics Book-4 93

Simplification
Rules of sign

To identify whether the given number is positive or negative, we write positive
sign (+) or negative sign (-) just before the number. The number with no sign
just before it is considered to have the (+) sign.

+5 is a positive number
-5 is a negative number.
5 is a positive number.

Rules of sign in addition and subtraction:

1. The addition of two numbers each with a positive sign gives the sum
with positive sign.

i.e. (+ number) + (+ number) = + sum

eg. + 6 + 8 = + 14 (Add the numbers and put + sign before the sum)

2. The addition of two numbers each with a negative sign each gives the
sum with negative sign.

i.e. (- number) + (- number) = - sum

(-5) + (-6) = -11 (Add the numbers and put – sign before the sum)

3. The addition of a number with positive sign and another number with
a negative sign gives the sum with positive or negative sign depending
upon the sign of the greater number.

i.e. (number ) + (- number) = ( + sum) If the greater number is positive.

(+ number) + ( - number) = ( - sum) If the greater number is negative.

eg. ( +8) + (-6) = (+2) Subtract the smaller number
\ 8>6 from the greater number and
(+6) + (-8) = (-2) put the sign of greater number
\ 8>6 before the result.

94 Oasis School Mathematics Book-4

Class Assignment b. (+5) + (-7) = c. (-6) + (+8)] =
f. (-6) + (-8) =
1. Find the sum: i. (-3) + (-3) =
a. (+5) + (+7) = l. (-9) + (+6) =

d. (-7) + (+2) = e. (-7) + (-2) =

g. (+6) + (-2) = h. (+9) + (-3) =

j. (+7) + (+9) = k. (+7) + (-9) =

Teacher's Signature

Rules of sign in multiplication and division:

1. The multiplication of two numbers both with a positive sign gives the product
with a positive sign.

i.e. (+number) × ( + number) = ( + product)
eg. (+6) × (+4) = +24 Multiply the numbers and put positive sign.

2. The multiplication of two numbers both with a negative sign gives the product
with a positive sign.

i.e. (- number) × (- number) = (+ product)

(- 4) × (- 6) = (+24) Multiply the numbers and put positive sign before the product.

3. The multiplication of two numbers with a positive and a negative sign gives the
product with a negative sign. i.e. (- number) × (+ number) = (- product)

eg. (+5) × (-6) = (-30)

(-6) × (+8) = (-48) Multiply the numbers and put negative sign.

4. The division of two numbers each with a positive sign gives the quotient with a
positive sign.

i.e. (+ number) ÷ (+ number) = (+quotient) (+24) ÷ (+6) = (+4)

5. The division of two numbers with a negative sign each gives the quotient with
a positive sign.

i.e. (-number) ÷ (- number) = (+quotient) (-32) ÷ (-4) = (+8)

6. The division of one number with positive sign and another number with
negative sign gives the quotient with a negative sign.

i.e. (+ number) ÷ (-number) = (-quotient)

(- number) ÷ (+ number) = (-quotient)

eg. (+20) ÷ (-4) = -5 (-24) ÷ (+6) = - 4

Oasis School Mathematics Book-4 95

Class Assignment

1. Find the product or quotient:

a. (-6) × (+8) = b. (+6) × (+3) = c. (-8) × (-3) =
f. (-7) × (-2) =
d. (+3) × (-7) = e. (+7) × (-4) = i. (-9) × (+2) =
l. (+18) ÷ (+6) =
g. (+9) × (+3) = h. (-3) × (+4) = o. (-18) ÷ (-6) =

j. (-12) ÷ (+4) = k. (-6) ÷ (-2) =

m. (+27) ÷ (-3) = n. (+21) ÷ (-7) =

Teacher's Signature

Order of operations

If a mathematical expression contains two or more than two mathematical
operations, such expression can be solved using mathematical operations in a
proper order.
Four fundamental operations are addition, subtraction, multiplication and
division. The simplification can be done using DMAS rule. Where D, M, A and
S give the order of simplification.
Where,

D means division (÷)

M means multiplication (×)

A means addition (+), and

S means subtraction (-)

Order of simplification 1st operation
Division 2nd operation
Multiplication 3rd operation
Addition 4th operation
Subtraction

96 Oasis School Mathematics Book-4

Example 1: Example 2:

Simplify: 19 – 9 × 5 + 30 20 ÷ 2 – 10 × 2 + 20 ÷ 5 + 20
Solution : Solution:
19 – 9 × 5 + 30 20 ÷ 2 – 10 × 2 + 20 ÷ 5 + 20
= 19 – 45 + 30 (multiply first) = 10 – 10 × 2 + 4 + 20 (divide first)
= 49 – 45 (add next) = 10 – 20 + 4 + 20 (multiply next)
= 4 (subtract at last) = 34 – 20 (add next)
= 14 (subtract)
Example 3:

Divide 63 by 7 and add the quotient with the product of 5 and 7.
Solution:
The mathematical expression for the given statement is:
63 ÷ 7 + 5 × 7
= 9 + 5 × 7
= 9 + 35
= 44

Exercise 3.12

1. Simplify:

a. 9 + 8 – 7 b. 12 – 13 + 16 c. -16 + 12 + 6

d. -25 + 13 – 16 e. 35 – 6 -7 f. 42 – 7 – 18 + 14

2. Simplify:

a. 12 × 5 + 52 – 73 b. 82 – 4 × 16 – 18 c. 12 – 5 × 4 + 3 × 2 – 8

d. 3 × 8 – 6 × 2 + 7 e. -10 + 6 × 3 – 7 f. -30 + 4 × 6 + 7

3. Simplify

a. 28 ÷ 7 – 4 × 8 + 6 – 12 b. 35 – 42 ÷ 7 + 6 × 2

c. 7 × 8 – 21 + 18 × 85 ÷ 17 d. 120 ÷ 10 + 6 × 4

e. 84 ÷ 12 × 4 – 3 × 60 ÷ 20 f. 32 + 91 ÷ 13 – 6 × 4 – 7

g. 44 + 9 × 3 – 35 × 6 0 ÷ 20 h. 66 ÷ 11 × 5 + 45 ÷ 9 - 30

i. 16 × 4 – 3 × 6 + 8 ÷ 4 j. 21 – 3 × 18 ÷ 6 + 22

k. 12 ÷ 3 + 15 ÷ 5 × 6 – 3 + 18 l. 15 – 2 × 3 + 48 ÷ 4 - 15

Oasis School Mathematics Book-4 97

4. Make mathematical expression and simplify:
a. The product of 3 and 7 is added to 20.
b. The product of 14 and 2 is subtracted from 50.
c. 25 is subtracted from the product of 7 and 4.
d. The quotient of 30 divided by 3 is added to 7.
e. The quotient of 52 divided by 13 is multiplied by 2.
f. The quotient of 70 divided by 5 is subtracted from 20.
g. The quotient of 56 divided by 8 is added to the product of 7 and 3.

1. a. 10 b. 15 c. 2 d. -28 e. 22 f. 31
2. a. 39 b. 0 c. -10 d. 19 e. 1 f. 1
3. a. -34 b. 41 c. 125 d. 36 e. 19 f. 8 g. -34 h. 5 i. 48 j. 34
k. 37 l. 6 4. a. 41 b. 22 c. 3 d. 17 e. 8 f. 6 g. 28

Uses of brackets in simplification
The brackets used in simplification are:
( ) small brackets
{ } curly brackets
[ ] square brackets
The order of operation inside the brackets are:
( ) First in order
{ } Second in order
[ ] Third in order
In the whole process of simplification, we have to use BODMAS rule.
Where, BODMAS gives the order of simplification.
‘B’ stands for removal of brackets. [{( )}]
i.e. simplification of the parts inside the brackets
‘O’ stands for ‘of’
‘D’ stands for ‘Division’ (÷)
‘M’ stands for ‘Multiplication’ (×)
‘A’ stands for ‘Addition’ (+)
‘S’ stands for ‘Subtraction’ (-)

98 Oasis School Mathematics Book-4

Example:
Simplify: 15 – [20 ÷ {30 – 4(10 – 5)}]

Solution:

15 – [20 ÷ {30 – 4(10 – 5)}]

= 15 – [20 ÷ {30 – 4 × 5}] Subtraction inside ( )

= 15 – [20 ÷ {30-20}] Multiplication inside { }

= 15 – [20 ÷ 10] Subtraction inside { }

= 15 – 2 Division inside [ ]

= 13

Exercise 3.13

1. Simplify:

a. 25 + (10 – 5 + 2) b. 25 – (6 × 3 + 5)

c. 18 + 3 (5 + 2 – 3) d. 27 + 2(15 – 2 × 6)

e. 3 × (7 + 6) – 12 f. (3 + 7) × (4 – 2)

g. (7 × 6) ÷ 14 + 7 h. (15 × 3 ) ÷ (4 + 5)

i. (10 ÷ 2 × 3) – (4 × 5 – 6)

2. Simplify:

a. {12 + (15 – 3) ÷ 4} b. 27 + {16 – (12 + 3)}

c. {(12 + 7) – (30 – 22)} × 8 ÷ 4 d. 3 {4 + 3 (6 – 2)}

e. 10 – 2 {8 ÷ (16 ÷ 4)} f. 28 ÷ [3 + 16 ÷ {2 + 8 ÷ (1 + 3)}]

g. 16 – [20 ÷ { 20 – 3 ( 10 – 5)}] h. 30 + {12 – (16 – 56 ÷ 8)}

i. 18 + [9 + {13 + 3 + (4 × 16 ÷ 8 )}] j. 30 – [11 + {18 – (48 ÷ 8 × 2 )}]

k. 28 + 3 [{12 + 2 (14 – 4) ÷ 5} – 6]÷ 3

1. a. 32 b. 2 c. 30 d. 33 e. 27 f. 20 g. 10 h. 5 i. 1
2. a. 15 b. 28 c. 22 d. 48 e. 6 f. 4 g. 12 h. 33 i. 51 j. 4 k. 38

Oasis School Mathematics Book-4 99

Worksheet

Solve the given problems. Identify the hidden name using the respective letter
code of answer.

11345 2592 17 6 22 36 9362 43685 2 300 49

AD E GH IM P RS V

3 7 2 8 6 2 4 1 2 3 2 4 88 ÷ 4 =
+ 5 6 3 4 –51067 ×8
9362

M

6316 + 5100 – 71 = XLIX =

Successor of the number 43684 = 2 2 6 9
×5

Only one prime which is even =

Rounding off the number 392 to its nearest ten.

Predecessor of the number 11346

6 4 8 H.C.F. of 12 and 18
×4

11 × 3 + 7 – 18 = LCM of 9 and 12

IX CCCLXII Dividend with divisor 5
22 – 9 × 5 + 25 = Quotient 7 and Remainder 1

Quotient of 90 divided by 6 added to 2

\ The name of the poet is
...................................... ...................................... ......................................

100 Oasis School Mathematics Book-4