- RD Sharma Solutions for Class 6 Maths Chapter 4 - Overview
- RD Sharma Solutions for Class 6 Maths Chapter 4 Operations on Whole Numbers
- Download RD Sharma Solutions for Class 6 Maths Chapter 4 – Operations on Whole Numbers PDF Here
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RD Sharma Solutions for Class 6 Maths Chapter 4 Operations on Whole Numbers
By Maitree Choube
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Updated on 28 Apr 2025, 11:43 IST
RD Sharma Solutions for Class 6 Maths Chapter 4: If you are looking for clear and reliable help with Class 6 Maths Chapter 4 Operations on Whole Numbers, the RD Sharma Solutions are a perfect choice. Designed by expert teachers, these solutions are carefully created to match the latest CBSE syllabus, ensuring you stay fully prepared for your school exams.
Each question is explained with step-by-step solutions, making it easier for young learners to build a strong conceptual understanding. Whether you want to boost your problem-solving skills or simply need a quick guide for practice and revision, RD Sharma Solutions offer simple and smooth explanations that make learning fun and easy. With these solutions, students can confidently master the important concepts of operations on whole numbers and achieve better results in their exams.
RD Sharma Solutions for Class 6 Maths Chapter 4 - Overview
| Aspects | Details |
| Class | Class 6 |
| Subject | Mathematics / Maths |
| Book | RD Sharma |
| Chapter Number | 4 |
| Name of Chapter | Operations on Whole Numbers |
| Study Material Here | RD Sharma Class 6 Maths Chapter 4 Operations on Whole Numbers Solutions |
| RD Sharma Solutions of All Chapters of This Class | RD Sharma Class 6 Solutions |
| All RD Sharma Solutions PDF Available | Yes |
RD Sharma Solutions for Class 6 Maths Chapter 4 Operations on Whole Numbers
RD Sharma Solutions for Class 6 Maths Chapter 4 Operations on Whole Numbers are the helpful resource for students who want to understand how to add, subtract, multiply, and divide whole numbers confidently. Designed by subject experts, these RD Sharma Solutions for Class 6 Chapter 4 provide simple, step-by-step solutions that match the latest CBSE syllabus, ensuring students build a strong and clear understanding from the basics.
Students who want extra practice can also work on Class 6 Maths worksheets along with the RD Sharma Solutions. These worksheets are a great way to strengthen problem-solving skills and improve accuracy in calculations. For easy and fast revision, RD Sharma Class 6 Maths PDF download options are available, helping students to access learning materials anytime, anywhere.
To further support their studies, students can also check out NCERT Solutions for Class 6 Maths, which offer another layer of explanation and extra practice opportunities. If you want a complete learning package, the RD Sharma Class 6 full book PDF brings all the chapters together, making it easier to study and revise the entire syllabus.
Using RD Sharma Solutions for Class 6 Maths Chapter 4 will definitely help students master the concept of operations on whole numbers and prepare well for exams with full confidence and clarity.
Don't Miss:
- RD Sharma Solutions for Class 6 Maths Chapter 1
- RD Sharma Solutions for Class 6 Maths Chapter 2
- RD Sharma Solutions for Class 6 Maths Chapter 3
- RD Sharma Solutions for Class 6 Maths Chapter 4
Download RD Sharma Solutions for Class 6 Maths Chapter 4 – Operations on Whole Numbers PDF Here
RD Sharma Solutions for Class 6 Maths Chapter 4 – Operations on Whole Numbers PDF is designed to help students understand how to perform basic operations like addition, subtraction, multiplication, and division with whole numbers. This chapter is very important for building a strong foundation in mathematics and includes the following exercises:
👉 Exercise 4.1 – Addition of Whole Numbers
👉 Exercise 4.2 – Subtraction of Whole Numbers
👉 Exercise 4.3 – Multiplication of Whole Numbers
👉 Exercise 4.4 – Division of Whole Numbers
We have provided an easy-to-download RD Sharma Class 6 Maths Chapter 4 PDF to make practice and revision super simple for students. The Operations on Whole Numbers PDF offers clear, step-by-step solutions that strictly follow the CBSE syllabus, making it perfect for mastering the basic concepts and improving accuracy.
These detailed solutions are created by subject experts to strengthen students' problem-solving skills and boost their confidence for school exams. With the RD Sharma Solutions PDF, students can easily study anytime, revise quickly, and develop a deeper conceptual understanding of operations on whole numbers.
Also Check: Properties of Whole Numbers
Access Answers to RD Sharma Solutions for Class 6 Maths Chapter 4 Operations on Whole Numbers
Exercise 4.1 PAGE: 4.4
- Fill in the blanks to make each of the following a true statement:
(i) 359 + 476 = 476 + …..
(ii) …. + 1952 = 1952 + 2008
(iii) 90758 + 0 = ….
(iv) 54321 + (489 + 699) = 489 + (54321 + …..)
Solution:
(i) 359 + 476 = 476 + 359 using commutativity
(ii) 2008 + 1952 = 1952 + 2008 using commutativity
(iii) 90758 + 0 = 90758 using the additive identity
(iv) 54321 + (489 + 699) = 489 + (54321 + 699) using associativity
- Add each of the following and check by reversing the order of addends:
(i) 5628 + 39784
(ii) 923584 + 178
(iii) 15409 + 112
(iv) 2359 + 641
Solution:
(i) We get
5628 + 39784 = 45412
By reversing the order of addends
39784 + 5628 = 45412
(ii) We get
923584 + 178 = 923762
By reversing the order of addends
178 + 923584 = 923762
(iii) We get
15409 + 112 = 15521
By reversing the order of addends
112 + 15409 = 15521
(iv) We get
2359 + 641 = 3000
By reversing the order of addends
641 + 2359 = 3000
- Determine the sum by suitable rearrangements:
(i) 953 + 407 + 647
(ii) 15409 + 178 + 591 + 322
(iii) 2359 + 10001 + 2641 + 9999
(iv) 1 + 2 + 3 + 4 + 1996 + 1997 + 1998 + 1999
(v) 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20
Solution:
(i) 953 + 407 + 647
We know that
53 + 47 = 100
It can be written as
(953 + 647) + 407 = 1600 + 407
On further calculation
(953 + 647) + 407 = 2007
(ii) 15409 + 178 + 591 + 322
We know that
409 + 91 = 500 and 78 + 22 = 100
It can be written as
(15409 + 591) + (178 + 322) = 16000 + 500
On further calculation
(15409 + 591) + (178 + 322) = 16500
(iii) 2359 + 10001 + 2641 + 9999
We know that
59 + 41 = 100 and 99 + 01 = 100
It can be written as
(2359 + 2641) + (10001 + 9999) = 5000 + 20000
On further calculation
(2359 + 2641) + (10001 + 9999) = 25000
(iv) 1 + 2 + 3 + 4 + 1996 + 1997 + 1998 + 1999
We know that
99 + 1 = 100, 98 + 2 = 100, 97 + 3 = 100 and 96 + 4 = 100
It can be written as
(1 + 1999) + (2 + 1998) + (3 + 1997) + (4 + 1996) = 2000 + 2000 + 2000 + 2000
On further calculation
(1 + 1999) + (2 + 1998) + (3 + 1997) + (4 + 1996) = 8000
(v) 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20
We know that
10 + 20 = 30, 1 + 9 = 10, 2 + 8 = 10, 3 + 7 = 10 and 4 + 6 = 10
It can be written as
(10 + 20) + (11 + 19) + (12 + 18) + (13 + 17) + (14 + 16) = 30 + 30 + 30 + 30 + 30 + 15
On further calculation
(10 + 20) + (11 + 19) + (12 + 18) + (13 + 17) + (14 + 16) = 150 + 15 = 165
- Which of the following statements are true and which are false:
(i) The sum of two odd numbers is an odd number.
(ii) The sum of two odd numbers is an even number.
(iii) The sum of two even numbers is an even number.
(iv) The sum of two even numbers is an odd number.
(v) The sum of an even number and an odd number is an odd number.
(vi) The sum of an odd number and an even number is an even number.
(vii) Every whole number is a natural number.
(viii) Every natural number is a whole number.
(ix) There is a whole number which when added to a whole number, gives that number.
(x) There is a natural number which when added to a natural number, gives that number.
(xi) Commutativity and associativity are properties of whole numbers.
(xii) Commutativity and associativity are properties of addition of whole numbers.
Solution:
(i) False. We know that, 1 + 3 = 4 where 4 is an even number.
(ii) True. We know that, 5 + 7 = 12 where 12 is an even number.
(iii) True. We know that, 2 + 4 = 6 where 6 is an even number.
(iv) False. We know that, 4 + 6 = 10 where 10 is an even number.
(v) True. We know that, 2 + 1 = 3 where 3 is an odd number.
(vi) False. We know that, 3 + 2 = 5 where 5 is an odd number.
(vii) False. Whole number starts from 0 whereas natural numbers start from 1.
(viii) True. All the natural numbers are also whole number.
(ix) True. We know that, 1 + 0 = 1 where 1 is a whole number.
(x) False. We know that 2 + 1 = 3 which is not that number.
(xi) False. Commutativity and associativity are not properties of whole numbers.
(xii) True. Commutativity and associativity are properties of addition of whole numbers.
Exercise 4.3 page: 4.14
- Fill in the blanks to make each of the following a true statement:
(i) 785 × 0 = …..
(ii) 4567 × 1 = …..
(iii) 475 × 129 = 129 × …..
(iv) ….. × 8975 = 8975 × 1243
(v) 10 × 100 × …. = 10000
(vi) 27 × 18 = 27 × 9 + 27 × ….. + 27 × 5
(vii) 12 × 45 = 12 × 50 – 12 × …..
(viii) 78 × 89 = 78 × 100 – 78 × ….. + 78 × 5
(ix) 66 × 85 = 66 × 90 – 66 × ….. – 66
(x) 49 × 66 + 49 × 34 = 49 × (….. + …..)
Solution:
(i) 785 × 0 = 0
(ii) 4567 × 1 = 4567 based on multiplicative identity
(iii) 475 × 129 = 129 × 475 based on commutativity
(iv) 1243 × 8975 = 8975 × 1243 based on commutativity
(v) 10 × 100 × 10 = 10000
(vi) 27 × 18 = 27 × 9 + 27 × 4 + 27 × 5
(vii) 12 × 45 = 12 × 50 – 12 × 5
(viii) 78 × 89 = 78 × 100 – 78 × 16 + 78 × 5
(ix) 66 × 85 = 66 × 90 – 66 × 4 – 66
(x) 49 × 66 + 49 × 34 = 49 × (66 + 34)
- Determine each of the following products by suitable rearrangements:
(i) 2 × 1497 × 50
(ii) 4 × 358 × 25
(iii) 495 × 625 × 16
(iv) 625 × 20 × 8 × 50
Solution:
(i) 2 × 1497 × 50
It can be written as
2 × 1497 × 50 = (2 × 50) × 1497
= 100 × 1497
= 149700
(ii) 4 × 358 × 25
It can be written as
4 × 358 × 25 = (4 × 25) × 358
= 100 × 358
= 35800
(iii) 495 × 625 × 16
It can be written as
495 × 625 × 16 = (625 × 16) × 495
= 10000 × 495
= 4950000
(iv) 625 × 20 × 8 × 50
It can be written as
625 × 20 × 8 × 50 = (625 × 8) × (20 × 50)
= 5000 × 1000
= 5000000
- Using distributivity of multiplication over addition of whole numbers, find each of the following products:
(i) 736 × 103
(ii) 258 × 1008
(iii) 258 × 1008
Solution:
(i) 736 × 103
It can be written as
= 736 × (100 + 3)
By using distributivity of multiplication over addition of whole numbers
= (736 × 100) + (736 × 3)
On further calculation
= 73600 + 2208
We get
= 75808
(ii) 258 × 1008
It can be written as
= 258 × (1000 + 8)
By using distributivity of multiplication over addition of whole numbers
= (258 × 1000) + (258 × 8)
On further calculation
= 258000 + 2064
We get
= 260064
(iii) 258 × 1008
It can be written as
= 258 × (1000 + 8)
By using distributivity of multiplication over addition of whole numbers
= (258 × 1000) + (258 × 8)
On further calculation
= 258000 + 2064
We get
= 260064
- Find each of the following products:
(i) 736 × 93
(ii) 816 × 745
(iii) 2032 × 613
Solution:
(i) 736 × 93
It can be written as
= 736 × (100 – 7)
By using distributivity of multiplication over subtraction of whole numbers
= (736 × 100) – (736 × 7)
On further calculation
= 73600 – 5152
We get
= 68448
(ii) 816 × 745
It can be written as
= 816 × (750 – 5)
By using distributivity of multiplication over subtraction of whole numbers
= (816 × 750) – (816 × 5)
On further calculation
= 612000 – 4080
We get
= 607920
(iii) 2032 × 613
It can be written as
= 2032 × (600 + 13)
By using distributivity of multiplication over addition of whole numbers
= (2032 × 600) + (2032 × 13)
On further calculation
= 1219200 + 26416
We get
= 1245616
- Find the values of each of the following using properties:
(i) 493 × 8 + 493 × 2
(ii) 24579 × 93 + 7 × 24579
(iii) 1568 × 184 – 1568 × 84
(iv) 15625 × 15625 – 15625 × 5625
Solution:
(i) 493 × 8 + 493 × 2
It can be written as
= 493 × (8 + 2)
By using distributivity of multiplication over addition of whole numbers
= 493 × 10
On further calculation
= 4930
(ii) 24579 × 93 + 7 × 24579
It can be written as
= 24579 × (93 + 7)
By using distributivity of multiplication over addition of whole numbers
= 24579 × 100
On further calculation
= 2457900
(iii) 1568 × 184 – 1568 × 84
It can be written as
= 1568 × (184 – 84)
By using distributivity of multiplication over subtraction of whole numbers
= 1568 × 100
On further calculation
= 156800
(iv) 15625 × 15625 – 15625 × 5625
It can be written as
= 15625 × (15625 – 5625)
By using distributivity of multiplication over subrtaction of whole numbers
= 15625 × 10000
On further calculation
= 156250000
- Determine the product of:
(i) the greatest number of four digits and the smallest number of three digits.
(ii) the greatest number of five digits and the greatest number of three digits.
Solution:
(i) We know that
Largest four digit number = 9999
Smallest three digit number = 100
Product of both = 9999 × 100 = 999900
Hence, the product of the greatest number of four digits and the smallest number of three digits is 999900.
(ii) We know that
Largest five digit number = 99999
Largest three digit number = 999
Product of both = 99999 × 999
It can be written as
= 99999 × (1000 – 1)
By using distributivity of multiplication over addition subrtaction of whole numbers
= (99999 × 1000) – (99999 × 1)
On further calculation
= 99999000 – 99999
We get
= 99899001
- In each of the following, fill in the blanks, so that the statement is true:
(i) (500 + 7) (300 – 1) = 299 × …..
(ii) 888 + 777 + 555 = 111 × …..
(iii) 75 × 425 = (70 + 5) (….. + 85)
(iv) 89 × (100 – 2) = 98 × (100 – …..)
(v) (15 + 5) (15 – 5) = 225 – …..
(vi) 9 × (10000 + …..) = 98766
Solution:
(i) By considering LHS
(500 + 7) (300 – 1)
We get
= 507 × 299
By using commutativity
= 299 × 507
(ii) By considering LHS
888 + 777 + 555
We get
= 111 (8 + 7 + 5)
By using distributivity
= 111 × 20
(iii) By considering LHS
75 × 425
We get
= (70 + 5) × 425
It can be written as
= (70 + 5) (340 + 85)
(iv) By considering LHS
89 × (100 – 2)
We get
= 89 × 98
It can be written as
= 98 × 89
By using commutativity
= 98 × (100 – 11)
(v) By considering LHS
(15 + 5) (15 – 5)
We get
= 20 × 10
On further calculation
= 200
It can be written as
= 225 – 25
(vi) By considering LHS
9 × (10000 + 974) = 98766
- A dealer purchased 125 colour television sets. If the cost of each set is Rs 19820, determine the cost of all sets together.
Solution:
It is given that
Cost of each television set = Rs 19820
So we get
Cost of 125 television sets = 19820 × 125
It can be written as
= 19820 × (100 + 25)
By using distributivity of multiplication over addition of whole numbers
= (19820 × 100) + (19820 × 25)
On further calculation
= 1982000 + 495500
So we get
= Rs 2477500
- The annual fee charged from a student of class VI in a school is Rs 8880. If there are, in all, 235 students in class VI, find the total collection.
Solution:
Annual fee per student = Rs 8880
So we get
Annual fee charged for 235 students = 8880 × 235 = 2086800
Therefore, the total collection is Rs 2086800.
- A group housing society constructed 350 flats. If the cost of construction for each flat is Rs 993570, what is the total cost of construction of all the flats.
Solution:
Cost of construction for each flat = Rs 993570
Number of flats constructed = 350
So we get
Cost of construction of 350 flats = 993570 × 350 = Rs 347749500
Therefore, the total cost of construction of all the flats is Rs 347749500.
- The product of two whole numbers is zero. What do you conclude?
Solution:
The product of two whole numbers is zero, which means that at least one number or both of them are zero.
- What are the whole numbers which when multiplied with itself gives the same number?
Solution:
Two numbers when multiplied with itself gives the same number.
For example: 0 × 0 = 0 and 1 × 1 = 1
- In a large housing complex, there are 15 small buildings and 22 large building. Each of the large buildings has 10 floors with 2 apartments on each floor. Each of the small buildings has 12 floors with 3 apartments on each floor. How many apartments are there in all.
Solution:
It is given that
No. of large buildings = 22
No. of small buildings = 15
No. of floors in 1 large building = 10
No. of apartments on 1 floor = 2
So total apartment in 1 large building = 10 × 2 = 20
The same way
No. of apartments in 1 small building = 12 × 3 = 36
So the total apartment in entire housing complex = (22 × 20) + (15 × 36) = 440 + 540 = 980
Therefore, there are 980 apartments in all.
RD Sharma Solutions for Class 6 Maths Chapter 4 FAQs
What is covered in RD Sharma Solutions for Class 6 Maths Chapter 4?
RD Sharma Solutions for Class 6 Maths Chapter 4 covers addition, subtraction, multiplication, and division of whole numbers. It explains properties like closure, commutativity, associativity, and distributivity in a clear and simple way.
Why is practicing operations on whole numbers important in Class 6?
Practicing operations on whole numbers builds a strong foundation for higher-level mathematics. It improves calculation speed, accuracy, and problem-solving skills essential for school exams.
How do RD Sharma Solutions for Chapter 4 help in school exam preparation?
RD Sharma Solutions for Chapter 4 offer step-by-step explanations and follow the latest CBSE syllabus. They make concepts easy to understand and help students prepare effectively for exams with extra practice and quick revision.
What kind of problems can I expect in RD Sharma Class 6 Chapter 4?
In RD Sharma Class 6 Chapter 4, you can expect problems related to simple operations, property-based questions, word problems, and pattern-based exercises on whole numbers.
Can I download the RD Sharma Solutions for Chapter 4 PDF for free?
Yes, students can easily access and download RD Sharma Solutions for Class 6 Maths Chapter 4 PDF online. It helps in practicing anytime and revising concepts anywhere.
Are RD Sharma Solutions enough for mastering operations on whole numbers?
RD Sharma Solutions provide a strong and detailed explanation, but combining them with Class 6 Maths NCERT Solutions and worksheets ensures complete mastery of operations on whole numbers.
Is the RD Sharma Class 6 Chapter 4 PDF aligned with the CBSE syllabus?
Yes, the RD Sharma Class 6 Maths Chapter 4 PDF is completely aligned with the latest CBSE syllabus, making it a reliable and updated resource for students.