NCERT Solutions Class 6 Maths Chapter 2
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NCERT Solutions For Class 6 Mathematics Chapter 2: Whole Numbers
In continuation with the previous chapter, NCERT Solutions for Class 6 Mathematics Chapter 2 aims to give students a deeper understanding of numbers. After learning about numbers in the first chapter, students will feel more confident about understanding Whole Numbers. NCERT Solutions for Class 6 Mathematics Chapter 2 by Extramarks is a wellorganised, verified, and structured resource for students.
NCERT Solutions for Class 6 Mathematics Chapter 2  Whole Numbers
Our educational experts developed NCERT solutions to ensure that you comprehend all of the major topics covered in this chapter. It includes solutions to the problems in each exercise and essential equations, Mathematical concepts, and smart approach to problemsolving.
Access NCERT Solutions for Mathematics Chapter 2 – Whole Numbers
Extramarks' NCERT Solutions for Class 6 Mathematics Chapter 2 is essential for exam preparation since it contains stepbystep solutions and easy shortcuts and effective strategies to inculcate mathematical thinking. NCERT Solutions for Class 6 Mathematics are available on the official website of Extramarks. The solutions are prepared following the latest CBSE curriculum and the best part about these solutions is that they are available for all the subjects, not just mathematics.
Exercise (2.1)
It contains 8 questions and solutions.
Exercise (2.2)
It has 7 questions and solutions.
Exercise (2.3)
This exercise has 5 questions and solutions.
NCERT Solutions for Class 6 Mathematics Chapter 2  Whole Numbers
NCERT Mathematics Solutions for Class 6 are available on Extramarks website. Students can access solutions and practise questions as per their convenience. Furthermore, the solutions can be saved for future use, such as to prepare for term and final exams as well as for olympiads .
2.1 Introduction
Chapter 2 Whole Numbers of Class 6 discusses numbers and explains the principles of successor and predecessor in natural numbers. Every whole number has a successor and every whole number except zero has a predecessor. If the student has a good understanding of the term predecessor and successor, students can learn them quickly.
2.2 Whole Numbers
All the natural numbers are whole numbers but all whole numbers are not a natural number. Aryabhatta discovered the digit zero to elucidate the situation. Whole numbers are made up of allnatural numbers, including zero. In short, Zero is a whole number but not a natural number. The NCERT Solutions for Chapter 2 of the Mathematics book have detailed explanations.
2.3 Number Line
The number line is a new concept for students of Class 6 Mathematics Chapter 2. Students will learn how to represent numbers on a straight line, unit distance and number line for whole numbers. This topic also includes addition, subtraction and multiplication on a number line. NCERT solutions will definitely help the students to solve more problems with the help of examples.
2.4 Properties of Whole Numbers
Having now learnt about whole numbers, students will get to know about the properties of whole numbers. The property of whole number includes:
 Closure Property: The Whole Numbers can hold the closure condition for both addition and multiplication.
 We can multiply two whole numbers in any order.
 The division of a whole number by zero is not defined.
 Property of Association: For both addition and multiplication, the Whole Numbers meet the associated property.
 Multiplication Distributive Over Addition: It indicates that Whole Numbers can only meet the distributivity of multiplication on additions which means multiplying a number by sum or difference
 Any number multiplied by the whole number zero, becomes zero.
 The result of the division of two whole numbers is not always a whole number.
2.5 Patterns in Whole Numbers
NCERT Solutions for Chapter 2 Mathematics Class 6 Next students about the patterns seen in Whole Numbers in a very simplified manner. The trend refers to a particular number that is represented or appears. It is something that students may learn with zeal because the numbers can take the form of a straight line, square, rectangle, triangle, or any other shape.
All Exercises in the Chapter Given Below
Extramarks provides all the questions and answers in NCERT Solutions for Class 6 Mathematics and helps students to learn the concepts and principles of mathematics, properties of whole numbers and new topics introduced in the chapter. .
Key Takeaways of the NCERT Solutions of Class 6 Mathematics Chapter 2: Whole Numbers 3
NCERT solutions on Extramarks gives a wealth of information to build a solid foundation in the formative years. NCERT solutions provide the following advantages:
 The explanations are simple and give clear understanding at the student's level.
 It includes enough unresolved questions to test the proficiency level of the student in each topic.
Q.1 Write the next three natural numbers after 10999.
Ans
The next three natural numbers after 10999 are:
10999 + 1 = 11000
11000 + 1 = 11001
11001 + 1 = 11002
Q.2 Write the three whole numbers occurring just before 10001.
Ans
Three whole numbers occurring just before 10001 are:
10001 – 1 = 10000
10000 – 1 = 9999
9999 – 1 = 9998
Q.3 Which is the smallest whole number?
Ans
0 is the smallest whole number.
Q.4 How many whole numbers are there between 32 and 53?
Ans
Number of whole numbers between 32 and 53 = 53 – 32 = 21
Q.5 Write the successor of:
(a) 2440701 (b) 100199
(c) 1099999 (d) 2345670
Ans
(a) Successor of 2440701 = 2440702
(b) Successor of 100199 = 100200
(c) Successor of 1099999 = 1100000
(d) Successor of 2345670 = 2345671
Q.6 Write the predecessor of:
(a) 94 (b) 10000
(c) 208090 (d) 7654321
Ans
(a) Predecessor of 94 = 93
(b) Predecessor of 10000 = 9999
(c) Predecessor of 208090 = 208089
(d) Predecessor of 7654321 = 7654320
Q.7 In each of the following pairs of numbers, state which whole number is on the left of the other number on the number line. Also write them with the appropriate sign (>, <) between them.
(a) 530, 503
(b) 370, 307
(c) 98765, 56789
(d) 9830415, 10023001
Ans
(a) 530 > 503
So, 503 appear on left side of 530 on a number line.
(b) 370 > 307
So, 307 appear on left side of 370 on a number line.
(c) 98765 > 56789
So, 56789 appear on left side of 98765 on a number line.
(d) 9830415 < 10023001
So, 9830415 appear on left side of 10023001 on a number line.
Q.8 Which of the following statements are true (T) and which are false (F)?
(a) Zero is the smallest natural number.
(b) 400 is the predecessor of 399.
(c) Zero is the smallest whole number.
(d) 600 is the successor of 599.
(e) All natural numbers are whole numbers.
(f) All whole numbers are natural numbers.
(g) The predecessor of a two digit number is never a single digit number.
(h) 1 is the smallest whole number.
(i) The natural number 1 has no predecessor.
(j) The whole number 1 has no predecessor.
(k) The whole number 13 lies between 11 and 12.
(l) The whole number 0 has no predecessor.
(m) The successor of a two digit number is always a two digit number.
Ans
(a) Zero is the smallest natural number. (F)
(b) 400 is the predecessor of 399. (F)
(c) Zero is the smallest whole number. (T)
(d) 600 is the successor of 599. (T)
(e) All natural numbers are whole numbers. (T)
(f) All whole numbers are natural numbers. (F)
(g) The predecessor of a two digit number is never a single digit number. (F)
(h) 1 is the smallest whole number. (F)
(i) The natural number 1 has no predecessor.(T)
(j) The whole number 1 has no predecessor. (F)
(k) The whole number 13 lies between 11 and 12. (F)
(l) The whole number 0 has no predecessor. (T)
(m) The successor of a two digit number is always a two digit number. (F)
Q.9 Find the sum by suitable rearrangement:
(a) 837 + 208 + 363
(b) 1962 + 453 + 1538 + 647
Ans
(a) 837 + 208 + 363 = (837 + 363) + 208
= 1200 + 208
= 1408
(b) 1962 + 453 + 1538 + 647
= (1962 + 1538) + (453 + 647)
= 3500 + 1100
= 4600
Q.10 Find the product by suitable rearrangement:
(a) 2 × 1768 × 50
(b) 4 × 166 × 25
(c) 8 × 291 × 125
(d) 625 × 279 × 16
(e) 285 × 5 × 60
(f) 125 × 40 × 8 × 25
Ans
(a) 2 × 1768 × 50 = 1768 ×(2 × 50)
= 1768 × 100
= 176800
(b) 4 × 166 × 25 = 166 × (4 × 25)
= 166 × 100
= 16600
(c) 8 × 291 × 125 = 291 × (8 × 125)
= 291 × 1000 = 291000
(d) 625 × 279 × 16 = 279 × (625 × 16)
= 279 × 10000
= 2790000
(e) 285 × 5 × 60 = 285 × (5 × 60)
= 285 × 300
= 85500
(f) 125 × 40 × 8 × 25 = (125 × 8) × (40 × 25)
= 1000 × 1000
= 1000000
Q.11 Find the value of the following:
(a) 297 × 17 + 297 × 3
(b) 54279 × 92 + 8 × 54279
(c) 81265 × 169 – 81265 × 69
(d) 3845 × 5 × 782 + 769 × 25 × 218
Ans
(a) 297 × 17 + 297 × 3 = 297 × (17 + 3)
= 297 × 20
= 5,940
(b) 54279 × 92 + 8 × 54279
= 54279 × (92 + 8)
= 54279 × 100
= 54,27,900
(c) 81265 × 169 – 81265 × 69
= 81265 × (169 – 69)
= 81265 × 100
= 81,26,500
(d) 3845 × 5 × 782 + 769 × 25 × 218
= 19225 × 782 + 19225 × 218
= 19225 × (782 + 218)
= 19225 × 1000
= 1,92,25,000
Q.12 Find the product using suitable properties.
(a) 738 × 103 (b) 854 × 102
(c) 258 × 1008 (d) 1005 × 168
Ans
(a) 738 × 103 = 738 × (100 + 3)
= (738 × 100) + (738 × 3)
[By the distributivity of multiplication over addition]
= 73800 + 2214
= 76014
(b) 854 × 102 = 854 × (100 + 2)
= (854 × 100) + (854 × 2)
[By the distributivity of multiplication over addition]
= 85400 + 1708
= 87,108
(c) 258 × 1008 = 258 × (1000 + 8)
= (258 × 1000) + (258 × 8)
[By the distributivity of multiplication over addition]
= 258000 + 2064
= 2,60,064
(d) 1005 × 168 = 168 × (1000 + 5)
= (168 × 1000) + (168 × 5)
[By the distributivity of multiplication over addition]
= 168000 + 840
= 1,68,840
Q. 13 Which of the following will not represent zero:(a)1+0 (b)0×0 (c) 02 (d)10102
Ans
Option (a) will not represent zero.
Q.14 Match the following:
(i) 425 × 136 = 425 × (6 + 30 +100) (a) Commutativity under multiplication.
(ii) 2 × 49 × 50 = 2 × 50 × 49 (b) Commutativity under addition.
(iii) 80 + 2005 + 20 = 80 + 20 + 2005 (c) Distributivity of multiplication over addition.
Ans
The matched pairs are:
(i) → (c)
(ii) → (a)
(iii) → (b)
Q.15 If the product of two whole numbers is zero, can we say that one or both of them will be zero? Justify through examples.
Ans
Yes, product of two whole numbers is zero if one or both of them will be zero.
Example: 0 × 2 = 0
0 × 0 = 0
Q.16 If the product of two whole numbers is 1, can we say that one or both of them will be 1? Justify through examples.
Ans
Product of two whole numbers will be 1 if and only if both of them will be 1. For example: 1 × 1 = 1.
Q.17 Find using distributive property:
(a) 728 × 101 (b) 5437 × 1001
(c) 824 × 25 (d) 4275 × 125
(e) 504 × 35
Ans
(a) 728 × 101 = 728 × (100 + 1)
= 728 × 100 + 728 × 1
= 72800 + 728
= 73,528
(b) 5437 × 1001 = 5437 × (1000 + 1)
= 5437 × 1000 + 5437 × 1
= 5437000 + 5437
= 54,42,437
(c) 824 × 25 = 824 × (20 + 5)
= 824 × 20 + 824 × 5
= 16480 + 4120
= 20,600
(d) 4275 × 125 = 4275 × (100 + 20 + 5)
= 4275 × 100 + 4275 × 20 + 4275 × 5
= 427500 + 85500 + 21375
= 5,34,375
(e) 504 × 35 = 35 × (500 + 4)
= 35 × 500 + 35 × 4
= 17500 + 140
= 17,640
Q.18 Study the pattern:
1 × 8 + 1 = 9 1234 × 8 + 4 = 9876
12 × 8 + 2 = 98 12345 × 8 + 5 = 98765
123 × 8 + 3 = 987
Write the next two steps. Can you say how the pattern works?
Ans
The next two steps are:
123456 × 8 + 6 = 987654
1234567 × 8 + 7 = 9876543
The working pattern is as given below:
1 × 8 + 1 = 9
12 × 8 + 2 = (11 + 1) × 8 + 2
= 11 × 8 + 1 × 8 + 2
= 88 + 9 + 1
= 88 + first term + 1
= 98
123 × 8 + 3 = (111 + 11 + 1) × 8 + 3
= 111 × 8 + 11 × 8 + 1 × 8 + 3
= 888 + 88 + 8 + 2 + 1
= 888 + second term + 1
= 987
1234 × 8 + 4 = (1111 + 111 + 11 + 1) × 8 + 4
= 1111 × 8 + 111 × 8 + 11 × 8 + 1 × 8 + 4
= 8888 + 888 + 88 + 8 + 3 + 1
= 8888 + third term + 1
= 9876
12345 × 8 + 5 = (11111 + 1111 + 111 + 11 + 1) × 8 + 5
= 11111 × 8 +1111 × 8 + 111 × 8 +11 × 8
+ 1 × 8 + 4 + 1
= 88888 + 8888 + 888 + 88 + 8 + 4 + 1
= 88888 + fourth term + 1
= 98765
In this way we can follow this pattern to find next terms.
Q.19 A taxi driver filled his car petrol tank with 40 litres of petrol on Monday. The next day, he filled the tank with 50 litres of petrol. If the petrol costs ₹44 per litre, how much did he spend in all on petrol?
Ans
The cost of one litre petrol = ₹44
Petrol filled in tank on Monday = 40 litres
Petrol filled in tank on Tuesday = 50 litres
Driver spend money on petrol = 44 × (40+50)
= 44 × 90
= ₹3960
Thus, a taxi driver spent ₹3960 on petrol.
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FAQs (Frequently Asked Questions)
The numbers that include natural numbers and zero. Not a fraction or decimal. Examples of whole numbers are 0,1,2,3,4,5, etc. The notion of whole numbers includes all natural numbers that begin with 1. Whole numbers contain both positive integers and zero. Visit the Extramarks site for more information and help. Even if you think the problems in this chapter are trivial, you must practise them thoroughly. Exam preparation is essential!
You can find the solutions on Extramarks. The answer to all the in text exercises as well as chapter end exercises are available for the students. In case any student wants further practice to become more confident about the underlying concept besides and be thorough with new concepts, a range of modules and sample papers are all available on Extramarks website and mobile app.
Number zero is a nonnegative integer, a whole number, and a real number. It’s neither a counting, odd, positive natural or negative whole number,nor a complex number. In numerous respects, zero is difficult to categorise. However, it can be employed in a variety of complexvalued equations.
Zero is a nonnegative integer categorised as a whole number, a real number, or a fraction. It is neither an undercounting number, nor an odd, positive natural, or negative whole number, nor a complex number. It isn’t easy to divide zero into different categories. It can, however, be used in a variety of complexnumber equations.
Definitely. You must practise all questions because they cover a wide range of topics and concepts and will provide you with a thorough idea about the types of questions that may be asked and the overall pattern of the question paper. Every exercise should be reworked diligently.
The numbers have six features in general. There will, however, be no requirement that all members satisfy all of the properties. Some of the qualities were met, while others were not. The following are the attributes available:
 Closure property
 Associative property
 Commutative property
 Distributive property
 Identity property
 Inverse property etc.
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