NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers

# NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers

Edited By Ramraj Saini | Updated on Nov 29, 2023 12:44 PM IST

NCERT solutions for class 6 maths chapter 2 Whole Numbers provided here that are created by expert team at careers360. These solutions are developed considering the latest syllabus of CBSE. Students looking for NCERT Class 6 solution for Maths chapter 2 pdf can access the solutions for all topics in this article. The Whole Number chapter is a part of the number system unit. Referring to NCERT 6 Maths solution chapter 2 aid to clear doubts and prepare well. CBSE NCERT solutions for whole numbers class 6 are covering the solutions for the questions from every concept of NCERT. Representation and properties of whole numbers are important topics.

Class 6 NCERT Maths solutions of chapter 2 are prepared as per questions available in NCERT Class 6 Syllabus. There are 3 exercises in this chapter containing 38 questions. NCERT Class 6 maths solution chapter 2 contains the answer to all the 38 problems. CBSE NCERT solutions for Class 6 Maths chapter 2 Whole Numbers are answering every problem related to the whole numbers. Along with all these, you can click on the link given to get NCERT Solutions for other classes and subjects.

## NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers - Important Formula

1. Whole numbers: 0, 1, 2, 3, and so on.

2. A factor of a number is a number that divides the other number without leaving a remainder.

3. A multiple of a number is a number that is exactly divisible by the given number.

4. The number 1 is a factor of every number and has only one factor.

5. Even numbers are those that are divisible by 2, while odd numbers are not divisible by 2.

6. Divisibility rules:

• A number is divisible by 2 if its unit digit is 0, 2, 4, 6, or 8.

• A number is divisible by 3 if the sum of its digits is divisible by 3.

• A number is divisible by 4 if the digits in its tens and units place are divisible by 4.

• A number is divisible by 5 if its unit digit is 0 or 5.

• A number is divisible by 6 if it satisfies the divisibility rules for both 2 and 3.

• A number is divisible by 8 if the number formed by its hundreds, tens, and units placed is divisible by 8.

• A number is divisible by 9 if the sum of its digits is divisible by 9.

• A number is divisible by 10 if its units place digit is 0.

• A number is divisible by 11 if the difference between the sum of its digits in odd places and the sum of its digits in even places is either 0 or divisible by 11.

7. The LCM (Least Common Multiple) of two numbers, a and b, is the smallest positive integer that is divisible by both a and b.

8. The HCF (Highest Common Factor) of two numbers, a and b, is the largest positive integer that divides both a and b.

9. LCM(A, B) HCF(A, B) = AB

10. Properties of whole numbers:

• Closure Property of Addition: a + b is a whole number

• Closure Property of Multiplication: a × b is a whole number

(a + b) + c = a + (b + c)

• Associativity of Multiplication:
a × (b × c) = (a × b) × c

• Distributive Property of Multiplication over Addition:
a × (b + c) = a × b + a × c

• Distributive Property of Multiplication over Subtraction:
a × (b - c) = a × b - a × c.

• Existence of Multiplicative Identity:
a + 0 = a = 0 + a

• Existence of Multiplicative Identity:
a × 0 = 0 = 0 × a

• Unit Multiplication:
a × 1 = a = 1 × a

## NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers - Important Topics

Natural Numbers: The numbers 1, 2, 3,... which we use for counting are known as natural numbers. Natural numbers represent the set of positive integers starting from 1 and extending infinitely. They are the most basic numbers used for counting and ordering objects.

Successor and Predecessor: If you add 1 to a natural number, we get its successor. If you subtract 1 from a natural number, you get its predecessor. Successor and predecessor are terms used to denote the number that comes after or before a given number in the natural number sequence.

Existence of Successor and Predecessor: Every natural number has a successor, meaning that we can always find the next number in the sequence by adding 1. Every natural number except 1 has a predecessor, which is the number that comes before it in the sequence.

Whole Numbers: If we add the number zero to the collection of natural numbers, we get the collection of whole numbers. Whole numbers include 0 along with the set of natural numbers. Thus, the numbers 0, 1, 2, 3,... form the collection of whole numbers.

Successor and Predecessor of Whole Numbers: Every whole number has a successor, meaning there is always a next number in the sequence. Every whole number except zero has a predecessor, which is the number that comes before it in the sequence.

Relationship between Natural Numbers and Whole Numbers: All natural numbers are whole numbers because they are included in the set of whole numbers. However, all whole numbers are not natural numbers since zero is a whole number but not a natural number.

Number Line Representation: A number line is a line marked with points at equal intervals, starting from 0 and extending in both positive and negative directions. It provides a visual representation of numbers. Natural and whole numbers can be represented on a number line, allowing for easy visualization and operations such as addition, subtraction, and multiplication.

Operations on Number Line: Addition on a number line corresponds to moving to the right, whereas subtraction corresponds to moving to the left. Multiplication corresponds to making jumps of equal distance starting from zero. Number lines aid in understanding and performing these operations.

Closure Property: Adding two whole numbers always gives a whole number. Similarly, multiplying two whole numbers always gives a whole number. This property is known as closure, which means that a particular operation performed on two whole numbers results in another whole number. However, whole numbers are not closed under subtraction and division.

Division by Zero: Division by zero is not defined. It is considered undefined because it leads to contradictory or nonsensical results. Dividing any number by zero is not permissible in mathematics.

Free download NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers PDF for CBSE Exam.

## NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers (Intext Questions and Exercise)

NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Topic: Predecessor and Successor

Answer: The predecessor and successor of:

• 19:
• Predecessor : 18
• Successor : 20
• 1997:
• Predecessor : 1996
• Successor : 1998
• 12000
• Predecessor : 11999
• Successor : 12001
• 49:
• Predecessor : 48
• Successor : 50
• 100000:
• Predecessor : 99999
• Successor : 100001

Answer: Every natural number has a predecessor. Although, it is interesting to know that the predecessor of 1 is not a natural number.

Answer: Every natural number has a successor. There is no last natural number. There are infinite natural numbers.

NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Topic: Whole Numbers

Answer: Yes, all the natural numbers are whole numbers. But, all whole numbers are not natural numbers.
Natural numbers = 1, 2, 3, 4, ....

Whole numbers= 1, 2, 3, 4, ....

Answer: No, all whole numbers are not natural numbers. 0 is a whole number, but it is not a natural number.

Answer: There are infinite whole numbers. Hence, there is no greatest whole number. Every whole number you can think of has a successor, which is greater than than the number.

NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Exercise: 2.1

The next three natural numbers are:
10999 + 1 = 11000
10999 + 2 = 11001
10999 + 3 = 11002

Three whole numbers occurring just before are:
11001 - 1 = 10000
11001 - 2 = 9999
11001 - 3 = 9998

Answer: The smallest whole number is 0. It has no whole number predecessor.

Answer: Given numbers are: 32 and 53

Number of whole numbers between 32 and 53 = (53-32) - 1 = 21 - 1 = 20

There are 20 whole numbers between 32 and 53

Answer: The successor of following numbers are:

• (a) 2440701
• $2440701+1 = 2440702$
• (b) 100199
• $100199 +1 = 100200$
• (c) 1099999
• $1099999+1 = 1100000$
• (d) 2345670
• $2345670+1 = 2345671$

Answer: The predecessor of the following numbers are:

• (a) 94
• $94-1=93$
• (b) 10000
• $10000 -1=9999$
• (c) 208090
• $208090 -1=208089$
• (d) 7654321
• $7654321-1=7654320$

Answer: The number on the left on the number line is smaller than the number that is on the right on the number line.

• (a) 530, 503
• $503$ is on the left.
• $\therefore 530 > 503$
• (b) 370, 307
• $307$ is on the left.
• $\therefore 370> 307$
• (c) 98765, 56789
• $56789$ is on the left.
• $\therefore 98765> 56789$
• (d) 9830415, 10023001
• $9830415$ is on the left.
• $\therefore 9830415< 10023001$

(a) Zero is the smallest natural number. - False. 0 is not a natural number.
(b) 400 is the predecessor of 399. - False. 400 is the successor of 399.
(c) Zero is the smallest whole number. - True.
(d) 600 is the successor of 599. - True
(e) All natural numbers are the whole numbers.- True.
(f) All whole numbers are natural numbers.- False. 0 is a whole number but not a natural number.
(g) The predecessor of a two-digit number is never a single-digit number.- False. The predecessor of 10 is 9.
(h) 1 is the smallest whole number. - False. 0 is the smallest whole number.
(i) The natural number 1 has no predecessor. - True.
(j) The whole number 1 has no predecessor. - False. The whole number 1 has 0 as its predecessor.
(k) The whole number 13 lies between 11 and 12.- False. The whole number 13 lies on the right side of 12 on the number line.
(l) The whole number 0 has no predecessor.- True.
(m) The successor of a two-digit number is always a two-digit number- False. The successor of 99 is 100.

NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Topic: Properties of Whole Numbers

Answer: 7 + 18 + 13; 16 + 12 + 4

NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Exercise: 2.2

• (a) 837 + 208 + 363
$837 + 208 + 363 =837 + 363+ 208$
$= (837 + 363)+ 208$
$= 1200 + 208$
$= 1408$
• (b) 1962 + 453 + 1538 + 647
$1962 + 453 + 1538 + 647 = 1962+ 1538 + 453 + 647$
$= (1962+ 1538) + (453 + 647)$
$= 3500 + 1100$
$= 4600$

Answer: The product of the following by suitable rearrangement are:

• (a) $2 \times 1768 \times 50$
$\\ = 2 \times 50\times 1768 \\ = (2 \times 50)\times 1768 \\ = 100\times 1768 \\ = 176800$
• (b) $4 \times 166 \times 25$
$\\ = 4\times 25 \times 166 \\ = (4\times 25 )\times 166 \\ = 100 \times 166 \\ = 16600$
• (c) $8 \times 291 \times125$
$\\ = 8 \times125 \times 291 \\ =( 8 \times125) \times 291 \\ = 1000\times 291 \\ = 291000$
• (d) $625 \times 279 \times 16$
$\\ = 625 \times 16\times 279 \\ = (625 \times 16)\times 279 \\ = 10000\times 279 \\ = 2790000$
• (e) $285 \times 5 \times 60$
$\\ = 285 \times (5 \times 60) \\ = 285 \times 300 \\ = 85500$
• (f) $125 \times 40 \times 8 \times 25$
$\\ = 125\times 8 \times 40 \times 25 \\ = (125\times 8 )\times (40 \times 25) \\ = 1000 \times 1000 \\ = 1000000$

• (a) $297 \times 17 + 297 \times 3$
Using Distributive law.
$\\ = 297 \times (17 +3) \\ = 297 \times 20 \\ = 5940$
• (b) $54279 \times 92 + 8 \times 54279$
Using Commutative under multiplication
$54279 \times 92 + 54279 \times 8$
Using Distributive law.
$\\ = 54279 \times( 92 +8) \\ = 54279 \times 100 \\ = 5427900$
• (c) $81265 \times 169 - 81265 \times 69$
Using Distributive law.
$\\ = 81265 \times (169 - 69) \\ = 81265 \times 100 \\ = 8126500$
• (d) $3845 \times 5 \times 782 + 769 \times 25 \times 218$
$\\ = (3845 \times 5) \times 782 + (769 \times 25) \times 218 \\ = 19225 \times 782 + 19225\times 218$
Using distributive law.
$\\ = 19225 \times( 782 + 218) \\ = 19225 \times1000 \\ = 19225000$

Answer: The product of the folllowing using suitable properties are:

• (a) $738 \times 103$
$\\ = 738 \times (100+3)$
Using distributive law.
$\\ = 738 \times 100+738 \times3 \\ = 73800+2214 \\ = 76014$
• (b) $854 \times 102$
$\\ = 854 \times (100+2)$
Using distributive law.
$\\ = 854 \times 100+854 \times 2 \\ = 85400+1708 \\ = 87108$
• (c) $258 \times 1008$
$\\ = 258 \times (1000+8)$
Using Distributive law.
$\\ = 258 \times 1000+258 \times8 \\ = 258000+2064 \\ = 260064$
• (d) $1005 \times 168$
$\\ = (1000+5) \times 168$
Using Distributive law.
$\\ = 1000\times 168+5 \times 168 \\ = 168000+840 \\ = 168840$

Answer: Amount of petrol filled on Monday = $40\ litres$
Amount of petrol filled on Tuesday = $40\ litres$
$\therefore$ Total amount of petrol = $(40+40)\ litres = 80\ litres$
Cost of 1 litre of petrol = $Rs.\ 44$
$\therefore$ Cost of $80\ litres$ of petrol = $Rs.\ (44\times80)$
$= Rs.\ 3520$

Answer: Amount of milk supplied in the morning = $32\ litres$
Amount of milk supplied in the evening = $68\ litres$
$\therefore$ Total amount of petrol = $(32+68)\ litres = 100\ litres$
Cost of 1 litre of milk = $Rs.\ 45$
$\therefore$ Cost of $100\ litres$ of milk = $Rs.\ (45\times100)$
$= Rs.\ 4500$

 (i) $425 \times 136 = 425 \times (6 + 30 +100)$ (c) Distributivity of multiplication over addition. (ii) $2 \times 49 \times 50 = 2 \times 50 \times 49$ (a) Commutativity under multiplication. (iii) $80 + 2005 + 20 = 80 + 20 + 2005$ (b) Commutativity under addition.

NCERT solutions for class 6 maths chapter 2 Whole Numbers Topic: Patterns in Whole Numbers

Answer: $1,5,7,11,13$ can be shown only as a line. They cannot be shown as a rectangle or square or triangle.

Answer: $4$ and $9$ can be shown as squares.
4: 2 rows and 2 columns.
9: 3 rows and 3 columns

Answer: $4, 6, 8,10, 12$ can be shown as rectangles. (Note: We are not counting squares as rectangles here)

Answer: 3, 6, 10, 15, 21, 28, 36.

Give at least five other such examples.

Answer: We can represent a number by two rectangles. for example 12 = 3 x 4 or 2 x 6

five other such examples are :

• 24 = 12 x 2 or 24 = 6 x 4
• 18 = 9 x 2 or 18 = 3 x 6
• 15 = 15 x 1 or 15 = 3 x 5
• 30 = 10 x 3 or 30 = 5 x 6
• 40 = 10 x 4 or 40 = 5 x 8.

NCERT solutions for class 6 maths chapter 2 Whole Numbers Exercise: 2.3

• (a) 1 + 0
It does not represent zero.
• (b) 0 × 0
It represents zero.
• (c) $\frac{0}{2}=0$
It represents zero.
• (d) $\frac{10-10}{2}=0$
It represents zero.

Answer: If the product of 2 whole numbers is zero, then one of them is definitely zero.
For example, 0 x 2 = 0 and 17 x 0 = 0
If the product of 2 whole numbers is zero, then both of them may be zero.
0 x 0 = 0
However, 2 x 3 = 6 (Since numbers to be multiplied are not equal to zero, the result of the product will also be non-zero.)

Answer: If the product of 2 numbers is 1, then both the numbers have to equal to 1.
For example, 1 x 1 = 1
However, 1 x 6 = 6
Clearly, the product of two whole numbers will be 1 in the situation when both numbers to be multiplied are 1.

• (a) 728 $\dpi{100} \times$ 101= 728 $\dpi{100} \times$ (100 + 1)
• = 728 $\dpi{100} \times$ 100 + 728 $\dpi{100} \times$ 1
• = 72800 + 728
• = 73528
• (b) 5437 $\dpi{100} \times$ 1001 = 5437 $\dpi{100} \times$ (1000 + 1)
• = 5437 $\dpi{100} \times$ 1000 + 5437 $\dpi{100} \times$ 1
• = 5437000 + 5437
• = 5442437
• (c) 824 $\dpi{100} \times$ 25 (800 + 24) $\dpi{100} \times$ 25 = (800 + 25 - 1) 25
• =800 $\dpi{100} \times$ 25+25 x 25-1 $\dpi{100} \times$ 25
• = 20000 + 625 - 25
• = 20000 + 600
• = 20600
• (d) 4275 $\dpi{100} \times$ 125 = (4000 + 200 + 100 - 25) $\dpi{100} \times$ 125
• = 4000 $\dpi{100} \times$ 125 + 200 $\dpi{100} \times$ 125 + 100 $\dpi{100} \times$ 125 - 25 $\dpi{100} \times$ 125
• = 500000 + 25000 + 12500 - 3125
• = 534375
• (e) 504 $\dpi{100} \times$ 35 = (500 + 4) $\dpi{100} \times$ 35
• = 500 x 35 +4 $\dpi{100} \times$ 35
• = 17500 + 140
• = 17640

Answer: 123456 $\dpi{100} \times$ 8 + 6 = 987648 + 6 = 987654
1234567 $\dpi{100} \times$ 8 + 7 = 9876536 + 7 = 9876543
Yes, the pattern works.
As 123456 = 111111 + 11111 + 1111 + 111 + 11 + 1,
123456 $\dpi{100} \times$ 8 = (111111 + 11111 + 1111 + 111 + 11 + 1) $\dpi{100} \times$ 8
= 111111 $\dpi{100} \times$ 8 + 11111 $\dpi{100} \times$ 8 + 1111 $\dpi{100} \times$ 8 + 111 $\dpi{100} \times$ 8 + 11 $\dpi{100} \times$ 8 + 1 $\dpi{100} \times$ 8
= 888888 + 88888 + 8888 + 888 + 88 + 8
= 987648
And,
123456 $\dpi{100} \times$ 8 + 6 = 987648 + 6 = 987648

## NCERT Solutions for Class 6 Subject wise

 NCERT Solutions for Class 6 Maths NCERT Solutions for Class 6 Science

## NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers - Topics

2.1 Identities and Special Properties of Whole Numbers.

2.2 Introduction to Whole Numbers.

2.3 Patterns in Whole Numbers.

2.4 Properties of Whole Numbers.

## NCERT Solutions for Class 6 Mathematics Chapter wise

 Chapters No. Chapters Name Chapter - 1 Chapter - 2 Whole Numbers Chapter - 3 Chapter - 4 Chapter - 5 Chapter - 6 Chapter - 7 Chapter - 8 Chapter - 9 Chapter -10 Chapter -11 Chapter -12 Chapter -13 Symmetry Chapter -14

## Key Features Of NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers

Comprehensive Coverage: The class 6 chapter 2 provides comprehensive coverage of the topic of whole numbers, ensuring that all important concepts and subtopics are included. It covers the definition of whole numbers, their properties, representation on a number line, and operations such as addition, subtraction, and multiplication.

Conceptual Clarity: The chapter focuses on providing conceptual clarity to students. It explains the fundamental concepts of whole numbers in a clear and concise manner, enabling students to grasp the underlying principles and theories effectively.

Problem-solving Approach: The chapter adopts a problem-solving approach, offering numerous examples and practice exercises to reinforce understanding. It encourages students to apply the concepts learned to solve real-life problems and mathematical problems related to whole numbers.

Step-by-step Solutions: The chapter provides step-by-step solutions to exercises and examples, guiding students in the correct approach to solving problems. This systematic approach assists students in understanding the process of solving mathematical problems related to whole numbers.

Practice Exercises: The chapter includes a variety of practice exercises, ranging from basic to challenging, to reinforce the understanding of whole numbers. These exercises allow students to practice and apply their knowledge, enabling them to develop proficiency in working with whole numbers.

Also Check -

NCERT Books and NCERT Syllabus

Keep Working hard and happy learning!

1. What is a number line as per Chapter 2 of NCERT Solutions for Class 6 Maths?

According to NCERT Solutions for Class 6 Maths Chapter 2, a number line is a graphical representation of a straight and horizontal line that is graduated, and numbers are marked on it. Numbers written on the left-hand side of the number line are smaller, while numbers written on the right-hand side are greater.

2. What is the meaning of multiplicative identity in Chapter 2 of NCERT Solutions for Class 6 Maths?

As per NCERT Solutions for Class 6 Maths Chapter 2, the multiplicative identity is a number that, when multiplied by another whole number, gives the same whole number. In this context, the number 1 is the multiplicative identity, as a × 1 = a (where 'a' is any whole number). NCERT Solutions provide students with clear definitions and explanations of mathematical concepts in line with the CBSE syllabus. Several examples presented in the solutions will assist students in solving problems related to mathematical relations with ease.

3. What are the topic covered in NCERT Class 6 Maths Chapter 2

The following topics are covered in the NCERT syllabus Class 6 Maths chapter 2

• Whole Numbers
• The Number Line
• Properties Of Whole Numbers
• Patterns in Whole Numbers
4. Explain the properties of whole numbers in NCERT Solutions for Class 6 Maths Chapter 2.

The properties of whole numbers, as covered in NCERT Solutions for Class 6 Maths Chapter 2, are as follows:

1. Addition and multiplication of any two whole numbers result in a whole number.
2. Subtraction and division of any two whole numbers may or may not result in a whole number.

Students are provided with solved examples before each exercise-wise problem to help them understand the problem-solving methods efficiently. By practicing problems from the NCERT textbook, students can enhance their conceptual understanding, which is essential for performing well in the annual exams.

## Upcoming School Exams

#### National Means Cum-Merit Scholarship

Application Date:22 July,2024 - 31 August,2024

Exam Date:19 September,2024 - 19 September,2024

Exam Date:20 September,2024 - 20 September,2024

Exam Date:26 September,2024 - 26 September,2024

Application Date:30 September,2024 - 30 September,2024

Get answers from students and experts

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9