# NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers

NCERT solutions for class 6 maths chapter 2 Whole Numbers This chapter is a part of the number system unit. It will specifically deal with the whole numbers. The whole number is a new topic and it is important to learn also. It starts with the introduction of whole numbers and revision of natural numbers and then goes on with discussions of the predecessor and successor of whole numbers and natural numbers. CBSE NCERT solutions for class 6 maths chapter 2 Whole Numbers are covering the solutions for the questions from every concept. The other important topics of this chapter are a representation of whole numbers in the number line, properties of whole numbers and patterns in whole numbers. Before coming to this chapter you must ensure that you are comfortable with the natural numbers. The natural numbers are the numbers 1, 2, 3, 4, 5... which you use for counting. The natural numbers along with zero forms the collection of the whole numbers. CBSE NCERT solutions for class 6 maths chapter 2 Whole Numbers are covering in-depth solutions to every problem related to the whole numbers. In this particular chapter, there are 3 exercises containing 38 questions. Solutions of NCERT for class 6 maths chapter 2 Whole Numbers contains all the 38 problems' solution to make your preparation better. Along with all these, you can click on the link given to get NCERT solutions for other classes and subjects.

The solutions for following exercise are given

Exercise 2.1

Exercise 2.2

Exercise 2.3

## NCERT solutions for class 6 maths chapter 2 Whole Numbers Topic: Predecessor and Successor

The predecessor and successor are:

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

Every natural number has a predecessor.

Although it is interesting to know that the predecessor of 1 is not a natural number.

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

Yes, all natural numbers are whole numbers.

But, all whole numbers are not natural numbers.

Natural numbers = $1, 2, 3, 4, .....$

Whole numbers= $\mathbf{0},1,2,3,4, ....$

No, all whole numbers are not natural numbers.

0 is a whole number, but it is not a natural number.

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.

## Solutions for NCERT class 6 maths chapter 2 Whole Numbers Exercise: 2.1

Given, $10999$

The next three natural numbers are:

$10999+1= 11000$

$10999+2= 11001$

$10999+3= 11002$

Given,

$10001$

Three whole numbers occurring just before are:

$10001-1 = 10000$

$10001-2 = 9999$

$10001-3 = 9998$

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

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 $32\ and\ 53$

The successor of following numbers are:

(a) 2440701

$2440701+1 = 2440702$

(b) 100199

$100199 +1 = 100200$

(c) 1099999

$1099999+1 = 1100000$

(d) 2345670

$2345670+1 = 2345671$

The predecessor of the following numbers are:

(a) 94

$94-1=93$

(b) 10000

$10000 -1=9999$

(c) 208090

$208090 -1=208089$

(d) 7654321

$7654321-1=7654320$

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

7 + 18 + 13; 16 + 12 + 4

## NCERT solutions for class 6 maths chapter 2 Whole Numbers Exercise: 2.2

Sum by suitable rearrangement:

(a) 837 + 208 + 363

$837 + 208 + 363 =837 + 363+ 208$

$= (837 + 363)+ 208$

$= 1200 + 208$

$= 1408$

(b) 1962 + 453 + 1538 + 647

$1962 + 453 + 1538 + 647 = 1962+ 1538 + 453 + 647$

$= (1962+ 1538) + (453 + 647)$

$= 3500 + 1100$

$= 4600$

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.

$\\ = 19225 \times( 782 + 218) \\ = 19225 \times1000 \\ = 19225000$

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.

$\\ = 1000\times 168+5 \times 168 \\ = 168000+840 \\ = 168840$

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 = $Rs.\ (44\times80)$

$= Rs.\ 3520$

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

$1,5,7,11,13$ can be shown only as a line.

They cannot be shown as rectangle or square or triangle.

$4$ and $9$ can be shown as squares.

4: 2 rows and 2 columns.

9: 3 rows and 3 columns

$4, 6, 8,10, 12$ can be shown as rectangles.

(Note: We are not counting squares as recatangles here)

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

Give at least five other such examples.

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.

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.)

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

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 mathematics chapter wise

 Chapters No. Chapters Name Chapter - 1 NCERT solutions for class 6 maths chapter 1 Knowing Our Numbers Chapter - 2 Solutions of NCERT for class 6 maths chapter 2 Whole Numbers Chapter - 3 CBSE NCERT solutions for class 6 maths chapter 3 Playing with Numbers Chapter - 4 NCERT solutions for class 6 maths chapter 4 Basic Geometrical Ideas Chapter - 5 Solutions of NCERT for class 6 maths chapter 5 Understanding Elementary Shapes Chapter - 6 CBSE NCERT solutions for class 6 maths chapter 6 Integers Chapter - 7 NCERT solutions for class 6 maths chapter 7 Fractions Chapter - 8 Solutions of NCERT for class 6 maths chapter 8 Decimals Chapter - 9 CBSE NCERT solutions for class 6 maths chapter 9 Data Handling Chapter -10 NCERT solutions for class 6 maths chapter 10 Mensuration Chapter -11 Solutions of NCERT for class 6 maths chapter 11 Algebra Chapter -12 CBSE NCERT solutions for class 6 maths chapter 12 Ratio and Proportion Chapter -13 NCERT solutions for class 6 maths chapter 13 Symmetry Chapter -14 Solutions of NCERT for class 6 maths chapter 14 Practical Geometry

## NCERT solutions for class 6 subject wise

 NCERT Solutions for class 6 maths Solutions of NCERT for class 6 science

## How to use NCERT solutions for class 6 maths chapter 2 Whole Numbers?

• Please ensure you must have done the previous chapter.

• Go through the concepts and observations given in the NCERT textbook.

• Take a look through some examples to understand the pattern to answer a specific question.

• Implement all the learning while doing the exercise problems.

• While doing the exercise problem you can assist yourself by using NCERT solutions for class 6 maths chapter 2 Whole Numbers.

Keep Working hard and happy learning!

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