# NCERT Solutions For Class 6 Maths Chapter 2 : Whole Numbers

Contents

NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers are complete study materials for students preparing for the Class 6 math test. Knowledge Glow specialists created these NCERT Solutions to help students understand the fundamental topics covered in this chapter. It includes solved questions for each exercise, as well as crucial formulas, quick suggestions, and the best approaches for solving the problems.

Students are thus recommended to tackle NCERT textbook problems in order to adequately prepare for their final test. Solving more problems would offer students an idea of the types of questions that might be given in the chapter ‘Whole Numbers’ exam.

NCERT Solutions for Class 6 Chapter 2: Whole Numbers are available.

## Exercise 2.1

1. Write the next three natural numbers after 10999.

Solutions:

The next three natural numbers after 10999 are 11000, 11001 and 11002

2. Write the three whole numbers that come before 10001.

Solutions:

The three whole numbers occurring just before 10001 are 10000, 9999 and 9998

3. Which of the following is the smallest whole number?

Solutions:

The smallest whole number is 0

4. What are the total number of whole numbers between 32 and 53?

Solutions:

The whole numbers between 32 and 53 are

(33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52)

Hence, there are 20 whole numbers between 32 and 53

5. Write the successor of:

(a) 2440701 (b) 100199 (c) 1099999 (d) 2345670

Solutions:

The successors are

(a) 2440701 + 1 = 2440702

(b) 100199 + 1 = 100200

(c) 1099999 + 1 = 1100000

(d) 2345670 + 1 = 2345671

6. Write the predecessor of:

(a) 94 (b) 10000 (c) 208090 (d) 7654321

Solutions:

The predecessors are

(a) 94 – 1 = 93

(b) 10000 – 1 = 9999

(c) 208090 – 1 = 208089

(d) 7654321 – 1 = 7654320

7. Identify which whole number is to the left of the other number on the number line in each of the following pairs of numbers. Also, provide the proper indication (>, ) between them.

(a) 530, 503 (b) 370, 307 (c) 98765, 56789 (d) 9830415, 10023001

Solutions:

(a) Since, 530 > 503

Hence, 503 is on the left side of 530 on the number line

(b) Since, 370 > 307

Hence, 307 is on the left side of 370 on the number line

(c) Since, 98765 > 56789

Hence, 56789 is on the left side of 98765 on the number line

(d) Since, 9830415 < 10023001

Hence, 9830415 is on the left side of 10023001 on the number line

8. Which of the following statements is correct (T) and which is incorrect (F)?

(a) The smallest natural number is zero.

Solutions:

False

The number 0 is not a natural number.

(b) 400 is the predecessor of 399.

Solution:

False

The predecessor of 399 is 398 Since, (399 – 1 = 398)

(c) The smallest entire number is zero.

Solution:

True

The smallest whole number is zero.

(d) 600 is 599’s successor.

Solution:

True

Since (599 + 1 = 600)

(e) All natural numbers are integers.

Solution:

True

Natural numbers are all entire numbers.

(f) Natural numbers are all whole numbers.

Solution:

False

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

(g) A two-digit number’s predecessor is never a single digit number.

Solution:

False

For instance, the precursor of 10 is 9.

(h) The smallest whole number is 1.

Solution:

False

The smallest entire number is 0.

(i) There is no precedence for the natural number 1.

True

0 is the predecessor of 1, however it is not a natural number.

(j) There is no precedence for the full number 1.

Solution:

False

0 is a full number that comes before 1.

## Exercise 2.2

1. Determine the sum using appropriate rearrangement:

(a) 837 + 208 + 363

(b) 1962 + 453 + 1538 + 647

Solutions:

(a) Given 837 + 208 + 363

= (837 + 363) + 208

= 1200 + 208

= 1408

(b) Given 1962 + 453 + 1538 + 647

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

= 3500 + 1100

= 4600

2. Find the product through appropriate 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

Solutions:

(a) Given 2 × 1768 × 50

= 2 × 50 × 1768

= 100 × 1768

= 176800

(b) Given 4 × 166 × 25

= 4 × 25 × 166

= 100 × 166

= 16600

(c) Given 8 × 291 × 125

= 8 × 125 × 291

= 1000 × 291

= 291000

(d) Given 625 × 279 × 16

= 625 × 16 × 279

= 10000 × 279

= 2790000

(e) Given 285 × 5 × 60

= 285 × 300

= 85500

(f) Given 125 × 40 × 8 × 25

= 125 × 8 × 40 × 25

= 1000 × 1000

= 1000000

3. 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

Solutions:

(a) Given 297 × 17 + 297 × 3

= 297 × (17 + 3)

= 297 × 20

= 5940

(b) Given 54279 × 92 + 8 × 54279

= 54279 × 92 + 54279 × 8

= 54279 × (92 + 8)

= 54279 × 100

= 5427900

(c) Given 81265 × 169 – 81265 × 69

= 81265 × (169 – 69)

= 81265 × 100

= 8126500

(d) Given 3845 × 5 × 782 + 769 × 25 × 218

= 3845 × 5 × 782 + 769 × 5 × 5 × 218

= 3845 × 5 × 782 + 3845 × 5 × 218

= 3845 × 5 × (782 + 218)

= 19225 × 1000

= 19225000

4. Find the product using suitable properties.

(a) 738 × 103

(b) 854 × 102

(c) 258 × 1008

(d) 1005 × 168

Solutions:

(a) Given 738 × 103

= 738 × (100 + 3)

= 738 × 100 + 738 × 3 (using distributive property)

= 73800 + 2214

= 76014

(b) Given 854 × 102

= 854 × (100 + 2)

= 854 × 100 + 854 × 2 (using distributive property)

= 85400 + 1708

= 87108

(c) Given 258 × 1008

= 258 × (1000 + 8)

= 258 × 1000 + 258 × 8 (using distributive property)

= 258000 + 2064

= 260064

(d) Given 1005 × 168

= (1000 + 5) × 168

= 1000 × 168 + 5 × 168 (using distributive property)

= 168000 + 840

= 168840

5. On Monday, a taxi driver filled his car’s gas tank with 40 gallons. He filled the tank with 50 litres of gasoline the next day. How much did he spend on gasoline overall if it costs 44 cents a litre?

Solutions:

Monday’s gasoline consumption was 40 litres.

Tuesday’s gasoline consumption was 50 litres.

Total amount of gasoline filled = (40 + 50) litres

Petrol price per litre = 44

Total amount spent = 44 (40 + 50).

= 44 × 90

= ₹ 3960

6. A seller provides a hotel with 32 litres of milk in the morning and 68 litres of milk in the evening. How much money is owed to the seller per day if the milk costs 45 cents per litre?

Solutions:

The amount of milk delivered in the morning is 32 litres.

The amount of milk delivered in the evening is 68 litres.

Milk price per litre = 45

Total cost of milk per day = 45 × (32 + 68)

= 45 × 100

= ₹ 4500

As a result, the vendor is owed \$4500 per day.

7. 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.

Solutions:

(i) 425 × 136 = 425 × (6 + 30 + 100) (c) Distributivity of multiplication over addition.

Hence (c) is the correct answer

(ii) 2 × 49 × 50 = 2 × 50 × 49 (a) Commutativity under multiplication

Hence, (a) is the correct answer

(iii) 80 + 2005 + 20 = 80 + 20 + 2005 (b) Commutativity under addition

Hence, (b) is the correct answer

## Exercise 2.3 PAGE NO: 43

1. Which of the following will not represent zero:

(a) 1 + 0

(b) 0 × 0

(c) 0 / 2

(d) (10 – 10) / 2

Solutions:

(a) 1 + 0 = 1

Hence, it does not represent zero

(b) 0 × 0 = 0

Hence, it represents zero

(c) 0 / 2 = 0

Hence, it represents zero

(d) (10 – 10) / 2 = 0 / 2 = 0

Hence, it represents zero

2. Can we argue that if the product of two whole numbers is zero, one or both of them will be zero? Exemplification is required.

Solutions:

If the product of two whole numbers is zero, one of them must be zero.

For example, 0 3 = 0 and 15 0 = 0

If the product of two whole numbers is zero, then both may be zero.

For instance, 0 0 Equals 0.

Yes, if the product of two whole numbers is 0, they are both zero.

3. Can we argue that if the product of two whole numbers is 1, one or both of them will be 1? Exemplification is required.

Solutions:

If the product of two whole numbers is one, both numbers should be one.

Example: 1 × 1 = 1

But 1 × 5 = 5

As a result, it is obvious that the product of two whole integers will be 1 only when both numbers to be multiplied are 1.

4. Determine using the distributive property:

(a) 728 × 101

(b) 5437 × 1001

(c) 824 × 25

(d) 4275 × 125

(e) 504 × 35

Solutions:

(a) Given 728 × 101

= 728 × (100 + 1)

= 728 × 100 + 728 × 1

= 72800 + 728

= 73528

(b) Given 5437 × 1001

= 5437 × (1000 + 1)

= 5437 × 1000 + 5437 × 1

= 5437000 + 5437

= 5442437

(c) Given 824 × 25

= (800 + 24) × 25

= (800 + 25 – 1) × 25

= 800 × 25 + 25 × 25 – 1 × 25

= 20000 + 625 – 25

= 20000 + 600

= 20600

(d) Given 4275 × 125

= (4000 + 200 + 100 – 25) × 125

= (4000 × 125 + 200 × 125 + 100 × 125 – 25 × 125)

= 500000 + 25000 + 12500 – 3125

= 534375

(e) Given 504 × 35

= (500 + 4) × 35

= 500 × 35 + 4 × 35

= 17500 + 140

= 17640

5. 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?

(Hint: 12345 = 11111 + 1111 + 111 + 11 + 1)

Solutions:

123456 × 8 + 6 = 987654

1234567 × 8 + 7 = 9876543

Given 123456 = (111111 + 11111 + 1111 + 111 + 11 + 1)

123456 × 8 = (111111 + 11111 + 1111 + 111 + 11 + 1) × 8

= 111111 × 8 + 11111 × 8 + 1111 × 8 + 111 × 8 + 11 × 8 + 1 × 8

= 888888 + 88888 + 8888 + 888 + 88 + 8

= 987648

123456 × 8 + 6 = 987648 + 6

= 987654

Yes, here the pattern works

1234567 × 8 + 7 = 9876543

Given 1234567 = (1111111 + 111111 + 11111 + 1111 + 111 + 11 + 1)

1234567 × 8 = (1111111 + 111111 + 11111 + 1111 + 111 + 11 + 1) × 8

= 1111111 × 8 + 111111 × 8 + 11111 × 8 + 1111 × 8 + 111 × 8 + 11 × 8 + 1 × 8

= 8888888 + 888888 + 88888 + 8888 + 888 + 88 + 8

= 9876536

1234567 × 8 + 7 = 9876536 + 7

= 9876543

Yes, here the pattern works

## Frequently Asked Questions on NCERT Solutions for Class 6 Maths Chapter 2

1. According to Chapter 2 of NCERT Solutions for Class 6 Maths, what is a number line?

A number line, according to the NCERT Solutions for Class 6 Maths Chapter 2, is a drawing of a graduated straight and horizontal line on which numbers are inscribed. A number printed on the left side of the number line is smaller, whereas a number written on the right side of the number line is larger.

In NCERT Solutions for Class 6 Maths Chapter 2, explain the qualities of the whole number.

Whole numbers have the following properties: 1. Addition and multiplication of any two whole numbers result in a whole number.

2. The addition and division of any two whole numbers may or may not result in a whole number.

Students are given solved examples before each exercise-based problem to help them understand how to solve problems in a shorter amount of time. Students will strengthen their conceptual understanding by solving problems from the NCERT textbook, which is required to score well in the yearly exams.

3. What does multiplicative identity mean in Chapter 2 of NCERT Solutions for Class 6 Maths?

When one whole number is multiplied by another whole number, the result is the same whole number. As a 1 = a, 1 is the Multiplicative identity (a is any whole number). According to the CBSE syllabus, NCERT Solutions provide students with accurate definition and understanding of relations. Several examples provided in the solutions will assist students in tackling relational problems with ease.