**RBSE Solutions for Class 6 MATHS Chapter 2** Welcome to the world of RBSE Class 6 Maths! As students embark on their mathematical journey, Chapter 2 is a crucial milestone. This article aims to provide a comprehensive guide to RBSE solutions for Class 6 Maths Chapter 2, ensuring a solid foundation in the subject. Whether you’re a student seeking clarity or a teacher searching for effective teaching aids, this article has got you covered.

## Table of Contents

Understanding the Importance of Chapter 2: Chapter 2 of RBSE Class 6 Maths is designed to introduce students to the fundamental concepts of whole numbers. It lays the groundwork for more complex mathematical concepts that students will encounter in higher classes. A strong grasp of these basics is essential for future academic success, making Chapter 2 a pivotal point in the learning journey.

Access RBSE Solutions for Class 6 Chapter 2: Whole Numbers

## RBSE Solutions for Class 6 MATHS Chapter 2 Exercise 2.1 PAGE No.: 31

**1. Write the next three natural numbers after 10999.**

**Solutions:**

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

170

**2. Write the three whole numbers occurring just before 10001.**

**Solutions:**

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

**3. Which is the smallest whole number?**

**Solutions:**

The smallest whole number is 0.

**4. How many whole numbers are there between 32 and 53?**

**Solutions:**

The whole numbers between 32 and 53 are as follows:

(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

94

(d) 7654321 – 1 = 7654320

**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**

**Solutions:**

(a) 530 > 503

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

(b) 370 > 307

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

(c) 98765 > 56789

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

(d) 9830415 < 10023001

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

**8. Which of the following statements are true (T) and which are false (F)?**

**(a) Zero is the smallest natural number.**

**Solution:**

False

0 is not a natural number.

**(b) 400 is the predecessor of 399.**

**Solution:**

False

The predecessor of 399 is 398 because (399 – 1 = 398)

**(c) Zero is the smallest whole number.**

**Solution:**

True

Zero is the smallest whole number.

**(d) 600 is the successor of 599.**

**Solution:**

True

Since (599 + 1 = 600)

**(e) All natural numbers are whole numbers.**

**Solution:**

True

All natural numbers are whole numbers.

**(f) All whole numbers are natural numbers.**

**Solution:**

False

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

**(g) The predecessor of a two-digit number is never a single-digit number.**

**Solution:**

False

For example, the predecessor of 10 is 9.

**(h) 1 is the smallest whole number.**

**Solution:**

False

0 is the smallest whole number.

**(i) The natural number 1 has no predecessor.**

True

The predecessor of 1 is 0, but it is not a natural number.

**(j) The whole number 1 has no predecessor.**

**Solution:**

False

0 is the predecessor of 1 and is a whole number.

**(k) The whole number 13 lies between 11 and 12.**

**Solution:**

False

13 does not lie between 11 and 12.

**(l) The whole number 0 has no predecessor.**

**Solution:**

True

The predecessor of 0 is -1 and is not a whole number.

**(m) The successor of a two-digit number is always a two-digit number.**

**Solution:**

False

For example, the successor of 99 is 100

## RBSE Solutions for Class 6 MATHS Chapter 2 Exercise 2.2 PAGE No.: 40

**1. Find the sum by suitable 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 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**

**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

110

(e) Given 285 × 5 × 60

= 285 × 300

= 85500

(f) Given 125 × 40 × 8 × 25

= 125 × 8 × 40 × 25

= 1000 × 1000

= 1000000

81

**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. 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?**

**Solutions:**

Petrol quantity filled on Monday = 40 litres

Petrol quantity filled on Tuesday = 50 litres

Total petrol quantity filled = (40 + 50) litre

Cost of petrol per litre = ₹ 44

Total money spent = 44 × (40 + 50)

= 44 × 90

= ₹ 3960

113

**6. A vendor supplies 32 litres of milk to a hotel in the morning and 68 litres of milk in the evening. If the milk costs ₹ 45 per litre, how much money is due to the vendor per day?**

**Solutions:**

Milk quantity supplied in the morning = 32 litres

Milk quantity supplied in the evening = 68 litres

Cost of milk per litre = ₹ 45

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

= 45 × 100

= ₹ 4500

Hence, the money due to the vendor per day is ₹ 4500

**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

## RBSE Solutions for Class 6 MATHS Chapter 2 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. If the product of two whole numbers is zero, can we say that one or both of them will be zero? Justify through examples.**

**Solutions:**

If the product of two whole numbers is zero, definitely one of them is zero

Example: 0 × 3 = 0 and 15 × 0 = 0

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

Example: 0 × 0 = 0

Yes, if the product of two whole numbers is zero, then both of them will be zero.

**3. If the product of two whole numbers is 1, can we say that one or both of them will be 1? Justify through examples.**

**Solutions:**

If the product of two whole numbers is 1, both numbers should be equal to 1

Example: 1 × 1 = 1

But 1 × 5 = 5

Hence, it’s clear that the product of two whole numbers will be 1, only in situations when both numbers to be multiplied are 1.

**4. Find using 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

88

(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.

RBSE Class 6 Maths Chapter 2 is the gateway to a world of mathematical understanding. By providing LEARNING-friendly solutions with comprehensive explanations, we aim to support students and educators in mastering these foundational concepts. Let’s empower the next generation with the tools they need to excel in mathematics and beyond. Happy learning!