RD Sharma Solutions for Class 6 Maths Chapter 4 Exercise 4.6

RD Sharma Solutions for Class 6 Maths Chapter 4 Exercise 4.6 provides detailed explanations and solutions for multiple-choice questions related to whole numbers.

This exercise helps students understand important concepts like the smallest whole number, even and odd numbers, predecessors and successors, divisibility, and basic arithmetic operations using properties like distributive law.

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The questions also cover practical applications such as finding the number of whole numbers between two given numbers and calculating dividends from quotient and remainder. These solutions guide students step-by-step to solve each problem effectively, making it easier for them to grasp key concepts and prepare well for exams.

RD Sharma Solutions for Class 6 Maths Chapter 4 Exercise 4.6 Overview

Exercise 4.6 focuses on various problems related to whole numbers and their properties. It includes multiple-choice questions that test students' understanding of important concepts like identifying the smallest whole number, even and odd numbers, predecessors and successors of numbers, and basic operations such as addition, subtraction, multiplication, and division.

The exercise also covers how to find the number of whole numbers between two given numbers and applies multiplication properties such as distributive law for simplifying calculations.

This exercise helps students practice solving real-life math problems involving whole numbers, develop their problem-solving skills, and improve their understanding of how numbers work in different contexts. Each question is designed to reinforce basic math concepts and prepare students for exams by providing clear, step-by-step solutions.

RD Sharma Solutions for Class 6 Maths Operations on Whole Numbers Chapter 4 Exercise 4.6

Here are the detailed solutions for RD Sharma Class 6 Maths Chapter 4 Exercise 4.6 on  Operations on Whole Numbers. 

Mark the correct alternative in each of the following:

1. Which one of the following is the smallest whole number?
(a) 1                 (b) 2                     (c) 0                                 (d) None of these

Solution:

Option (c) is the correct answer.

We know that the set of whole numbers is {0, 1, 2, 3, 4 …}.

Hence, the smallest whole number is 0.

2. Which one of the following is the smallest even whole number?
(a) 0                           (b) 1                               (c) 2                                  (d) None of these

Solution:

Option (c) is the correct answer.

We know that the natural numbers, along with 0, form the collection of whole numbers.

Hence, the numbers 0, 1, 2, 3, 4 … form the collection of whole numbers.

So the number which is divisible by 2 is an even number, and 2 is the smallest even number.

3. Which one of the following is the smallest odd whole number?
(a) 0                           (b) 1                               (c) 3                                  (d) 5

Solution:

Option (b) is the correct answer.

We know that the natural numbers, along with 0, form the collection of whole numbers.

Hence, the numbers 0, 1, 2, 3, and 4 … form the collection of whole numbers.

So the natural number which is not divisible by 2 is called an odd whole number, and 1 is the smallest odd whole number.

4. How many whole numbers are between 437 and 487?
(a) 50                           (b) 49                               (c) 51                                  (d) None of these

Solution:

Option (b) is the correct answer.

We know that the whole numbers between 437 and 487 are 438, 439, 440, 441, …, 484, 485 and 486.

In order to find the required number of whole numbers, subtract 437 from 487 and then subtract 1 again.

Hence, there are (487 − 437) − 1 whole number lying between 437 and 487.

So we get (487 − 437) − 1 = 50 − 1 = 49

5. The product of the successor of 999 and the predecessor of 1001 is
(a) one lakh                        (b) one billion                           (c) one million                            (d) one crore

Solution:

Option (c) is the correct answer.

We know that the successor of 999 = 999 + 1 = 1000

So the predecessor of 1001 = 1001 − 1 = 1000

It can be written as

Product of them = (Successor of 999) × (Predecessor of 1001)

By substituting the values

Product of them = 1000 × 1000 = 1000000 = one million

6. Which one of the following whole numbers does not have a predecessor?
(a) 1                        (b) 0                          (c) 2                           (d) None of these

Solution:

Option (b) is the correct answer.

We know that the numbers 0, 1, 2, 3, 4 … form the collection of whole numbers.

Hence, the smallest whole number is 0, which does not have a predecessor.

7. The number of whole numbers between the smallest whole number and the greatest 2-digit number is
(a) 101                        (b) 100                          (c) 99                           (d) 98

Solution:

Option (d) is the correct answer.

We know that the smallest whole number = 0

So the greatest 2-digit whole number = 99

Whole numbers which lie between 0 and 99 are 1, 2, 3, 4,…, 97, 98.

In order to find the number of whole numbers between 0 and 99, first subtract 1 from the difference of 0 and 99.

So the number of whole numbers between 0 and 99 = (99 − 0) − 1 = 99 − 1 = 98

8. If n is a whole number such that n + n = n, then n =?
(a) 1                        (b) 2                          (c) 3                           (d) None of these

Solution:

Option (d) is the correct answer.

We know that 0 + 0 = 0, 1 + 1 = 2, 2 + 2 = 4….

Hence, the statement n + n = n is true only when n = 0.

9. The predecessor of the smallest 3-digit number is
(a) 999                        (b) 99                          (c) 100                           (d) 101

Solution:

Option (b) is the correct answer.

We know that the smallest 3-digit number = 100

So the predecessor of 3 digit number = 100 − 1 = 99

10. The least number of 4-digits which is exactly divisible by 9 is
(a) 1008                        (b) 1009                          (c) 1026                           (d) 1018

Solution:

Option (a) is the correct answer.

We know that the least 4-digit number = 1000

Hence, the least 4-digits which is exactly divisible by 9 is 1000 + (9 − 1) = 1008

11. The number which when divided by 53 gives 8 as quotient and 5 as remainder is
(a) 424             (b) 419                   (c) 429                          (d) None of these

Solution:

Option (c) is the correct answer.

It is given that

Divisor = 53, Quotient = 8 and Remainder = 5.

By using the relation, we get

Dividend = Divisor × Quotient + Remainder

By substituting the values

Dividend = 53 × 8 + 5 = 424 + 5 =429

Hence, the required number is 429.

12. The whole number n satisfying n + 35 = 101 is
(a) 65             (b) 67                   (c) 64                          (d) 66

Solution:

Option (d) is the correct answer.

It is given that

n + 35 = 101

By adding − 35 on both sides

n + 35 + (− 35) = 101 + (− 35)

On further calculation

n + 0 = 66

So we get

n = 66

13. The 4 × 378 × 25 is
(a) 37800             (b) 3780                   (c) 9450                          (d) 30078

Solution:

Option (a) is the correct answer.

We can write it as

4 × 378 × 25 = 4 × 25 × 378

On further calculation

4 × 378 × 25 = 100 × 378 = 37800

14. The value of 1735 × 1232 − 1735 × 232 is
(a) 17350             (b) 173500                   (c) 1735000                          (d) 173505

Solution:

Option (c) is the correct answer.

By using the distributive law of multiplication over subtraction

1735 × 1232 − 1735 × 232 = 1735(1232 − 232)

On further calculation

1735 × 1232 − 1735 × 232 = 1735 × 1000 = 1735000

15. The value of 47 × 99 is
(a) 4635             (b) 4653                   (c) 4563                          (d) 6453

Solution:

Option (b) is the correct answer.

It can be written as

99 = 100 − 1

So we get

47 × 99 = 47 × (100 − 1)

On further calculation

47 × 99 = 47× 100 – 47 = 4700 – 47 = 4653

Hence, the value of 47 × 99 is 4653.

RD Sharma Solutions for Class 6 Maths Chapter 4 Exercise 4.6 PDF

RD Sharma Solutions for Class 6 Maths Chapter 4 Exercise 4.6 PDF provides comprehensive and easy-to-understand answers to all questions in the exercise.

These solutions help students grasp important concepts related to whole numbers, their properties, and arithmetic operations with clear explanations and step-by-step methods. Students can download the PDF from the link below to study anytime and improve their problem-solving skills effectively.

Key Features of RD Sharma Solutions for Class 6 Maths Chapter 4 Operations on Whole Numbers Exercise 4.6

• Step-by-step answers for all questions in Exercise 4.6 to help students understand each problem clearly.

• Detailed explanations of addition, subtraction, multiplication, and division of whole numbers for better conceptual understanding.

• Includes a variety of problems to practice different types of operations on whole numbers.

• Solutions are prepared keeping the exam pattern in mind to help students score well.

• Uses simple and easy-to-understand language suitable for Class 6 students.

• Ideal for completing homework and assignments accurately and on time.

• Strengthens arithmetic skills, boosting confidence in solving similar questions independently.