Open App

Class 6 Exam > Class 6 Notes > Mathematics (Maths) Class 6 > RD Sharma Solutions: Operations on Whole Numbers (Exercise 4.2)

Operations on Whole Numbers (Exercise 4.2) RD Sharma Solutions | Mathematics (Maths) Class 6 PDF Download

Download, print and study this document offline

Please wait while the PDF view is loading

 Page 1


 
 
 
 
 
 
Exercise 4.2                                                                     PAGE: 4.8 
1. A magic square is an array of numbers having the same number of rows and columns and the sum of 
numbers in each row, column or diagonal being the same. Fill in the blank cells of the following magic 
squares: 
(i)  
 
(ii)  
 
Solution: 
 
(i) We know that  
Considering diagonal values 13 + 12 + 11 = 36 
So we get 
No. in the first cell of the first row = 36 – (8 + 13) = 15 
No. in the first cell of the second row = 36 – (15 + 11) = 10 
No. in the third cell of the second row = 36 – (10 + 12) = 14 
No. in the second cell of the third row = 36 – (8 + 12) = 16 
No. in the third cell of the third row = 36 – (11 + 16) = 9 
 
 
(ii) We know that 
Considering diagonal values 20 + 19 + 18 + 17 + 16 = 90 
So we get 
No. in the second cell of the first row = 90 – (22 + 6 + 13 + 20) = 29 
No. in the first cell of the second row = 90 – (22 + 9 + 15 + 16) = 28 
No. in the fifth cell of the second row = 90 – (28 + 10 + 12 + 19) = 21 
No. in the fifth cell of the third row = 90 – (9 + 11 + 18 + 25) = 27 
No. in the fifth cell of the fourth row = 90 – (15 + 17 + 24 + 26) = 8 
No. in the second cell of the fifth row = 90 – (29 + 10 + 11 + 17) = 23 
No. in the third cell of the fifth row = 90 – (6 + 12 + 18 + 24) = 30 
Page 2


 
 
 
 
 
 
Exercise 4.2                                                                     PAGE: 4.8 
1. A magic square is an array of numbers having the same number of rows and columns and the sum of 
numbers in each row, column or diagonal being the same. Fill in the blank cells of the following magic 
squares: 
(i)  
 
(ii)  
 
Solution: 
 
(i) We know that  
Considering diagonal values 13 + 12 + 11 = 36 
So we get 
No. in the first cell of the first row = 36 – (8 + 13) = 15 
No. in the first cell of the second row = 36 – (15 + 11) = 10 
No. in the third cell of the second row = 36 – (10 + 12) = 14 
No. in the second cell of the third row = 36 – (8 + 12) = 16 
No. in the third cell of the third row = 36 – (11 + 16) = 9 
 
 
(ii) We know that 
Considering diagonal values 20 + 19 + 18 + 17 + 16 = 90 
So we get 
No. in the second cell of the first row = 90 – (22 + 6 + 13 + 20) = 29 
No. in the first cell of the second row = 90 – (22 + 9 + 15 + 16) = 28 
No. in the fifth cell of the second row = 90 – (28 + 10 + 12 + 19) = 21 
No. in the fifth cell of the third row = 90 – (9 + 11 + 18 + 25) = 27 
No. in the fifth cell of the fourth row = 90 – (15 + 17 + 24 + 26) = 8 
No. in the second cell of the fifth row = 90 – (29 + 10 + 11 + 17) = 23 
No. in the third cell of the fifth row = 90 – (6 + 12 + 18 + 24) = 30 
 
 
 
 
 
 
 
 
2. Perform the following subtractions and check your results by performing corresponding additions: 
(i) 57839 – 2983 
(ii) 92507 – 10879 
(iii) 400000 – 98798 
(iv) 5050501 – 969696 
(v) 200000 – 97531 
(vi) 3030301 – 868686 
Solution: 
 
(i) 57839 – 2983 
We know that  
57839 – 2983 = 54856 
By addition 
54856 + 2983 = 57839 
 
(ii) 92507 – 10879 
We know that 
92507 – 10879 = 81628 
By addition 
81628 + 10879 = 92507 
 
(iii) 400000 – 98798 
We know that 
400000 – 98798 = 301202 
By addition 
301202 + 98798 = 400000 
 
(iv) 5050501 – 969696 
We know that 
5050501 – 969696 = 4080805 
By addition 
4080805 + 969696 = 5050501 
 
(v) 200000 – 97531 
We know that 
200000 – 97531 = 102469 
By addition 
102469 + 97531 = 200000 
 
(vi) 3030301 – 868686 
We know that 
Page 3


 
 
 
 
 
 
Exercise 4.2                                                                     PAGE: 4.8 
1. A magic square is an array of numbers having the same number of rows and columns and the sum of 
numbers in each row, column or diagonal being the same. Fill in the blank cells of the following magic 
squares: 
(i)  
 
(ii)  
 
Solution: 
 
(i) We know that  
Considering diagonal values 13 + 12 + 11 = 36 
So we get 
No. in the first cell of the first row = 36 – (8 + 13) = 15 
No. in the first cell of the second row = 36 – (15 + 11) = 10 
No. in the third cell of the second row = 36 – (10 + 12) = 14 
No. in the second cell of the third row = 36 – (8 + 12) = 16 
No. in the third cell of the third row = 36 – (11 + 16) = 9 
 
 
(ii) We know that 
Considering diagonal values 20 + 19 + 18 + 17 + 16 = 90 
So we get 
No. in the second cell of the first row = 90 – (22 + 6 + 13 + 20) = 29 
No. in the first cell of the second row = 90 – (22 + 9 + 15 + 16) = 28 
No. in the fifth cell of the second row = 90 – (28 + 10 + 12 + 19) = 21 
No. in the fifth cell of the third row = 90 – (9 + 11 + 18 + 25) = 27 
No. in the fifth cell of the fourth row = 90 – (15 + 17 + 24 + 26) = 8 
No. in the second cell of the fifth row = 90 – (29 + 10 + 11 + 17) = 23 
No. in the third cell of the fifth row = 90 – (6 + 12 + 18 + 24) = 30 
 
 
 
 
 
 
 
 
2. Perform the following subtractions and check your results by performing corresponding additions: 
(i) 57839 – 2983 
(ii) 92507 – 10879 
(iii) 400000 – 98798 
(iv) 5050501 – 969696 
(v) 200000 – 97531 
(vi) 3030301 – 868686 
Solution: 
 
(i) 57839 – 2983 
We know that  
57839 – 2983 = 54856 
By addition 
54856 + 2983 = 57839 
 
(ii) 92507 – 10879 
We know that 
92507 – 10879 = 81628 
By addition 
81628 + 10879 = 92507 
 
(iii) 400000 – 98798 
We know that 
400000 – 98798 = 301202 
By addition 
301202 + 98798 = 400000 
 
(iv) 5050501 – 969696 
We know that 
5050501 – 969696 = 4080805 
By addition 
4080805 + 969696 = 5050501 
 
(v) 200000 – 97531 
We know that 
200000 – 97531 = 102469 
By addition 
102469 + 97531 = 200000 
 
(vi) 3030301 – 868686 
We know that 
 
 
 
 
 
 
3030301 – 868686 = 2161615 
By addition 
2161615 + 868686 = 3030301 
 
3. Replace each * by the correct digit in each of the following: 
(i)  
 
(ii)  
 
(iii)  
 
(iv)  
 
(v)  
 
(vi)  
 
Solution: 
 
(i) We know that in the units digit  
6 - * = 7 where the value of * is 9 as 1 gets carried from 7 at tens place to 6 at units place  
6 at the units place becomes 16 so 16 – 9 = 7 
When 7 is reduced by 1 it gives 6 so 6 – 3 = 3 
We know that 
8 - * = 6 so we get * value as 2 
 
 
(ii) We know that in the units digit 
Page 4


 
 
 
 
 
 
Exercise 4.2                                                                     PAGE: 4.8 
1. A magic square is an array of numbers having the same number of rows and columns and the sum of 
numbers in each row, column or diagonal being the same. Fill in the blank cells of the following magic 
squares: 
(i)  
 
(ii)  
 
Solution: 
 
(i) We know that  
Considering diagonal values 13 + 12 + 11 = 36 
So we get 
No. in the first cell of the first row = 36 – (8 + 13) = 15 
No. in the first cell of the second row = 36 – (15 + 11) = 10 
No. in the third cell of the second row = 36 – (10 + 12) = 14 
No. in the second cell of the third row = 36 – (8 + 12) = 16 
No. in the third cell of the third row = 36 – (11 + 16) = 9 
 
 
(ii) We know that 
Considering diagonal values 20 + 19 + 18 + 17 + 16 = 90 
So we get 
No. in the second cell of the first row = 90 – (22 + 6 + 13 + 20) = 29 
No. in the first cell of the second row = 90 – (22 + 9 + 15 + 16) = 28 
No. in the fifth cell of the second row = 90 – (28 + 10 + 12 + 19) = 21 
No. in the fifth cell of the third row = 90 – (9 + 11 + 18 + 25) = 27 
No. in the fifth cell of the fourth row = 90 – (15 + 17 + 24 + 26) = 8 
No. in the second cell of the fifth row = 90 – (29 + 10 + 11 + 17) = 23 
No. in the third cell of the fifth row = 90 – (6 + 12 + 18 + 24) = 30 
 
 
 
 
 
 
 
 
2. Perform the following subtractions and check your results by performing corresponding additions: 
(i) 57839 – 2983 
(ii) 92507 – 10879 
(iii) 400000 – 98798 
(iv) 5050501 – 969696 
(v) 200000 – 97531 
(vi) 3030301 – 868686 
Solution: 
 
(i) 57839 – 2983 
We know that  
57839 – 2983 = 54856 
By addition 
54856 + 2983 = 57839 
 
(ii) 92507 – 10879 
We know that 
92507 – 10879 = 81628 
By addition 
81628 + 10879 = 92507 
 
(iii) 400000 – 98798 
We know that 
400000 – 98798 = 301202 
By addition 
301202 + 98798 = 400000 
 
(iv) 5050501 – 969696 
We know that 
5050501 – 969696 = 4080805 
By addition 
4080805 + 969696 = 5050501 
 
(v) 200000 – 97531 
We know that 
200000 – 97531 = 102469 
By addition 
102469 + 97531 = 200000 
 
(vi) 3030301 – 868686 
We know that 
 
 
 
 
 
 
3030301 – 868686 = 2161615 
By addition 
2161615 + 868686 = 3030301 
 
3. Replace each * by the correct digit in each of the following: 
(i)  
 
(ii)  
 
(iii)  
 
(iv)  
 
(v)  
 
(vi)  
 
Solution: 
 
(i) We know that in the units digit  
6 - * = 7 where the value of * is 9 as 1 gets carried from 7 at tens place to 6 at units place  
6 at the units place becomes 16 so 16 – 9 = 7 
When 7 is reduced by 1 it gives 6 so 6 – 3 = 3 
We know that 
8 - * = 6 so we get * value as 2 
 
 
(ii) We know that in the units digit 
 
 
 
 
 
 
9 – 4 = 5 
Tens digit 8 – 3 = 5 
So the missing blank can be found by subtracting 3455 from 8989 
Difference between them = 3455 
So the answer is 
 
 
(iii) We know that in units digit  
17 – 8 = 9  
Tens digit = 9 – 7 = 2 
So we get 
Hundreds place 10 – 9 = 1 
Thousands place 9 – 8 = 1 
So the addend difference = 5061129 
Subtract 5061129 from 6000107 to get addend 
 
So the answer is 
 
 
(iv) We know that in units digit  
10 – 1 = 9 
Lakhs place 9 – 0 = 9 
So the addend difference = 970429 
Subtract 970429 from 1000000 to get the addend 
 
So the correct answer is 
 
 
(v) We know that in units digit  
13 – 7 = 6 
Tens digit 9 – 8 = 1 
Hundreds place 9 – 9 = 0 
Thousands place 10 – 6 = 4 
So the addend difference = 4844016 
Page 5


 
 
 
 
 
 
Exercise 4.2                                                                     PAGE: 4.8 
1. A magic square is an array of numbers having the same number of rows and columns and the sum of 
numbers in each row, column or diagonal being the same. Fill in the blank cells of the following magic 
squares: 
(i)  
 
(ii)  
 
Solution: 
 
(i) We know that  
Considering diagonal values 13 + 12 + 11 = 36 
So we get 
No. in the first cell of the first row = 36 – (8 + 13) = 15 
No. in the first cell of the second row = 36 – (15 + 11) = 10 
No. in the third cell of the second row = 36 – (10 + 12) = 14 
No. in the second cell of the third row = 36 – (8 + 12) = 16 
No. in the third cell of the third row = 36 – (11 + 16) = 9 
 
 
(ii) We know that 
Considering diagonal values 20 + 19 + 18 + 17 + 16 = 90 
So we get 
No. in the second cell of the first row = 90 – (22 + 6 + 13 + 20) = 29 
No. in the first cell of the second row = 90 – (22 + 9 + 15 + 16) = 28 
No. in the fifth cell of the second row = 90 – (28 + 10 + 12 + 19) = 21 
No. in the fifth cell of the third row = 90 – (9 + 11 + 18 + 25) = 27 
No. in the fifth cell of the fourth row = 90 – (15 + 17 + 24 + 26) = 8 
No. in the second cell of the fifth row = 90 – (29 + 10 + 11 + 17) = 23 
No. in the third cell of the fifth row = 90 – (6 + 12 + 18 + 24) = 30 
 
 
 
 
 
 
 
 
2. Perform the following subtractions and check your results by performing corresponding additions: 
(i) 57839 – 2983 
(ii) 92507 – 10879 
(iii) 400000 – 98798 
(iv) 5050501 – 969696 
(v) 200000 – 97531 
(vi) 3030301 – 868686 
Solution: 
 
(i) 57839 – 2983 
We know that  
57839 – 2983 = 54856 
By addition 
54856 + 2983 = 57839 
 
(ii) 92507 – 10879 
We know that 
92507 – 10879 = 81628 
By addition 
81628 + 10879 = 92507 
 
(iii) 400000 – 98798 
We know that 
400000 – 98798 = 301202 
By addition 
301202 + 98798 = 400000 
 
(iv) 5050501 – 969696 
We know that 
5050501 – 969696 = 4080805 
By addition 
4080805 + 969696 = 5050501 
 
(v) 200000 – 97531 
We know that 
200000 – 97531 = 102469 
By addition 
102469 + 97531 = 200000 
 
(vi) 3030301 – 868686 
We know that 
 
 
 
 
 
 
3030301 – 868686 = 2161615 
By addition 
2161615 + 868686 = 3030301 
 
3. Replace each * by the correct digit in each of the following: 
(i)  
 
(ii)  
 
(iii)  
 
(iv)  
 
(v)  
 
(vi)  
 
Solution: 
 
(i) We know that in the units digit  
6 - * = 7 where the value of * is 9 as 1 gets carried from 7 at tens place to 6 at units place  
6 at the units place becomes 16 so 16 – 9 = 7 
When 7 is reduced by 1 it gives 6 so 6 – 3 = 3 
We know that 
8 - * = 6 so we get * value as 2 
 
 
(ii) We know that in the units digit 
 
 
 
 
 
 
9 – 4 = 5 
Tens digit 8 – 3 = 5 
So the missing blank can be found by subtracting 3455 from 8989 
Difference between them = 3455 
So the answer is 
 
 
(iii) We know that in units digit  
17 – 8 = 9  
Tens digit = 9 – 7 = 2 
So we get 
Hundreds place 10 – 9 = 1 
Thousands place 9 – 8 = 1 
So the addend difference = 5061129 
Subtract 5061129 from 6000107 to get addend 
 
So the answer is 
 
 
(iv) We know that in units digit  
10 – 1 = 9 
Lakhs place 9 – 0 = 9 
So the addend difference = 970429 
Subtract 970429 from 1000000 to get the addend 
 
So the correct answer is 
 
 
(v) We know that in units digit  
13 – 7 = 6 
Tens digit 9 – 8 = 1 
Hundreds place 9 – 9 = 0 
Thousands place 10 – 6 = 4 
So the addend difference = 4844016 
 
 
 
 
 
 
Subtract 4844016 from 5001003 to get the addend 
 
So the answer is  
 
 
(vi) We know that units digit  
11 – 9 = 2 
So the addend difference = 54322 
Subtract 54322 from 111111 to get the addend 
 
So the answer is 
 
 
4. What is the difference between the largest number of five digits and the smallest number of six digits? 
Solution: 
 
99999 is the largest number of five digits 
100000 is the largest number of six digits 
Difference = 100000 – 99999 = 1 
 
Therefore, 1 is the difference between the largest number of five digits and smallest number of six digits. 
 
5. Find the difference between the largest number of 4 digits and the smallest number of 7 digits. 
Solution: 
 
9999 is the largest number of 4 digits 
1000000 is the smallest number of 6 digits 
Difference = 1000000 – 9999 = 990001 
 
Therefore, 990001 is the difference between the largest number of 4 digits and the smallest number of 7 digits. 
 
6. Rohit deposited Rs 125000 in his savings bank account. Later he withdrew Rs 35425 from it. How much 
money was left in his account? 
Solution: 
 
Money deposited in savings bank account = Rs 125000 
Money withdrawn = Rs 35425

	Mathematics (Maths) Class 6 94 videos\|347 docs\|54 tests

Mathematics (Maths) Class 6

94 videos|347 docs|54 tests

Join Course for Free

Top Courses for Class 6

View all

Related Exams

Class 6

About this Document

	4.87/5 Rating
	Oct 24, 2024 Last updated

Document Description: RD Sharma Solutions: Operations on Whole Numbers (Exercise 4.2) for Class 6 2024 is part of Mathematics (Maths) Class 6 preparation. The notes and questions for RD Sharma Solutions: Operations on Whole Numbers (Exercise 4.2) have been prepared according to the Class 6 exam syllabus. Information about RD Sharma Solutions: Operations on Whole Numbers (Exercise 4.2) covers topics like and RD Sharma Solutions: Operations on Whole Numbers (Exercise 4.2) Example, for Class 6 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for RD Sharma Solutions: Operations on Whole Numbers (Exercise 4.2).

Introduction of RD Sharma Solutions: Operations on Whole Numbers (Exercise 4.2) in English is available as part of our Mathematics (Maths) Class 6 for Class 6 & RD Sharma Solutions: Operations on Whole Numbers (Exercise 4.2) in Hindi for Mathematics (Maths) Class 6 course. Download more important topics related with notes, lectures and mock test series for Class 6 Exam by signing up for free. Class 6: Operations on Whole Numbers (Exercise 4.2) RD Sharma Solutions | Mathematics (Maths) Class 6

Description

Full syllabus notes, lecture & questions for Operations on Whole Numbers (Exercise 4.2) RD Sharma Solutions | Mathematics (Maths) Class 6 - Class 6 | Plus excerises question with solution to help you revise complete syllabus for Mathematics (Maths) Class 6 | Best notes, free PDF download

Information about RD Sharma Solutions: Operations on Whole Numbers (Exercise 4.2)

In this doc you can find the meaning of RD Sharma Solutions: Operations on Whole Numbers (Exercise 4.2) defined & explained in the simplest way possible. Besides explaining types of RD Sharma Solutions: Operations on Whole Numbers (Exercise 4.2) theory, EduRev gives you an ample number of questions to practice RD Sharma Solutions: Operations on Whole Numbers (Exercise 4.2) tests, examples and also practice Class 6 tests

	Mathematics (Maths) Class 6 94 videos\|347 docs\|54 tests

Mathematics (Maths) Class 6

94 videos|347 docs|54 tests

Join Course for Free

Download as PDF

Explore Courses for Class 6 exam

Top Courses for Class 6

Explore Courses

Signup for Free!

Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.

Start learning for Free

10M+ students study on EduRev

practice quizzes

Important questions

Sample Paper

Objective type Questions

mock tests for examination

MCQs

Viva Questions

Operations on Whole Numbers (Exercise 4.2) RD Sharma Solutions | Mathematics (Maths) Class 6

Exam

Operations on Whole Numbers (Exercise 4.2) RD Sharma Solutions | Mathematics (Maths) Class 6

Extra Questions

Summary

past year papers

pdf

study material

video lectures

Operations on Whole Numbers (Exercise 4.2) RD Sharma Solutions | Mathematics (Maths) Class 6

Free

ppt

Semester Notes

Previous Year Questions with Solutions

shortcuts and tricks

;

Additional Information about RD Sharma Solutions: Operations on Whole Numbers (Exercise 4.2) for Class 6 Preparation

RD Sharma Solutions: Operations on Whole Numbers (Exercise 4.2) Free PDF Download

The RD Sharma Solutions: Operations on Whole Numbers (Exercise 4.2) is an invaluable resource that delves deep into the core of the Class 6 exam. These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective. With the help of these notes, you can grasp complex subjects quickly, revise important points easily, and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner, allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material, or toppers' notes, this PDF has got you covered. Download the RD Sharma Solutions: Operations on Whole Numbers (Exercise 4.2) now and kickstart your journey towards success in the Class 6 exam.

Importance of RD Sharma Solutions: Operations on Whole Numbers (Exercise 4.2)

The importance of RD Sharma Solutions: Operations on Whole Numbers (Exercise 4.2) cannot be overstated, especially for Class 6 aspirants. This document holds the key to success in the Class 6 exam. It offers a detailed understanding of the concept, providing invaluable insights into the topic. By knowing the concepts well in advance, students can plan their preparation effectively. Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.

RD Sharma Solutions: Operations on Whole Numbers (Exercise 4.2) Notes

RD Sharma Solutions: Operations on Whole Numbers (Exercise 4.2) Notes offer in-depth insights into the specific topic to help you master it with ease. This comprehensive document covers all aspects related to RD Sharma Solutions: Operations on Whole Numbers (Exercise 4.2). It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation. Practice papers and question papers enable you to assess your progress effectively. Additionally, the paper analysis provides valuable tips for tackling the exam strategically. Access to Toppers' notes gives you an edge in understanding complex concepts. Whether you're a beginner or aiming for advanced proficiency, RD Sharma Solutions: Operations on Whole Numbers (Exercise 4.2) Notes on EduRev are your ultimate resource for success.

RD Sharma Solutions: Operations on Whole Numbers (Exercise 4.2) Class 6 Questions

The "RD Sharma Solutions: Operations on Whole Numbers (Exercise 4.2) Class 6 Questions" guide is a valuable resource for all aspiring students preparing for the Class 6 exam. It focuses on providing a wide range of practice questions to help students gauge their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation. The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level. Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.

Study RD Sharma Solutions: Operations on Whole Numbers (Exercise 4.2) on the App

Students of Class 6 can study RD Sharma Solutions: Operations on Whole Numbers (Exercise 4.2) alongwith tests & analysis from the EduRev app, which will help them while preparing for their exam. Apart from the RD Sharma Solutions: Operations on Whole Numbers (Exercise 4.2), students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc. The EduRev App will make your learning easier as you can access it from anywhere you want. The content of RD Sharma Solutions: Operations on Whole Numbers (Exercise 4.2) is prepared as per the latest Class 6 syllabus.

Education Revolution