RD Sharma Solutions for Class 6 Chapter 4: Operations on Whole Numbers Exercise 4.2

RD Sharma Class 6 Solutions Chapter 4 Ex 4.2 PDF Free Download

RD Sharma Solutions for Class 6 Maths Exercise 4.2 PDF

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RD Sharma Solutions for Class 6 Chapter 4: Operations on Whole Numbers Exercise 4.2 Download PDF

 

rd sharma solution class 6 maths chapter 4 ex 2
rd sharma solution class 6 maths chapter 4 ex 2
rd sharma solution class 6 maths chapter 4 ex 2
rd sharma solution class 6 maths chapter 4 ex 2
rd sharma solution class 6 maths chapter 4 ex 2
rd sharma solution class 6 maths chapter 4 ex 2

 

Access RD Sharma Solutions for Class 6 Chapter 4: Operations on Whole Numbers Exercise 4.2

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)

RD Sharma Solutions Class 6 Chapter 4 Ex 4.2 Image 1

(ii)

RD Sharma Solutions Class 6 Chapter 4 Ex 4.2 Image 2

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

RD Sharma Solutions Class 6 Chapter 4 Ex 4.2 Image 3

(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

RD Sharma Solutions Class 6 Chapter 4 Ex 4.2 Image 4

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)

RD Sharma Solutions Class 6 Chapter 4 Ex 4.2 Image 5

(ii)

RD Sharma Solutions Class 6 Chapter 4 Ex 4.2 Image 6

(iii)

RD Sharma Solutions Class 6 Chapter 4 Ex 4.2 Image 7

(iv)

RD Sharma Solutions Class 6 Chapter 4 Ex 4.2 Image 8

(v)

RD Sharma Solutions Class 6 Chapter 4 Ex 4.2 Image 9

(vi)

RD Sharma Solutions Class 6 Chapter 4 Ex 4.2 Image 10

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

RD Sharma Solutions Class 6 Chapter 4 Ex 4.2 Image 11

(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

RD Sharma Solutions Class 6 Chapter 4 Ex 4.2 Image 12

(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

RD Sharma Solutions Class 6 Chapter 4 Ex 4.2 Image 13

So the answer is

RD Sharma Solutions Class 6 Chapter 4 Ex 4.2 Image 14

(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

RD Sharma Solutions Class 6 Chapter 4 Ex 4.2 Image 15

So the correct answer is

RD Sharma Solutions Class 6 Chapter 4 Ex 4.2 Image 16

(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

RD Sharma Solutions Class 6 Chapter 4 Ex 4.2 Image 17

So the answer is

RD Sharma Solutions Class 6 Chapter 4 Ex 4.2 Image 18

(vi) We know that units digit

11 – 9 = 2

So the addend difference = 54322

Subtract 54322 from 111111 to get the addend

RD Sharma Solutions Class 6 Chapter 4 Ex 4.2 Image 19

So the answer is

RD Sharma Solutions Class 6 Chapter 4 Ex 4.2 Image 20

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

So the money which is left out in his account = 125000 – 35425 = Rs 89575

Hence, Rs 89575 is left in his account.

7. The population of a town is 96209. If the number of men is 29642 and that of women is 29167, determine the number of children.

Solution:

Population of a town = 96209

No. of men = 29642

No. of women = 29167

Total number of men and women = 29642 + 29167 = 58809

So the number of children = Population of a town – Total number of men and women

Number of children = 96209 – 58809 = 37400

Hence, there are 37400 children.

8. The digits of 6 and 9 of the number 36490 are interchanged. Find the difference between the original number and the new number.

Solution:

It is given that

Number = 39460

Number after interchanging 6 and 9 = 36490

Difference between them = 39460 – 36490 = 2790

Therefore, the difference between the original number and new number is 2970.

9. The population of a town was 59000. In one year it was increased by 4536 due to new births. However, 9218 persons died or left the town during the year. What was the population at the end of the year?

Solution:

Population of a town = 59000

Population increase = 4536

Population decrease = 9218

So the population at the end of year = 59000 + 4536 – 9218 = 54318

Therefore, the population at the end of the year is 54318.


Access other exercise solutions of Class 6 Maths Chapter 4: Operations on Whole Numbers

Exercise 4.1 Solutions 4 Questions

Exercise 4.3 Solutions 13 Questions

Exercise 4.4 Solutions 9 Questions

Exercise 4.5 Solutions 8 Questions

Objective Type Questions

RD Sharma Solutions Class 6 Maths Chapter 4 – Operations on Whole Numbers Exercise 4.2

RD Sharma Solutions Class 6 Maths Chapter 4 Operations on Whole Numbers Exercise 4.2 explains properties of subtraction and problem solving methods based on CBSE syllabus.

Key features of RD Sharma Solutions for Class 6 Maths Chapter 4: Operations on Whole Numbers Exercise 4.2

  • The solutions prepared by subject experts improve conceptual knowledge among the students.
  • Huge number of problems present before each exercise help students in understanding the methods of solving them.
  • The exercise wise problems are made available in PDF format which can be downloaded for reference purpose.
  • Self study methods are improved among students which significantly contributes to their success in higher classes.

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