# Rd Sharma Solutions Class 6 Chapter 4

## Rd Sharma Solutions Class 6 Chapter 4 Operation on Whole Numbers

Welcome to NCTB Solution. Here with this post we are going to help 6th class students for the Solutions of Rd Sharma Class 6 Mathematics, Chapter 4, Operation on Whole Numbers. Here students can easily find Exercise wise solution for chapter 4, Operation on Whole Numbers. Students will find proper solutions for Exercise 4.1, 4.2, 4.3, 4.4 and 4.5. Our teachers solved every problem with easily understandable methods so that every students can understand easily. Here all solutions are based on the CBSE latest curriculum.

Operation on Whole Numbers Exercise 4.1 Solution

Question no – (1)

Solution :

(i) 359 + 476 = 476 + 359 …(Commutative law)

(ii) 2008 + 1952 = 1952 + 2008 …(Commutative)

(iii) 90758 + 0 = 9078 …(Additive identity)

(iv) 54321 + (489 + 699) 489 + (54321 + 699) …(Associative)

Question no – (2)

Solution :

(i) 5628 + 39784

= 45412

Now, by reversing,

= 39784+ 5628

= 45412

(ii) 923584 + 178

= 923762

Now, by reversing,

178 + 923584

= 923762

(iii) 15409 + 112

= 15521

Now, by reversing,

= 112 + 15409

= 15521

(iv) 2359 + 641

= 3000

Now, by reversing,

641 + 2359

= 3000

Question no – (4)

Solution :

(i) Given statement is False.

(ii) Given statement is True.

(ii) Given statement is True.

(iv) Given statement is False.

(v) Given statement is True.

(vi) Given statement is False.

(vii) Given statement is False.

(viii) Given statement is True.

(ix) Given statement is True.

(x) Given statement is False.

(xi) Given statement is False.

(xii) Given statement is True.

Operation on Whole Numbers Exercise 4.2 Solution

Question no – (1)

Solution :

(i)

 15 8 13 10 12 14 11 16 9

Diagonal values,

= 13 + 12 + 11

= 36

(ii)

 22 29 6 13 20 28 10 12 19 21 9 11 18 25 27 15 17 24 26 8 16 23 30 7 14

Consider the diagonal values,

= 20 + 15 + 18 + 17 + 16

= 90

Question no – (2)

Solution :

(i) 57839 – 2983 = 54856

= 54856 + 2983

= 57839

(ii) 92507- 10879 = 81628

= 81628 + 10879

= 92507

(iii) 400000 – 98798 = 301202

301202 + 98798

= 400000

(iv) 5050501 – 969696 = 4080805

= 4080805 + 969696

= 5050501

(v) 200000 – 97531 = 1024469

= 102469 + 97351

= 200000

Question no – (4)

Solution :

We know that,

Largest no of 5 digits = 99999

Largest no of 6 digits = 100000

Difference is,

= (100000 – 99999)

= 1

Therefore, the difference will be 1.

Question no – (5)

Solution :

As we know that,

Largest number of 4 digits = 9999

Smallest number of 7 digits = 1000000

Difference is,

= (1000000 – 9999)

= 990001

Therefore, the difference will be 990001.

Question no – (6)

Solution :

In the given question,

Rohit deposited = Rs 125000

He withdrew = Rs 35425

Money left = ?

Money left in his account,

= (125000 – 25425)

= 89575 Rs

Therefore, Rs 89575 was left in Rohit’s account.

Question no – (7)

Solution :

According to the question,

Population of a town is 96209

Men = 29642

Women = 29167

Children = ?

Total no of number + women,

= (29642 + 29167)

= 58809

No of children,

= (96209 – 58809)

= 37400

Thus, the the number of children will be 37400.

Question no – (8)

Solution :

Original no = 39460

After interchanging 6 and 9

= 36490 (new number)

Difference is,

= (39460 – 36490)

= 2790

Therefore, the difference will be 2790.

Question no – (9)

Solution :

In the given question,

population of a town = 59000.

Increased due to new births = 4536

persons died or left the town = 9218

Population at the end of the year,

= 59000 + 4536 – 9218

= 54318

Therefore, the population at the end of the year will be 54318.

Operation on Whole Numbers Exercise 4.3 Solution

Question no – (1)

Solution :

(i) 785 × 0 = 0

(ii) 4567 × 1 = 4567

(iii) 475 × 129 = 129 × 475

(iv) 1243 × 8975 × 1243

(v) 10 × 100 × = 10 = 10000

(vi) 27 × 18 = 27 × 9 + 27 × 4 + 27 × 5

(vii) 12 × 45 = 12 × 50 – 12 × 5

(viii) 78 × 89 = 78 × 100 – 78 × 16 + 78 × 5

(ix) 66 × 85 = 66 × 90 – 66 × 4 – 66

(x) 49 × 66 + 49 × 34 = 49 × (66 + 34)

Question no – (2)

Solution :

(i) 2 × 1497 × 50

= (2 × 50) × 1497

= 100 × 1497

= 149700

(ii) 4 × 358 × 25

= (4 × 25) × 358

= 100 × 358

= 35800

(iii) 495 × 625 × 16

= (625 × 16) × 495

= 10000 × 495 ‘

= 4950000

(iv) 625 × 20 × 8 × 50

= 625 × 20 × 8 × 50

= (625 × 8) (20 × 500

= 5000000

Question no – (3)

Solution :

(i) 736 × 103

= 736 (100 + 3)

= (736 × 100) + (736 × 3)

= 73600 + (736 × 3)

= 788808

(ii) 258 × 1008

= 258 (1000 + 8)

= (258 × 1000) × (258 × 8)

= 258000 + 2064

= 260064

Question no – (4)

Solution :

(i) 736 × 93

= 736 × 93 = 736 (100 – 7)

= (736 × 100) – (736 × 7)

= 73600 – 5752

= 68448

So, the product will be 68448

(ii) 816 × 745

= 816 × 745 = 816 × (750 – 6)

= (816 × 750) – (816 × 5)

= 612000 – 4080

= 607920

So, the product will be 607920

(iii) 2032 × 613

= 2032 × 613 = 2032 (600 + 13)

= (2032 × 600) + (2032 × 13)

= 1219200 + 26416

= 1245616

So, the product will be 1245616

Question no – (8)

Solution :

In the question,

Dealer purchased = 125 colour television

Cost of each set = Rs 19820

Determine cost of all sets together = ?

Cost of 125 television sets,

= (19820 × 125) Rs

= 2477500 Rs

Therefore, the cost of all sets together will be 2477500 Rs.

Question no – (9)

Solution :

In the given question,

The annual fee of class VI = Rs 8880.

Students in class VI = 235

Annual fee changed for 235 students,

= (8880 × 235)

= 208600 Rs

Therefore, the total collection is 208600 Rs.

Question no – (10)

Solution :

In the question we get,

Constructed flats= 350

Cost of construction for each flat = Rs 993570

Cost of Constriction,

= (993570 × 350) Rs

= 347749500 Rs

Hence, total cost of construction of all the flats will be 347749500 Rs.

Question no – (11)

Solution :

I conclude, The product of two whole numbers is Zero when at least one number or both now are Zero,

Question no – (12)

Solution :

Two numbers when multiplied with gives the same number 0 × 0 = 0, 1 × 1 = 1

Question no – (13)

Solution :

Total apartment in 1 large building,

= 10 × 2

= 20

1 small building, no of apartment,

= 12 × 3

= 36

Total apartment,

= (22 × 20) + (15 × 36)

= 440 + 540

= 980

Hence, there will be total 980 apartments.

Operation on Whole Numbers Exercise 4.4 Solution

Question no – (1)

Solution :

Yes, there exist a whole number we know, the whole number is 1,

Where 1 ÷ 1 = 1.

Question no – (2)

Solution :

(i) 23457 ÷ 1

= 23457

(ii) 0 ÷ 97

= 0

(iii) 476 + (840 ÷ 84)

= 476 + 10

= 486

(iv) 964 – (425 ÷ 425)

= 964 – 1

= 963

(v) (2758 ÷ 2758) – (2758 ÷ 2758)

= 1 – 1

= 0

Question no – (3)

Solution :

(i) 10 ÷ (5×2) = (10÷5) × (10÷2)

= L.H.S 10 ÷ (5×2)

= 10 ÷ 10

= 1

R.H.S (10÷5) × (10÷2)

= 2 × 5

= 10

Therefore, L.H.S ≠ R.H.S

(ii) (53–14) ÷ 7 = 35 ÷ 7 – 14 ÷ 7

= L.H.S (35–14) ÷ 7

= 21 ÷ 7

= 3

R.H.S, 35 ÷ 7 – 14 ÷ 7

= 5 – 2

= 3

Therefore, L.H.S = R.H.S (Proved)

(iii) 35 – 14 ÷ 7 = 35 ÷ 7 – 14 ÷ 7

= L.H.S, 25 – 14 ÷ 7

= 35 – 2

= 33

R.H.S, 35 ÷ 7 – 14 ÷ 7

= 5 ÷ 2

= 3

Therefore, L.H.S ≠ R.H.S

(iv) (20 – 5) ÷ 5 = 20 ÷ 5 – 5

= L.H.S (20–5) ÷ 5

= 15 ÷ 5

= 3

R.H.S, 20 ÷ 5 – 5

= 4 – 5

= – 1

Therefore, L.H.S ≠ R.H.S

(v) 12×(14 ÷ 7) = (12×14) ÷ (12×7)

= L.H.S, 12 (14÷7)

= 12×2

= 24

R.H.S, (12×14) ÷ (12÷7)

= 168 ÷ 84

= 2

Therefore, L.H.S ≠ R.H.S

Question no – (4)

Solution :

(i) Given, 7772 ÷ 58

∴ Quotient = 134,

Remainder = 0

For check, We, know,

Divided = Divisor × Quotient + Remainder.

7772 = 58 × 134 + 0

= 7772

L.H.S = R.H.S (proved)

(ii) Given, 6906 ÷ 35

Hence,

Quotient = 197

Remainder = 11

(iii) Given, 16135 ÷ 875

Here,

Quotient = 18,

Remainder = 385

Question no – (5)

Solution :

As we know that,

Divided = Divisor × Quotient + Remainder

35 × 20 + 18

= 700 + 18

= 718

Therefore, the required number is 718.

Question no – (6)

Solution :

As we know,

Divided = Divisor × Quotient + Remainder

Divided = 58 × 40 + 31

= 2320 + 31

= 2351

Therefore, the required number is 2351.

Question no – (7)

Solution :

In the question,

Product of two numbers = 504347

One number = 1591

Other number = ?

Other number is,

= 504347/1591

= 317

Hence, the other number will be 317.

Question no – (8)

Solution :

As we know,

Divided = Divisor × Quotient + Remainder

59761 = Divisor × 189 + 37

= 59761 – 37 = 189 Divisor

Now the required divisor,

= 59724/189

= 316

Therefore, the Divisor will be 316.

Question no – (9)

Solution :

As we know that,

Divided = Divisor × Quotient + Remainder

55390 = 299 × Quotient + 75

= 299× Quotient = 55390 – 75

Now the required quotient,

= 55315/299

= 185

Hence, the quotient will be 185.

Operation on Whole Numbers Exercise 4.5 Solution

Question no – (1)

Solution :

(i) 10th square number

= 10th square number (10×10)

= (10)2

= 100

Therefore, 10th square number is 100.

(ii) 6th triangular number,

= 6 × (6 + 1)/2

= 6 × 7/2

= 42/2

= 21

Hence, the 6th triangular number is 21.

Question no – (2)

Solution :

(i) Yes, a rectangular number also be a square number.

(ii) Yes, a triangular number also be a square number.

Question no – (3)

Solution :

Here, we know that,

1 × 5 = 5

2 × 6 = 12

3 × 7 = 21

4 × 8 = 32

First four products of two numbers with difference 4.

= 5 – 1 = 4; 6 – 2 = 4

= 7 – 3 = 4; 8 – 4 = 4

Question no – (4)

Solution :

9 × 9 + 7 = 88

98 × 9 + 6 = 888

987 × 9 + 5 = 8888

9876 × 9 + 5 = 88888

98765 × 9 + 3 = 8888888

987654 × 9 + 2 = 8888888

9876543 × 9 + 1 = 88888888

Question no – (5)

Solution :

6 × 2 – 5 = 7

7 × 3 – 12 = 9

8 × 4 – 21 = 11

9 × 5 – 32 = 13

10 × 6 – 45 = 15

11 × 7 – 60 = 17

12 × 8 – 77 = 19

Question no – (6)

Solution :

By observing the above pattern,

(i) 1 + 3 + 5 + 7 + 9 + 11

= 6 × 6

= 36

(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15

= 8 × 8

= 64

(iii) 21 + 23 + 25 + __ + 51

= 262 – 102

= 676 – 100

= 576

Question no – (7)

Solution :

By observing the above pattern,

(i) 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10

= (10×11)/2

= 110/2

= 55

(ii) 50 + 51 + 52 + __ + 100

= 100 × 100/2 – 49 × 50/2

= 5050 – 1225

= 3825

(iii) 2 + 4 + 6 + 8+ 10 + __ + 100

= 2 (1 + 2 + 3 + 4 + 5 + ……)

= 2 × (50 ×57/2)

= 2550

Previous Chapter Solution :

Updated: June 7, 2023 — 2:04 pm