NCERT Solutions For Class 6 Maths Chapter 2

NCERT Solutions Class 6 Maths Whole Numbers

NCERT NCERT Solutions for class 6 Maths Chapter 2 Whole Numbers is a comprehensive study material for students preparing for class 6 mathematics. These solutions are prepared by our experts to promote proper understanding of basic concepts included in this chapter. It includes exercise wise solved questions along with important formulas, shortcut tips and more. These class 6 Maths NCERT Solutions Chapter 2 Whole Numbers comprises of detailed explanation of all the concepts such as Natural and whole numbers, associative property, commutative property, distributive property, Additive identity, Number Line and Addition, Subtraction and Multiplication on Number Line, patterns in whole numbers, etc. The chapter 2 of class 6 maths has the following sections

Section Number Topic
2.1 Introduction
2.2 Whole Numbers
2.3 The Number Line
2.4 Properties of Whole Numbers
2.5 Patterns in Whole Numbers

Students can also download NCERT Solutions for class 6 Maths Chapter 2 pdf for better assistance and preparation.Solutions For Class 6 Maths Chapter 2

NCERT Solutions Class 6 Maths Chapter 2 Exercises

Exercise 2.1

Question 1:

List out any four natural numbers after 11999.

Answer:

Three natural numbers after 11999 are 12000, 12001, and 12002.

Question 2:

List out any four whole numbers, which come just before 10001?

Answer:

The numbers, which come before 10001, are 10000, 9999, and 9998

Question 3:

The smallest whole number is:

Answer:

The smallest whole number is zero.

Question 4:

What are the number of whole numbers lying between 42 and 65?

Answer:

The number of whole numbers lying between 42 and 65 is 22 (65 – 42 – 1 = 22).

They are as follows 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, and 64.

Question 5:

What is the predecessor of?

(a) 95 (b) 1000 (c) 302780 (d) 6754300

Answer:

(a) 95 – 1 = 94

(b) 1000 – 1 = 999

(c) 302780 – 1 = 302779

(d) 6754300 – 1 = 6754299

Question 6:

What is the successor of?

(a) 88 (b) 9999 (c) 84729 (d) 63513

Answer:

(a) 88 + 1 = 89

(b) 9999 + 1 = 10000

(c) 84729 + 1 = 84730

(d) 63513 + 1 = 63514

Question 7:

In the following given pair of numbers, find the whole number which lies on the left side of the other number in on the number line, Also put the appropriate sign (<,>) between them.

(a) 115, 151 (b) 435, 345

(c) 9876567, 9876576 (d) 1234567, 1233567

Answer:

(a) 115 < 151

115 lie on the left side of 151 on the number line.

(b) 435 > 345

345 lie on the left side of 435 on the number line.

 

(c) 9876567 < 9876576

9876567 lie on to the left of 9876576 on the number line.

 

(d) 1234567 > 1233567

1233567 lie on the left side of 1234567 on the number line.

 

Question 8:

Mark True (T) or False (F) for the given statements.

(a) Zero is the smallest natural number

(b) 400 is the successor of 399

(c) Smallest whole number is Zero

(d) 500 is the predecessor of 499

(e) All natural numbers are whole numbers

(f) All whole numbers are natural numbers

(g) Predecessor of a two-digit numbers can never be a single digit number

(h) The smallest whole number is 1

(i) The natural 1 number has no predecessor

(j) The whole number 1 has no predecessor

(k) The whole number 21 lies between 25 and 27

(l) The whole number 0 have a predecessor

(m) The successor of a three digit number can never be four digit number

Answer:

(a) False, 0 does not come under natural number

(b) True, as 399 + 1 = 400

(c) True

(d) False, predecessor of 499 is 499 -1 = 488

(e) True

(f) False, 0 is a whole number but it is not a natural number

(g) False, as 10 is a 2 digit number but its predecessor is 9 which is a single digit number

(h) False, the smallest whole number is 0

(i) True, since 0 is the successor of 1 but it’s not a natural number

(j) False, 0 is a whole number

(k) False

(l) False, 0 have a predecessor -1 which is not a whole number

(m) False, 999 is a three digit number and its successor is 10000 and it’s a four digit number

 

EXERCISE- 2.2

Question 1:

Frame the numbers in proper order and find the sum

a) 820 + 200 +325

b) 1960 + 448 + 1520 + 638

Answer:

a) (800 + 325) + 200 = 1325

b) (1960 +1520) + (448 + 638) = 3480 + 1086 = 4566

 

Question 2:

Find the product by framing it in suitable order

a) 2\(\times\) 1750\(\times\) 45

b) 3\(\times\) 165\(\times\)25

c) 6\(\times\)290\(\times\)120

d) 600\(\times\)286\(\times\)18

e) 286\(\times\)8\(\times\)70

f) 130\(\times\)45\(\times\)6\(\times\)30

Answer:

a) 2\(\times\)1750\(\times\)45

= (2\(\times\)45) \(\times\)1750 = 90\(\times\)1750 = 157,500

b) 3\(\times\)165\(\times\)25

= (3\(\times\)25) \(\times\)165 = 75\(\times\)165 = 12,375

c) 6\(\times\)290\(\times\)120

= (6\(\times\)120) \(\times\)290 = 720\(\times\)290 = 208,800

d) 600\(\times\)286\(\times\)18

= (600\(\times\)18) \(\times\)286 = 10800\(\times\)286 = 3,088,800

e) 286\(\times\)8\(\times\)70

= 286\(\times\) (8\(\times\)70) = 286\(\times\)560 = 160,160

f) 130\(\times\)45\(\times\)6\(\times\)30

= (130\(\times\)6) \(\times\) (45\(\times\)30) = 780\(\times\)1350 = 1,053,000

 

Question 3:

Find the solution of the following

a) 290\(\times\)15 + 290\(\times\)5

Answer:

Since 290 is repeated, we can take it as a common term

290\(\times\) (15+5) = 290\(\times\)20 = 5800

 

b) 54270\(\times\)95 + 6\(\times\)54270

Answer:

Since 54270 is repeated, we can take it as a common term

54270\(\times\) (95+6) = 54270\(\times\)101 = 5,481,270

 

c) 8250\(\times\)62 – 8250\(\times\)50

Answer:

Since 8250 is repeated, we can take it as a common term

8250\(\times\) (62-50) = 8250\(\times\)12 = 99,000

 

d) 3845\(\times\)5\(\times\)720 + 769\(\times\)25\(\times\)220

Answer:

= 3845\(\times\)5\(\times\)720 + 769\(\times\)5\(\times\)5\(\times\)220

= 3845\(\times\)5\(\times\)720 + (769\(\times\)5)5\(\times\)220

= 3845\(\times\)5\(\times\)720 + 3845\(\times\)5\(\times\)220

Since 3845 and 5 is repeated, we can take it as a common term

= 3845\(\times\)5\(\times\) (720 +220)

= 19,225\(\times\)940

= 18,071,500

 

Question 4:

Use suitable properties to find out the product

a) 642\(\times\)105

Answer:

= 642\(\times\) (100+5)

By distributive law

A\(\times\) (B+C) = A\(\times\)B + A\(\times\)C

= 642\(\times\)100 + 642\(\times\)5

= 64200 + 3210

= 67,410

 

b) 850\(\times\)103

Answer:

= 850\(\times\) (100+3)

By distributive law,

A\(\times\) (B+C) = A\(\times\)B + A\(\times\)C

= 850\(\times\)100 + 850\(\times\)3

= 85000 + 2550

= 87,550

 

c) 260\(\times\)1006

Answer:

= 260\(\times\) (1000+6)

By distributive law

A\(\times\) (B+C) = A\(\times\)B + A\(\times\)C

= 260\(\times\)1000 + 260\(\times\)6

= 260000 + 1560

= 261,560

 

d) 1008\(\times\)168

Answer:

= (1000\(\times\)8) \(\times\)168

By distributive law

A\(\times\) (B+C) = A\(\times\)B + A\(\times\)C

= 1000\(\times\)168 + 8\(\times\)168

= 168000 + 1344

= 169,344

 

Question 5:

A cab driver filled his vehicle tank with 35litres of diesel on Tuesday. The next day, he filled the tank with 45litres of diesel. If the diesel costs Rs.32 per litre. How much did he spend on diesel?

Answer:

The quantity of diesel filled on Tuesday = 35L

The quantity of diesel filled on Wednesday = 45L

The total quantity filled = (35+45)L

Cost of diesel (per L) = Rs.32

The total money spent on diesel = 32\(\times\) (35+45)

= 32\(\times\)80

= Rs.2560

The total amount spent on diesel = Rs.2560

Question 6:

A milkman supplies 30 liters of milk to a teashop in the morning and 60 liters of milk in the evening. If the cost of milk is 12 per liter, how much does the milkman earn a day?

Answer:

The quantity of milk supplied in the morning = 30L

The quantity of milk supplied in the evening = 60L

The total quantity filled = (30+60)L

Cost of milk (per L) = Rs.12

(L represents litres)

The total money spent on diesel = 12\(\times\) (30+60)

= 12\(\times\)90

= Rs.1080

The total amount spent on diesel = Rs.1080

Question 7:

Write the answers using correct property

a) 424\(\times\)128

Answer:

424\(\times\)128 = 424\(\times\) (100+20+8)

This is distributive property of multiplication over addition

 

b) 2\(\times\)49\(\times\)50

Answer:

2\(\times\)48\(\times\)51 = 2\(\times\)51\(\times\)48

This is commutative property under multiplication

 

c) 82+2000+20

Answer:

82+2000+20 = 82+20+2000

This is commutative property under addition

 

EXERCISE- 2.3

Question 1:

Which out of the following questions will represent zero;

a) 2+0

b) 1\(\times\)0

c) \(\frac{0}{2}\)

d) \(\frac{20-20}{2}\)

Answer:

a) 2 + 0 = 2

It does not represent zero, since the answer is 2.

 

b) 1*0 = 0

It represents zero, since number multiplied with anything is zero.

 

c) \(\frac{0}{2}\) = 0

It represents zero, since number divided with anything is zero.

 

d) \(\frac{20-20}{2}\) = 0

It represents zero.

 

Question 2:

The product of two numbers is zero; can we say that one or both of them will be zero or non-zero? State reasons.

Answer:

First case

The product of 2 whole numbers will be zero only when one of the number is multiplied with zero

Ex)

5\(\times\)0 = 0

18\(\times\)0 = 0

Second case

The product of 2 whole numbers will be zero when both the numbers are multiplied are zero

0\(\times\)0 = 0

Third case

The product of 2 numbers will not be zero, if both are multiplied with a non-zero number

4\(\times\)3 = 12

5\(\times\)4 = 20

Question 3:

The product of two whole numbers is 1, can we say that one or both of them will be 1? State reasons.

Answer:

First case

The product of 2 whole numbers will be one, if both the numbers are multiplied with 1.

Ex)

1\(\times\)1 =1

Second case

The product of 2 whole numbers will not be 1, if it is multiplied by numbers other than 1.

6\(\times\)5 = 30

So, it is clear that the product of two numbers will be 1 only when the numbers are multiplied with 1.

 

Question 4:

Find the solution using distributive property.

a) 725\(\times\)102

Answer:

By distributive law

A\(\times\) (B+C) = A\(\times\)B + A\(\times\)C

= 728\(\times\) (100+2)

= 728\(\times\)100 + 728\(\times\)2

= 72800 + 1456

= 74,256

 

b) 5450\(\times\)1004

Answer:

By distributive law,

A\(\times\) (B+C) = A\(\times\)B + A\(\times\)C

= 5450\(\times\) (1000+4)

= (5450\(\times\)1000) + (5450\(\times\)4)

= 5450000 + 21800

= 5471800

 

c) 724\(\times\)25

Answer:

By distributive law,

A\(\times\) (B+C) = A\(\times\)B + A\(\times\)C

= (700+24) \(\times\)25

= (700 + 25 – 1) \(\times\)25

= 700\(\times\)25 + 25\(\times\)25 – (1\(\times\)25)

= 17500 + 625 – 25

= 18100

 

d) 4225\(\times\)125

Answer:

By distributive law,

A\(\times\) (B+C) = A\(\times\)B + A\(\times\)C

= (4000 + 200 +100 -75) \(\times\)125

= (4000\(\times\)125) + (200\(\times\)125) + (100\(\times\)125) – (75\(\times\)125)

= (500000) + (25000) + (12500) – (9375)

= 528125

 

e) 508\(\times\)35

Answer:

By distributive law,

A\(\times\) (B+C) = A\(\times\)B + A\(\times\)C

= (500 + 8) \(\times\)35

= (500\(\times\)35) + (8\(\times\)35)

= 17500 + 280

= 17780

 

Question 5:

Analyze the pattern

(1\(\times\)8) + 1 = 91234\(\times\)8 + 4 = 9876

(12\(\times\)8) +2 = 981234\(\times\) 8 + 5 = 98765

(123\(\times\)8) + 3 = 987

Follow the next two steps and explain the working of the pattern

(Note: 12345 = 11111 + 1111 + 111 + 11 + 1)

Answer:

123456\(\times\)8 + 6 = 987648 + 6 = 987654

1234567\(\times\)8 + 7 = 9876536 + 7 =9876543

The answer is yes and the pattern works

W.K.T

123456 = (111111 + 11111 + 1111 + 111 + 11 + 1)

123456\(\times\)8 = (111111 + 11111 + 1111 + 111 + 11 +1) \(\times\)8

= (111111\(\times\)8) + (11111\(\times\)8) + (1111\(\times\)8) + (111\(\times\)8) + (11\(\times\)8) + (1\(\times\)8)

= 888888 + 88888 + 8888 + 888 + 88 + 8

= 987648

So,(123456\(\times\)8) + 6 = 987648 + 6 = 987654

This is an interesting and essential chapter to learn. A few of the important topics in the chapter are

  • Distributive Property of Whole Numbers
  • Mathematical Operations on Number Line
  • Additive and Multiplicative Identities for Whole Numbers


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