**Whole Numbers**- As the name suggests this chapter explains the

**whole numbers**.

- The
**natural numbers**along with**zero**form the collection of**whole numbers.**

**predecessor**and

**successor**followed by concept of

**whole numbers.**

- If you add 1 to a
**natural number**, we get its**successor**. If you subtract 1 from a**natural number**, you get its**predecessor.** - Every
**natural number**has a**successor**. - Every
**natural number**except 1 has a**predecessor**. - Every
**whole number**has a**successor**. - Every
**whole number**except zero has a**predecessor**. - All
**natural numbers**are**whole numbers**, but all**whole numbers are not natural numbers**.

The topic

**number line**is discussed in detail along with the operations like

**addition, subtraction and multiplication**that can be performed on them.

This is followed by

**Properties of whole numbers**. Various properties associated with whole numbers are explained in this chapter with examples.

**Closure property**- Division of a whole number by 0 is not defined.
**Commutativity**of addition and multiplication**Associativity**of addition and multiplication**Distributivity of multiplication over addition****Zero**is called an**identity**for addition of whole numbers or**additive identity**for whole number**s.****Whole number 1**is the**identity for multiplication**of whole numbers.

**Patterns in whole numbers**are the last topic that is discussed in this chapter-

**Whole numbers**. These patterns are formed using numbers and

**arrangement of dots**. This section contains 5 questions in the column titled '

**Try These**'. Students must try these questions as they are not only fun to attempt but also help in building the concept.

A total of 3 exercises are given in the chapter. Ample number of solved examples are given for reference to solve the unsolved questions.

Important points are mentioned at the end of the chapter in the form of a summary.

#### Page No 31:

#### Question 1:

Write the next three natural numbers after 10999.

#### Answer:

Next three natural numbers after 10999 are

11000, 11001, 11002

#### Page No 31:

#### Question 2:

Write the three whole numbers occurring just before 10001.

#### Answer:

3 whole numbers just before 10001 are

10000, 9999, 9998

#### Page No 31:

#### Question 3:

Which is the smallest whole number?

#### Answer:

The smallest whole number is 0.

#### Page No 31:

#### Question 4:

How many whole numbers are there between 32 and 53?

#### Answer:

Whole numbers between 32 and 53 = 20 (53 − 32 − 1 = 20)

(33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52)

#### Page No 31:

#### Question 5:

Write the successor of:

(a) 2440701 (b) 100199

(c) 1099999 (d) 2345670

#### Answer:

(a) 2440701 + 1 = 2440702

(b) 100199 + 1 = 100200

(c) 1099999 + 1 = 1100000

(d) 2345670 + 1 = 2345671

#### Page No 31:

#### Question 6:

Write the predecessor of:

(a) 94 (b) 10000

(c) 208090 (d) 7654321

#### Answer:

(a) 94 − 1 = 93

(b) 10000 − 1 = 9999

(c) 208090 − 1 =208089

(d) 7654321 − 1 = 7654320

#### Page No 31:

#### Question 7:

In each of the following pairs of numbers, state which whole number is on the left of the other number on the number line. Also write them with the appropriate sign (>, <) between them.

(a) 530, 503 (b) 370, 307

(c) 98765, 56789 (d) 9830415, 10023001

#### Answer:

(a) 530, 503

As 530 > 503,

503 is on the left side of 530 on the number line.

(b) 370, 307

As 370 > 307,

307 is on the left side of 370 on the number line.

(c) 98765, 56789

As 98765 > 56789,

56789 is on the left side of 98765 on the number line.

(d) 9830415, 10023001

Since 98, 30, 415 < 1, 00, 23, 001,

98,30,415 is on the left side of 1,00,23,001 on the number line.

#### Page No 31:

#### Question 8:

Which of the following statements are true (T) and which are false (F)?

(a) Zero is the smallest natural number.

(b) 400 is the predecessor of 399.

(c) Zero is the smallest whole number.

(d) 600 is the successor of 599.

(e) All natural numbers are whole numbers.

(f) All whole numbers are natural numbers.

(g) The predecessor of a two digit number is never a single digit number.

(h) 1 is the smallest whole number.

(i) The natural number 1 has no predecessor.

(j) The whole number 1 has no predecessor.

(k) The whole number 13 lies between 11 and 12.

(l) The whole number 0 has no predecessor.

(m) The successor of a two digit number is always a two digit number.

#### Answer:

(a) False, 0 is not a natural number.

(b) False, as predecessor of 399 is 398 (399 − 1 = 398).

(c) True

(d) True, as 599 + 1 = 600

(e) True

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

(g) False, as predecessor of 10 is 9.

(h) False, 0 is the smallest whole number.

(i) True, as 0 is the predecessor of 1 but it is not a natural number.

(j) False, as 0 is the predecessor of 1 and it is a whole number.

(k) False, 13 does not lie in between 11 and 12.

(l) True, predecessor of 0 is −1, which is not a whole number.

(m) False, as successor of 99 is 100.

#### Page No 40:

#### Question 1:

Find the sum by suitable rearrangement:

(a) 837 + 208 + 363 (b) 1962 + 453 + 1538 + 647

#### Answer:

(a) 837 + 208 + 363 = (837 + 363) + 208

= 1200 + 208 = 1408

(b) 1962 + 453 + 1538 + 647 = (1962 + 1538) + (453 + 647)

= 3500 + 1100 = 4600

#### Page No 40:

#### Question 2:

Find the product by suitable rearrangement:

(a) 2 × 1768 × 50 (b) 4 × 166 × 25

(c) 8 × 291 × 125 (d) 625 × 279 × 16

(e) 285 × 5 × 60 (f) 125 × 40 × 8 × 25

#### Answer:

(a) 2 × 1768 × 50 = 2 × 50 × 1768

= 100 × 1768 = 176800

(b) 4 × 166 × 25 = 4 × 25 × 166

= 100 × 166 = 16600

(c) 8 × 291 × 125 = 8 × 125 × 291

= 1000 × 291 = 291000

(d) 625 × 279 × 16 = 625 × 16 × 279

= 10000 × 279 = 2790000

(e) 285 × 5 × 60 = 285 × 300 = 85500

(f) 125 × 40 × 8 × 25 = 125 × 8 × 40 × 25

= 1000 × 1000 = 1000000

#### Page No 40:

#### Question 3:

Find the value of the following:

(a) 297 × 17 + 297 × 3 (b) 54279 × 92 + 8 × 54279

(c) 81265 × 169 − 81265 × 69 (d) 3845 × 5 × 782 + 769 × 25 × 218

#### Answer:

(a) 297 × 17 + 297 × 3 = 297 × (17 + 3)

= 297 × 20 = 5940

(b) 54279 × 92 + 8 × 54279 = 54279 × 92 + 54279 × 8

= 54279 × (92 + 8)

= 54279 × 100 = 5427900

(c) 81265 × 169 − 81265 × 69 = 81265 × (169 − 69)

= 81265 × 100 = 8126500

(d) 3845 × 5 × 782 + 769 × 25 × 218

= 3845 × 5 × 782 + 769 × 5 × 5 × 218

= 3845 × 5 × 782 + 3845 × 5 × 218

= 3845 × 5 × (782 + 218)

= 19225 × 1000 = 19225000

##### Video Solution for knowing our numbers (Page: 40 , Q.No.: 3)

NCERT Solution for Class 6 math - knowing our numbers 40 , Question 3

#### Page No 40:

#### Question 4:

Find the product using suitable properties.

(a) 738 × 103 (b) 854 × 102

(c) 258 × 1008 (d) 1005 × 168

#### Answer:

(a) 738 × 103 = 738 × (100 + 3)

= 738 × 100 + 738 × 3 (Distributive property)

= 73800 + 2214

= 76014

(b) 854 × 102 = 854 × (100 + 2)

= 854 × 100 + 854 × 2 (Distributive property)

= 85400 + 1708 = 87108

(c) 258 × 1008 = 258 × (1000 + 8)

= 258 × 1000 + 258 × 8 (Distributive property)

= 258000 + 2064 = 260064

(d) 1005 × 168 = (1000 + 5) × 168

= 1000 × 168 + 5 × 168 (Distributive property)

= 168000 + 840 = 168840

#### Page No 40:

#### Question 5:

A taxi driver filled his car petrol tank with 40 litres of petrol on Monday. The next day, he filled the tank with 50 litres of petrol. If the petrol costs Rs 44 per litre, how much did he spend in all on petrol?

#### Answer:

Quantity of petrol filled on Monday = 40 *l*

Quantity of petrol filled on Tuesday = 50 *l*

Total quantity filled = (40 + 50) *l*

Cost of petrol (per *l*) = Rs 44

Total money spent = 44 × (40 + 50)

= 44 × 90 = Rs 3960

##### Video Solution for knowing our numbers (Page: 40 , Q.No.: 5)

NCERT Solution for Class 6 math - knowing our numbers 40 , Question 5

#### Page No 41:

#### Question 6:

A vendor supplies 32 litres of milk to a hotel in the morning and 68 litres of milk in the evening. If the milk costs Rs 15 per litre, how much money is due to the vendor per day?

#### Answer:

Quantity of milk supplied in the morning = 32 *l*

Quantity of milk supplied in the evening = 68* l*

Total of milk per litre = (32 + 68) *l*

Cost of milk per litre = Rs 15

Total cost per day = 15 × (32 + 68)

= 15 × 100 = Rs 1500

##### Video Solution for knowing our numbers (Page: 41 , Q.No.: 6)

NCERT Solution for Class 6 math - knowing our numbers 41 , Question 6

#### Page No 41:

#### Question 7:

Match the following:

(i) 425 × 136 = 425 × (6 + 30 + 100)

(a) Commutativity under multiplication

(ii) 2 × 49 × 50 = 2 × 50 × 49

(b) Commutativity under addition

(iii) 80 + 2005 + 20 = 80 + 20 + 2005

(c) Distributivity of multiplication over addition

#### Answer:

(i) 425 × 136 = 425 × (6 + 30 + 100) [Distributivity of multiplication over addition]

Hence, (c)

(ii) 2 × 49 × 50 = 2 × 50 × 49 [Commutativity under multiplication]

Hence, (a)

(iii) 80 + 2005 + 20 = 80 + 20 + 2005 [Commutativity under addition]

Hence, (b)

#### Page No 43:

#### Question 1:

Which of the following will not represent zero?

(a) 1 + 0 (b) 0 × 0

(c) (d)

#### Answer:

(a) 1 + 0 = 1

It does not represent zero.

(b) 0 × 0 = 0

It represents zero.

(c)

It represents zero.

(d)

It represents zero.

#### Page No 43:

#### Question 2:

If the product of two whole numbers is zero, can we say that one or both of them will be zero? Justify through examples.

#### Answer:

If the product of 2 whole numbers is zero, then one of them is definitely zero.

For example, 0 × 2 = 0 and 17 × 0 = 0

If the product of 2 whole numbers is zero, then both of them may be zero.

0 × 0 = 0

However, 2 × 3 = 6

(Since numbers to be multiplied are not equal to zero, the result of the product will also be non-zero.)

#### Page No 44:

#### Question 3:

If the product of two whole numbers is 1, can we say that one of both of them will be 1? Justify through examples.

#### Answer:

If the product of 2 numbers is 1, then both the numbers have to be equal to 1.

For example, 1 × 1 = 1

However, 1 × 6 = 6

Clearly, the product of two whole numbers will be 1 in the situation when both numbers to be multiplied are 1.

#### Page No 44:

#### Question 4:

Find using distributive property:

(a) 728 × 101 (b) 5437 × 1001

(c) 824 × 25 (d) 4275 × 125

(e) 504 × 35

#### Answer:

(a) 728 × 101= 728 × (100 + 1)

= 728 × 100 + 728 × 1

= 72800 + 728 = 73528

(b) 5437 × 1001 = 5437 × (1000 + 1)

= 5437 × 1000 + 5437 × 1

= 5437000 + 5437 = 5442437

(c) 824 × 25 = (800 + 24) × 25

= (800 + 25 − 1) × 25

= 800 × 25 + 25 × 25 − 1 × 25

= 20000 + 625 − 25

= 20000 + 600 = 20600

(d) 4275 × 125 = (4000 + 200 + 100 − 25) × 125

= 4000 × 125 + 200 × 125 + 100 × 125 − 25 × 125

= 500000 + 25000 + 12500 − 3125

= 534375

(e) 504 × 35 = (500 + 4) × 35

= 500 × 35 + 4 × 35

= 17500 + 140 = 17640

#### Page No 44:

#### Question 5:

Study the pattern:

1 × 8 + 1 = 9 1234 × 8 + 4 = 9876

12 × 8 + 2 = 98 12345 × 8 + 5 = 98765

123 × 8 + 3 = 987

Write the next two steps. Can you say how the pattern works?

(Hint: 12345 = 11111 + 1111 + 111 + 11 + 1).

#### Answer:

123456 × 8 + 6 = 987648 + 6 = 987654

1234567 × 8 + 7 = 9876536 + 7 = 9876543

Yes, the pattern works.

As 123456 = 111111 + 11111 + 1111 + 111 + 11 + 1,

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

= 111111 × 8 + 11111 × 8 + 1111 × 8 + 111 × 8 + 11 × 8 + 1 × 8

= 888888 + 88888 + 8888 + 888 + 88 + 8 = 987648

123456 × 8 + 6 = 987648 + 6 = 987654

##### Video Solution for knowing our numbers (Page: 44 , Q.No.: 5)

NCERT Solution for Class 6 math - knowing our numbers 44 , Question 5

View NCERT Solutions for all chapters of Class 6