Contents

NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 6 |

Subject |
Maths |

Chapter |
Chapter 2 |

Chapter Name |
Whole Numbers |

Exercise |
Ex 2.1, Ex 2.2, Ex 2.3 |

Number of Questions Solved |
20 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers

### Chapter 2 Whole Numbers Exercise 2.1

**Question 1.**

Write the next three natural numbers after 10999.

**Solution:**

The next three natural numbers after 10999 are 10999 +1,10999 + 2 and 10999 + 3

i.e.,11000, 11001 and 11002.

**Question 2.**

Write the three whole numbers occurring just before 10001.

**Solution:**

The three whole numbers occurring just before 10001 are 10001 – 1, 10001 – 2 and 10001 – 3

i.e., 10000, 9999 and 9998.

**Question 3.**

Which is the smallest whole number?

**Solution:**

The smallest whole number is 0.

**Question 4.**

How many whole numbers are there between 32 and 53?

**Solution:**

There are (53 – 32) -1 = 21 – 1= 20 whole numbers between 32 and 53.

**Question 5.**

Write the successor of:

**(a)** 2440701

**(b)** 100199

**(c)** 1099999

**(d)** 2345670

**Solution:**

**(a)** Successor of 2440701 = 2440701 + 1 = 2440702

**(b)** Successor of 100199 = 100199 + 1 = 100200

**(c)** Successor of 1099999 =1099999 +1 = 1100000

**(d)** Successor of 2345670 = 2345670 + 1 = 2345671

**Question 6.**

Write the predecessor of:

**(a)** 94

**(b)** 10000

**(c)** 208090

**(d)** 7654321

**Solution:**

**(a)** Predecessor of 94 = 94 – 1 = 93

**(b)** Predecessor of 10000 = 10000 – 1 = 9999

**(c)** Predecessor of 208090 = 208090 – 1 = 208089

**(d)** Predecessor of 7654321 =7654321 – 1 = 7654320

**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

**Solution:**

**(a)** Here 503 lies on the left of 530 on the number line.

∴ 530 > 503

**(b)** Here 307 lies on the left of 370 on the number line.

∴ 370 > 307

**(c)** Here 56789.lies on the left of 98765 on the number line.

∴ 98765 > 56789

**(d)** Here 9830415 lies on the left of 10023001

∴ 9830415 > 10023001

**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.

**Solution:**

**(a)** False

**(b)** False

**(c)** True

**(d)** True

**(e)** True

**(f)** False

**(g)** False

**(h)** False

**(i)** True

**(j)** False

**(k)** False

**(l)** True

**(m)** False

### Chapter 2 Whole Numbers Exercise 2.2

**Question 1.**

Find the sum by suitable rearrangement:

**(a)** 837 + 208 + 363

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

**Solution:**

**Question 2.**

Find the product by suitable rearrangement:

**(a)** 2 x 1768 x 50

**(b)** 4 x 166 x 25

**(c)** 8 x 291 x 125

**(d)** 625 x 279 x 16

**(e)** 285 x 5 x 60

**(f)** 125 x 40 x 8 x 25

**Solution:**

**(a)** 2 x 1768 x 50 = (2 x 50) x 1768

= 100 x 1768 = 176800

**(b)** 4 x 166 x 25 = (4 x 25) x 166

= 100 x 166 = 16600

**(c)** 8 x 291 x 125 = (8 x 125) x 291

= 1000 x 291 = 291000

**(d)** 625 x 279 x 16 = (625 x 16) x 279

= 10000 x 279 = 2790000

**(e)** 285 x 5 x 60 = 285 x (5 x 60)

= 285 x 300 = 85500

**(f)** 125 x 40 x 8 x 25 = (125 x 8) x (40 x 25)

= 1000 x 1000

= 1000000

**Question 3.**

Find the value of the following:

**(a)** 297 x 17 + 297 x 3

**(b)** 54279 x 92 + 8 x 54279

**(c)** 81265 x 169 – 81265 x 69

**(d)** 3845 x 5 x 782 + 769 x 25 x 218

**Solution:**

**(a)** 297 x 17 + 297 x 3 = 297 x (17 + 3)

= 297 x 20

= 5949

**(b)** 54279 x 92 + 8 x 54279

= 54279 x (92 + 8)

= 54279 x 100 = 5427900

**(c)** 81265 x 169 – 81265 x 69

= 81265 x (169 -69)

= 81265 x 100

= 8126500

**(d)** 3845 x 5 x 782 + 769 x 25 x 218

= 3845 x 5 x 782 + (769 x 5) x 5 x 218

= 3845x5x782 + 3845x5x218

= (3845 x 5) x (782+ 218)

=19225 x 1000

=19225000

**Question 4.**

Find the product using suitable properties :

**(a)** 738 x 103

**(b)** 854 x 102

**(c)** 258 x 1008

**(d)** 1005 x 168

**Solution:**

**(a)** 738 x 103 = 738 x (100 + 3)

= 738 x 100 + 738 x 3 = 73800 + 2214 = 76014

**(b)** 854 x 102 = 854 x (100 + 2)

= 854 x 100 + 854 x 2 = 85400 + 1708 = 87108

**(c)** 258 x 1008 = 258 x (1000 + 8)

= 258 x 1000 + 258 x 8 = 258000 + 2064 = 260064

**(d)** 1005 x 168 = (1000 + 5) x 168

= 1000 x 168 + 5 x 168

= 168000 + 840 = 168840

**Question 5.**

A taxidriver 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 ? 44 per litre, how much did he spend in all on petrol?

**Solution:**

Petrol filled on Monday and Tuesday

= 40 litres + 50 litres

= 90 litres

Cost of petrol @ ₹ 44 per litre

= 90 x ₹ 44 = ₹ 3960

**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 ? 15 per litre, how much money is due to the vendor per day?

**Solution:**

Milk supplied to a hotel in the morning and evening

= 32 litres + 68 litres = 100 litres

Money due to vendor per day.

= 100 x ₹ 15 = ₹ 1500

**Question 7.**

Match the following:

**(i)** 425 x 136 = 425 x (6 + 30 +100) (a) Commutativity under multiplication

**(ii)** 2 x 49 x 50 = 2 x 50 x 49 (b) Commutativity under addition

**(iii)** 80 + 2005 + 20 = 80 + 20 + 2005 (c) Distributivity of multiplication over addition

**Solution:**

Matching is as under :

**(i)** → (c)

**(ii)** → (a)

**(iii)** → (b)

### Chapter 2 Whole Numbers Exercise 2.3

**Question 1.**

Which of the following will not represent zero :

**(a)** a + 0

**(b)** 0 x 0

**(c)**

**(d)**

**Solution:**

**(a)** 1 + 0 = 1 ≠ 0

**(b)** 0 x 0 = 0

**(c)**

**(d)**

Thus, only (a) does not represent zero.

**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.

**Solution:**

We know that the product of any whole number and zero is always zero i. e., a x 0 = 0, where a is any whole number.

**Question 3.**

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

**Solution:**

We know that if a is any whole number, then

a x l = a -1 x a

∴ Clearly a must be = 1

Thus, both the numbers should be 1 i. e., 1 x 1 = 1

**Question 4.**

Find by using distributive property :

**(a)** 728 x 101

**(b)** 5437 x 1001

**(c)** 824 x 25

**(d)** 4275 x 125

**(e)** 504 x 35

**Solution:**

**Question 5.**

Study the pattern :

1 x 8 +1 = 9

12 x 8 + 2 = 98

123 x 8 + 3 = 987

1234 x 8 + 4 = 9876

12345 x 8 + 5 = 98765

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

**Solution:**

Write the next two steps :

123456 x 8 + 6 = 987654

1234567×8 + 7 = 9876543

How the pattern works?

1 x8 + 1= 9

(11 + 1) x 8 + 2 = 12 x 8 + 2 = 98

(111 + 11 + 1) x 8 + 3 = 123 x 8 + 3 = 987 .

(1111 + 111 + 11 + 1) x 8 + 4 =1234 x 8 + 4 =9876

(11111 + 1111 + 111 + 11 + 1) x 8 + 5

= 12345 x 8 + 5=98765

(111111 + 11111 + 1111 + 111 + 11 + 1) x 8+ 6

= 123456 x 8 + 6 = 987654

and (1111111 + 111111 + 11111 +1111 +111 +11 +1) x 8 + 7

= 1234567 x 8 + 7 = 9876543

We hope the NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers, drop a comment below and we will get back to you at the earliest.

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