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NCERT solutions for class 6 maths chapter 2 Whole Numbers

NCERT Solutions for Class 6 Maths offer comprehensive explanations for the questions found within the NCERT textbooks endorsed by the Central Board of Secondary Education (CBSE). Orchids the international school provides these NCERT Class 6 Maths Solutions on a chapter-by-chapter basis, aiming to assist students in resolving any uncertainties and acquiring a profound comprehension of the subject matter. These resources, including NCERT Solutions, are conveniently accessible in PDF format, allowing students to download them for offline learning.

NCERT Solutions for Maths Whole Numbers

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Access Answers to NCERT solutions for class 6 maths chapter 2 Whole Numbers

Introduction

Question 1 :

 Write the next three natural numbers after 10999.

 

Answer :

The next three natural numbers after 10999 are 11000, 11001 and 11002.


Question 2 :

Write the three whole numbers occurring just before 10001.

 

Answer :

The three whole numbers occurring just before 10001 are 10000, 9999 and 9998.

 


Question 3 :

Which is the smallest whole number?

 

Answer :

The smallest whole number is 0.


Question 4 :

How many whole numbers are there between 32 and 53?

 

Answer :

 

The whole numbers between 32 and 53 are (33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52).

 

Hence, there are 20 whole numbers between 32 and 53.

 


Question 5 :

Write the predecessor of:

 

(a) 94

 

 (b) 10000

 

 (c) 208090

 

 (d) 7654321

 

Answer :

The predecessors are

 

(a) 94 – 1 = 93

 

(b) 10000 – 1 = 9999

 

(c) 208090 – 1 = 208089

 

(d) 7654321 – 1 = 7654320

 


Question 6 :

 Write the successor of:

 

(a) 2440701

 

(b) 100199

 

(c) 1099999

 

(d) 2345670

 

Answer :

The successors are

 

(a) 2440701 + 1 = 2440702

 

(b) 100199 + 1 = 100200

 

(c) 1099999 + 1 = 1100000

 

(d) 2345670 + 1 = 2345671

 


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

 

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

 

(b) 370 > 307

 

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

 

(c) 98765 > 56789

 

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

 

(d) 9830415 < 10023001

 

Hence, 9830415 is on the left side of 10023001 on the number line.

 


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.(l) The whole number 0 has no predecessor.

 

(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

The predecessor of 399 is 398 since (399 – 1 = 398).

 

(c).True

Zero is the smallest whole number.

 

(d).True

Since (599 + 1 = 600).

 

(e).True

All natural numbers are whole numbers.

 

(f).False

0 is a whole number but is not a natural number.

 

(g).False

For example, the predecessor of 10 is 9.

 

(h).False

0 is the smallest whole number.

 

(i).True

The predecessor of 1 is 0, but it is not a natural number.

 

(j).False

0 is the predecessor of 1 and is a whole number.

 

(k).False

13 does not lie between 11 and 12.

 

(l).True

The predecessor of 0 is -1 and is not a whole number.

 

(m).False

As the successor of 99 is 100.

 


Whole Numbers

Question 1 :

 Find the sum by suitable rearrangement:

 

(a) 837 + 208 + 363

 

(b) 1962 + 453 + 1538 + 647

 

Answer :

(a) Given 837 + 208 + 363

= (837 + 363) + 208

= 1200 + 208

= 1408

(b) Given 1962 + 453 + 1538 + 647

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

= 3500 + 1100

= 4600

 


Question 2 :

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) Given 297 × 17 + 297 × 3

= 297 × (17 + 3)

= 297 × 20

= 5940

(b) Given 54279 × 92 + 8 × 54279

= 54279 × 92 + 54279 × 8

= 54279 × (92 + 8)

= 54279 × 100

= 5427900

(c) Given 81265 × 169 – 81265 × 69

= 81265 × (169 – 69)

= 81265 × 100

= 8126500

(d) Given 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

 


Question 3 :

Find the product using suitable properties.

(a) 738 × 103

 

(b) 854 × 102

 

(c) 258 × 1008

 

(d) 1005 × 168

 

Answer :

(a) Given 738 × 103

= 738 × (100 + 3)

= 738 × 100 + 738 × 3 (using distributive property)

= 73800 + 2214

= 76014

(b) Given 854 × 102

= 854 × (100 + 2)

= 854 × 100 + 854 × 2 (using distributive property)

= 85400 + 1708

= 87108

(c) Given 258 × 1008

= 258 × (1000 + 8)

= 258 × 1000 + 258 × 8 (using distributive property)

= 258000 + 2064

= 260064

(d) Given 1005 × 168

= (1000 + 5) × 168

= 1000 × 168 + 5 × 168 (using distributive property)

= 168000 + 840

= 168840

 


Question 4 :

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?

 

Answer :

Petrol quantity filled on Monday = 40 litres

Petrol quantity filled on Tuesday = 50 litres

Total petrol quantity filled = (40 + 50) litre

Cost of petrol per litre = ₹ 44

Total money spent = 44 × (40 + 50)

= 44 × 90

= ₹ 3960

 


Question 5 :

 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 ₹ 45 per litre, how much money is due to the vendor per day?

 

Answer :

Milk quantity supplied in the morning = 32 litres

Milk quantity supplied in the evening = 68 litres

Cost of milk per litre = ₹ 45

Total cost of milk per day = 45 × (32 + 68)

= 45 × 100

= ₹ 4500

Hence, the money due to the vendor per day is ₹ 4500

 


Question 6 :

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) Given 2 × 1768 × 50

= 2 × 50 × 1768

= 100 × 1768

= 176800

(b) Given 4 × 166 × 25

= 4 × 25 × 166

= 100 × 166

= 16600

(c) Given 8 × 291 × 125

= 8 × 125 × 291

= 1000 × 291

= 291000

(d) Given 625 × 279 × 16

= 625 × 16 × 279

= 10000 × 279

= 2790000

(e) Given 285 × 5 × 60

= 285 × 300

= 85500

(f) Given 125 × 40 × 8 × 25

= 125 × 8 × 40 × 25

= 1000 × 1000

= 1000000

 


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)  (c) Distributivity of multiplication over addition. Hence, (c) is the correct answer
(ii) 2 × 49 × 50 = 2 × 50 × 49  (a) Commutativity under multiplication Hence, (a) is the correct answer

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

(b) Commutativity under addition

Hence, (b) is the correct answer

 

 

 

 

 

 


The Number Line

Question 1 :

Which of the following will not represent zero?

(a) 1 + 0

(b) 0 × 0

(c) 0 / 2

(d) (10 – 10) / 2

 

Answer :

(a) 1 + 0 = 1

Hence, it does not represent zero

(b) 0 × 0 = 0

Hence, it represents zero

(c) 0 / 2 = 0

Hence, it represents zero

(d) (10 – 10) / 2 = 0 / 2 = 0

Hence, it represents 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.

 

Answer :

If the product of two whole numbers is zero, definitely one of them is zero

Example: 0 × 3 = 0 and 15 × 0 = 0

If the product of two whole numbers is zero, both of them may be zero

Example: 0 × 0 = 0

Yes, if the product of two whole numbers is zero, then both of them will be zero

 


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.

 

Answer :

If the product of two whole numbers is 1, both numbers should be equal to 1

Example: 1 × 1 = 1

But 1 × 5 = 5

Hence, it is clear that the product of two whole numbers will be 1, only in situations when both numbers to be multiplied are 1.

 


Question 4 :

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 = 987654

1234567 × 8 + 7 = 9876543

Given 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

Yes, here the pattern works

1234567 × 8 + 7 = 9876543

Given 1234567 = (1111111 + 111111 + 11111 + 1111 + 111 + 11 + 1)

1234567 × 8 = (1111111 + 111111 + 11111 + 1111 + 111 + 11 + 1) × 8

= 1111111 × 8 + 111111 × 8 + 11111 × 8 + 1111 × 8 + 111 × 8 + 11 × 8 + 1 × 8

= 8888888 + 888888 + 88888 + 8888 + 888 + 88 + 8

= 9876536

1234567 × 8 + 7 = 9876536 + 7

= 9876543

Yes, here the pattern works

 


Question 5 :

 Find using distributive property:

 

(a) 728 × 101

 

 

(b) 5437 × 1001

 

 

(c) 824 × 25

 

 

(d) 4275 × 125

 

 

(e) 504 × 35

 

 

Answer :

(a) Given 728 × 101

= 728 × (100 + 1)

= 728 × 100 + 728 × 1

= 72800 + 728

= 73528

(b) Given 5437 × 1001

= 5437 × (1000 + 1)

= 5437 × 1000 + 5437 × 1

= 5437000 + 5437

= 5442437

(c) Given 824 × 25

= (800 + 24) × 25

= (800 + 25 – 1) × 25

= 800 × 25 + 25 × 25 – 1 × 25

= 20000 + 625 – 25

= 20000 + 600

= 20600

(d) Given 4275 × 125

= (4000 + 200 + 100 – 25) × 125

= (4000 × 125 + 200 × 125 + 100 × 125 – 25 × 125)

= 500000 + 25000 + 12500 – 3125

= 534375

(e) Given 504 × 35

= (500 + 4) × 35

= 500 × 35 + 4 × 35

= 17500 + 140

= 17640

 


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