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 chapterbychapter 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.
The NCERT Solutions For Class 6 Maths Chapter 2  Whole Numbers are tailored to help the students master the concepts that are key to success in their classrooms. The solutions given in the PDF are developed by experts and correlate with the CBSE syllabus of 20232024. These solutions provide thorough explanations with a stepbystep approach to solving problems. Students can easily get a hold of the subject and learn the basics with a deeper understanding. Additionally, they can practice better, be confident, and perform well in their examinations with the support of this PDF.
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Students can access the NCERT Solutions For Class 6 Maths Chapter 2  Whole Numbers. Curated by experts according to the CBSE syllabus for 2023–2024, these stepbystep solutions make Maths much easier to understand and learn for the students. These solutions can be used in practice by students to attain skills in solving problems, reinforce important learning objectives, and be wellprepared for tests.
Write the next three natural numbers after 10999.
The next three natural numbers after 10999 are 11000, 11001 and 11002.
Write the three whole numbers occurring just before 10001.
The three whole numbers occurring just before 10001 are 10000, 9999 and 9998.
Which is the smallest whole number?
The smallest whole number is 0.
How many whole numbers are there between 32 and 53?
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.
Write the predecessor of:
(a) 94
(b) 10000
(c) 208090
(d) 7654321
The predecessors are
(a) 94 – 1 = 93
(b) 10000 – 1 = 9999
(c) 208090 – 1 = 208089
(d) 7654321 – 1 = 7654320
Write the successor of:
(a) 2440701
(b) 100199
(c) 1099999
(d) 2345670
The successors are
(a) 2440701 + 1 = 2440702
(b) 100199 + 1 = 100200
(c) 1099999 + 1 = 1100000
(d) 2345670 + 1 = 2345671
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
(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.
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 twodigit number is never a singledigit 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 twodigit number is always a twodigit number.
(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.
Find the sum by suitable rearrangement:
(a) 837 + 208 + 363
(b) 1962 + 453 + 1538 + 647
(a) Given 837 + 208 + 363
= (837 + 363) + 208
= 1200 + 208
= 1408
(b) Given 1962 + 453 + 1538 + 647
= (1962 + 1538) + (453 + 647)
= 3500 + 1100
= 4600
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
(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
Find the product using suitable properties.
(a) 738 × 103
(b) 854 × 102
(c) 258 × 1008
(d) 1005 × 168
(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
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?
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
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?
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
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
(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
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. 
(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

Which of the following will not represent zero?
(a) 1 + 0
(b) 0 × 0
(c) 0 / 2
(d) (10 – 10) / 2
(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
If the product of two whole numbers is zero, can we say that one or both of them will be zero? Justify through examples.
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
If the product of two whole numbers is 1, can we say that one or both of them will be 1? Justify through examples.
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.
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)
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
Find using distributive property:
(a) 728 × 101
(b) 5437 × 1001
(c) 824 × 25
(d) 4275 × 125
(e) 504 × 35
(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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The NCERT solution for Class 6 Chapter 2: Whole Numbers is important as it provides a structured approach to learning, ensuring that students develop a strong understanding of foundational concepts early in their academic journey. By mastering these basics, students can build confidence and readiness for tackling more difficult concepts in their further education.
Yes, the NCERT solution for Class 6 Chapter 2: Whole Numbers is quite useful for students in preparing for their exams. The solutions are simple, clear, and concise allowing students to understand them better. They can solve the practice questions and exercises that allow them to get examready in no time.
You can get all the NCERT solutions for Class 6 Maths Chapter 2 from the official website of the Orchids International School. These solutions are tailored by subject matter experts and are very easy to understand.
Yes, students must practice all the questions provided in the NCERT solution for Class 6 Maths Chapter 2: Whole Numbers as it will help them gain a comprehensive understanding of the concept, identify their weak areas, and strengthen their preparation.
Students can utilize the NCERT solution for Class 6 Maths Chapter 2 effectively by practicing the solutions regularly. Solve the exercises and practice questions given in the solution.