Chapter 2 of NCERT Solutions for Class 6 Maths focuses on introducing and developing a strong foundation in understanding whole numbers, paving the way for more complex mathematical concepts. Here, you find class 6 maths chapter 2 all exercises solutions.
Maths Class 6th Chapter 1 Solutions:
Exercise 2.1 PAGE No.: 31
1. Write the next three natural numbers after 10999.
Solutions:
The next three natural numbers after 10999 are 11000, 11001 and 11002.
2. Write the three whole numbers occurring just before 10001.
Solutions:
The three whole numbers occurring just before 10001 are 10000, 9999 and 9998.
3. Which is the smallest whole number?
Solutions:
The smallest whole number is 0.
4. How many whole numbers are there between 32 and 53?
Solutions:
The whole numbers between 32 and 53 are as follows:
(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
5. Write the successor of:
(a) 2440701 (b) 100199 (c) 1099999 (d) 2345670
Solutions:
The successors are
(a) 2440701 + 1 = 2440702
(b) 100199 + 1 = 100200
(c) 1099999 + 1 = 1100000
(d) 2345670 + 1 = 2345671
6. Write the predecessor of:
(a) 94 (b) 10000 (c) 208090 (d) 7654321
Solutions:
The predecessors are
(a) 94 – 1 = 93
(b) 10000 – 1 = 9999
(c) 208090 – 1 = 208089
(d) 7654321 – 1 = 7654320
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
Solutions:
(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
8. Which of the following statements are true (T) and which are false (F)?
(a) Zero is the smallest natural number.
Solution:
False
0 is not a natural number.
(b) 400 is the predecessor of 399.
Solution:
False
The predecessor of 399 is 398 because (399 – 1 = 398)
(c) Zero is the smallest whole number.
Solution:
True
Zero is the smallest whole number.
(d) 600 is the successor of 599.
Solution:
True
Since (599 + 1 = 600)
(e) All natural numbers are whole numbers.
Solution:
True
All natural numbers are whole numbers.
(f) All whole numbers are natural numbers.
Solution:
False
0 is a whole number but is not a natural number.
(g) The predecessor of a two-digit number is never a single-digit number.
Solution:
False
For example, the predecessor of 10 is 9.
(h) 1 is the smallest whole number.
Solution:
False
0 is the smallest whole number.
(i) The natural number 1 has no predecessor.
True
The predecessor of 1 is 0, but it is not a natural number.
(j) The whole number 1 has no predecessor.
Solution:
False
0 is the predecessor of 1 and is a whole number.
(k) The whole number 13 lies between 11 and 12.
Solution:
False
13 does not lie between 11 and 12.
(l) The whole number 0 has no predecessor.
Solution:
True
The predecessor of 0 is -1 and is not a whole number.
(m) The successor of a two-digit number is always a two-digit number.
Solution:
False
For example, the successor of 99 is 100
Exercise 2.2 PAGE No.: 40
1. Find the sum by suitable rearrangement:
(a) 837 + 208 + 363
(b) 1962 + 453 + 1538 + 647
Solutions:
(a) Given 837 + 208 + 363
= (837 + 363) + 208
= 1200 + 208
= 1408
(b) Given 1962 + 453 + 1538 + 647
= (1962 + 1538) + (453 + 647)
= 3500 + 1100
= 4600
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
Solutions:
(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
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
Solutions:
(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
4. Find the product using suitable properties.
(a) 738 × 103
(b) 854 × 102
(c) 258 × 1008
(d) 1005 × 168
Solutions:
(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
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 ₹ 44 per litre, how much did he spend in all on petrol?
Solutions:
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
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 ₹ 45 per litre, how much money is due to the vendor per day?
Solutions:
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
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.
Solutions:
(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
Exercise 2.3 PAGE No.: 43
1. Which of the following will not represent zero?
(a) 1 + 0
(b) 0 × 0
(c) 0 / 2
(d) (10 – 10) / 2
Solutions:
(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.
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.
Solutions:
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.
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.
Solutions:
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’s clear that the product of two whole numbers will be 1, only in situations when both numbers to be multiplied are 1.
4. Find using distributive property:
(a) 728 × 101
(b) 5437 × 1001
(c) 824 × 25
(d) 4275 × 125
(e) 504 × 35
Solutions:
(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
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)
Solutions:
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
Fundamentals of Whole Numbers
Whole numbers, a set of numbers comprising zero and all positive integers, play a vital role in arithmetic. These numbers possess unique properties that distinguish them from other numerical sets. For instance, they are closed under addition, subtraction, multiplication, and division.
Basic operations involving whole numbers, such as addition, subtraction, multiplication, and division, are essential skills covered in this chapter. Real-life examples illustrating these operations help in better comprehension.
Place Value and Face Value
Understanding the place value and face value of digits within a number is crucial. NCERT Solutions break down these concepts, offering methods to determine place values and face values, supplemented with exercises for practice.
Properties of Whole Numbers
Properties like closure, commutativity, associativity, and distributivity are inherent to whole numbers and are explored in detail. Understanding these properties aids in solving complex mathematical problems efficiently.
Prime and Composite Numbers
Identification and differentiation between prime and composite numbers form a significant part of this chapter. Students engage in exercises to grasp the importance of prime numbers in mathematics.
Factors and Multiples
The concept of factors and multiples is elucidated with examples. Finding factors and multiples of numbers and their application in problem-solving are highlighted.
Understanding Prime Factorization
Prime factorization, a critical concept in simplifying fractions, is explained thoroughly. Methods to find prime factors and their significance in mathematics are emphasized.
Common Factors and Common Multiples
Understanding common factors and multiples and applying them to solve problems is an essential skill taught in this section.
Applications of Whole Numbers in Daily Life
The chapter delves into practical applications of whole numbers in everyday scenarios, showcasing their relevance beyond the confines of mathematical exercises.
Tips to Master NCERT Solutions for Class 6 Maths Chapter 2
Effective study techniques, strategies, and practice resources are recommended to aid in mastering the concepts covered in this chapter.
Importance of NCERT Solutions
NCERT Solutions act as a guiding light, aiding students in understanding complex mathematical concepts with ease. The structured approach and comprehensive explanations provided by NCERT Solutions make learning mathematics an enjoyable experience.
Challenges and How to Overcome Them
Some students may face challenges in grasping the concepts of whole numbers. However, with dedication and practice, these difficulties can be overcome.
