NCERT Solutions for Class 6 Maths Chapter 2 — Whole Numbers
Whole numbers are the building blocks of arithmetic. Starting from zero and going upward, they help us count, measure, and calculate. This chapter covers the number line, properties of whole numbers, and patterns that make calculations easier. Every solution below is worked out step by step.
Key Concepts
| Concept | What It Means | Example |
|---|---|---|
| Whole Numbers | The collection of numbers 0, 1, 2, 3, 4, 5, ... | 0 is the smallest whole number; there is no largest |
| Number Line | A straight line with numbers marked at equal intervals, starting from 0 | Moving right = addition; moving left = subtraction |
| Predecessor | The number that comes just before a given number | Predecessor of 7 is 6; 0 has no predecessor in whole numbers |
| Successor | The number that comes just after a given number | Successor of 7 is 8 |
| Closure Property | The result of an operation on whole numbers is also a whole number | 5 + 3 = 8 (whole); 5 − 3 = 2 (whole), but 3 − 5 is not a whole number |
| Commutative Property | Changing the order does not change the result | 3 + 7 = 7 + 3 = 10; 4 × 6 = 6 × 4 = 24 |
| Associative Property | Grouping does not change the result | (2 + 3) + 4 = 2 + (3 + 4) = 9 |
| Distributive Property | Multiplication distributes over addition | 7 × (20 + 3) = (7 × 20) + (7 × 3) = 140 + 21 = 161 |
| Identity Elements | Additive identity = 0; Multiplicative identity = 1 | a + 0 = a; a × 1 = a |
| Patterns in Whole Numbers | Numbers can be arranged as dots in rows and columns, revealing shapes | 6 = 2 × 3 (rectangle), 9 = 3 × 3 (square) |
Exercise 2.1 — Introduction to Whole Numbers and Number Line
Question 1: Write the next three natural numbers after 10999
- Step 1: The successor of 10999 is 10999 + 1 = 11000
- Step 2: The successor of 11000 is 11000 + 1 = 11001
- Step 3: The successor of 11001 is 11001 + 1 = 11002
Question 2: Write the three whole numbers occurring just before 10001
- Step 1: Predecessor of 10001 = 10001 − 1 = 10000
- Step 2: Predecessor of 10000 = 10000 − 1 = 9999
- Step 3: Predecessor of 9999 = 9999 − 1 = 9998
Question 3: Which is the smallest whole number?
- Whole numbers start from 0 and go as 0, 1, 2, 3, ...
- The number 0 has no predecessor among whole numbers.
Question 4: How many whole numbers are there between 32 and 53?
- Step 1: Whole numbers between 32 and 53 means: 33, 34, 35, ..., 52
- Step 2: Number of whole numbers = 52 − 33 + 1 = 20
- Alternative: 53 − 32 − 1 = 20
Question 5: Write the successor of: (a) 2440701 (b) 100199 (c) 1099999 (d) 2345670
- (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: Fill in the blanks
(a) Every natural number is a whole number.
(b) Every whole number is a natural number.
Question 7: Represent the following on the number line: (a) 3 + 4 (b) 7 − 2
Question 8: Find: (a) 15 + 0 (b) 0 × 27 (c) 1 × 0 (d) 0 ÷ 5
- (a) 15 + 0 = 15 (0 is the additive identity: a + 0 = a)
- (b) 0 × 27 = 0 (any number multiplied by 0 is 0)
- (c) 1 × 0 = 0 (any number multiplied by 0 is 0)
- (d) 0 ÷ 5 = 0 (0 divided by any non-zero number is 0)
Exercise 2.2 — Properties of Whole Numbers
Question 1: Find the sum by suitable rearrangement
- Step 1: Rearrange to group numbers with convenient sums: 837 + 363 + 208
- Step 2: 837 + 363 = 1200
- Step 3: 1200 + 208 = 1408
- (We used the commutative and associative properties to rearrange.)
(b) 1962 + 453 + 1538 + 647
- Step 1: Rearrange: 1962 + 1538 + 453 + 647
- Step 2: 1962 + 1538 = 3500
- Step 3: 453 + 647 = 1100
- Step 4: 3500 + 1100 = 4600
Question 2: Find the product by suitable rearrangement
- Step 1: Rearrange using commutative property: 8 × 125 × 291
- Step 2: 8 × 125 = 1000
- Step 3: 1000 × 291 = 2,91,000
(b) 25 × 5678 × 4
- Step 1: Rearrange: 25 × 4 × 5678
- Step 2: 25 × 4 = 100
- Step 3: 100 × 5678 = 5,67,800
(c) 2 × 1785 × 50
- Step 1: Rearrange: 2 × 50 × 1785
- Step 2: 2 × 50 = 100
- Step 3: 100 × 1785 = 1,78,500
Question 3: Find the product using suitable properties
Question 4: Which property is demonstrated?
(b) (4 + 5) + 6 = 4 + (5 + 6)
(c) 7 × (8 + 2) = 7 × 8 + 7 × 2
Question 5: Study the pattern and fill in the blanks
- Step 1: Observe the pattern: The result increases by one more digit each time, all 9s except the last digit.
- Step 2: 1 × 8 + 1 = 9
- Step 3: 12 × 8 + 2 = 98
- Step 4: 123 × 8 + 3 = 987
- Step 5: 1234 × 8 + 4 = 9876
- Verification: 1234 × 8 = 9872; 9872 + 4 = 9876 ✓
(b) 1 × 9 + 2 = 11; 12 × 9 + 3 = 111; 123 × 9 + 4 = 1111; 1234 × 9 + 5 = _______
- Step 1: The pattern gives all 1s, increasing by one digit each time.
- Step 2: 1 × 9 + 2 = 11
- Step 3: 12 × 9 + 3 = 111
- Step 4: 123 × 9 + 4 = 1111
- Step 5: 1234 × 9 + 5 = 11111
- Verification: 1234 × 9 = 11106; 11106 + 5 = 11111 ✓
Question 6: Are the following statements true or false?
Question 7: Verify the following
- LHS: 25 + 35 = 60
- RHS: 35 + 25 = 60
- LHS = RHS → Verified
(b) 6 × 9 = 9 × 6 (Commutative property of multiplication)
- LHS: 6 × 9 = 54
- RHS: 9 × 6 = 54
- LHS = RHS → Verified
(c) (3 + 4) + 5 = 3 + (4 + 5) (Associative property of addition)
- LHS: (3 + 4) + 5 = 7 + 5 = 12
- RHS: 3 + (4 + 5) = 3 + 9 = 12
- LHS = RHS → Verified
(d) (4 × 5) × 6 = 4 × (5 × 6) (Associative property of multiplication)
- LHS: (4 × 5) × 6 = 20 × 6 = 120
- RHS: 4 × (5 × 6) = 4 × 30 = 120
- LHS = RHS → Verified
Question 8: Find using distributive property
- Step 1: Break 48 as 50 − 2
- Step 2: 125 × 48 = 125 × (50 − 2) = (125 × 50) − (125 × 2)
- Step 3: = 6250 − 250 = 6,000
(b) 4275 × 125 + 4275 × 75
- Step 1: Factor out 4275 using distributive property: 4275 × (125 + 75)
- Step 2: 125 + 75 = 200
- Step 3: 4275 × 200 = 8,55,000
(c) 342 × 101
- Step 1: Break 101 as 100 + 1
- Step 2: 342 × 101 = 342 × (100 + 1) = (342 × 100) + (342 × 1)
- Step 3: = 34200 + 342 = 34,542
Patterns in Whole Numbers
Question 1: Express 9 as a pattern of dots in different ways
- As a line: 9 dots in a single row: ● ● ● ● ● ● ● ● ● (1 × 9)
- As a rectangle: 3 rows of 3 dots each (3 × 3 — this forms a square)
- As a rectangle: 1 row of 9 dots (1 × 9)
So 9 can be arranged as:
- 1 × 9 (a line)
- 9 × 1 (a line, transposed)
- 3 × 3 (a square)
9 is a square number because it forms a perfect square pattern.
Question 2: Which numbers from 1 to 10 can be arranged as squares?
- A number n can form a square pattern if n = k × k for some whole number k.
- 1 = 1 × 1 ✓
- 4 = 2 × 2 ✓
- 9 = 3 × 3 ✓
Question 3: Which numbers from 1 to 10 can be arranged as rectangles with more than one row and more than one column?
- We need n = a × b where a > 1 and b > 1.
- 4 = 2 × 2 ✓
- 6 = 2 × 3 ✓
- 8 = 2 × 4 ✓
- 9 = 3 × 3 ✓
- 10 = 2 × 5 ✓
Question 4: Which numbers from 1 to 10 can only be arranged as a line (one row)?
- Numbers that have only 1 as a factor (besides themselves) can only be arranged as a line.
- These are prime numbers: 2, 3, 5, 7
Question 5: Observe the pattern and find the next line
1 = 1
1 + 3 = 4
1 + 3 + 5 = 9
1 + 3 + 5 + 7 = 16
1 + 3 + 5 + 7 + 9 = ?
- Step 1: Each sum equals a perfect square: 1 = 1², 4 = 2², 9 = 3², 16 = 4²
- Step 2: The next line should equal 5² = 25
- Step 3: 1 + 3 + 5 + 7 + 9 = 25
- This pattern shows that the sum of the first n odd numbers equals n².
Question 6: Observe the pattern and find the next line
1 = 1
2 + 3 + 4 = 9
3 + 4 + 5 + 6 + 7 = 25
4 + 5 + 6 + 7 + 8 + 9 + 10 = 49
- Step 1: Each sum equals a perfect square: 1 = 1², 9 = 3², 25 = 5², 49 = 7²
- Step 2: The number of terms in each sum equals the starting number: 1 term, 3 terms, 5 terms, 7 terms
- Step 3: The next starting number is 5, and there should be 9 terms
- Step 4: 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 = 81 (= 9²)
Common Mistakes
1. Confusing natural numbers and whole numbers: Natural numbers start from 1 (1, 2, 3, ...). Whole numbers start from 0 (0, 1, 2, 3, ...). Zero is a whole number but NOT a natural number.
2. Saying whole numbers are closed under subtraction: This is false. For example, 3 − 5 = −2, which is not a whole number. Whole numbers are only closed under addition and multiplication.
3. Forgetting division properties: Whole numbers are NOT closed under division. Also, dividing by zero is undefined — never write a ÷ 0.
4. Misapplying the distributive property: Remember: a × (b + c) = a × b + a × c. Do NOT distribute addition over multiplication, i.e., a + (b × c) is NOT equal to (a + b) × (a + c).
5. Confusing additive and multiplicative identities: Zero (0) is the identity for addition (a + 0 = a). One (1) is the identity for multiplication (a × 1 = a). Do not mix them up.
6. Incorrect number line operations: On a number line, addition means moving right, subtraction means moving left. Drawing the arrow in the wrong direction leads to wrong answers.
Exam Checklist
- [ ] I can define whole numbers and distinguish them from natural numbers (whole numbers include 0).
- [ ] I can identify the successor and predecessor of any whole number (0 has no predecessor).
- [ ] I can represent addition and subtraction on a number line.
- [ ] I can apply the closure property — whole numbers are closed under addition and multiplication but NOT under subtraction and division.
- [ ] I can apply the commutative property — addition and multiplication are commutative; subtraction and division are not.
- [ ] I can apply the associative property — addition and multiplication are associative; subtraction and division are not.
- [ ] I can use the distributive property to simplify calculations: a × (b + c) = a × b + a × c.
- [ ] I know the identity elements: 0 for addition, 1 for multiplication.
- [ ] I can use patterns in whole numbers to simplify mental calculations.
- [ ] I can arrange numbers as dot patterns (line, rectangle, square) and identify which numbers form each shape.
Frequently Asked Questions
Whole numbers are 0, 1, 2, 3, 4, ... They include all natural numbers plus zero.