# NCERT Class 6 Maths Solutions for Chapter 2 Whole Numbers

## Exercise 2.1 (Page 31)

**1. Write the next three natural numbers after 10999.**

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

**2. Write the three whole numbers occurring just before 10001.**

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

**3. Which is the smallest whole number?**

The smallest whole number is 0.

**4. How many whole numbers are there between 32 and 53?**

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

(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

(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

(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.
- (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.

(a) False – 0 is not a natural number.

(b) False – The predecessor of 399 is 398 because (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 – For example, the successor of 99 is 100.

## Exercise 2.2 (Page 40)

**1. 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

**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

(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

(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

(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?**

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?**

Milk supplied in the morning = 32 litres

Milk supplied in the evening = 68 litres

Total milk supplied = (32 + 68) litres

Cost of milk per litre = ₹ 45

Total money due to the vendor = 45 × (32 + 68)

= 45 × 100

= ₹ 4500

## Exercise 2.3 (Page 42)

**1. 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 (not zero)

(b) 0 × 0 = 0

(c) 0 ÷ 2 = 0

(d) (10 – 10) ÷ 2 = 0 ÷ 2 = 0

Therefore, option (a) will not represent 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.**

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

For example:

0 × 4 = 0 (one number is zero)

0 × 0 = 0 (both numbers are zero)

This shows that if the product is zero, at least one of the numbers must be zero.

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

Yes, if the product of two whole numbers is 1, then one or both of them must be 1.

For example:

1 × 1 = 1 (both numbers are 1)

This shows that if the product is 1, both numbers must be 1.

**4. Find the using distributive property:**

- (a) 728 × 101
- (b) 5437 × 1001
- (c) 824 × 25
- (d) 4275 × 125

(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

= 824 × (20 + 5)

= 824 × 20 + 824 × 5

= 16480 + 4120

= 20600

(d) Given 4275 × 125

= 4275 × (100 + 25)

= 4275 × 100 + 4275 × 25

= 427500 + 106875

= 534375

**5. Study the pattern:**

- 4 × 6 = 24
- 4 × 66 = 264
- 4 × 666 = 2664
- 4 × 6666 = 26664
- 4 × 66666 = 266664

Observe the number of sixes and number of digits in the product.

**6. Using distributive property, solve the following:**

- (a) 258 × 1008
- (b) 1005 × 168

(a) Given 258 × 1008

= 258 × (1000 + 8)

= 258 × 1000 + 258 × 8

= 258000 + 2064

= 260064

(b) Given 1005 × 168

= (1000 + 5) × 168

= 1000 × 168 + 5 × 168

= 168000 + 840

= 168840