## NCERT SOLUTIONS FOR CLASS 6TH CLASS MATHS CHAPTER 2, WHOLE NUMBERS SOLUTIONS

Chapter 2 class 6 Maths NCERT Solutions for Whole Numbers are given.

You should Study the textbook lesson Whole numbers very well.

You should also observe the example problems and solutions given in the textbook.

**You can also see the solutions of some chapters for 6th class mathematics.**

NCERT solutions for class 6th chapter 1

NCERT solutions for class 6th chapter 3

Some chapters solutions links are given below.

Whole Numbers

Exercise 2.2 (Eliminated for 2023 – 2024)

Exercise 2.3 (Eliminated for 2023 – 2024)

### WHOLE NUMBERS class 6 NCERT maths

**EXERCISE 2.1**

**Problem 1**

1. Write the next three natural numbers after 10999.

**Solution:**

10,999

10,999 + 1=11,000

11,000 +1 = 11,001

11,001 + 1 = 11,002

Therefore, the next three natural numbers after 10,999 are 11,000, 11,001, 11,002.

**Problem 2**

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

**Solution:**

10.001

10,001 – 1 = 10,000

10,000 – 1 =9,999

9,999 – 1 = 9,998

Therefore, the three whole numbers occurring just before are 10,000, 9,999, 9,998.

**Problem 3**

3. Which is the smallest whole number?

**Solution:**

Zero (0) is the smallest whole number.

**Problem 4**

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

**Solution:**

53 – 32 = 21 -1 =20 (1 is zero)

**Problem 5**

5.Write the successor of:

a. 2440701, b. 100199, c, 1099999, d.2345670

**Solution:**

The successor of 24,40,701 is 24,40,702

The successor of 1,00,199 is 1,00,200

The successor of 10,99,999 is 1,10,000

The successor of 23,45,670 is 23,45,671

**Problem 6**

6. Write the predecessor of:

a. 94, b. 10000, c. 208090, d. 7654321

**Solutions:**

a. The predecessor of 94 is 94 – 1 = 93

b. The predecessor of 10,000 is 10,000 – 1 = 9,999

c. The predecessor of 2,08,090 is 2,08,090 – 1 = 2,08,089

The predecessor of 76,54,321 is 76,54,321 – 1 = 7,654,320

**Problem 7**

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. 503 is on the left of 530, 530 > 503.

b. 307 is on the left of 370, 370 > 307.

c. 56,789 is on the left of 98,765, 98,765 > 56,789.

d. 9,830,415 is on the left of 1,00,23,001, 9830415 < 1,00,23,001.

**Problem 8**

8. Which of the following statements are true (T) and which are false (F)?

a. Zero is the smallest natural number.

**Solution:** False

b. 400 is predecessor of 399.

**Solution:** False

c. Zero is the smallest whole number.

**Solution:** True

d. 600 is the successor of 599.

**Solution:** True

e. All natural numbers are whole numbers.

**Solution:** True

f. All whole numbers are natural numbers.

**Solution: False**

g. The predecessor of a two-digit number is never a single digit number.

**Solution:** False

h. I is the smallest whole number.

**Solution: False**

i. The natural number 1 has no predecessor.

**Solution:** True

j. The whole number 1 has no predecessor.

**Solution:** False

k. The whole number 13 lies between 11 and 12.

**Solution:** False

l. The whole number ‘ o ‘ has no predecessor.

**Solution:** True

m. The successor of a two-digit number is always a two-digit number.

**Solution:** False

#### NCERT SOLUTIONS FOR MATHS 6TH CLASS CHAPTER 2, WHOLE NUMBERS

**EXERCISE 2.2 (Eliminated for 2023 – 2024)**

**Problem 1**

*1. Find the sum by suitable rearrangement:*

*a. 837 + 208 + 363*

*b. 1962 + 453 + 1538 + 647*

**Solutions :**

837 + 208 + 363

= 837 + 363 + 208

= (837 + 363) + 208

= 1200 + 208 = 1,400

b. 1962 + 453 + 1538 + 647

= 1962 + 1538 + 453+ 647

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

= 3,500 + 1,100 = 4,600

**Problem 2**

*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. 2 × 1768 × 50

= 2 × 50 × 1768

= ( 2 × 50 ) ×1768

= 100 × 1768 = 1,76,800

b. 4 × 166 × 25

= 4 × 25 ×166

= ( 4 ×25 ) ×166

= 100 × 166 = 16,600

c. 8 × 291 × 125

8 × 125 × 291

= ( 8 × 125 ) × 291

= 1,000 × 291 = 2,91,000

d. 625 × 279 × 16

= 625 × 16 × 279

= ( 625 × 16 ) × 279

= 10,000 × 279 = 27,90,000

e. 285 × 5 × 60

= 285 × ( 5 × 60 )

= 285 × 300 = 85,500

f. 125 × 40 × 8 ×25

= 125 × 8 × 40 × 25

= ( 125 × 8 ) × ( 40 × 25 )

= 1,000 × 1,000 = 10,00,000

**Problem 3**

*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. 297 × 17 + 297 × 3

= 297 × ( 17 + 3 )

= 297 × 20 = 5,940

b. 54279 × 92 + 8 × 54279

= 54279 × ( 92 + 8 )

= 54,279 × 100 = 54,27,900

c. 81265 ×169 – 81265 × 69

= 81265 × ( 169 – 69 )

= 81265 × 100 = 81,26,500

d. 3845 × 5 × 782 + 769 × 25 × 218

= 3845 × 5 × 782 + 769 × 5 × 5 × 218

= 3845 × 5 × 782 + 3845 × 5 × 218

= ( 3845 × 5 ) × 782 + ( 3845 × 5 ) ×218

= 19225 × 782 + 19225 × 218

= 19225 × ( 782 + 218 )

= 19225 × 1000 = 1,92,25,000

**Problem 4**

*4. Find the product using suitable properties.*

*a. 738 × 103, b. 854 × 102*

*c. 258 × 1008, d. 1005 × 168*

**Solutions** :

*a. 738 × 103*

**= ***738 × ( 100 + 3 ) *

*= 738 × 100 + 738 × 3*

*= 73,800 + 2,214 = 76,014*

*b. 854 × 102*

*= 854 × ( 100 + 2 )*

*= 854 × 100 + 854 × 2*

*= 85,400 + 1,708 = 87,108*

*c. 258 × 1008*

*= 258 × (1,000 + 8 )*

*= 258 × 1,000 + 258 × 8*

*= 2,58,000 + 2064 = 2,60,064*

*d. 1,005 × 168*

*= ( 1,000 + 5 ) × 168*

*= 1,000 × 168 + 5 × 168*

*= 1,68,000 + 840 = 1,68,840*

**Problem 5**

*5. A taxi driver filled his car petrol tank with 40 liters of petrol on Monday. The next day , he filled the tank with 50 liters of the petrol costs Rs. 44 per liters, how many did he spend in all on petrol?*

**Solution** :

The petrol filled on Monday = 40 liters

The petrol filled on the next day = 50 liters

The total petrol filled on both the day = 40 + 50 = 90 liters

The cost petrol for 1 liter = Rs. 44

The cost of petrol for 90 liters = 44 × Rs.90 = Rs.3960

Therefore,the taxi driver spent Rs.3960 on petrol.

**Problem 6**

*6. A vendor supplies 32 liters of milk to a hotel in the morning and 68 liters of milk in the evening. If the milk costs Rs 15 per liter,how much money is due to the vendor per day ?*

**Solution :**

The supply of milk in morning to a hotel = 32 liters

The supply of milk in evening to the hotel = 68 liters

The total supply of milk per day = 32 + 68 = 100 liters

The cost of milk for 1 litre = Rs.15

The cost of milk for 100 litres = 15 × 100 = Rs.1500

Therefore, Rs.1500 is due to the vendor per day.

**Problem 7**

*7. Match the following :*

*i. 425 × 136 = 425 × ( 6+ 30 + 100 )*

*ii. 2 × 49 × 50 = 2 × 50 × 49*

*iii. 80 + 2005 + 20 = 80 + 20 × 2005*

*a. Commutativity under multiplication.*

*b. Commutativity under addition.*

*c. Distributive of multiplication over addition.*

**Solutions :**

*i – c*

*ii***– ***a*

*iii – b*

##### SOLUTIONS OF WHOLE NUMBERS VI CLASS MATHS CBSE

##### NCERT SOLUTIONS FOR CLASS 6TH MATHS CHAPTER 2

**EXERCISE 2.3 (Eliminated for 2023 – 2024)**

**Problem 1**

*1. Which of the following will not represent zero:*

*a. 1 + 0 *

*b. 0 × 0*

*c. 0/2 *

*d. 10 × 0/2*

**Solution :**

a.( 1 + 0 = 1 )

**Problem 2**

*2. If the product of two whole numbers is zero, can we say that one or both of them will be zero ? Justify through example.*

**Solution :**

Yes.

If we multiply any number with zero, the product will be zero.

Example : 1 × 0 = 0, 3 × 0 = 0

If both numbers are zero, the product also be zero. Example : 0 × 0 = 0

**Problem 3**

*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.*

**Solution :**

If we multiply any number with 1 the product can not be 1.

Example : 3 × 1 = 3, 5 × 1 = 5

If both numbers are 1, then the product is 1. Example : 1 × 1 = 1

**Problem 4**

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

*a. 728 × 101 *

*b. 5437 × 1001*

*c. 824 × 25 *

*d. 4275 × 125*

**Solutions :**

a. 728 × 101

= 728 × ( 100 + 1 )

= 728 × 100 + 728 × 1

= 72,800 + 728 = 73,528

b. 5437 × 1001

= 5437 × ( 1000 + 1 )

= 5437 × 1000 + 5437 × 1

= 54,37,000 + 5437 = 54,42,437

c. 824 × 25

= 824 × 25 × 4/4

= 824 × 100 /4

= 82400/4 = 20,600

d. 4275 × 125

= 4275 × 125 × 8/8

= 4275 × 1000/8

= 42,75,000/8 = 5,34,375

e. 504 × 35

= 504 × 35 × 2/2

= 504 × 70 / 2

= 35280 /2 = 17,640

**Problem 5**

*5. Study the pattern :*

*1 × 8 + 1 = 9*

*12 × 8 +2 = 98*

*123 × 8 +3 = 987*

*1234 x 8 + 4 = 9876*

*12345 × 8 +5 = 98765*

*Write the two new steps. Can you say the pattern works ?*

*( Hint : 12345 = 11111 + 1111 + 111 + 11 + 1 )*.

**Solution :**

1 × 8 + 1 = 9

12 × 8 +2 = 98

123 × 8 +3 = 987

1234 x 8 + 4 = 9876

12345 × 8 +5 = 98765

123456 × 8 +6 = 987654

1234567 × 8 + 7 = 9876543

###### CBSE SOLUTIONS FOR MATHS CLASS 6TH CHAPTER 2, WHOLE NUMBERS

**Note:**** **Observe the solutions of whole numbers and try them in your own methods.

**Chapter 4 Basic Geometrical Ideas**

**Integers**

Ncert solutions for class 6th chapter 6

**Fractions**

Exercise 7.1

Exercise 7.2

Exercise 7.3

**Chapter 14 Practical Geometry class 6**

Ncert solutions for class 7 th maths some chapters.

Maths solutions class 10 real numbers

You can also see

Ncert solutions for class 7 th maths chapter 4 simple equations.

Ncert solutions for class 8 maths chapter 1 rational numbers.

Intermediate maths trigonometry solutions

Ncert solutions for maths class 6 some chapters

Ncert solutions for maths class 7 some chapters