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

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

Study the textbook lesson very well.

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 soutions for class 6th chapter 3

Ncert solutions for class6th chapter 6

**Chapter 4 Basic Geometrical Ideas**

**Chapter 14 Practical Geometry class 6**

Exercise 14.2

Exercise 14.3

Exercise 14.4

Exercise 14.5

Exercise 14.6

Ncert solutions for class 7 th maths some chapters.

Maths solutions class 10 real numbers

### WHOLE NUMBERS

**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

Therefor, 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 and which are false ?*

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**

**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**

**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.

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