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 maths NCERT Solutions
EXERCISE 2.1
Std 6 maths textbook chapter 2 Whole Numbers problem 1 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
Std 6 maths textbook chapter 2 Whole Numbers problem 2 exercise 2.1
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
Std 6 maths textbook chapter 2 Whole Numbers problem 3 exercise 2.1
3. Which is the smallest whole number?
Solution:
Zero (0) is the smallest whole number.
Problem 4
Std 6 maths textbook chapter 2 Whole Numbers problem 4 exercise 2.1
4. How many whole numbers are there between 32 and 53?
Solution:
53 – 32 = 21 -1 =20 (1 is zero)
Problem 5
Chapter 2 Whole Numbers problem 5 exercise 2.1 class 6 maths NCERT solutions
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
Chapter 2 Whole Numbers problem 6 exercise 2.1 class 6 maths NCERT solutions
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
Chapter 2 Whole Numbers problem 7 exercise 2.1 class 6 maths NCERT solutions
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
Chapter 2 Whole Numbers problem 8 exercise 2.1 class 6 maths NCERT solutions
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.
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