Check below the NCERT Solutions for Class 6 Maths Chapter 2 - Whole Numbers:
Content:
1. Book PDF
2. Exercise 2.1
3. Exercise 2.2
4. Exercise 2.3
Book PDF:1. Book PDF
2. Exercise 2.1
3. Exercise 2.2
4. Exercise 2.3
NCERT Books for class 6 Maths Chapter 2: Whole Numbers in English Medium is given below to download in PDF form free. There are total 14 chapters in grade 6 Maths. All chapters are compulsory for the school test or terminal exams.The first chapter is Knowing Our Numbers and the last chapter is Practical Geometry. All the chapters are equally important for exams. Solutions of each chapter in English Medium or Hindi Medium is available on LEAF EDU91 website.
Exercise 2.1:
Q. Write the next three natural numbers after 10999.
Ans. The next three natural numbers after 10999 are 10999 +1,10999 + 2 and 10999 + 3
i.e.,11000, 11001 and 11002.
Ans. The next three natural numbers after 10999 are 10999 +1,10999 + 2 and 10999 + 3
i.e.,11000, 11001 and 11002.
Q. Write the three whole numbers occurring just before 10001.
Ans. The three whole numbers occurring just before 10001 are 10001 – 1, 10001 – 2 and 10001 – 3
i.e., 10000, 9999 and 9998.
Ans. The three whole numbers occurring just before 10001 are 10001 – 1, 10001 – 2 and 10001 – 3
i.e., 10000, 9999 and 9998.
Q. Which is the smallest whole number?
Ans. The smallest whole number is 0.
Ans. The smallest whole number is 0.
Q. How many whole numbers are there between 32 and 53?
Ans. There are (53 – 32) -1 = 21 – 1= 20 whole numbers between 32 and 53.
Ans. There are (53 – 32) -1 = 21 – 1= 20 whole numbers between 32 and 53.
Q. Write the successor of:
(a) 2440701
(b) 100199
(c) 1099999
(d) 2345670
Ans.
(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
(a) 2440701
(b) 100199
(c) 1099999
(d) 2345670
Ans.
(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
Q.Write the predecessor of:
(a) 94
(b) 10000
(c) 208090
(d) 7654321
Ans.
(a) Predecessor of 94 = 94 – 1 = 93
(b) Predecessor of 10000 = 10000 – 1 = 9999
(c) Predecessor of 208090 = 208090 – 1 = 208089
(d) Predecessor of 7654321 =7654321 – 1 = 7654320
(a) 94
(b) 10000
(c) 208090
(d) 7654321
Ans.
(a) Predecessor of 94 = 94 – 1 = 93
(b) Predecessor of 10000 = 10000 – 1 = 9999
(c) Predecessor of 208090 = 208090 – 1 = 208089
(d) Predecessor of 7654321 =7654321 – 1 = 7654320
Q.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
Ans.
(a) Here 503 lies on the left of 530 on the number line.
∴ 530 > 503
(b) Here 307 lies on the left of 370 on the number line.
∴ 370 > 307
(c) Here 56789.lies on the left of 98765 on the number line.
∴ 98765 > 56789
(d) Here 9830415 lies on the left of 10023001
∴ 9830415 > 10023001
(a) 530, 503
(b) 370, 307
(c) 98765, 56789
(d) 9830415, 10023001
Ans.
(a) Here 503 lies on the left of 530 on the number line.
∴ 530 > 503
(b) Here 307 lies on the left of 370 on the number line.
∴ 370 > 307
(c) Here 56789.lies on the left of 98765 on the number line.
∴ 98765 > 56789
(d) Here 9830415 lies on the left of 10023001
∴ 9830415 > 10023001
Which of the following statements are true (T) and which are false (F)?
Q. Zero is the smallest natural number.
Ans. False
Q. 400 is the predecessor of 399.
Ans. False
Q. Zero is the smallest whole number.
Ans. True
Q. 600 is the successor of 599.
Ans. True
Q. All natural numbers are whole numbers.
Ans. True
Q. All whole numbers are natural numbers.
Ans. False
Q. The predecessor of a two digit number is never a single digit number.
Ans. False
Q. 1 is the smallest whole number.
Ans. False
Q. The natural number 1 has no predecessor.
Ans. True
Q. The whole number 1 has no predecessor.
Ans. False
Q. The whole number 13 lies between 11 and 12.
Ans. False
Q. The whole number 0 has no predecessor.
Ans. True
Q. The successor of a two digit number is always a two digit number.
Ans. False
Q. Zero is the smallest natural number.
Ans. False
Q. 400 is the predecessor of 399.
Ans. False
Q. Zero is the smallest whole number.
Ans. True
Q. 600 is the successor of 599.
Ans. True
Q. All natural numbers are whole numbers.
Ans. True
Q. All whole numbers are natural numbers.
Ans. False
Q. The predecessor of a two digit number is never a single digit number.
Ans. False
Q. 1 is the smallest whole number.
Ans. False
Q. The natural number 1 has no predecessor.
Ans. True
Q. The whole number 1 has no predecessor.
Ans. False
Q. The whole number 13 lies between 11 and 12.
Ans. False
Q. The whole number 0 has no predecessor.
Ans. True
Q. The successor of a two digit number is always a two digit number.
Ans. False
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Exercise 2.2:
Q. Find the sum by suitable rearrangement:
(a) 837 + 208 + 363
(b) 1962 + 453 +1538 + 647
Ans.
(a) 837 + 208 + 363
(b) 1962 + 453 +1538 + 647
Ans.
Q.Find the product by suitable rearrangement:
(a) 2 x 1768 x 50
(b) 4 x 166 x 25
(c) 8 x 291 x 125
(d) 625 x 279 x 16
(e) 285 x 5 x 60
(f) 125 x 40 x 8 x 25
Ans.
(a) 2 x 1768 x 50 = (2 x 50) x 1768
= 100 x 1768 = 176800
(b) 4 x 166 x 25 = (4 x 25) x 166
= 100 x 166 = 16600
(c) 8 x 291 x 125 = (8 x 125) x 291
= 1000 x 291 = 291000
(d) 625 x 279 x 16 = (625 x 16) x 279
= 10000 x 279 = 2790000
(e) 285 x 5 x 60 = 285 x (5 x 60)
= 285 x 300 = 85500
(f) 125 x 40 x 8 x 25 = (125 x 8) x (40 x 25)
= 1000 x 1000 = 1000000
(a) 2 x 1768 x 50
(b) 4 x 166 x 25
(c) 8 x 291 x 125
(d) 625 x 279 x 16
(e) 285 x 5 x 60
(f) 125 x 40 x 8 x 25
Ans.
(a) 2 x 1768 x 50 = (2 x 50) x 1768
= 100 x 1768 = 176800
(b) 4 x 166 x 25 = (4 x 25) x 166
= 100 x 166 = 16600
(c) 8 x 291 x 125 = (8 x 125) x 291
= 1000 x 291 = 291000
(d) 625 x 279 x 16 = (625 x 16) x 279
= 10000 x 279 = 2790000
(e) 285 x 5 x 60 = 285 x (5 x 60)
= 285 x 300 = 85500
(f) 125 x 40 x 8 x 25 = (125 x 8) x (40 x 25)
= 1000 x 1000 = 1000000
Q.Find the value of the following:
(a) 297 x 17 + 297 x 3
(b) 54279 x 92 + 8 x 54279
(c) 81265 x 169 – 81265 x 69
(d) 3845 x 5 x 782 + 769 x 25 x 218
Ans.
(a) 297 x 17 + 297 x 3 = 297 x (17 + 3)
= 297 x 20
= 5949
(b) 54279 x 92 + 8 x 54279
= 54279 x (92 + 8)
= 54279 x 100 = 5427900
(c) 81265 x 169 – 81265 x 69
= 81265 x (169 -69)
= 81265 x 100
= 8126500
(d) 3845 x 5 x 782 + 769 x 25 x 218
= 3845 x 5 x 782 + (769 x 5) x 5 x 218
= 3845x5x782 + 3845x5x218
= (3845 x 5) x (782+ 218)
=19225 x 1000
=19225000
(a) 297 x 17 + 297 x 3
(b) 54279 x 92 + 8 x 54279
(c) 81265 x 169 – 81265 x 69
(d) 3845 x 5 x 782 + 769 x 25 x 218
Ans.
(a) 297 x 17 + 297 x 3 = 297 x (17 + 3)
= 297 x 20
= 5949
(b) 54279 x 92 + 8 x 54279
= 54279 x (92 + 8)
= 54279 x 100 = 5427900
(c) 81265 x 169 – 81265 x 69
= 81265 x (169 -69)
= 81265 x 100
= 8126500
(d) 3845 x 5 x 782 + 769 x 25 x 218
= 3845 x 5 x 782 + (769 x 5) x 5 x 218
= 3845x5x782 + 3845x5x218
= (3845 x 5) x (782+ 218)
=19225 x 1000
=19225000
Q. Find the product using suitable properties :
(a) 738 x 103
(b) 854 x 102
(c) 258 x 1008
(d) 1005 x 168
Ans.
(a) 738 x 103 = 738 x (100 + 3)
= 738 x 100 + 738 x 3 = 73800 + 2214 = 76014
(b) 854 x 102 = 854 x (100 + 2)
= 854 x 100 + 854 x 2 = 85400 + 1708 = 87108
(c) 258 x 1008 = 258 x (1000 + 8)
= 258 x 1000 + 258 x 8 = 258000 + 2064 = 260064
(d) 1005 x 168 = (1000 + 5) x 168
= 1000 x 168 + 5 x 168
= 168000 + 840 = 168840
(a) 738 x 103
(b) 854 x 102
(c) 258 x 1008
(d) 1005 x 168
Ans.
(a) 738 x 103 = 738 x (100 + 3)
= 738 x 100 + 738 x 3 = 73800 + 2214 = 76014
(b) 854 x 102 = 854 x (100 + 2)
= 854 x 100 + 854 x 2 = 85400 + 1708 = 87108
(c) 258 x 1008 = 258 x (1000 + 8)
= 258 x 1000 + 258 x 8 = 258000 + 2064 = 260064
(d) 1005 x 168 = (1000 + 5) x 168
= 1000 x 168 + 5 x 168
= 168000 + 840 = 168840
Q. A taxidriver 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?
Ans. Petrol filled on Monday and Tuesday
= 40 litres + 50 litres
= 90 litres
Cost of petrol @ ₹ 44 per litre
= 90 x ₹ 44 = ₹ 3960
Ans. Petrol filled on Monday and Tuesday
= 40 litres + 50 litres
= 90 litres
Cost of petrol @ ₹ 44 per litre
= 90 x ₹ 44 = ₹ 3960
Q. 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 ? 15 per litre, how much money is due to the vendor per day?
Ans. Milk supplied to a hotel in the morning and evening
= 32 litres + 68 litres = 100 litres
Money due to vendor per day.
= 100 x ₹ 15 = ₹ 1500
Ans. Milk supplied to a hotel in the morning and evening
= 32 litres + 68 litres = 100 litres
Money due to vendor per day.
= 100 x ₹ 15 = ₹ 1500
Q. Match the following:
Ans.
| (i) 425 x 136 = 425 x (6 + 30 +100) | (c) Distributivity of multiplication over addition |
| (ii) 2 x 49 x 50 = 2 x 50 x 49 | (a) Commutativity under multiplication |
| (iii) 80 + 2005 + 20 = 80 + 20 + 2005 | (b) Commutativity under addition |
Exercise 2.3:
Q. Which of the following will not represent zero :
(a) a + 0
(b) 0 x 0
(c) 0⁄2
(d) 10-10⁄2
Ans.
(a) 1 + 0 = 1 ≠ 0
(b) 0 x 0 = 0
(c) 0⁄2 = 0 x 1⁄2 = 0
(d) 10-10⁄2 = 0⁄2 = 0 x 1⁄2 = 0
Thus, only (a) does not represent zero.
(a) a + 0
(b) 0 x 0
(c) 0⁄2
(d) 10-10⁄2
Ans.
(a) 1 + 0 = 1 ≠ 0
(b) 0 x 0 = 0
(c) 0⁄2 = 0 x 1⁄2 = 0
(d) 10-10⁄2 = 0⁄2 = 0 x 1⁄2 = 0
Thus, only (a) does not represent zero.
Q. If the product of two whole numbers is zero, can we say that one or both of them will be zero? Justify through examples.
Ans. We know that the product of any whole number and zero is always zero i. e., a x 0 = 0, where a is any whole number.
Ans. We know that the product of any whole number and zero is always zero i. e., a x 0 = 0, where a is any whole number.
Q. If the product of two whole numbers is 1, can we say that one or both of them will be 1? Justify through examples.
A. We know that if a is any whole number, then
a x l = a -1 x a
∴ Clearly a must be = 1
Thus, both the numbers should be 1 i. e., 1 x 1 = 1
A. We know that if a is any whole number, then
a x l = a -1 x a
∴ Clearly a must be = 1
Thus, both the numbers should be 1 i. e., 1 x 1 = 1
Q. Find by using distributive property :
(a) 728 x 101
(b) 5437 x 1001
(c) 824 x 25
(d) 4275 x 125
(e) 504 x 35
Ans.
(a) 728 x 101
(b) 5437 x 1001
(c) 824 x 25
(d) 4275 x 125
(e) 504 x 35
Ans.
Q. Study the pattern :
1 x 8 +1 = 9
12 x 8 + 2 = 98
123 x 8 + 3 = 987
1234 x 8 + 4 = 9876
12345 x 8 + 5 = 98765
Write the next two steps. Can you say how the pattern works?
Ans. Write the next two steps :
123456 x 8 + 6 = 987654
1234567×8 + 7 = 9876543
How the pattern works?
1 x8 + 1= 9
(11 + 1) x 8 + 2 = 12 x 8 + 2 = 98
(111 + 11 + 1) x 8 + 3 = 123 x 8 + 3 = 987 .
(1111 + 111 + 11 + 1) x 8 + 4 =1234 x 8 + 4 =9876
(11111 + 1111 + 111 + 11 + 1) x 8 + 5
= 12345 x 8 + 5=98765
(111111 + 11111 + 1111 + 111 + 11 + 1) x 8+ 6
= 123456 x 8 + 6 = 987654
and (1111111 + 111111 + 11111 +1111 +111 +11 +1) x 8 + 7
= 1234567 x 8 + 7 = 9876543
1 x 8 +1 = 9
12 x 8 + 2 = 98
123 x 8 + 3 = 987
1234 x 8 + 4 = 9876
12345 x 8 + 5 = 98765
Write the next two steps. Can you say how the pattern works?
Ans. Write the next two steps :
123456 x 8 + 6 = 987654
1234567×8 + 7 = 9876543
How the pattern works?
1 x8 + 1= 9
(11 + 1) x 8 + 2 = 12 x 8 + 2 = 98
(111 + 11 + 1) x 8 + 3 = 123 x 8 + 3 = 987 .
(1111 + 111 + 11 + 1) x 8 + 4 =1234 x 8 + 4 =9876
(11111 + 1111 + 111 + 11 + 1) x 8 + 5
= 12345 x 8 + 5=98765
(111111 + 11111 + 1111 + 111 + 11 + 1) x 8+ 6
= 123456 x 8 + 6 = 987654
and (1111111 + 111111 + 11111 +1111 +111 +11 +1) x 8 + 7
= 1234567 x 8 + 7 = 9876543
Download all the NCERT Solutions for Class 6 Maths Chapter 2 from the following link:
NCERT Solutions for Class 6 Maths Chapter 2 - Whole Numbers
Also Read:
Class 6 Maths NCERT Book
Class 6 Maths NCERT Solutions
Class 6 Science NCERT Book
Class 6 Science NCERT Solutions
Class 6 Maths NCERT Book
Class 6 Maths NCERT Solutions
Class 6 Science NCERT Book
Class 6 Science NCERT Solutions
