NCERT Class 6 Maths Solutions for Chapter 2 Whole Numbers
Exercise 2.1 (Page 31)
The next three natural numbers after 10999 are 11000, 11001, and 11002.
The three whole numbers occurring just before 10001 are 10000, 9999, and 9998.
The smallest whole number is 0.
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.
- (a) 2440701
- (b) 100199
- (c) 1099999
- (d) 2345670
(a) 2440701 + 1 = 2440702
(b) 100199 + 1 = 100200
(c) 1099999 + 1 = 1100000
(d) 2345670 + 1 = 2345671
- (a) 94
- (b) 10000
- (c) 208090
- (d) 7654321
(a) 94 – 1 = 93
(b) 10000 – 1 = 9999
(c) 208090 – 1 = 208089
(d) 7654321 – 1 = 7654320
- (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.
- (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)
- (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
- (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
- (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
- (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
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
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)
- (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.
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.
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.
- (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
- 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.
- (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
