## NCERT Solutions Class 6 Maths## Chapter - 2: Whole Numbers## Exercise 2.1
10999 +1 = 11000 11000 + 1 = 11001 11001 + 1 = 11002 Thus, the next three natural numbers are
numbers. 0, 1, 2, 3, .. and so on are the whole numbers. The three whole numbers occuring just before 10001 are: 10001 - 1 = 10000 10000 - 1 = 9999 9999 - 1 = 9998
The whole numbers between 32 and 53 are: 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, and 52. These are the total
For example, 2 is the successor of 1
2440701+ 1 = 2440702
100199 + 1 = 100200
1099999 + 1 = 1100000
2345670 + 1 =2345671
For example, 3 is the predecessor of 2
94 - 1 = 93
10000 - 1 = 9999
208090 - 1 = 208089
7654321 - 1 = 7654320
503 is less than 530 Hence,
307 is less than 370 Hence,
56789 is less than 98765 Hence, 56789 is on the left of 98765 on the number line.
9830415 is less than 10023001 Hence,
(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.
For example, Predecessor of 10 is 9. 10 - 1 = 9 (h) 1 is the smallest whole number.
Hence, 0 is the smallest whole number. (i) The natural number 1 has no predecessor.
(j) The whole number 1 has no predecessor.
1 - 1 = 0 0 is also a whole number. (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.
For example, Successor of 99 is 100. 100 is a three digit number. 99 + 1 = 100 ## Exercise 2.2## 1. Find the sum by suitable rearrangement:
= (837 + 363) + 208 = 1200 + 208 = 1408
= (1962 + 1538) + (453 + 647) = 3500 + 1100 = 4600 ## 2. Find the product by suitable rearrangement:
= 1768 × (2 × 50) = 1768 × 100 = 176800
= 166 × (4 × 25) = 166 × 100 = 16600
= 291 × (8 × 125) = 291 × (1000) = 291000
= 279 × (625 × 16) = 279 × 10000 = 2790000
= 285 × (5 × 60) = 285 × 300 = 85500
= (125 × 8) × (40 × 25) = 1000 × 1000 = 1000000 ## 3. Find the value of the following:
= 297 × (17 + 3) = 297 × 20 = 5940
= 54279 × (92 + 8) = 54279 × 100 = 5427900
= 81265 × (169 - 69) = 81265 × 100 = 8126500
(3845 × 5) × 782 + (769 × 25) × 218 = 19225 × 782 + 19225 × 218 = 19225 × (782 + 218) = 19225 × 1000 = 19225000 ## 4. Find the product using suitable properties.
= 738 × (100 + 3) = 738 × 100 + 738 × 3 = 73800 + 2214 = 76014
854 × (100 + 2) = 854 × 100 + 854 × 2 = 85400 + 1708 = 87108
= 258 × (1000 + 8) = 258 × 1000 + 258 × 8 = 258000 + 2064 = 260064
168 × (1000 + 5) = 168 × 1000 + 168 × 5 = 168000 + 840 = 168840
Petrol filled on the next day = 50 litres Total petrol filled on both the days = 40 + 50 = 90 litres Cost of petrol per litre = 44 Total cost of petrol for 90 litres = 44 × 90 = 3960 Thus, he spend rupees
Milk supplied in the evening = 68 litres Total milk supplied on the same day = 32 + 68 = 100 litres Cost of milk per litre = 45 Cost of 100 litres of milk = 45 × 100 = 4500 Thus, rupees
(i) 425 × 136 = 425 × (6 + 30 +100) - (c) Distributivity of multiplication (ii) 2 × 49 × 50 = 2 × 50 × 49 - (a) Commutativity under multiplication. (iii) 80 + 2005 + 20 = 80 + 20 + 2005 - (b) Commutativity under addition. over addition. ## Exercise 2.3
1 + 0 = 1
0 × 0 = 0
0/2 = 0
(10 -10) = 0 0/2 = 0 Thus, only option
For example, 0 × 1 = 0 2 × 0 = 0 5 × 0 = 0 0 × 6 = 0 Thus, if any one number is 0, the product will always be 0. 0 × 0 = 0 Thus, if both numbers are 0, the product will always be 0.
For example, 1 × 1 = 1 Thus, if both numbers are 1, the product will always be 1. 2 × 1 = 2 5 × 1 = 5 1 × 3 = 3 Thus, if any one number is 1, the product will not be 1.
= 728 × (100 + 1) = 728 × 100 + 728 × 1 = 72800 + 728 = 73528
= 5437 × (1000 + 1) = 5437 × 1000 + 5437 × 1 = 5437000 + 5437 = 5442437
= 824 × (20 + 5) = 824 × 20 + 824 × 5 = 16480 + 4120 = 20600
= 4275 × (100 + 20 + 5) = 4275 × 100 + 4275 × 20 + 4275 × 5 = 427500 + 85500 + 21375 = 534375
= 504 × (30 + 5) = 504 × 30 + 504 × 5 = 15120 + 2520 = 17640
1 × 8 + 1 = 9 12 × 8 + 2 = 98 123 × 8 + 3 = 987 1234 × 8 + 4 = 9876 12345 × 8 + 5 = 98765 Write the next two steps. Can you say how the pattern works?
123456 × 8 + 6 = 987654 1234567 × 8 + 7 = 9876543
1 × 8 + 1 = 9 (11 + 1) × 8 + (1 + 1) = 98 (111 + 11 + 1) × 8 + (1 + 1 + 1) = 987 (1111 + 111 + 11 + 1) × 8 + (1 + 1 + 1 + 1) = 9876 (11111 + 1111 + 111 + 11 + 1) × 8 + (1 + 1 + 1 + 1 + 1) = 98765 (111111 + 11111 + 1111 + 111 + 11 + 1) × 8 + (1 + 1 + 1 + 1 + 1 + 1) = 987654 (1111111 + 111111 + 11111 + 1111 + 111 + 11 + 1) × 8 + (1 + 1 + 1 + 1 + 1 + 1 + 1) = 9876543 Next TopicClass 6 Maths Chapter 3 |