Kseeb Class 6 Mathematics Chapter 2 Whole Numbers Solutions
Welcome to NCTB Solutions. Here with this post we are going to help 6th class students by providing Solutions for KSEEB Class 6 Mathematics chapter 2 Whole Numbers. Here students can easily find all the solutions for Whole Numbers Exercise 2.1, 2.2 and 2.3. Also here our Expert Mathematic Teacher’s solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 2 solutions. Here all the solutions are based on Karnataka State Board latest syllabus.
Whole Numbers Exercise 2.1 Solutions :
(1) Write the next three natural numbers after 10999.
Solution :
The next there natural number after 10999 are
= 10999 +1
= 11000, 11000 + 1
= 11001 and 11001 + 1
= 11002
(2) Write the three whole numbers occurring just before 10001.
Solution :
The three whole numbers occurring just before 10001 are
10001 – 1 = 10000,
10000 – 1 = 9999,
9999 – 1 = 9998
(3) Which is the smallest whole number?
Solution :
The smallest whole number is 0
(4) How many whole numbers are there between 32 and 53?
Solution :
There are 20 whole numbers between 32 and 53.
Such that 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 50, 51, 52
(5) Write the successor of :
(a) 2440701
(b) 100199
(c) 1099999
(d) 2345670
Solution :
(a) The successor of 2440701 is
= 2440701 + 1
= 2440702
(b) The successor of 100199 is
= 100199 + 1
= 100200
(c) The successor of 1099999 is
= 1099999 + 1
= 1100000
(d) The successor of 2345670 is
= 2345670 + 1
= 2345671
(6) Write the predecessor of :
(a) 94
(b) 10000
(c) 208090
(d) 7654321
Solution :
(a) The predecessor of 94 is
= 94 – 1
= 93
(b) The predecessor of 10000 is
= 10000 – 1
= 9999
(c) The predecessor of 10000 is
= 10000 – 1
= 9999
(d) The predecessor of 7654321 is
= 7654321 – 1
= 7654320
(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
Solution :
(a) 503 is the left of 530.
= 503 < 530
(b) 307 is the left of 370.
= 307< 370
(c) 56789 is the left of 98765.
= 56789 < 98765
(d) 9830415 is the left of 10023001.
= 9830415 < 10023001
(8) Which of the following statements are true (T) and which are false (F) ?
(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.
Solution :
(a) False, since zero is the smallest whole number.
(b) False. since the predecessor of 399 is 399 – 1 = 398
(c) True
(d) True
(e) True
(f) False
(g) False, because the predecessor of a two digit number is a single digit number or two digit number.
(i) False, 0 is the smallest whole number.
(j) False, The whole number 1 has 0 predecessor
(k) False. The whole number 13 lies between 12 and 13.
(l) True
(m) False, since the successor of 99 is 100 i.e. three digit number.
Whole Numbers Exercise 2.2 Solutions :
(1) Find the sum by suitable rearrangement:
(a) 837 + 208 + 363
(b) 1962 + 453 + 1538 + 647
Solution :
(a) 837 + 208 + 363
(837 + 363) + 208
1200 + 208
=1408
837 + 208 + 363 = 1408
(b) 1962 + 453 + 1538 + 647
(1962 + 1538) + (453 + 647)
3500 + 1100
= 4600
1962 + 453 + 1538 + 647 = 4600
(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
Solution :
(a) 2 × 1768 × 50
= (2 × 50) ×1768
= 100 × 1768
= 176800
∴ 2 × 1768 × 50 = 176800.
(b) 4 × 166 × 25
= (4 × 25) × 166
= 100 × 166
= 16600
∴ 4 × 166 × 25 = 16600
(c) 8 × 291 × 125
= (8 × 125) × 291
= 1000 × 291
= 291000
∴ 8 × 291 × 125 =291000
(d) 625 × 279 × 16
= (625 × 16) × 279
= 10,000 × 279
= 2790000
∴ 625 × 279 × 16 =2790000.
(e) 285 × 5 × 60
= 285 × (5 × 60)
= 285 × 300
= 85500
∴ 285 × 5 × 60 = 85500
(f) 125 × 40 × 8 × 25
= (125 × 8)× ( 25× 40)
= 1000 × 1000
= 1,000,000
∴ 125 × 40 × 8 × 25 = 1,000,000
(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
Solution :
(a) 297 × 17 + 297 × 3
= 297 × 17 + 297 × 3
= 297 × (17 + 3)
= 297 × 20
= 5940
= 297 × 17 + 297 × 3= 5940
(b) 54279 × 92 + 8 × 54279
= 54279 × 92 + 8 × 54279
= 54279 × (92 + 8)
= 54279 × 100
= 5427900
= 54279 × 92 + 8 × 54279=5427900
(c) 81265 × 169 – 81265 × 69
= 81265 × 169 – 81265 × 69
= 81265 × (169 – 69)
= 81265 × 100
= 8126500
= 81265 × 169 – 81265 × 69=8126500
(d) 3845 × 5 × 782 + 769 × 25 × 218
= (3845 × 5 × 782)+ (769 × 25 × 218)
= 1, 50, 33,950 + 41, 91,050
= 1, 92, 25,000
= 3845 × 5 × 782 + 769 × 25 × 218 = 1, 92, 25,000
(4) Find the product using suitable properties.
(a) 738 × 103
(b) 854 × 102
(c) 258 × 1008
(d) 1005 × 168
Solution :
(a) 738 × 103
= 738 × 103
= (738 × 100) +(738 × 3)
= 73800 + 2214
= 76014
= 738 × 103= 76014
(b) 854 × 102
= 854 × 102
= (854 × 100) +(854 × 2)
= 85400 + 1708
= 87108
854 × 102=87108
(c) 258 × 1008
= 258 × 1008
= (258 × 1000) + (258 × 8)
= 258000 + 2064
= 2, 60,064
= 258 × 1008 = 2, 60,064
(d) 1005 × 168
= 1005 × 168
= (1000 × 168) + (5 × 168)
= 168000 + 840
= 168840
= 1005 × 168 = 1, 68,840
(5) A taxi driver 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?
Solution :
A taxi driver filled his car petrol tank with 40 litres of petrol on Monday.
The next day, he filled the tank with 50 litres of petrol.
Total petrol filled
= 40 + 50
= 90 Litre.
The petrol costs 44 per litre.
Amount spend to fill petrol
= 40 × 90
= 3600.
(6) 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 ₹ 45 per litre, how much money is due to the vendor per day?
Solution :
The cost of milk is RS 45 per litre.
Due money for the morning is for 32 litres of milk is = 45 × 32
Due money for the evening is for 68 litres of milk is = 45 × 68
Therefore total due money,
= 45 × 32 + 45 × 68
= 45 (32 + 68)
(Distributivity of multiplication over addition)
∴ Rs. 4500 is due to the vendor per day.
(7) Match the following:
(i) 425 × 136 = 425 × (6 + 30 + 100) (a) Commutativity under multiplication.
(ii) 2 × 49 × 50 = 2 × 50 × 49 (b) Commutativity under addition.
(iii) 80 + 2005 + 20 = 80 + 20 + 2005 (c) Distributivity of multiplication over addition.
Solution :
(i) 425 × 136 = 425 × (6 + 30 +100) – (a) Distributivity of multiplication over addition
(ii) 2 × 49 × 50 = 2 × 50 × 49 – (b) Commutativity under multiplication
(iii) 80 + 2005 + 20 = 80 + 20 + 2005 – (c) Commutativity under addition.
Whole Numbers Exercise 2.3 Solutions :
(1) Which of the following will not represent zero
(a) 1 + 0
(b) 0 × 0
(c) 0/2
(D) 10-10/2
Solution :
1 + 0 = 1,
0 × 0 = 0,
0/2 = 0,
10 – 10/2 = 0/2 = 0
∴ 1 + 0 does not/will not represent zero.
(2) If the product of two whole numbers is zero, can we say that one or both of them will be zero? Justify through examples.
Solution :
Example :
(i) 2 × 0 = 0
(ii) 0 × 0 = 0
(iii) 3 × 4 = 12
After observing the above examples in example
(i) 2 and 0 are whole numbers in example
(ii) 0 and 0 are whole number and their product is zero, but in example
(iii) 3 and 4 are whole numbers but their product is not equal to zero.
We have come to a connection that if the product of two whole number is zero we can say that one or both of them will be zero.
(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 :
Example :
(i) 0 × 1 = 0
(ii) 1 × 2 = 2
(iii) 1 × 1 = 1
In above given example all are product of whole number i.e. 0, 2 and 1 after observing all of them we get 1 from example (iii) is 1 X 1 = 1 From this we can conclude that if the product of two whole numbers is 1, we can say that both of them will be 1.
(4) Find using distributive property
(a) 728 × 101
(b) 5437 × 1001
(c) 824 × 25
(d) 4275 × 125
(e) 504 × 35
Solution :
(a) 728 × 101
728 × 101
(728 × 100) +(728 × 1)
= 72800 + 728
= 73528
728 × 101= 73528
(b) 5437 × 1001
(5437 × 1000) + (5437 × 1)
5437000 + 5437
= 54, 42,437
5437 × 1001 =54, 42,437
(c) 824 × 25
(824 × 20) + (824 × 5)
16480 + 4120
= 20,600
824 × 25= 20,600
(d) 4275 × 125
4275 × 125
(4275 × 100) + (4275 × 25)
427500 + 1, 06,875
= 5, 34,375
4275 × 125= 5, 34,375
(e) 504 × 35
(504 × 30) + (504 × 5)
15120 + 2520
= 17640
504 × 35 =17640
(5) Study the pattern :
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?
Solution :
1 × 8 + 1 = 9
12 × 8 + 2 = 98
123 × 8 + 3 = 9876
1234 × 8 + 4 = 9876
12345 × 8 + 5 = 98765
123456 × 8 + 6 = 987654
1234567 × 8 + 7 = 9876543
12345678 × 8 = 8 = 98765432
123456789 × 9 + 9 = 987654321
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