# Class 6 Maths Chapter 2 Whole Numbers

Exercise 2.1

Ex 2.1 Class 6 Maths Question 1.
Write the next three natural numbers after 10999.
Solution:
The next three natural numbers after 10999 are 11000, 11001 and 11002.

Ex 2.1 Class 6 Maths Question 2.
Write three whole numbers occurring just before 10001.
Solution:
10001 – 1 = 10000
10000 – 1 = 9999
9999 – 1 = 9998
Hence, three whole numbers just before 10001 are 10000, 9999 and 9998.

Ex 2.1 Class 6 Maths Question 3.
Which is the smallest whole number?
Solution:
0 is the smallest whole number.

Ex 2.1 Class 6 Maths Question 4.
How many whole numbers are there between 32 and 53?
Solution:
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, 52.

Ex 2.1 Class 6 Maths Question 5.
Write the successor of:
(a) 2440701
(b) 100199
(c) 1099999
(d) 2345670
Solution:
(a) Successor of 244070 is 244070 + 1 = 244071
Hence, successor of 244070 is 244071.
(b) Successor of 100199 is 100199 + 1 = 100200
Hence, successor of 100199 is 100200.
(c) Successor of 1099999 is 1099999 + 1 = 1100000
Hence, successor of 1099999 is 1100000.
(d) Successor of 2345670 is 2345670 + 1 = 2345671
Hence, successor of 2345670 is 2345671

Ex 2.1 Class 6 Maths Question 6.
Write the predecessor of:
(a) 94
(b) 10000
(c) 208090
(d) 7654321
Solution:
(a) Predecessor of 94 is 94 – 1 = 93
Hence, predecessor of 94 is 93.
(b) Predecessor of 1000 is 10000 – 1 = 9999
Hence, predecessor of 10000 is 9999.
(c) Predecessor of 208090 is 208090 -1 = 208089
Hence, predecessor of 208090 is 208089.
(d) Predecessor of 7654321 is 7654321 – 1 = 7654320
Hence, predecessor of 7654321 is 7654320.

Ex 2.1 Class 6 Maths Question 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:
We know that the smaller number is always on the left side of the greater number on number line.
(a) 530, 503
Clearly 503 is smaller than 530.
Hence, 503 will be on left side of 530 on number line.
Expression: 503 < 530 or 530 > 503

(b) 307 < 370
Clearly 307 is smaller than 370.
Hence, 307 will be on the left side of 370 on number line.
Expression: 307 < 370 or 370 > 307.

(c) 98765, 56789
Clearly 56789 is smaller than 98765.
Hence, 56789 will be on left side of 98765 on number line.
Expression: 56789 < 98765 or 98765 > 56789.

(d) 9830415, 10023001
Clearly, 9830415 is smaller than 10023001
Hence, 9830415 will be on the left side of 10023001 on the number line.
Expression: 9830415 < 10023001 or 10023001 > 9830415.

Ex 2.1 Class 6 Maths Question 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) This statement is false (F)
(b) This statement is false (F)
(c) This statement is true (T)
(d) This statement is true (T)
(e) This statement is true (T)
(f) This statement is false (F)
(g) This statement is false (F)
(h) This statement is false (F)
(i) This statement is true (T)
(J) This statement is false (F)
(k) This statement is false (F)
(l) This statement is true (T)
(m) This statement is false (F).

Exercise 2.2

Ex 2.2 Class 6 Maths Question 1.
Find the sum by suitable arrangement:
(a) 837 + 208 + 363
(b) 1962 + 453,+ 1538 + 647
Solution:
(a) 837 + 208 + 363 = (837 + 363) + 208
= 1200 + 208 [Using associative property]
= 1408

(b) 1962 + 453 + 1538 + 647
= (1962 + 1538) + (453 + 647)
= 3500 + 1100 = 4600

Ex 2.2 Class 6 Maths Question 2.
Find the product by suitable arrangement:
(а) 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
Solution:
(a) 2 x 1768 x 50 = (2 x 50) x 1768 = 176800
(b) 4 x 166 x 25 = 166 x (25 x 4) = 166 x 100 = 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 = (300 – 15)x 300 = 300 x 300 – 15 x 300 = 90000 – 4500 = 85500
(f) 125 x 40 x 8 x 25 = (125 x 8) x (40 x 25) = 1000 X 1000 = 1000000

Ex 2.2 Class 6 Maths Question 3.
Find the value of the following:
(а) 297 x 17 + 297 x 3
(б) 54279 x 92 + 8 x 54279
(c) 81265 x 169 – 81265 x 69
(d) 3845 x 5 x 782 + 769 x 25 x 218
Solution:
(a) 297 x 17 x 297 x 3 = 297 x (17 + 3)
= 297 x 20 = 297 x 2 x 10
= 594 x 10 = 5940

(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
= 3845 x 5 x 782 + (769 x 5) x 5 x 218
= 3845 x 5 x 782 + 3845 x 5 x 218
= 3845 x 5 x 782 + 3845 x 5 x 218
= 3845 x 5 x (782 + 218)
= 3845 x 5 x 1000
= 19225 x 1000
= 19225000

Ex 2.2 Class 6 Maths Question 4.
Find the product using suitable properties.
(a) 738 x 103
(b) 854 x 102
(c) 258 x 1008
(d) 1005 x 168
Solution:
(a) 738 x 103 = 738 x (100 + 3)
= 738 x 100 + 738 x 3 [Using distributive property]
= 73800 + 2214 = 76014

(b) 854 x 102 = 854 x (100 + 2)
= 854 x 100 + 854 x 2 [Using distributive property]
= 85400 + 1708 = 87108

(c) 258 x 1008 = 258 x (1000 + 8)
= 258 x 1000 + 258 x 8 [Using distributive property]
= 258000 + 2064 = 260064

(d) 1005 x 168 = (1000 + 5) x 168
= 1000 x 168 + 5 x 168 [Using distributive property]
= 168000 + 840 = 168840

Ex 2.2 Class 6 Maths Question 5.
A taxidriver filled his car petrol tank with 40 litres of petrol on Monday. The next day, he filled the tank with 50 litre of petrol. If the petrol cost ₹44 per litre, how much did he spend in all on petrol?
Solution:
Petrol filled on Monday = 40 litres
Cost of petrol = ₹44 per litre
Petrol filled on Tuesday = 50 litre
Cost of petrol = ₹44 pet litre
∴ Total money spent in all
= ₹(40 x 44 + 50 x 44)
= ₹(40 + 50) x 44 = ₹90 x 44 = ₹3960

Ex 2.2 Class 6 Maths Question 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 ₹15 per litre, how much money is due to the vendor per day?
Solution:
Milk supplied in the morning = 32 litres
Cost of milk = ₹15 per litre
Milk supplied in the evening = 68 litres
Cost of milk = ₹15 per litre

Ex 2.2 Class 6 Maths Question 7.
Match the following:
(i) 425 x 136 = 425 x (6 + 30 + 100)
(ii) 2 x 49 x 50 = 2 x 50 x 49
(iii) 80 + 2005 + 20 = 80 + 20 + 2005
Hence (i) ↔ (c), (ii) ↔ (a) and (iii) ↔ (b)
∴ Money paid to the vendor
= ₹ (32 x 15 + 68 x 15)
= ₹(32 + 68) x 15
= ₹100 x 15
= ₹1500
(a) Commutativity under multiplication
(c) Distributivity of multiplication over addition

Exercise 2.3

Ex 2.3 Class 6 Maths Question 1.
Which of the following will not represent zero:
(a) 1 + 0
(b) 0 x 0
(c) 02
(d) 10−102
Solution:
(a) 1 + 0 = 1 ≠ 0, does not represent zero.
(b) 0 x 0 = 0, represents zero
(c) 02 = 0, represents zero.
(d) 10−102 = 02 = 0 represents zero.

Ex 2.3 Class 6 Maths Question 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:
Yes, Examples:
5 x 0 = 0
0 x 8 = 0
0 x 0 = 0

Ex 2.3 Class 6 Maths Question 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:
This is only true, when each of the number are 1.
1 x 1 = 1

Ex 2.3 Class 6 Maths Question 4.
Find using distributive property:
(а) 728 x 101
(b) 5437 x 1001
(c) 824 x 25
(d) 4275 x 125
(e) 504 x 35
Solution:
(a) 728 x 101 = 728 x (100 + 1)
= 728 x 100 + 728 x 1
= 72800 + 728
= 73528

(b) 5437 x 1001 = 5437 x (1000 + 1)
= 5437 x 1000 + 5437 x 1
= 5437000 + 5437
= 5442437

(c) 824 x 25 = 824 x (20 + 5)
= 824 x 20 + 824 x 5
= 16480 + 4120
= 20600

(d) 4275 x 125 = 4275 x (100 + 20 + 5)
= 4275 x 100 + 4275 x 20 + 4275 x 5
= 427500 + 85500 + 21375
= 534375

(e) 504 x 35 = (500 + 4) x 35
= 500 x 35 + 4 x 35
= 17500 + 140
= 17640

Ex 2.3 Class 6 Maths Question 5.
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?
Solution:
Step I: 123456 x 8 + 6 = 987654
Step II: 1234567 x 8 + 7 = 9876543

Working pattern:
(1) x 8 + 1 = 9
(12) x 8 + 2 = (11 + 1) x 8 + 2 = 98
(123) x 8 + 3 = (111 + 11 + 1) x 8 + 3 = 987
(1234) x 8 + 4 = (1111 + 111 + 11 + 1) x 8 + 4 = 9876
(12345) x 8 + 5 = (11111 + 1111 + 111 + 11 + 1) x 8 + 5 = 98765

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