Whole Numbers
Exercise 2.2
Ex 2.1 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 [by using associative property]
= 1408
(b) 1962 + 453 + 1538 + 647
= (1962 + 1538) + (453 + 647)
= 3500 + 1100 = 4600
Ex 2.2 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 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 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 [By using distributive property]
= 73800 + 2214 = 76014
(b) 854 x 102 = 854 x (100 + 2)
= 854 x 100 + 854 x 2 [By using distributive property]
= 85400 + 1708 = 87108
(c) 258 x 1008 = 258 x (1000 + 8)
= 258 x 1000 + 258 x 8 [By using distributive property]
= 258000 + 2064 = 260064
(d) 1005 x 168 = (1000 + 5) x 168
= 1000 x 168 + 5 x 168 [By using distributive property]
= 168000 + 840 = 168840
Ex 2.2 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
Petrol filled on Tuesday = 50 litre
Total petrol filled = 40+50 = 90 litres
Cost of 1 litre petrol = ₹44 per litre
Cost of 90 litres petrol = 44×90
= 44x (100-10)
=44×100 – 44 x10
=4400 – 440
= ₹3960
(Another Method-)
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 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 morning = 32 litres
Milk supplied in evening = 68 litres
Total supply = 32+68 = 100 litres
Cost of 1-litre milk = ₹15 per litre
Cost of 100 litres milk = 15 x 100 = ₹1500
(Another Method-)
Supply of milk in morning = 32 litres
Cost of milk = ₹15 per litre
Supply of milk in evening = 68 litres
Cost of milk = ₹15 per litre
∴ Money paid to the vendor
= ₹ (32 x 15 + 68 x 15)
= ₹(32 + 68) x 15
= ₹100 x 15
= ₹1500
Ex 2.2 Question 7.
Match the following:
(i) 425 x 136 = 425 x (6 + 30 + 100) (a) Commutativity under multiplication
(ii) 2 x 49 x 50 = 2 x 50 x 49 (b) Commutativity under addition
(iii) 80 + 2005 + 20 = 80 + 20 + 2005 (c) Distributivity of multiplication over addition
(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
