Whole Numbers Class 6 Ex. 2.2
Ncert Class 6 Math Free Solution.
Exercise 2.2
Question 1 :- Find the sum by suitable rearrangement:
(a) 837 + 208 + 363
(b) 1962 + 453 + 1538 + 647
Solution 1:-
(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.
Question 2 :- Find the product by suitable rearrangement:
(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 2:-
(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 = 16,600
(c) 8 x 291 x 125
(8 x 125) x 291 = 1000 x 291 = 2,91,000
(d) 625 x 279 x 16
(625 x 16) x 279 = 10,000 x 279 = 27,90,000
(e) 285 x 5 x 60
285 X (5 x 60) = 285 x 300
= (300 – 15)x 300 = 300 x 300 – 15 x 300
= 90,000 – 4,500 = 85,500
( f ) 125 X 40 X 8 X 25
(125 x 8) X (40 x 25) = 1,000 X 1,000 = 10,00,000
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 3:-
(a) 297 x 17 + 297 x 3 = 297 x (17 + 3)
= 297 X 20 = 297 X 2 X 10
= 594 X 10 = 5,940.
(b) 54279 x 92 + 8 x 54279 = 54279 x (92 + 8)
= 54279 x 100 = 54,27,900.
(c) 81,265 x 169 – 81,265 x 69
= 81265 x (169 – 69)
= 81265 x 100 = 81,26,500.
(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 + 218)
= 3845 x 5 x 1000
= 19225 x 1000
= 1,92,25,000.
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 4:-
(a) 738 x 103 = 738 x (100 + 3)
= 738 x 100 + 738 x 3 [Using distributive property]
= 73800 + 2214 = 76,014.
(b) 854 x 102 = 854 x (100 + 2)
= 854 x 100 + 854 x 2 [Using distributive property]
= 85400 + 1708 = 87,108.
(c) 258 x 1008 = 258 x (1000 + 8)
= 258 x 1000 + 258 x 8 [Using distributive property]
= 258000 + 2064 = 2,60,064.
(d) 1005 x 168 = (1000 + 5) x 168
= 1000 x 168 + 5 x 168 [Using distributive property]
= 168000 + 840 = 1,68,840.
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 litres, how much did he spend in all on petrol?
Solution 5:-
Petrol filled on Monday = 40 litres
Petrol filled on Tuesday = 50 litres
Cost of petrol = ₹44 pet litre
So, Total money spent in all = ₹ (40 x 44 + 50 x 44)
= ₹(40 + 50) x 44 = ₹90 x 44 = ₹ 3,960.
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 ₹45 per litre, how much money is due to the vendor per day?
Solution 6 :-
Supply of the milk in the morning = 32 litres
Supply of the milk in the evening = 68 litres
Cost of milk = ₹45 per litre
So, Total supplies of milk = (32 + 68) litre = 100 litre
Cost of 100 litres milk = 45 x 100 = ₹4500
So, the amount ₹4500 is due to the vendor per day.
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
Solution 7:-
(i) ↔ (c),
(ii) ↔ (a)
(iii) ↔ (b).