NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Exercise 2.2<\/h2>\n
Question 1.
\nFind the sum by suitable rearrangement:
\n(a) 837 + 208 + 363
\n(b) 1962 + 453 + 1538 + 647
\nAnswer:
\n(a) 837 + 208 + 363
\n= (837 +363)+208
\n= 1200 + 208 = 1408<\/p>\n
(b) 1962 + 453 + 1538 + 647
\n= (1962+ 1538)+ (453+ 647)
\n= 3500 + 1100 = 4600<\/p>\n
![\"NCERT](\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?resize=170%2C17&ssl=1\")
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Question 2.
\nFind the product by suitable rearrangement:
\n(a) 2 \u00d7 1768 \u00d7 50
\n(b) 4 \u00d7 166 \u00d7 25
\n(c) 8 \u00d7 291 \u00d7 125
\n(d) 625 \u00d7 279 \u00d7 16
\n(e) 285 \u00d7 5 \u00d7 60
\n(f) 125 \u00d7 40 \u00d7 8 \u00d7 25
\nAnswer:
\n(a) 2 \u00d7 1768 \u00d7 50 = (2 \u00d7 50) \u00d7 1768
\n= 100 \u00d7 1768 = 176800
\n(b) 4 \u00d7 166 \u00d7 25 = (4 \u00d7 25) \u00d7 166
\n= 100 \u00d7 166 = 16600
\n(c) 8 \u00d7 291 \u00d7 125 = (8 \u00d7 125) \u00d7 291
\n= 1000 \u00d7 291 = 291000
\n(d) 625 \u00d7 279 \u00d7 16 = (625 \u00d7 16) \u00d7 279
\n= 10000 \u00d7 279 = 2790000
\n(e) 285 \u00d7 5 \u00d7 60 = 285 \u00d7 (5 \u00d7 60)
\n= 285 \u00d7 300 = 85500
\n(f) 125 \u00d7 40 \u00d7 8 \u00d7 25
\n= (125 \u00d7 8) \u00d7 (40 \u00d7 25)
\n= 1000 \u00d7 1000 = 1000000<\/p>\n
Question 3.
\nFind the value of the following:
\n(a) 297 \u00d7 17 + 297 \u00d7 3
\n(b) 54279 \u00d7 92 + 8 \u00d7 54279
\n(c) 81265 \u00d7 169 – 81265 \u00d7 69
\n(d) 3845 \u00d7 5 \u00d7 782 + 769 \u00d7 25 \u00d7 218
\nAnswer:
\n(a) 297 \u00d7 17 + 297 \u00d7 3
\n= 297 \u00d7 (17 + 3)
\n= 297 \u00d7 20 = 5940<\/p>\n
(b) 54279 \u00d7 92 + 8 \u00d7 54279
\n= 54279 \u00d7 (92 + 8)
\n= 54279 \u00d7 100 = 5427900<\/p>\n
(c) 81265 \u00d7 169 – 81265 \u00d7 69
\n= 81265 \u00d7 (169 – 69)
\n= 81265 \u00d7 100 = 8126500<\/p>\n
(d) 3845 \u00d7 5 \u00d7 782 + 769 \u00d7 25 \u00d7 218
\n= 3845 \u00d7 5 \u00d7 782 + 769 \u00d7 5 \u00d7 5 \u00d7 218
\n= 3845 \u00d7 5 \u00d7 782 + 3845 \u00d7 5 \u00d7 218
\n= 3845 \u00d7 5 \u00d7 (782 + 218)
\n= 3845 \u00d7 5 \u00d7 1000 = 19225000<\/p>\n
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Question 4.
\nFind the product using suitable properties.
\n(a) 738 \u00d7 103
\n(b) 854 \u00d7 102
\n(c) 258 \u00d7 1008
\n(d) 1005 \u00d7 168
\nAnswer:
\n(a) 738 \u00d7 103 = 738 \u00d7 (100 + 3)
\n= 738 \u00d7 100 + 738 \u00d7 3
\n(Using distributive property)
\n= 73800 + 2214 = 76014<\/p>\n
(b) 854 \u00d7 102 = 854 \u00d7 (100 + 2)
\n= 854 \u00d7 100 + 854 \u00d7 2
\n(Using distributive property)
\n= 85400 + 1708 = 87108<\/p>\n
(c) 258 \u00d7 1008 = 258 \u00d7 (1000 + 8)
\n= 258 \u00d7 1000 + 258 \u00d7 8
\n= 258000 + 2064 = 260064<\/p>\n
(d) 1005 \u00d7 168 = (1000 +5) \u00d7 168
\n= 1000 \u00d7 168 + 5 \u00d7 168
\n= 168000 + 840 = 168840<\/p>\n
Question 5.
\nA taxi driver filled his car petrol tank with 40 litres of petrol on Monday. The next day, he tilled the tank with 50 litres of petrol. If the petrol costs \u20b9 44 per litre, how much did he spend in all on petrol?
\nAnswer:
\nPetrol filled on Monday = 40 litres
\nPetrol filled on next day (i.e. Tue.) = 50 litres
\nTotal petrol filled = 90 litres
\nNow, Cost of 1 litre petrol = \u20b9 44
\nCost of 90 litres petrol = 44 \u00d7 90 = 44 \u00d7 (100 – 10)
\n= 44 \u00d7 100 – 44 \u00d7 10 = 4400 – 440
\n= \u20b9 3960
\nTherefore, he spent ? 3960 on petrol.<\/p>\n
<\/p>\n
Question 6.
\nA vendor supplies 32 litres of milk to a hotel in the morning and 68 litres of milk in the evening. If the milk costs ? 41 per litre, how much money is due to the vendor per day?
\nAnswer:
\nSupply of milk in morning the = 32 litres
\nSupply of milk in the evening = 68 litres
\nThe cost of milk = \u20b9 45 per litre
\n= \u20b9 45 \u00d7 [32+68]= \u20b9 45 \u00d7 100 = \u20b9 4500
\nTherefore, \u20b9 4500 is due to the vendor per day.<\/p>\n
Question 7.
\nMatch the following:<\/p>\n