{"id":16631,"date":"2019-12-30T17:56:16","date_gmt":"2019-12-30T12:26:16","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=16631"},"modified":"2021-07-02T15:33:26","modified_gmt":"2021-07-02T10:03:26","slug":"kseeb-solutions-for-class-6-maths-chapter-2-ex-2-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-for-class-6-maths-chapter-2-ex-2-2\/","title":{"rendered":"KSEEB Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.2"},"content":{"rendered":"

Students can Download Chapter 2 Whole Numbers Ex 2.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths<\/a> helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka State Syllabus Class 6 Maths Chapter 2 Whole Numbers Ex 2.2<\/h2>\n

Question 1.
\nFind the sum by suitable rearrangement:
\na. 837 + 208 + 363 = (837 + 363) + 208
\n= 1200 + 208
\n= 1408<\/p>\n

b. 1962 + 453 + 1538 + 647 = (1962 + 1538) + (453 + 647)
\n= 3500 + 1100 = 4600.<\/p>\n

Question 2.
\nFind the product by suitable rearrangement:
\na. 2 \u00d7 1768 \u00d7 50 = (2 \u00d7 50) \u00d7 1768
\n= 100 \u00d7 1768 = 176800<\/p>\n

b. 4 \u00d7 166 \u00d7 25 = (4 \u00d7 25) \u00d7 166
\n= 100 \u00d7 166 = 16600<\/p>\n

c. 8 \u00d7 291 \u00d7 125 = (8 \u00d7 125) \u00d7 291
\n= 1000 \u00d7 291 = 291000<\/p>\n

\"KSEEB<\/p>\n

d. 625 \u00d7 279 \u00d7 16 = (625 \u00d7 16) \u00d7 279
\n= 10000 \u00d7 279 = 2790000<\/p>\n

e. 285 \u00d7 5 \u00d7 60 = 285 \u00d7 5 \u00d7 60 = 285 \u00d7 (5 \u00d7 60) = 285 \u00d7 300 = 85,500<\/p>\n

f. 125 \u00d7 40 \u00d7 8 \u00d7 25 = 125 \u00d7 8 \u00d7 40 \u00d7 25 = (125 \u00d7 40) \u00d7 (8 \u00d7 25) (5000) \u00d7 (200)
\n= 1000 \u00d7 1000 = 10,00,000.<\/p>\n

Question 3.
\nFind the value of the following
\na. 297 \u00d7 17 + 297 \u00d7 3
\nb. 54279 \u00d7 92 + 8 \u00d7 54279
\nc. 81265 \u00d7 169 – 81265 \u00d7 69
\nd. 3845 \u00d7 5 \u00d7 782 + 769 \u00d7 25 \u00d7 218
\nSolution:
\na. 297 \u00d7 17 + 297 \u00d7 3 = 297 \u00d7 (17 + 3)
\n= 297 \u00d7 20 = 5,940<\/p>\n

b. 54279 \u00d7 92 + 8 \u00d7 54279 = 54279 \u00d7 92 + 54279 \u00d7 8
\n= 54279 \u00d7 (92 + 8)
\n= 54279 \u00d7 100 = 5427900<\/p>\n

\"KSEEB<\/p>\n

c. 81265 \u00d7 169 – 81265 \u00d7 69 = 81265 \u00d7 (169 – 169)
\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 + (769 \u00d7 5) \u00d7 5 \u00d7 218 = 3845 \u00d7 5 \u00d7 (782 + 218)
\n= 19225 \u00d7 1000 = 1,92,25,000<\/p>\n

Question 4.
\nFind the product using suitable properties.
\na. 738 \u00d7 103
\nb. 854 \u00d7 102
\nc. 258 \u00d7 1008
\nd. 1005 \u00d7 168
\nSolution:
\na. 738 \u00d7 103
\n= 738 \u00d7 (100 + 3)
\n= 738 \u00d7 100 + 738 \u00d7 3 (Distributive property)
\n= 73800 + 2214 = 76014<\/p>\n

\"KSEEB<\/p>\n

b. 854 \u00d7 102
\n= 854 \u00d7 (100 + 2)
\n854 \u00d7 100 + 854 \u00d7 2 (Distributive property)
\n= 85400 + 1708 = 87108<\/p>\n

c. 258 \u00d7 1008
\n= 258 \u00d7 ( 1000 + 8)
\n= 258 \u00d7 1000 + 258 \u00d7 8 (Distributive property )
\n= 258000 + 2064 = 260064<\/p>\n

d. 1005 \u00d7 168 = (1000 + 5) \u00d7 168
\n= 1000 \u00d7 168 + 5 \u00d7 168 (Distributive property )
\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 filled the tank with 50 litres of petrol. If the petrol costs Rs 44 per litre, how much did he spend in all on petrol?
\nSolution:
\nQuantity of petrol filled on monday = 40l
\nQuantity of petrol filled on tuesday = 50 l
\nTotal quantity filled = (40+ 50) l
\nCost of petrol (per l) = Rs 44
\nTotal money spent = 44 \u00d7 (40 + 50)
\n= 44 \u00d7 90 = Rs.3960<\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nA vendor supplies 32 litres of milk to a hotel in the morning and 68 liters of milk in the evening. If the milk costs Rs 15 per litre, how much money is due to the vendor per day?
\nSolution:
\nQuantity of milk supplied in the morning = 32
\nQuantity of milk supplied is the evening = 681
\nTotal of milk per litre = (32+ 68)
\nCost of milk per litre = Rs 15
\nTotal cost per day = 15 \u00d7 (32 + 68)
\n= 15 \u00d7 100 = Rs 1500<\/p>\n

\"KSEEB<\/p>\n

Question 7.
\nMatch the following
\n1. 425 \u00d7 136 = 425 \u00d7 (6 + 30 + 100) ____ Commutativity under multiplication
\n2. 2 \u00d7 49 \u00d7 50 = 2 \u00d7 50 \u00d7 49 ____ Commutativity under addition
\n3. 80 + 2005 + 20 = 80 + 20 + 2005 ____ Distributivity of multiplication over addition.
\nSolution:
\n1. 425 \u00d7 136 = 425 \u00d7 (6 + 30 + 100) ___ Distributivity of multiplication over addition
\n2. 2 \u00d7 49 \u00d7 50 = 2 \u00d7 50 \u00d7 49 ___ Commutativity under multiplication
\n3. 80 + 2005 + 20 = 80 + 20 + 2005 ___ Commutativity under addition<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Chapter 2 Whole Numbers Ex 2.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka State Syllabus Class 6 Maths Chapter 2 Whole Numbers Ex 2.2 Question 1. Find the sum by …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[83],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/16631"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=16631"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/16631\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=16631"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=16631"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=16631"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}