{"id":4847,"date":"2020-09-09T07:28:35","date_gmt":"2020-09-09T01:58:35","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=4847"},"modified":"2021-07-02T15:03:36","modified_gmt":"2021-07-02T09:33:36","slug":"kseeb-solutions-class-8-maths-chapter-5-ex-5-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-class-8-maths-chapter-5-ex-5-2\/","title":{"rendered":"KSEEB Solutions for Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Ex 5.2"},"content":{"rendered":"<p>Students can Download Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Ex 5.2 Questions and Answers, Notes Pdf, <a href=\"https:\/\/kseebsolutions.guru\/kseeb-solutions-class-8-maths\/\">KSEEB Solutions for Class 8 Maths<\/a> helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n<h2>Karnataka Board Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Ex 5.2<\/h2>\n<p>Question 1.<br \/>\nFind the sum 1 + 3 + 5 + _____ +51 (the sum fo all odd numbers from 1 to 51) without actually adding them.<br \/>\nAnswer:<br \/>\nThe sum of first n odd numbers is n<sup>2<\/sup><br \/>\n\u2234 1 +3 + 5+ _____ + 51 = 26<sup>2<\/sup> = 676.<\/p>\n<p>Question 2.<br \/>\nExpress 144 as a sum of 12 odd numbers.<br \/>\nAnswer:<br \/>\n144 = 122 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23.<\/p>\n<p>Question 3.<br \/>\nFind the 14<sup>th<\/sup> and 15<sup>th<\/sup> triangular numbers and find their sum verify the statement 8 for this sum.<br \/>\nAnswer:<br \/>\n14 th triangular number =1 + 2 + 3+4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 = 105.<br \/>\n15 th triangular number = 1 + 2 + 3 + 4 + ___ + 15 = 120<br \/>\nSum of 105 + 120 = 225<br \/>\nAccording to statement 8, the sum of 14th and 15th triangular numbers in equal to<br \/>\nn + (n + 1) = (n + 1)<sup>2<\/sup><br \/>\n105 + 120 = (14 + 1)<sup>2<\/sup><br \/>\n225 = 225.<\/p>\n<p>Question 4.<br \/>\nWhat are the remainders of a perfect square when divided by 5?<br \/>\nAnswer:<br \/>\nWhen perfect squares 9, 16, 25, 36, 49, 64 are divided by 5 the remainders are 4, 1, 0, 1, 4, 4 respectively.<br \/>\nTherefore the remainders of a perfect square when divided by 5 are 0, 1, and 4.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Students can Download Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Ex 5.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka Board Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots &hellip;<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/4847"}],"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=4847"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/4847\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=4847"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=4847"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=4847"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}