{"id":1194,"date":"2020-09-15T12:20:55","date_gmt":"2020-09-15T06:50:55","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=1194"},"modified":"2021-07-02T15:01:50","modified_gmt":"2021-07-02T09:31:50","slug":"kseeb-sslc-class-10-maths-solutions-chapter-8-ex-8-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-sslc-class-10-maths-solutions-chapter-8-ex-8-2\/","title":{"rendered":"KSEEB SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Ex 8.2"},"content":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Ex 8.2<\/strong> are part of KSEEB SSLC Class 10 Maths Solutions<\/a>. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Exercise 8.2.<\/p>\n

Karnataka SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Exercise 8.2<\/h2>\n

Question 1.
\nExpress each number as a product of its prime factors :
\n(i) 140
\n(ii) 156
\n(iii) 3825
\n(iv) 5005
\n(v) 7429
\nSolution:
\n\"KSEEB
\n\"KSEEB<\/p>\n

Question 2.
\nFind the LCM and HCF of the following pairs of integers and verify that LCM \u00d7 HCF = product of the two numbers.
\n(i) 26 and 91
\n(ii) 510 and 92
\n(iii) 336 and 54
\nSolution:
\n(i) 26 and 91
\na = 26, b = 91
\n\"KSEEB
\n26 = 2 \u00d7 13
\n91 = 7 \u00d7 13
\nLCM of (26, 91) = 2 \u00d7 13 \u00d7 7 = 182
\nHCF of (26, 91) = 13
\nVerification
\nLCM of (26, 91) \u00d7 H C F (26, 91) = 182 \u00d7 13 = 2366
\nproduct of two numbers = 26 \u00d7 91 = 2366
\nHence LCM \u00d7 HCF = Product of two numbers.<\/p>\n

(ii) 510 and 92
\na = 510, b = 92
\n\"KSEEB<\/p>\n

(iii) 336 and 54
\na = 336, b = 54
\n\"KSEEB<\/p>\n

Question 3.
\nFind the LCM and HCF of the following integers by applying the prime factorization method.
\n(i) 12, 15 and 21
\n(ii) 17, 23 and 29
\n(iii) 8, 9 and 25
\nSolution:
\n(i) 12, 15 and 21
\n12 = 2 \u00d7 2 \u00d7 3
\n15 = 3 \u00d7 5
\n21 = 3 \u00d7 7
\n\u2234 H.C.F. = 3
\nL.C.M. = 3 \u00d7 2 \u00d7 2 \u00d7 5 \u00d7 7 = 420
\n\u2234 HCF \u00d7 LCM = Product of three numbers
\n3 \u00d7 420 = 12 \u00d7 15 \u00d7 21
\n1260 = 1260<\/p>\n

ii) 17, 23 and 29
\n17, 23 and 29 are prime numbers
\nLCM of (17, 23, 29) = 17 \u00d7 23 \u00d7 29
\n= 391 \u00d7 29
\n= 11339
\nHCF of (17, 23, 29) = 1
\n(\u2235 17, 23 and 29 have no common Factor)<\/p>\n

(iii) 8, 9 and 25
\n8 = 2 \u00d7 2 \u00d7 2= 23
\n9 = 3 \u00d7 3 = 32
\n24 = 5 \u00d7 5 = 52
\n\u2234 H.C.F. = 1
\nL.C.M. = 23<\/sup> \u00d7 32<\/sup> \u00d7 52<\/sup> = 8 \u00d7 9 \u00d7 25 = 1800
\n\u2234 HCF \u00d7 LCM = Product of three numbers
\n1 \u00d7 1800= 8 \u00d7 9 \u00d7 25
\n1800 = 1800<\/p>\n

Question 4.
\nGiven that HCF (306, 657) = 9, find LCM (306, 657).
\nSolution:
\nH.C.F \u00d7 L.C.M = a \u00d7 b
\n(a, b) \u00d7 (a, b) = a \u00d7 b
\n9 \u00d7 x = 306 \u00d7 657
\nx = \\(\\frac{306 \\times 657}{9}\\)
\n\u2234 x = 22338
\n\u2234 L.C.M of (306, 657) = 22338<\/p>\n

Question 5.
\nCheck whether 6n<\/sup> can end with the digit 0 for any natural number ‘n’.
\nSolution:
\nIf the number ends with digit 0, then the prime factorization of the number must contain the factors of 10 as 2 \u00d7 5.
\nNow the prime factorization of 6n<\/sup> = (2 \u00d7 3)n<\/sup> = 2n<\/sup> \u00d7 3n<\/sup>
\nSince it does not contain the factors of 10 as 2 \u00d7 5, thus n<\/sup> cannot end with the digit 0, for any natural number n.<\/p>\n

Question 6.
\nExplain why 7 \u00d7 11 \u00d7 13 + 13 and 7 \u00d7 6 \u00d7 5 \u00d7 4 \u00d7 3 \u00d7 2 \u00d7 1 + 5 are composite numbers.
\nSolution:
\ni) 7 \u00d7 11 \u00d7 13 + 13 = 13(77 + 1)
\n= 13 \u00d7 78
\n\u2234 It is a composite number
\n[It is having more than two factors]<\/p>\n

ii) 7 \u00d7 6 \u00d7 5 \u00d7 4 \u00d7 3 \u00d7 2 \u00d7 1 + 5
\n=5[7 \u00d7 6 \u00d7 4 \u00d7 3 \u00d7 2 \u00d7 1 + 1]
\n= 5 (1008 + 1)
\n= 5 \u00d7 1009 = 5045
\n\u2234 It is a composite number.<\/p>\n

Question 7.
\nThere is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time and go in the same direction. After how many minutes will they meet again at the starting
\nSolution:
\nSonia takes 18 minutes to drive one round of the field.
\nRavi takes 12 minutes for the same.
\nSuppose they both start at the same point and at the same time.
\nTime to meet both again at the starting point,
\nFor this, we have to find out HCF of 18 and 12.
\n18 = 2 \u00d7 3 \u00d7 3 = 2 \u00d7 32<\/sup>
\n12 = 2 \u00d7 2 \u00d7 3 = 22<\/sup> \u00d7 3
\n\u2234 H.C.F. = 2 \u00d7 3 = 6
\n\u2234 After 6 minutes, they both meet at the same point.<\/p>\n

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Ex 8.2 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Exercise 8.2, drop a comment below and we will get back to you at the earliest.<\/p>\n","protected":false},"excerpt":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Ex 8.2 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Exercise 8.2. Karnataka SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Exercise 8.2 Question 1. Express each number as …<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/1194"}],"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\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=1194"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/1194\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=1194"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=1194"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=1194"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}