{"id":1137,"date":"2020-09-15T11:41:35","date_gmt":"2020-09-15T06:11:35","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=1137"},"modified":"2021-07-02T15:01:50","modified_gmt":"2021-07-02T09:31:50","slug":"kseeb-sslc-class-10-maths-solutions-chapter-8-ex-8-1","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-sslc-class-10-maths-solutions-chapter-8-ex-8-1\/","title":{"rendered":"KSEEB SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Ex 8.1"},"content":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Ex 8.1<\/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 8.1.<\/p>\n

Karnataka SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Exercise 8.1<\/h2>\n

Question 1.
\nUse Euclid\u2019s division algorithm to find the HCF of:
\n(i) 135 and 225
\n(ii) 196 and 38220
\n(iii) 867 and 255
\nSolution:
\n(i) 135 and 225
\n\"KSEEB
\n\u2234 90 = 45 \u00d7 2 + 0
\n\u2234 HCF of 135 and 225 is 45.<\/p>\n

(ii) 196 and 38220
\n\"KSEEB
\n\u2234 38220 = 196 \u00d7 195 + 0
\n\u2234 HCF of 196 and 38220 is 196.<\/p>\n

(iii) 867 and 2552
\n\"KSEEB
\n2 = 1 \u00d7 2 + 0
\n\u2234 HCF of 867 and 2552 is 1.<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\nShow that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
\nSolution:
\nLet us consider a positive odd integer as ‘a’.
\nOn dividing ‘a’ by 6, let q be the quotient and ‘r’ be the remainder.
\n\u2234 Using Euclid’s lemma, we get a = 6q + r
\nwhere 0 \u2264 r < 6 i.e., r = 0, 1, 2, 3, 4 or 5 i.e.,
\na = 6q + 0 = 6q or a = 6q + 1
\nor a = 6q + 2 or a = 6q + 3
\nor a = 6q + 4 or a = 6q + 5
\nBut, a = 6q, a = 6q + 2, a = 6q + 4 are even values of ‘a’.
\n[\u2235 6q = 2(3q) = 2m1<\/sub> 6q + 2 = 2(3q + 1) = 2m2<\/sub>,
\n6q + 4 = 2(3 q + 2) = 2m3<\/sub>]
\nBut \u2018a\u2019 being an odd integer, we have :
\na = 6q + 1, or a = 6q + 3, or a = 6q + 5<\/p>\n

Question 3.
\nAn army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march ?
\nSolution:
\nHCF of 32 and 616
\n\"KSEEB
\n\u2234 32 = 8 \u00d7 4 + 0
\n\u2234 HCF of 32 and 616 is 8.
\n\u2234 Both groups can march in 8 columns.<\/p>\n

Question 4.
\nUse Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
\n[Hint: Let x be any positive integer then it is of the form 3g, 3q + 1 or 3g + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m +1.]
\nSolution:
\nLet us consider an arbitrary positive integer as ‘x’ such that it is of the form
\n3q, (3q + 1) or (3q + 2)
\nFor x = 3q, we have x2<\/sup> = (3q)2<\/sup>
\n\u21d2 x2<\/sup> = 9q2<\/sup> = 3(3q2<\/sup>) = 3m ………. (1)
\nPutting 3q2<\/sup> = m, where m is an integer.
\nFor x = 3q + 1,
\nx2<\/sup> = (3q + 1)2<\/sup> = 9q2<\/sup> + 6q + 1
\n= 3(3q2<\/sup> + 2q) + 1 = 3m + 1 ………… (2)
\nPutting 3q2<\/sup> + 2q = m, where m is an integer.
\nFor x = 3q + 2,
\nx2<\/sup> = (3q + 2)2<\/sup>
\n= 9q2<\/sup> + 12q + 4 = (9q2<\/sup> + 12q + 3) + 1
\n= 3(3q2<\/sup> + 4q + 1) + 1 = 3m + 1 ……….. (3)
\nPutting 3q2<\/sup> + 4q +1 = m, where m is an integer.
\nFrom (1), (2) and (3),
\nx2<\/sup> = 3m or 3m + 1
\nThus, the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.<\/p>\n

\"KSEEB<\/p>\n

Question 5.
\nUse Euclid\u2019s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.
\nSolution:
\nLet a and b be two positive integers and a > b
\na = (b \u00d7 q) + r where q and r are positive integers and 0 \u2264 r < b.
\nLet b = 3 (if 9 is multiplied by 3 a perfect cube number is obtained)
\na = 3q + r where 0 \u2264 r < 3
\n(i) if r = 0, a = 3q
\n(ii) if r = 1, a = 3q + 1
\n(iii) if r = 2 a = 3q + 2
\nConsider cubes of these,
\ncase i) a = 3q
\na3<\/sup> = (3q)3<\/sup> = 27q3<\/sup> = 9(3q3<\/sup>)
\na3<\/sup> = 9m
\nwhere m= 3q3<\/sup> & m is an integer<\/p>\n

case ii) a = 3q + 1
\na3<\/sup> = (3q + 1)3<\/sup>
\n(a + b)3<\/sup> = a3<\/sup> + b3<\/sup> + 3a2<\/sup>b + 3ab2<\/sup>
\na3<\/sup> = 27q3<\/sup> + 1 + 27q3<\/sup> + 9q
\na3<\/sup> = 27 q3<\/sup> + 27 q2<\/sup> + 9q + 1
\na3<\/sup> = 9 (3q3<\/sup> + 3q2<\/sup> + q) + 1
\na3<\/sup> = 9m + 1
\nwhere m = 3q3<\/sup> + 3q2<\/sup> + q & m is an integer<\/p>\n

case iii) a = 3q + 2
\na3<\/sup> = (3q + 2)3<\/sup> = 27q3<\/sup> + 8 + 54q2<\/sup> + 36q
\na3<\/sup> = 27q3<\/sup> + 54q2<\/sup> + 36q + 8
\n= 9 (3q3<\/sup> + 6q2<\/sup> + 4q) + 8
\na3<\/sup> = 9m + 8
\nwhere m = 3q3<\/sup> + 6q2<\/sup> + 4q and m is an integer.
\n\u2234 Cube of any positive integer is either of the form 9m, 9m + 1 (or) 9m + 8 for some integer m.<\/p>\n

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Ex 8.1 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Exercise 8.1, 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 8.1 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Exercise 8.1. Karnataka SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Exercise 8.1 Question 1. Use Euclid\u2019s division algorithm …<\/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\/1137"}],"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=1137"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/1137\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=1137"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=1137"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=1137"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}