{"id":27580,"date":"2020-07-07T11:25:31","date_gmt":"2020-07-07T05:55:31","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=27580"},"modified":"2021-07-02T15:07:07","modified_gmt":"2021-07-02T09:37:07","slug":"kseeb-solutions-for-class-8-maths-chapter-10-additional-questions","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-for-class-8-maths-chapter-10-additional-questions\/","title":{"rendered":"KSEEB Solutions for Class 8 Maths Chapter 10 Exponents Additional Questions"},"content":{"rendered":"

Students can Download Maths Chapter 10 Exponents Additional Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths<\/a>\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka State Syllabus Class 8 Maths Chapter 10 Exponents Additional Questions<\/h2>\n

Question 1.
\nMark the correct option.
\n(a) The value of (3m<\/sup>)n<\/sup>, for every pair of integers (m, n) is
\nA. 3m+n
\n<\/sup>B. 3mn<\/sup>
\nC. 3mn<\/sup>
\nD. 3m<\/sup> + 3n<\/sup>
\nSolution:
\n(B) 3mn<\/sup><\/p>\n

(b) If x, y, 2x + \\(\\frac{\\mathbf{y}}{2}\\) are non-zero real numbers then \\(\\left(2 x+\\frac{y}{2}\\right)^{-1}\\left[(2 x)^{-1}+\\left(\\frac{y}{2}\\right)^{-1}\\right]\\) equals.
\nA. 1
\nB. x . y-1<\/sup>
\nC. x-1<\/sup> . y
\nD. x-1<\/sup> . y-1<\/sup>
\nSolution:
\n(D) x-1<\/sup> . y-1<\/sup><\/p>\n

(c) If 2x<\/sup> – 2x-2<\/sup> = 192 the value of x is
\nA. 5
\nB. 6
\nC. 7
\nD. 8
\nSolution:
\n(D) 8<\/p>\n

(d) The number\\(\\left[6^{\\left(6^{6}\\right)}\\right]^{\\frac{1}{6}}\\) is equal to
\nA. 66<\/sup>
\nB. 66 <\/sup>– 1
\nC. 6(65<\/sup>)
\nD. \\(6^{6^{5}}\\)
\nSolution:
\n(C) 6(65<\/sup>)<\/p>\n

\"KSEEB<\/p>\n

(e) The number of pairs of positive integers (m, n) such that mn<\/sup> = 25 is
\nA. 0
\nB. 1
\nC. 2
\nD. More than 2
\nSolution:
\n(B) 1<\/p>\n

Question 2.
\nThe diameter of the sun is 1.4 x 109 meters and that of the earth is about 1.2678 x 107<\/sup> meteres. Find the approximate ratio of the diameter of the sun to that of the earth.
\nSolution:
\n\"KSEEB
\n= 1.10427 x 109-7<\/sup>
\n= 1.10427 x 102<\/sup>
\n= 1.10427 x 100 = 110 . 427
\n\u2234 The ratio is 110 . 427 : 1<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nFind the value of each of the following expressions
\n(a) (-0.75)3<\/sup> + (0.3)-3<\/sup> – \\(\\left(-\\frac{3}{2}\\right)^{-3}\\)
\nSolution:
\n\"KSEEB<\/p>\n

(b) \"KSEEB
\nSolution:
\n\"KSEEB
\n\"KSEEB<\/p>\n

(c) \"KSEEB
\nSolution:
\n\"KSEEB<\/p>\n

Question 4.
\nHow many digits are there in the number 23<\/sup> x 55<\/sup> x 205<\/sup>?
\nSolution:
\n23<\/sup> x 55<\/sup> x 205<\/sup>
\n= 8 x 625 x 3200000 = 5000 x 32
\n= 16000000000
\n\u2234 The number has 11 digits.<\/p>\n

\"KSEEB<\/p>\n

Question 5.
\nIf a7 = 3 find the of
\n\"KSEEB
\nSolution:
\n\"KSEEB<\/p>\n

Question 6.
\nIf 2m<\/sup>. a2<\/sup> = 28<\/sup> where a, m are positive integers find all possible values of a + m.
\nSolution:
\n2m<\/sup> . a2<\/sup> = 28<\/sup>
\n2m<\/sup> . a2<\/sup> = 26<\/sup> . 22<\/sup>
\n\u2234m = 6 and a = 2
\na + m = 6 + 2 = 8<\/p>\n

Question 7.
\nSuppose 3k<\/sup> x b2<\/sup> = 64<\/sup> for some positive integers k and b. Find all possible values of k + b.
\nSolution:
\n3k<\/sup> x b2<\/sup> = 64<\/sup>
\n= (3 x 2)4<\/sup>
\n= 34<\/sup> x 24<\/sup> = 34<\/sup> x 22×2<\/sup> = 34 x (22<\/sup>)2<\/sup>
\n= 3k x b2<\/sup> = 34<\/sup> x 42<\/sup>
\n\u2234 k = 4 and b = 4
\nk + b = 4 + 4 = 8<\/p>\n

\"KSEEB<\/p>\n

Question 8.
\nFind the value of \\(\\frac{(625)^{6.25} \\times(25)^{2.60}}{(625)^{7.25} \\times(5)^{1.20}}\\)
\nSolution:
\n\"KSEEB
\n\"KSEEB<\/p>\n

Question 9.
\nA person had some rupees which is a power of 5. He gave a part of it to his friend which is also a power of 5. He was left with Rs. 500. How much did money he have?
\nSolution:
\nThe powers of 5 are 51<\/sup> = 5, 52<\/sup> = 25, 53<\/sup>
\n= 125, 54<\/sup> = 625,
\n55<\/sup> = 3125 etc.
\nWe find that 625 – 125 = 500
\n\u2234 The money with him was Rs. 625.<\/p>\n

\"KSEEB<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 10 Exponents Additional Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka State Syllabus Class 8 Maths Chapter 10 Exponents Additional Questions Question 1. Mark the correct option. (a) The value …<\/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\/27580"}],"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=27580"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/27580\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=27580"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=27580"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=27580"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}