{"id":1395,"date":"2020-09-15T11:05:11","date_gmt":"2020-09-15T05:35:11","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=1395"},"modified":"2021-07-02T15:01:50","modified_gmt":"2021-07-02T09:31:50","slug":"kseeb-sslc-class-10-maths-solutions-chapter-10-ex-10-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-sslc-class-10-maths-solutions-chapter-10-ex-10-2\/","title":{"rendered":"KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.2"},"content":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.2<\/strong> are part of KSEEB SSLC Class 10 Maths Solutions<\/a>. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Exercise 10.2.<\/p>\n

Karnataka SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Exercise 10.2<\/h2>\n

Question 1.
\nFind the roots of the following quadratic equations by factorisation:
\n(i) x2<\/sup> – 3x – 10 = 0 ,
\n(ii) 2x2<\/sup> + x – 6 = 0
\n(iii) \\(\\sqrt{2}\\)x2<\/sup> + 7x + \\(5 \\sqrt{2}\\) =0
\n(iv) 2x2<\/sup> – x + \\(\\frac{1}{8}\\) = 0
\n(v) 100x2<\/sup> – 20x + 1 = 0
\nSolution:
\n(i) x2<\/sup> – 3x – 10 = 0
\nx2<\/sup> – 5x + 2x – 10 = 0
\n\"KSEEB
\nx(x – 5) + 2 (x – 5) = 0
\n(x – 5) (x + 2) = 0
\nIf x + 2 = 0, then x = -2
\nIf x – 5 = 0, then x = 5
\n\u2234 x = -2 OR +5.<\/p>\n

(ii) 2x2<\/sup> + x – 6 = 0
\n2x2<\/sup> + 4x – 3x – 6 = 0
\n\"KSEEB
\n2x(x + 2) – 3(x + 2) = 0
\n(x + 2) (2x – 3) = 0
\nIf x + 2 = 0, then x = -2
\nIf 2x – 3 = 0, then 2x = 3 x = \\(\\frac{3}{2}\\)
\n\u2234 x = -2 OR \\(\\frac{3}{2}\\)<\/p>\n

\"KSEEB<\/p>\n

iv) 2x2<\/sup> – x + \\(\\frac{1}{8}\\) = 0 Multiply by 8
\n8(2x2<\/sup> -x + \\(\\frac{1}{8}\\)) = 0
\n16x2<\/sup> – 8x + \\(\\frac{8}{8}\\) = 0
\n16x2<\/sup> – 8x + 1 = 0
\n16x2<\/sup> – 4x – 4x + 1 = 0
\n4x (4x – 1) – 1 (4x – 1) =
\n(4x – 1) (4x – 1) = 0
\n4x – 1 = 0 or 4x – 1 = 0
\n4x = 1 or 4x = 1
\nx = \\(\\frac{1}{4}\\) or x = \\(\\frac{1}{4}\\)
\nx = \\(\\frac{1}{4}\\) are the roots of the equation 2x2<\/sup> – x + \\(\\frac{1}{8}\\) = 0<\/p>\n

(v) 100x2<\/sup> – 20x + 1 = 0
\n100x2<\/sup> – 10x – 10x + 1 = 0
\n10x(10x – 1) – 1 (10x – 1) = 0
\n(10x – 1) (10x – 1) = 0
\n(10x – 1)2<\/sup> = 0
\nIf 10x – 1 = 0, then
\n10x = 1
\n\u2234 x = \\(\\frac{1}{10}\\)
\n\u2234 Two roots are \\(\\frac{1}{10} , \\frac{1}{10}\\)<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\nSolve the problems:
\n(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.
\n(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was \u20b9 750. We would like to find out the number of toys produced on that day.
\nSolution:
\n(i) Let John had x marbles and Jivanti had (45 – x) marbles.
\nWhen both of them lost 5 marbles then equation becomes (x – 5) \u00d7 (45 – x – 5) = 124
\n\u21d2 (x – 5) \u00d7 (40 – x) = 124
\n\u21d2 x2<\/sup> – 45x + 324 = 0
\n\u21d2 x2<\/sup> – 9x – 36x + 324 = 0
\n\u21d2 x(x – 9) – 36(x – 9) = 0
\n\u21d2 (x – 9)(x – 36) = 0
\nEither x – 9 = 0 or x – 36 = 0
\nThus, x = 9 or x = 36
\n\u2234 If John had 9 marbles, then Jivanti had 45 – 9 = 36 marbles.
\nIf John had 36 marbles, then Jivanti had 45 – 36 = 9 marbles.<\/p>\n

(ii) Let the number of toys produced in a day be x.
\nThen cost of 1 toy = \\(\\frac{750}{x}\\)
\n\u21d2 \\(\\frac{750}{x}\\) = 55 – x
\n\u21d2 750 = 55x – x2<\/sup>
\n\u21d2 x2<\/sup> – 55x + 750 = 0
\n\u21d2 x2<\/sup> – 30x – 25x + 750 = 0
\n\u21d2 x(x – 30) – 25(x – 30) = 0
\n\u21d2 (x – 30)(x – 25) = 0
\nEither x – 30 = 0 or x – 25 = 0
\nx = 30 or x = 25<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nFind two numbers whose sum is 27 and the product is 182.
\nSolution:
\nSum of two numbers be 27
\nLet one number be ‘x’ other number be ’27 – x’
\nProduct of two numbers = 182
\nx(27 – x)= 182
\n27x – x2<\/sup> = 182
\nx2<\/sup> – 27x + 182 = 0
\nx2<\/sup> – 14x – 13x+ 182 = 0
\nx (x – 14) – 13 (x – 14) = 0
\n(x – 14) (x – 13) = 0
\nx + 14 = 0 (or) x – 13 = 0
\nx = – 14 or x = 13
\nx = 14, 13
\n\u2234 Two numbers are 14, 13 (or) 13, 14<\/p>\n

Question 4.
\nFind two consecutive positive integers, sum of whose squares is 365.
\nSolution:
\nIn that numbers, let one of the numbers be \u2019x’.
\nIt’s consecutive positive integer is (x + 1)
\nSum of their squares is 365.
\n\u2234 (x)2<\/sup> + (x + 1)2<\/sup> = 365
\nx2<\/sup> + x2<\/sup> + 2x + 1 = 365
\n2x2<\/sup> + 2x + 1 = 365
\n2x2<\/sup> + 2x + 1 – 365 = 0
\n2x2<\/sup> + 2x – 365 = 0
\nx2<\/sup> + x – 182 = 0
\n\"KSEEB
\nx2<\/sup> + 14x – 13x – 182 = 0
\nx(x + 14) – 13(x + 14) = 0
\n(x + 14) (x – 13) = 0
\nIf x + 14 = 0, then x = -14
\nIf x – 13 = 0, then x = 13
\nIn these positive integer is 13.
\n\u2234 One number, x = 13
\nIt consecutive number is, x + 1
\n= 13 + 1 = 14
\n\u2234 The Numbers are 14 and 13.<\/p>\n

\"KSEEB<\/p>\n

Question 5.
\nThe altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm., find the other two sides.
\nSolution:
\n\"KSEEB
\nIn \u22a5\u2206ABC,
\n\u2220ABC = 90\u00b0.
\nAB is altitude,
\nBC is base AC is Hypotenuse
\nLet Base, BC = x cm.
\nif so considered, :
\nAltitude AB = (x – 7) cm.
\nHypotenuse AC = 13 cm.
\nAs per Pythagoras theorem,
\nIn a right angled ABC,
\nAB2<\/sup> + BC2<\/sup> = AC2<\/sup>
\n(x – 7)2<\/sup> + (x)2<\/sup> = (13)2<\/sup>
\nx2<\/sup> – 14x + 49 + x2<\/sup>= 169
\n2x2<\/sup> – 14x + 49 = 169
\n2x2<\/sup> – 14x + 49 – 169 = 0
\n2x2<\/sup> – 14x – 120 = 0
\nx2<\/sup> – 7x – 60 = 0
\n\"KSEEB
\nx2<\/sup> – 12x + 5x – 6 0 = 0
\nx(x – 12) + 5(x – 12) = o
\n(x – 12) (x + 5) =0
\nIf x – 12 = 0, then x = 12
\nIf x + 5 = 0, then x = -5
\nPositive value, x = 12
\n\u2234 Base, BC = x = 12 cm.
\nAltitude, AB = x – 7 = 12 – 7 = 5 cm.<\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nA cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was \u20b9 90, find the number of articles produced and the cost of each article.
\nSolution:
\nLet the number of articles produced in a day = x
\n\u2234 Cost of production of each article = \u20b9 (2x + 3)
\nAccording to the condition,
\nTotal cost = \u20b9 90
\n\u21d2 x \u00d7 (2x + 3) = 90
\n\u21d2 2x2<\/sup> + 3x = 90
\n\u21d2 2x2<\/sup> + 3x – 90 = 0
\n\u21d2 2x2<\/sup> – 12x + 15x – 90 = 0
\n\u21d2 2x(x – 6) + 15(x – 6) = 0
\n\u21d2 (x – 6)(2x + 15) = 0
\nEither x – 6 = 0 or 2x + 15 = 0
\n\u21d2 x= 6 or x = \\(\\frac{-15}{2}\\)
\nBut the number of articles produced can never be negative.
\n\u21d2 x = \\(\\frac{-15}{2}\\) is rejected
\n\u2234 Cost of production of each article = \u20b9 (2 \u00d7 6 + 3) = \u20b9 15
\nThus, the required number of articles produced is 6 and the cost of each article is \u20b9 15.<\/p>\n

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.2 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Exercise 10.2, 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 10 Quadratic Equations Ex 10.2 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Exercise 10.2. Karnataka SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Exercise 10.2 Question 1. Find the roots of …<\/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\/1395"}],"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=1395"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/1395\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=1395"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=1395"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=1395"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}