{"id":1472,"date":"2020-09-15T11:59:23","date_gmt":"2020-09-15T06:29:23","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=1472"},"modified":"2021-07-02T15:01:50","modified_gmt":"2021-07-02T09:31:50","slug":"kseeb-sslc-class-10-maths-solutions-chapter-10-ex-10-3","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-sslc-class-10-maths-solutions-chapter-10-ex-10-3\/","title":{"rendered":"KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.3"},"content":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.3<\/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 10.3.<\/p>\n

Karnataka SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Exercise 10.3<\/h2>\n

Question 1.
\nFind the roots of the following quadratic equations, if they exist, by the method of completing the square:
\n(i) 2x2<\/sup> – 7x + 3 = 0.
\n(ii) 2x2<\/sup> + x – 4 = 0
\n(iii) 4x2<\/sup> + \\(4 \\sqrt{3} x\\) + 3 = 0
\n(iv) 2x2<\/sup> + x + 4 = 0
\nSolution:
\n(i) 2x2<\/sup> – 7x + 3 = 0
\nDividing all terms by 2,
\n\"KSEEB
\n\"KSEEB
\n\"KSEEB<\/p>\n

(ii) 2x2<\/sup> + x – 4 = 0
\nDividing all terms by 2,
\n\"KSEEB
\n\"KSEEB<\/p>\n

(iii) 4x2<\/sup> + \\(4 \\sqrt{3} x\\) + 3 = 0
\nDividing all terms by 4,
\n\"KSEEB<\/p>\n

(iv) 2x2<\/sup> + x + 4 = 0
\nDividing all terms of the equation by 2,
\n\"KSEEB
\nHere ‘x’ has no fixed value because \\(\\sqrt{-31}\\) is not a square root.<\/p>\n

Question 2.
\nFind the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.
\n(i) 2x2<\/sup> – 7x + 3 = 0.
\n(ii) 2x2<\/sup> + x – 4 = 0
\n(iii) 4x2<\/sup> + \\(4 \\sqrt{3} x\\) + 3 = 0
\n(iv) 2x2<\/sup> + x + 4 = 0
\nSolution:
\n(i) 2x2<\/sup> – 7x + 3 = 0.
\nHere, a = 2, b = -7, c = 3
\n\"KSEEB<\/p>\n

(ii) 2x2<\/sup> + x – 4 = 0
\nHere, a = 2, b = 1, c = 4
\n\"KSEEB<\/p>\n

(iii) 4x2<\/sup> + \\(4 \\sqrt{3} x\\) + 3 = 0
\nHere, a = 4, b = \\(4 \\sqrt{3} \\), c = 3
\n\"KSEEB
\n\"KSEEB<\/p>\n

(iv) 2x2<\/sup> + x + 4 = 0
\nHere, a = 2, b = 1, c = 4
\n\"KSEEB<\/p>\n

Question 3.
\nFind the roots of the following equations:
\n\"KSEEB
\nSolution:
\n\"KSEEB
\n\"KSEEB
\n(x – 2) (x – 1) = 0
\nIf x – 2 = 0, then x = 2
\nIf x – 1 = 0, then x = 1
\n\u2234 x = 2 OR x = 1.<\/p>\n

Question 4.
\nThe sum of the reciprocals of Rehman\u2019s ages, (in years) 3 years ago and 5 years from now is \\(\\frac{1}{3}\\). Find his present age.
\nSolution:
\nLet the present age of Rehman be ‘x’ years.
\nAfter 5 years his age will be (x + 5)
\n3 years back his age was (x – 3)
\n\"KSEEB
\n(x – 7) (x + 3) = 0
\nIf x – 7 = 0, then x = 7
\nIf x + 3 = 0, then x = -3
\n\u2234 x = 7 OR x = -3
\n\u2234 Present age of Rehman is 7 years.<\/p>\n

Question 5.
\nIn a class test, the sum of Shefali\u2019s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.
\nSolution:
\nLet Shefali’s marks in mathematics is Y and marks in English is (30 – x)
\nShefali’s gets 2 more marks in mathematics then x + 2 and 3 marks less in English is (30 – x – 3)
\nProduct of Shefali’s marks = 210
\n(x + 2) (30 – x – 3) = 210
\n(x + 2) (27 – x) = 210
\n27x + 54 – x2<\/sup> – 2x = 210
\n– x2<\/sup> + 25x + 54 = 210
\nx2<\/sup> – 25x + 210 – 54 = 0
\nx2<\/sup> – 25x + 156 = 0
\nx2<\/sup> – 13x – 12x + 156 = 0
\nx(x – 13) – 12 (x – 13) = 0
\n(x – 13) (x – 12) = 0
\nx – 13 = 0 (or) x – 12 = 0
\nx = 13 (or) x = 12
\nTherefore, Shefali’s marks in Mathematics =13
\nMarks in English = 30 – 13 = 17 (or) Shefali’s marks in Mathematics = 12 then.
\nMarks in English = 30 – 12 = 18<\/p>\n

Question 6.
\nThe diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.
\nSolution:
\n\"KSEEB
\nLet shorter side of rectangle ABCD be x metre,
\nThen, longer side of rectangle ABCD,
\nAB = (x + 30) m.
\nDiagonal AC = (x + 60) m.
\nIn \u22a5\u2206ABC, \u2220B = 90\u00b0.
\nAs per Pythagoras theorem,
\nAB2<\/sup> + BC2<\/sup> = AC2<\/sup>
\n(x + 30)2<\/sup> + (x)2<\/sup> = (x + 60)2<\/sup>
\nx2<\/sup> + 60x + 900 + x2<\/sup> = x2<\/sup> + 120x + 3600
\n2x2<\/sup> + 60x + 900 = x2<\/sup> + 120x + 3600
\n2x2<\/sup> – x2<\/sup> + 60x – 120x + 900 – 3600 = 0
\nx2<\/sup> – 60x – 2700 = 0
\n\"KSEEB
\nx2<\/sup> – 90x + 30x – 2700 = 0
\nx(x – 90) + 30 (x – 90) = 0
\n(x – 90) (x + 30) = 0
\nIf x – 90 = 0, then x = 90
\nIf x + 30 = 0, then x = -30
\nShorter side of rectangle, BC = x = 90 m.
\nLonger side of rectangle, AB = x + 30 = 90 + 30= 120 m.
\nDiagonal of rectangle, AC = x + 60 = 90 + 60 = 150 m.<\/p>\n

Question 7.
\nThe difference of squares of two numbers is 180. The square of the smellier number is 8 times the larger number. Find the two numbers.
\nSolution:
\nLet the small number be ‘x’.
\nThe Square of the smaller number is 8 times the larger number.
\n\u2234 x2<\/sup> = 8 \u00d7 Larger number.
\n\u2234 Larger number = \\(\\frac{1}{8}\\) x2<\/sup>.
\nThe difference between these numbers is 180.
\n\"KSEEB
\n\u2234 Smaller number is 12,
\nLarger number is 18.<\/p>\n

Question 8.
\nA train travels 360 km at a uniform speed. If the speed had been 5 km\/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
\nSolution:
\nLet initial speed of the train be x km\/h,
\nThen, time taken to cover 360 km
\nIf the speed had been 5 km\/h more,
\n\\(\\frac{360}{x}\\) Hr.
\nthen time taken to cover is \\(\\frac{360}{x+5}\\) Hr.
\n\"KSEEB
\nx(x + 365) = 360 (x + 5)
\nx2<\/sup> + 365x = 360x + 1800
\nx2<\/sup> + 365x – 360x – 1800 = 0
\nx2<\/sup> + 5x – 1800 = 0
\nx2<\/sup> + 45x – 40x – 1800 = 0
\nx(x + 45) – 40 (x + 45) = 0
\n(x + 45) (x – 40) = 0
\nIf x + 45 = 0, then x = – 45
\nIf x – 40 = 0, then x = 40
\n\u2234 Initial speed of a train is 40 km\/hr.<\/p>\n

Question 9.
\nTwo water taps together can fill a tank in \\(9 \\frac{3}{8}\\) hours. The tap of a larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank. .
\nSolution:
\nLet the time required to fill up the tank for tap having a larger diameter be ‘x’ Hr. For ‘x’ hour part of the tank filling is 1.
\nFor \\(9 \\frac{3}{8}\\) hr. part of the tank filling ………..?
\ni.e., \\(\\frac{75 \\times 1}{8 x}\\) part
\nTime required for smaller diameter is = (x – 10)Hr.
\nFor (x – 10)Hr part of the tank, filling is 1
\nFor \\(9 \\frac{3}{8}\\) Hr. part of the tank is …………?
\n\"KSEEB
\n\u2234Time required for larger diameter, x = 25 Hr.
\n\u2234 Time required for smaller diameter = x – 10 = 15 Hr.<\/p>\n

Question 10.
\nAn express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11km\/h more than that of the passenger train, find the average speed of the two trains.
\nSolution:
\nLet Average speed of passenger train be = x km\/h.
\n. But Average speed of Express train be = (x + 11) km\/h.
\nTime required for Passenger:
\nFor travelling x km time required is 1 Hr
\nFor travelling 132 km time required …….?
\n\\(\\frac{132}{x}\\) Hr.
\nTime required for Express train :
\nFor travelling (x + 11) km is 1 Hr.
\nFor travelling 132 km …………?
\ni.e., \\(\\frac{132}{(x+11)}\\) Hr
\nExpress train required 1 Hr. less comparing to Passenger train. .
\n\"KSEEB
\nx(x + 143) = 132 (x+ 11)
\nx2<\/sup> + 143x = 132x + 1452
\nx2<\/sup> + 143x – 132 x – 1452 = 0
\nx2<\/sup> + 11x – 1452 = 0
\nx2<\/sup> + 44x – 33x – 1452 = 0
\nx(x + 44) – 33(x + 44) = 0
\n(x + 44) (x – 33) = 0
\nIf x + 44 = 0, then x = -44
\nIf x – 33 = 0, then x = 33
\n\u2234 Average speed of passenger train is 33 km\/hr.
\n\u2234 Average speed of Express train = x + 11 = 33 + 11 = 44 km\/hr.<\/p>\n

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.3 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Exercise 10.3, 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 10.3 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Exercise 10.3. Karnataka SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Exercise 10.3 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\/1472"}],"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=1472"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/1472\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=1472"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=1472"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=1472"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}