{"id":25591,"date":"2021-06-23T14:10:34","date_gmt":"2021-06-23T08:40:34","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=25591"},"modified":"2021-06-23T14:10:34","modified_gmt":"2021-06-23T08:40:34","slug":"ncert-solutions-for-class-10-maths-chapter-4-ex-4-2","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-10-maths-chapter-4-ex-4-2\/","title":{"rendered":"NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.2"},"content":{"rendered":"

These NCERT Solutions for Class 10 Maths<\/a> Chapter Ex 4.2 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.2<\/h2>\n

\"NCERT<\/p>\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) \u221a2x2<\/sup> + 7x + 5\u221a2 = 0
\n(iv) 2x2<\/sup> – x + \\(\\frac { 1 }{ 8 }\\) = 0 8
\n(v) 100x2<\/sup> – 20 x + 1 = 0
\nSolution:
\n(i) We have,
\nx\u00b2 – 3x – 10 = 0
\nor x\u00b2 – 5x + 2x – 10 = 0
\nor x(x – 5) – 2(x – 5) = 0
\nor, (x – 5) (x + 2) = 0
\n\u2234 x – 5 = 0 or, x + 2 = 0
\n\u21d2 x = 5 or, x + 2 = 0
\nTherefore, the roots x\u00b2 – 2x -10 = 0 are – 5 and<\/p>\n

(ii) We have,
\n2x\u00b2 + x – 6 = 0
\nor, 2x\u00b2 + 4x – 3x – 6 = 0
\nor, 2x(x + 2) – 3(x + 2) = 0
\nor, (x + 2) (2x – 3) = 0
\n\u2234 (x + 2) = 0 or, (2x – 3) = 0
\n\u21d2 x = – 2 or, x = \\(\\frac { 3 }{ 2 }\\)
\nTherefore, the roots of 2x\u00b2 + x – 6 = 0 are – 2 and \\(\\frac { 3 }{ 2 }\\)<\/p>\n

(iii) We have,
\n\"NCERT
\nTherefore, the root of \\(\\sqrt{2}\\)2x\u00b2 + 7x + 5\\(\\sqrt{2}\\) = o are \\(\\frac{-5}{\\sqrt{2}}\\) and \\(\\sqrt{2}\\)<\/p>\n

(iv) We have,
\n\"NCERT
\nTherefore, the roots of 2x\u00b2 – x + \\(\\frac { 1 }{ 8 }\\) = 0 is \\(\\frac { 1 }{ 4 }\\) ; \\(\\frac { 1 }{ 4 }\\)<\/p>\n

(v) We have,
\n\"NCERT
\nTherefore, the roots of 100x\u00b2 – 20x + 1 = 0 is \\(\\frac { 1 }{ 10 }\\), \\(\\frac { 1 }{ 10 }\\).<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nSolve the problems given in Example 1.
\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.<\/p>\n

(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy in (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 number of marbles John have be x
\n\u2234 Number of marbles Jivanti have = 45 – x
\nThe number of marbles left with John when he lost 5 marbles = x – 5
\nThe number of marbles left with Jivanti when she lost 5 marbles = 45x – x – 5 = 40 – x
\nAccording to question,
\n(x – 5) (40 – x) = 124
\nor 40x – x\u00b2 – 200 + 5x = 124
\nor 45x – x\u00b2 – 200 = 124
\nor x\u00b2 – 45x + 324 = 0
\nor x\u00b2 – 36x – 9x + 324 = 0
\nor x(x – 36) – 9(x – 36) = 0
\nor (x – 36) (x – 9) = 0
\n\u2234 (x – 36) =0 or (x – 9) = 0
\n\u21d2 x = 36 or x = 9
\nTherefore, if number of marbles John have be 36
\nThen number of marbles Jivanti have = 45 – x = 45 – 36 = 9
\nAnd, if number of marbles John have be 9
\nThen, number of marbles Jivanti have = 45 – x = 45 – 9 = 36<\/p>\n

(ii) Let the number of toys produced on that day be x
\nTherefore; the cost of production (in rupees) of each toy that day = 55 – x
\nSo, the total cost of production (in rupees) that day = x(55 – x)
\nAccording to question,
\nx(55 – x) = 750
\nor 55x – x\u00b2 = 750
\nor x\u00b2 – 55x + 750 = 0
\nor x\u00b2 – 30x – 25x + 750 = 0
\nor x (x – 30) – 25 (x – 30) = 0
\n(x – 30) (x – 25) = 0
\n\u2234 (x – 30) = 0 or (x – 25) = 0
\n\u21d2 x = 30 or x = 25
\nTherefore, if the number of toys produced on that day is x = 30, then, the cost of production (in rupees) or each toy = 55 – x = 55 – 30 = \u20b9 25.
\nAnd, if the number of toys produced is x = 25,
\nthen, the cost of production = 55 – x = 55 – 25 = \u20b9 30.<\/p>\n

Question 3.
\nFind two numbers whose sum is 27, and product is 182.
\nSolution:
\nLet first number be x Second number be = 27 – x According to question,
\nx(27 – x) = 182 ‘
\nor 27x – x\u00b2 = 182
\nor x\u00b2 – 27x + 182 = 0
\nor x\u00b2 – 14x – 13 + 182 = 0
\nor x (x – 14) – 13 (x – 14) = 0
\nor (x – 14) (x – 13) = 0
\n\u2234 (x – 14) = 0 or, (x – 13) = 0
\n\u21d2 x = 14 or x = 13
\nTherefore, if first number is 14 then second number is
\n27 – x = 27 – 14 = 13
\nAnd, if first number is 13 then second number is
\n27 – x = 27 – 13 = 14<\/p>\n

\"NCERT<\/p>\n

Question 4.
\nFind two consecutive positive integers, sum of whose squares is 365.
\nSolution:
\nLet first consecutive positive integer be x
\n\u2234Second consecutive positive integer be x + 1
\nAccording to question,
\nx\u00b2 + (x + 1)\u00b2 = 365
\nor x\u00b2 + x\u00b2 + 2x + 1 = 365
\nor 2x\u00b2 + 2x + 1 – 365 = 0
\nor 2x\u00b2 + 2x – 364 = 0
\nor 2(x\u00b2 + x – 182) = 0
\nor x\u00b2 + x – 182 = 0
\nor x\u00b2 + 14x – 13x – 182 = 0
\nor x(x + 14) – 13 (x + 14) = 0
\nor (x + 14) (x- 13) = 0
\n\u2234 (x + 14) = 0 or (x – 13) = 0
\n\u21d2 x = – 14 or x = 13
\nBut, we has given that numbers are positive
\n\u2234 x = – 14 is neglected.
\nTherefore, x = 13
\n\u2234 First consecutive positive integer = x = 13
\nand second consecutive positive integer = 27 – x = 14<\/p>\n

Question 5.
\nThe altitude of a right traingle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
\nSolution:
\nLet the base of right angle triangle = x
\n\u2234 The height of right angle triangle = x – 7
\nAnd hypotenuse of right angle triangle = 13 cm
\nBy Pythagoras theorem, we have,
\n(height)\u00b2 + (base)\u00b2 = (hypotenuse)\u00b2
\n\u2234 (x – 7)\u00b2 + x\u00b2 = (13)\u00b2
\nor x\u00b2 – 14x + 49 + x\u00b2 = 169
\nor 2x\u00b2 – 14x + 49 – 169 = 0
\nor 2x\u00b2 -14x – 120 = 0
\nor 2(x\u00b2 – 7x – 60) = 0
\nor x\u00b2 – 7x – 60 = 0
\nor x\u00b2 – 12x + 5x – 60 = 0
\nor x (x – 12) + 5 (x – 12) = 0
\nor (x -12) (x + 5) = 0
\n\u2234 x – 12 = 0 or x + 5=0
\n\u21d2 x = 12 or x = – 5
\nBut, length cannot be in negative
\nSo, x = – 5 is neglected.
\n\u2234 x = 12
\nTherefore, base of the right angle triangle = x
\n\u2234 x = 12 cm
\nand height of the right angle triangle = x – 7 = 12 – 7 = 5 cm<\/p>\n

\"NCERT<\/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 numbr of articles produced on that day. It the total cost of production on that was \u20b9 90. Find the number of articles produced and the cost of each article.
\nSolution:
\nLet total number of articles produced = x
\nCost of production = 2x + 3
\nAccording to question,
\nx (2x + 3) = 90
\nor 2x\u00b2 + 3x = 90
\nor 2x\u00b2 + 3x – 90 = 0
\nor 2x\u00b2 + 15x – 12x – 90 = 0
\nor x(2x + 15) – 6 (2x + 15) = 0
\nor (2x +15) (x – 6) = 0
\n\u2234(2x + 15) = 0 or (x – 6) = 0
\nx = \\(\\frac { -15 }{ 2 }\\) or x = 6
\nBut, number of articles cannot be in negative.
\nSo, x = \\(\\frac { -15 }{ 2 }\\) is neglected.
\n\u2234 Total number of articles produced = x = 6
\nand cost of production = 2x + 3 = 2 x 6 + 3 = \u20b9 15.<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 10 Maths Chapter Ex 4.2 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.2 Question 1. Find the roots of the following quadratic equations by factorisation: (i) x2 – 3x – 10 = 0 (ii) 2×2 …<\/p>\n

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.2<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[2],"tags":[],"yoast_head":"\nNCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.2 - NCERT Solutions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-10-maths-chapter-4-ex-4-2\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.2 - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 10 Maths Chapter Ex 4.2 Questions and Answers are prepared by our highly skilled subject experts. 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NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.2 Question 1. 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