{"id":6034,"date":"2023-08-06T16:30:56","date_gmt":"2023-08-06T16:30:56","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=6034"},"modified":"2023-08-06T16:30:56","modified_gmt":"2023-08-06T16:30:56","slug":"ncert-solutions-class-10-maths-chapter-4-quadratic-equations-ex-4-2","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-4-quadratic-equations-ex-4-2\/","title":{"rendered":"NCERT Solutions Class 10 Maths Chapter 4 Quadratic Equations Ex 4.2"},"content":{"rendered":"<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">NCERT Solutions Class 10 Maths\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Chapter &#8211; 4 (Quadratic Equations)\u00a0<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">The NCERT Solutions in English Language for Class 10 Mathematics <strong>Chapter &#8211; 4 Quadratic Equations <\/strong>Exercise 4.2 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>Chapter : 4 Quadratic Equations<\/strong> <\/span><\/p>\n<ul>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-4-quadratic-equations-ex-4-1\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 4.1<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-4-quadratic-equations-ex-4-3\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 4.3<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-4-quadratic-equations-ex-4-4\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 4.4<\/a><\/span><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><span style=\"font-family: Georgia, Palatino;\"><strong><span style=\"color: #000000;\">Exercise &#8211; 4.2<\/span><\/strong><\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>1. Find the roots of the following quadratic equations by factorisation:<br \/>\n<\/strong><strong>(i)\u00a0<\/strong>x<sup>2<\/sup>\u00a0\u2013 3x\u00a0\u2013 10 = 0<br \/>\n<strong>(ii)<\/strong> 2x<sup>2<\/sup>\u00a0+\u00a0x\u00a0\u2013 6 = 0<br \/>\n<strong>(iii)<\/strong> \u221a2\u00a0x<sup>2<\/sup>\u00a0+ 7x\u00a0+ 5\u221a2\u00a0= 0<br \/>\n<strong>(iv)<\/strong> 2x<sup>2<\/sup>\u00a0\u2013\u00a0x\u00a0+1\/8 = 0<br \/>\n<strong>(v)<\/strong> 100x<sup>2<\/sup>\u00a0\u2013 20x\u00a0+ 1 = 0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solutions &#8211;<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(i)\u00a0<i>x<\/i><sup>2<\/sup>\u00a0\u2013 3<i>x<\/i> \u2013 10 = 0<\/strong><\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 <i>x<\/i><sup>2<\/sup>\u00a0&#8211; 5<i>x<\/i>\u00a0+ 2<i>x<\/i> &#8211; 10 = 0\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><i>\u21d2 x<\/i>(<i>x\u00a0<\/i>&#8211; 5)\u00a0+ 2(<i>x<\/i> &#8211; 5) = 0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 (<i>x<\/i>\u00a0&#8211; 5)(<i>x<\/i> + 2) = 0 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 x = 5 and\u00a0x = &#8211; 2<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Therefore, roots are : &#8211; 2, 5<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(ii) 2<i>x<\/i><sup>2<\/sup>\u00a0+\u00a0<i>x<\/i> \u2013 6 = 0\u00a0<\/strong><\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 2<i>x<\/i><sup>2<\/sup>\u00a0+ 4<i>x<\/i>\u00a0&#8211; 3<i>x<\/i> &#8211; 6 = 0\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 2<i>x<\/i>(<i>x<\/i>\u00a0+\u00a02) &#8211; 3(<i>x<\/i> + 2) = 0\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 (<i>x<\/i>\u00a0+ 2)(2<i>x<\/i> &#8211; 3) = 0\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 2x &#8211; 3 = 0 and x + 2 = 0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 2x = 3 and x = &#8211; 2<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 x = 3\/2 and x = &#8211; 2<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Therefore, roots are : 3\/2, -2<\/span><\/p>\n<div id=\"post27995201739301860783\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(iii) \u221a2\u00a0<i>x<\/i><sup>2<\/sup>\u00a0+ 7<i>x<\/i> + 5\u221a2 = 0\u00a0<\/strong><\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 \u221a2 <i>x<\/i><sup>2\u00a0<\/sup>+ 5<i>x<\/i>\u00a0+ 2<i>x<\/i> + 5\u221a2 = 0\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><i>\u21d2 x<\/i>\u00a0(\u221a2<i>x<\/i>\u00a0+ 5)\u00a0+\u00a0\u221a2(\u221a2<i>x<\/i> + 5) = 0\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 (\u221a2<i>x<\/i>\u00a0+ 5)(<i>x\u00a0<\/i>+ \u221a2) = 0 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 \u221a2x + 5 = 0 and x + \u221a2 = 0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 \u221a2x = &#8211; 5 and x = &#8211; \u221a2<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 x = &#8211; 5\/\u221a2 and x = &#8211; \u221a2<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Therefore, roots are : &#8211; 5\/\u221a2, &#8211; \u221a2<\/span><\/div>\n<div><\/div>\n<div><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(iv) 2<i>x<\/i><sup>2<\/sup>\u00a0\u2013\u00a0<i>x<\/i> + 1\/8 = 0\u00a0<\/strong><\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Multiplying both sides of the equation by 8:<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 2(8) x<sup>2<\/sup>\u00a0&#8211; 8(x) + (8)(1\/ 8) = (0)8<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 16x<sup>2<\/sup>\u00a0&#8211; 8x + 1 = 0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 16x<sup>2<\/sup>\u00a0&#8211; 4x &#8211; 4x + 1 = 0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 4x(4x<sup>\u00a0\u00a0<\/sup>&#8211; 1) -1(4x &#8211; 1) = 0\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 (4x &#8211; 1) (4x &#8211; 1) = 0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 (4x &#8211; 1)<sup>2<\/sup> = 0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 4x &#8211; 1 = 0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 x = 1\/4 and x = 1\/4<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Therefore, roots are : 1\/4, 1\/4<\/span><\/div>\n<div><\/div>\n<div><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(v) 100<i>x<\/i><sup>2<\/sup>\u00a0\u2013 20<i>x<\/i> + 1 = 0\u00a0<\/strong><\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 100<i>x<\/i><sup>2<\/sup>\u00a0\u2013 10<i>x<\/i>\u00a0&#8211; 10<i>x<\/i> + 1 = 0\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 10<i>x<\/i>(10<i>x<\/i>\u00a0&#8211; 1) -1(10<i>x<\/i> &#8211; 1) = 0\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 (10<i>x<\/i>\u00a0&#8211; 1)<sup>2\u00a0<\/sup>= 0 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 10x &#8211; 1 = 0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 x =1\/10 and x = 1\/10<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Therefore, roots are : 1\/10, 1\/10<\/span><\/div>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>2. Solve the problems given in Example 1.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Represent the following situations mathematically:<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(i) <\/strong>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.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solutions &#8211;<\/strong> Let the number of John&#8217;s marbles be\u00a0<i>x<\/i>.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Therefore, number of Jivanti&#8217;s marble = 45 &#8211;\u00a0<i>x<\/i><\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">After losing 5 marbles,<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Number of John&#8217;s marbles =\u00a0<i>x<\/i>\u00a0&#8211; 5<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Number of Jivanti&#8217;s marbles = 45 &#8211;\u00a0<i>x<\/i>\u00a0&#8211; 5 = 40 &#8211;<i>\u00a0x<\/i><\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">It is given that the product of their marbles is 124.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 (<i>x\u00a0<\/i>&#8211; 5)(40 &#8211;\u00a0<i>x<\/i>) = 124<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2\u00a0<i>x<\/i><sup>2<\/sup>\u00a0\u2013 45<i>x<\/i> + 324 = 0<br \/>\n\u21d2\u00a0<i>x<\/i><sup>2<\/sup>\u00a0\u2013 36<i>x<\/i>\u00a0&#8211; 9<i>x<\/i> + 324 = 0<br \/>\n\u21d2\u00a0<i>x<\/i>(<i>x<\/i>\u00a0&#8211; 36) -9(<i>x<\/i> &#8211; 36) = 0<br \/>\n\u21d2 (<i>x<\/i>\u00a0&#8211; 36)(<i>x<\/i> &#8211; 9) = 0<br \/>\nEither\u00a0<i>x<\/i>\u00a0&#8211; 36 = 0 or\u00a0<i>x<\/i> &#8211; 9 = 0<br \/>\n\u21d2\u00a0<i>x<\/i>\u00a0= 36 or\u00a0<i>x<\/i>\u00a0= 9<br \/>\nIf the number of John&#8217;s marbles = 36,<br \/>\nThen, number of Jivanti&#8217;s marbles = 45 &#8211; 36 = 9<br \/>\nIf number of John&#8217;s marbles = 9,<br \/>\nThen, number of Jivanti&#8217;s marbles = 45 &#8211; 9 = 36<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(ii) <\/strong>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\u00a0 750. We would like to find out the number of toys produced on that day.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solutions &#8211; <\/strong>Let the number of toys produced be\u00a0<i>x<\/i>.<br \/>\n\u2234 Cost of production of each toy = Rs (55 &#8211;\u00a0<i>x<\/i>)<br \/>\nIt is given that, total production of the toys = Rs 750<br \/>\n\u2234\u00a0<i>x<\/i>(55 &#8211;\u00a0<i>x<\/i>) = 750<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2\u00a0<i>x<\/i><sup>2<\/sup>\u00a0\u2013\u00a055<i>x<\/i>\u00a0+ 750 = 0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2\u00a0<i>x<\/i><sup>2<\/sup>\u00a0\u2013 25<i>x<\/i>\u00a0&#8211;\u00a030<i>x<\/i>\u00a0+ 750 = 0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2<i>\u00a0x<\/i>(<i>x<\/i>\u00a0&#8211; 25) -30(<i>x<\/i>\u00a0&#8211; 25) = 0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 (<i>x<\/i>\u00a0&#8211; 25)(<i>x<\/i>\u00a0&#8211; 30) = 0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Either,\u00a0<i>x<\/i>\u00a0-25 = 0 or\u00a0<i>x<\/i> &#8211; 30 = 0<br \/>\n\u21d2\u00a0<i>x<\/i>\u00a0= 25 or\u00a0<i>x<\/i>\u00a0= 30<br \/>\nHence, the number of toys will be either 25 or 30.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>3. Find two numbers whose sum is 27 and product is 182.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211; <\/strong>Let the first number be\u00a0<i>x<\/i>\u00a0and the second number is 27 &#8211;\u00a0<i>x<\/i>.<br \/>\nTherefore, their product =\u00a0<i>x<\/i>\u00a0(27 &#8211;\u00a0<i>x<\/i>)<br \/>\nIt is given that the product of these numbers is 182.<br \/>\nTherefore,\u00a0<i>x<\/i>(27 &#8211;\u00a0<i>x<\/i>) = 182<br \/>\n\u21d2\u00a0<i>x<\/i><sup>2<\/sup>\u00a0\u2013 27<i>x<\/i>\u00a0&#8211; 182 = 0<br \/>\n\u21d2\u00a0<i>x<\/i><sup>2<\/sup>\u00a0\u2013 13<i>x<\/i>\u00a0&#8211; 14<i>x<\/i>\u00a0+ 182 = 0<br \/>\n\u21d2\u00a0<i>x<\/i>(<i>x<\/i>\u00a0&#8211; 13) -14(<i>x<\/i>\u00a0&#8211; 13) = 0<br \/>\n\u21d2 (<i>x<\/i>\u00a0&#8211; 13)(<i>x<\/i>\u00a0-14) = 0<br \/>\nEither\u00a0<i>x<\/i>\u00a0= -13\u00a0= 0 or\u00a0<i>x<\/i>\u00a0&#8211; 14 = 0<br \/>\n\u21d2\u00a0<i>x<\/i>\u00a0= 13 or\u00a0<i>x<\/i>\u00a0= 14<br \/>\nIf first number = 13, then<br \/>\nOther number = 27 &#8211; 13 = 14<br \/>\nIf first number = 14, then<br \/>\nOther number = 27 &#8211; 14 = 13<br \/>\nTherefore, the numbers are 13 and 14.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>4. Find two consecutive positive integers, sum of whose squares is 365.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211; <\/strong>Let the consecutive positive integers be\u00a0<i>x<\/i>\u00a0and\u00a0<i>x<\/i>\u00a0+ 1.<br \/>\nTherefore,\u00a0<i>x<\/i><sup>2<\/sup>\u00a0+ (<i>x<\/i>\u00a0+ 1)<sup>2<\/sup>\u00a0= 365<br \/>\n\u21d2\u00a0<i>x<\/i><sup>2\u00a0<\/sup>+\u00a0<i>x<\/i><sup>2\u00a0<\/sup>+ 1 + 2<i>x<\/i>\u00a0= 365<br \/>\n\u21d2 2<i>x<\/i><sup>2<\/sup>\u00a0+ 2x &#8211; 364 = 0<br \/>\n\u21d2\u00a0<i>x<\/i><sup>2\u00a0<\/sup>+\u00a0<i>x\u00a0<\/i>&#8211; 182<i>\u00a0<\/i>= 0<br \/>\n\u21d2\u00a0<i>x<\/i><sup>2\u00a0<\/sup>+ 14<i>x<\/i>\u00a0&#8211; 13<i>x<\/i>\u00a0&#8211; 182 = 0<br \/>\n\u21d2\u00a0<i>x<\/i>(<i>x<\/i>\u00a0+\u00a014) -13(<i>x<\/i>\u00a0+\u00a014) = 0<br \/>\n\u21d2 (<i>x<\/i>\u00a0+ 14)(<i>x<\/i>\u00a0&#8211; 13) = 0<br \/>\nEither\u00a0<i>x<\/i>\u00a0+ 14 = 0 or\u00a0<i>x<\/i>\u00a0&#8211; 13 = 0,<br \/>\n\u21d2\u00a0<i>x<\/i>\u00a0= &#8211; 14 or\u00a0<i>x<\/i>\u00a0= 13<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Since the integers are positive,\u00a0<i>x<\/i>\u00a0can only be 13.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234\u00a0<i>x<\/i>\u00a0+ 1 = 13 + 1 = 14<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Therefore, two consecutive positive integers will be 13 and 14.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;\u00a0<\/strong>Let the base of the right triangle be\u00a0<i>x<\/i>\u00a0cm.<br \/>\nIts altitude = (<i>x<\/i>\u00a0&#8211; 7) cm<br \/>\nFrom Pythagoras theorem, we have<br \/>\nBase<sup>2<\/sup>\u00a0+ Altitude<sup>2<\/sup>\u00a0= Hypotenuse<sup>2<\/sup><br \/>\n\u2234\u00a0<i>x<\/i><sup>2\u00a0<\/sup>+ (<i>x<\/i>\u00a0&#8211; 7)<sup>2<\/sup>\u00a0= 132<br \/>\n\u21d2\u00a0<i>x<\/i><sup>2\u00a0<\/sup>+\u00a0<i>x<\/i><sup>2\u00a0<\/sup>+ 49 &#8211; 14<i>x<\/i>\u00a0= 169<br \/>\n\u21d2 2<i>x<\/i><sup>2\u00a0<\/sup>&#8211; 14<i>x<\/i>\u00a0&#8211; 120 = 0<br \/>\n\u21d2\u00a0<i>x<\/i><sup>2\u00a0<\/sup>&#8211; 7<i>x<\/i>\u00a0&#8211; 60 = 0<br \/>\n\u21d2\u00a0<i>x<\/i><sup>2\u00a0<\/sup>&#8211; 12<i>x\u00a0<\/i>+ 5<i>x<\/i>\u00a0&#8211; 60 = 0<br \/>\n\u21d2\u00a0<i>x<\/i>(<i>x<\/i>\u00a0&#8211; 12) + 5(<i>x\u00a0<\/i>&#8211; 12) = 0<br \/>\n\u21d2 (<i>x<\/i>\u00a0&#8211; 12)(<i>x<\/i>\u00a0+ 5) = 0<br \/>\nEither\u00a0<i>x<\/i>\u00a0&#8211; 12 = 0 or\u00a0<i>x<\/i>\u00a0+ 5 = 0,<br \/>\n\u21d2\u00a0<i>x<\/i>\u00a0= 12 or\u00a0<i>x<\/i>\u00a0= &#8211; 5<br \/>\nSince sides are positive,\u00a0<i>x<\/i>\u00a0can only be 12.<br \/>\nTherefore, the base of the given triangle is 12 cm and the altitude of this triangle will be (12 &#8211; 7) cm = 5 cm.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>6. A 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 Rs. 90, find the number of articles produced and the cost of each article.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;\u00a0 <\/strong>Let the number of articles produced be\u00a0<i>x<\/i>.<br \/>\nTherefore, cost of production of each article = Rs (2<i>x<\/i>\u00a0+ 3)<br \/>\nIt is given that the total production is Rs 90.\u2234\u00a0<i>x<\/i>(2<i>x<\/i>\u00a0+ 3) = 0<br \/>\n\u21d2 2<i>x<\/i><sup>2\u00a0<\/sup>+ 3<i>x<\/i>\u00a0&#8211; 90 = 0<br \/>\n\u21d2 2<i>x<\/i><sup>2\u00a0<\/sup>+ 15<i>x<\/i>\u00a0-12<i>x<\/i>\u00a0&#8211; 90 = 0<br \/>\n\u21d2\u00a0<i>x<\/i>(2<i>x<\/i>\u00a0+ 15) -6(2<i>x<\/i>\u00a0+ 15) = 0<br \/>\n\u21d2 (2<i>x<\/i>\u00a0+ 15)(<i>x<\/i>\u00a0&#8211; 6) = 0<br \/>\nEither 2<i>x<\/i>\u00a0+ 15 = 0 or\u00a0<i>x<\/i>\u00a0&#8211; 6 = 0<br \/>\n\u21d2\u00a0<i>x<\/i>\u00a0= -15\/2 or\u00a0<i>x<\/i> = 6 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">As the number of articles produced can only be a positive integer, therefore, x can only be 6. <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Hence, number of articles produced = 6 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Cost of each article = 2 \u00d7 6 + 3 = Rs 15.<\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"font-family: Georgia, Palatino;\"><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths\/\"><span style=\"color: #0000ff;\"><b><i>Go Back To Chapters<\/i><\/b><\/span><\/a><\/span><\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 10 Maths\u00a0 Chapter &#8211; 4 (Quadratic Equations)\u00a0 The NCERT Solutions in English Language for Class 10 Mathematics Chapter &#8211; 4 Quadratic Equations Exercise 4.2 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter : 4 Quadratic Equations NCERT Class 10 Maths Solution Ex &#8211; 4.1 [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1043],"tags":[1045,1069,1071,1044,1049,1048],"class_list":["post-6034","post","type-post","status-publish","format-standard","hentry","category-class-10-maths","tag-class-10-ncert-mathematics-solutions","tag-ncert-class-10-mathematics-chapter-4-quadratic-equations-solutions","tag-ncert-class-10-mathematics-exercise-4-2-solutions","tag-ncert-class-10-mathematics-solutions-class-10-ncert-solutions","tag-ncert-solutions-class-10-mathematics","tag-ncert-solutions-class-10-maths"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.5) - 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