{"id":126850,"date":"2021-09-13T17:13:01","date_gmt":"2021-09-13T11:43:01","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=126850"},"modified":"2021-09-13T17:13:04","modified_gmt":"2021-09-13T11:43:04","slug":"rd-sharma-class-10-solutions-chapter-8-exercise-8-13","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-8-exercise-8-13\/","title":{"rendered":"RD Sharma Class 10 Solutions Chapter 8 Quadratic Equations Exercise 8.13 (Updated for 2021-22)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-126891\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-8-Exercise-8.13.jpg\" alt=\"RD Sharma Class 10 Solutions Chapter 8 Exercise 8.13\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-8-Exercise-8.13.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-8-Exercise-8.13-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 8 Exercise 8.13:&nbsp;<\/strong>The main focus of this exercise is on various applications of quadratic equations. The <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-for-maths\/\"><strong>RD Sharma Class 10 Solutions<\/strong><\/a>&nbsp;created by Kopykitab experts is an excellent resource for students who want to clear up their doubts and focus on their weaker areas. Students can also refer to the <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-8-quadratic-equations\/\"><strong>RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations<\/strong><\/a> Exercise 8.13 PDF for any questions about the exercise problems.<\/p>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_47_1 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"ez-toc-toggle-icon-1\"><label for=\"item-69d5683a67e4f\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;direction:ltr;\"><svg style=\"fill: #000000;color:#000000\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" 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href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-8-exercise-8-13\/#download-rd-sharma-class-10-solutions-chapter-8-exercise-813-free-pdf\" title=\"Download RD Sharma Class 10 Solutions Chapter 8 Exercise 8.13 Free PDF\">Download RD Sharma Class 10 Solutions Chapter 8 Exercise 8.13 Free PDF<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-8-exercise-8-13\/#access-answers-to-rd-sharma-solutions-class-10-maths-chapter-8-exercise-813-important-question-with-answers\" title=\"Access answers to RD Sharma Solutions Class 10 Maths Chapter 8 Exercise 8.13- Important Question with Answers\">Access answers to RD Sharma Solutions Class 10 Maths Chapter 8 Exercise 8.13- Important Question with Answers<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-8-exercise-8-13\/#faqs-on-rd-sharma-class-10-solutions-chapter-8-exercise-813\" title=\"FAQs on RD Sharma Class 10 Solutions Chapter 8 Exercise 8.13\">FAQs on RD Sharma Class 10 Solutions Chapter 8 Exercise 8.13<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-8-exercise-8-13\/#how-is-rd-sharma-class-10-solutions-chapter-8-exercise-813-helpful-for-board-exams\" title=\"How is RD Sharma Class 10 Solutions Chapter 8 Exercise 8.13 helpful for board exams?\">How is RD Sharma Class 10 Solutions Chapter 8 Exercise 8.13 helpful for board exams?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-8-exercise-8-13\/#is-the-rd-sharma-class-10-solutions-chapter-8-exercise-813-available-on-the-kopykitab-website\" title=\"Is the RD Sharma Class 10 Solutions Chapter 8 Exercise 8.13 available on the Kopykitab website?\">Is the RD Sharma Class 10 Solutions Chapter 8 Exercise 8.13 available on the Kopykitab website?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-8-exercise-8-13\/#where-can-i-get-rd-sharma-class-10-solutions-chapter-8-exercise-813-free-pdf\" title=\"Where can I get RD Sharma Class 10 Solutions Chapter 8 Exercise 8.13 Free PDF?\">Where can I get RD Sharma Class 10 Solutions Chapter 8 Exercise 8.13 Free PDF?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-10-solutions-chapter-8-exercise-813-free-pdf\"><\/span>Download RD Sharma Class 10 Solutions Chapter 8 Exercise 8.13 Free PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<div id=\"example1\" style=\"text-align: justify;\">&nbsp;<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 800px;}<br \/>\n.pdfobject { border: 1px solid #666; }<br \/>\n<\/style><\/p>\n<p style=\"text-align: justify;\"><script src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solution-Class-10-Chapter-8-Exercise-8.13.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solution-Class-10-Chapter-8-Exercise-8.13.pdf\">RD Sharma Class 10 Solutions Chapter 8 Exercise 8.13<\/a><\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-rd-sharma-solutions-class-10-maths-chapter-8-exercise-813-important-question-with-answers\"><\/span>Access answers to RD Sharma Solutions Class 10 Maths Chapter 8 Exercise 8.13- Important Question with Answers<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br>A piece of cloth costs Rs. 35. If the piece were 4 m longer and each meter costs Re. one less, the cost would remain unchanged. How long is the piece?<br>Solution:<br>Let the length of a piece of cloth = x m<br>Total cost = Rs. 35<br>Cost of 1 m cloth = Rs.&nbsp;35x<br>According to the condition,<br><img src=\"https:\/\/farm2.staticflickr.com\/1721\/28739458808_13f2745af1_o.png\" alt=\"RD Sharma Class 10 Chapter 8 Quadratic Equations \" width=\"353\" height=\"364\"><br>\u21d2 x (x + 14) \u2013 10 (x + 14) = 0<br>\u21d2 (x + 14) (x \u2013 10) = 0<br>Either x + 14 = 0, then x = \u2013 14 which is not possible being negative<br>or x \u2013 10 = 0, then x = 10<br>Length of piece of cloth = 10 m<\/p>\n<p>Question 2.<br>Some students planned a picnic. The budget for food was Rs. 480. But eight of these failed to go and thus the cost of food for each member increased by Rs. 10. How many students attended the picnic?<br>Solution:<br>Let the number of students = x<br>and total budget = Rs. 480<br>Share of each student = Rs. 480x<br>According to the condition,<br><img src=\"https:\/\/farm2.staticflickr.com\/1755\/41712469485_aa557f2b58_o.png\" alt=\"Quadratic Equations Class 10 RD Sharma \" width=\"349\" height=\"286\"><br>\u21d2 x (x + 16) \u2013 24 (x + 16) = 0<br>\u21d2 (x + 16) (x \u2013 24) = 0<br>Either x + 16 = 0, then x = -16 which is not possible being negative<br>or x \u2013 24 = 0, then x = 24<br>Number of students = 24<br>and number of students who attended the picnic = 24 \u2013 8 = 16<\/p>\n<p>Question 3.<br>A dealer sells an article for Rs. 24 and gains as much percent as the cost price of the article. Find the cost price of the article.<br>Solution:<br>Let the cost price ofjfie the article = Rs. x<br>Selling price = Rs. 24<br>Gain = x %<br>According to the condition,<br><img src=\"https:\/\/farm2.staticflickr.com\/1754\/41712469715_d17f470f61_o.png\" alt=\"RD Sharma Class 10 Solutions Quadratic Equations \" width=\"247\" height=\"167\"><br><img src=\"https:\/\/farm2.staticflickr.com\/1721\/41712469535_7e83c67282_o.png\" alt=\"RD Sharma Class 10 Solutions Quadratic Equations Ex 8.13 \" width=\"357\" height=\"249\"><\/p>\n<p>Question 4.<br>Out of a group of swans,&nbsp;72&nbsp;times the square root of the total number are playing on the share of a pond. The two remaining ones are swinging in water. Find the total number of swans.<br>Solution:<br>Let the total number of swans = x<br>According to the condition,<br><img src=\"https:\/\/farm2.staticflickr.com\/1736\/41712469835_45ffb37f1d_o.png\" alt=\"RD Sharma Class 10 Solutions Chapter 8 Quadratic Equations \" width=\"350\" height=\"525\"><br>Number of total swans = 16<\/p>\n<p>Question 5.<br>If the list price of a toy is reduced by Rs. 2, a person can buy 2 toy mope for Rs. 360. Find the original price of the toy. (C.B.S.E. 2002C)<br>Solution:<br>The list price of the toy = Rs. x<br>Total amount = Rs. 360<br>Reduced price of each toy = (x \u2013 2)<br>According to the condition,<br><img src=\"https:\/\/farm2.staticflickr.com\/1750\/41712470185_bcfcbedd69_o.png\" alt=\"RD Sharma Class 10 Pdf Chapter 8 Quadratic Equations \" width=\"357\" height=\"318\"><br>\u21d2 x (x \u2013 20) + 18 (x \u2013 20) = 0<br>\u21d2 (x \u2013 20) (x + 18) = 0<br>Either x \u2013 20 = 0, then x = 20<br>or x + 18 = 0, then x = -18 which is not possible being negative<br>Price of each toy = Rs. 20<\/p>\n<p>Question 6.<br>Rs. 9000 were divided equally among a certain number of persons. Had there been 20 more persons, each would have got Rs. 160 less. Find the original number of persons.<br>Solution:<br>Total amount = Rs. 9000<br>Let number of persons = x<br>Then each share = Rs.&nbsp;9000x<br>Increased persons = (x + 20)<br>According to the condition,<br><img src=\"https:\/\/farm2.staticflickr.com\/1747\/41712470425_1c79ed1aa9_o.png\" alt=\"RD Sharma Solutions Class 10 Chapter 8 Quadratic Equations \" width=\"349\" height=\"371\"><br>Either x + 45 = 0, then x = -45 which is not possible being negative<br>or x \u2013 25 = 0, then x = 25<br>Number of persons = 25<\/p>\n<p>Question 7.<br>Some students planned a picnic. The budget for food was Rs. 500. But 5 of them failed to go and thus the cost of food for each number increased by Rs. 5. How many students attended the picnic? (C.B.S.E. 1999)<br>Solution:<br>Let the number of students = x<br>Total budget = Rs. 500<br>Share of each student = Rs.&nbsp;500x<br>No. of students failed to go = 5<br>According to the given condition,<br><img src=\"https:\/\/farm2.staticflickr.com\/1750\/28739459998_239e4dc356_o.png\" alt=\"Learncbse.In Class 10 Chapter 8 Quadratic Equations \" width=\"283\" height=\"179\"><br><img src=\"https:\/\/farm2.staticflickr.com\/1746\/41712470705_d7c6186e05_o.png\" alt=\"Class 10 RD Sharma Solutions Chapter 8 Quadratic Equations \" width=\"355\" height=\"288\"><\/p>\n<p>Question 8.<br>A pole has to be erected at a point on the boundary of a. circular park of diameter 13 meters in such a way that the difference, of its distances from two diametrically opposite fixed gates A and B on the boundary, is 7 meters. Is it possible to do so? If yes, at what distances from the two gates Should the pole be erected?<br>Solution:<br>In a circle, AB is the diameters and AB = 13 m<br><img src=\"https:\/\/farm2.staticflickr.com\/1722\/41712470935_8455c22d0b_o.png\" alt=\"RD Sharma Class 10 Pdf Free Download Full Book Chapter 8 Quadratic Equations \" width=\"356\" height=\"453\"><br>Either x + 12 = 0, then x = -12 which is not possible being negative<br>or x \u2013 5 = 0, then x = 5<br>P is at a distance of 5 m from B and 5 + 7 = 12 m from A<\/p>\n<p>Question 9.<br>In a class test, the sum of the marks obtained by P in Mathematics and Science is 28. Had he got 3 marks more in Mathematics and 4 marks less in Science. The product of their marks, would have been 180. Find his marks in the two subjects. (C.B.S.E. 2008)<br>Solution:<br>Sum of marks in Mathematics and Science = 28<br>Let marks in Math = x<br>Then marks in Science = 28 \u2013 x<br>According to the condition,<br>(x + 3) (28 \u2013 x \u2013 4) = 180<br>\u21d2 (x + 3) (24 \u2013 x) = 180<br>\u21d2 24x \u2013 x\u00b2 + 72 \u2013 3x = 180<br>\u21d2 21x \u2013 x\u00b2 + 72 \u2013 180 = 0<br>\u21d2 \u2013 x\u00b2 + 21x \u2013 108 = 0<br>\u21d2 x\u00b2 \u2013 21x + 108 = 0<br>\u21d2 x\u00b2 \u2013 9x \u2013 12x + 108 = 0<br>\u21d2 x (x \u2013 9) \u2013 12 (x \u2013 9) \u2013 0<br>\u21d2 (x \u2013 9)(x \u2013 12) = 0<br>Either x \u2013 9 = 0, then x = 9<br>or x \u2013 12 = 0, then x = 12<br>(i) If x = 9, then Marks in Maths = 9 and marks in Science = 28 \u2013 9 = 19<br>(ii) If x = 12, then Marks in Maths = 12 and marks in Science = 28 \u2013 12 = 16<\/p>\n<p>Question 10.<br>In 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 her marks would have been 210. Find her marks in two subjects. [NCERT]<br>Solution:<br>Sum of marks in Mathematics and English = 30<br>Let marks obtained in Mathematics = x<br>Then in English = 30 \u2013 x<br>According to the condition,<br>(x + 2) (30 \u2013 x \u2013 3) = 210<br>\u21d2 (x + 2) (27 \u2013 x) = 210<br>\u21d2 27x \u2013 x\u00b2 + 54 \u2013 2x \u2013 210 = 0<br>\u21d2 \u2013 x\u00b2 + 25x \u2013 156 = 0<br>\u21d2 x\u00b2 \u2013 25x + 156 = 0<br>\u21d2 x\u00b2 \u2013 12x \u2013 13x +156 = 0<br>\u21d2 x (x \u2013 12) \u2013 13 (x \u2013 12) = 0<br>\u21d2 (x \u2013 12) (x \u2013 13) = 0<br>Either x \u2013 12 = 0, then x = 12<br>or x \u2013 13 = 0, then x = 13<br>(i) If x = 12, then<br>Marks in Maths =12 and in English = 30 \u2013 12 = 18<br>(ii) If x = 13, then<br>Marks in Maths = 13 and in English = 30 \u2013 13 = 17<\/p>\n<p>Question 11.<br>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, And the number of articles produced and the cost of each article. [NCERT]<br>Solution:<br>Total cost = Rs. 90<br>Let number of articles = x<br>Then price of each articles = 2x + 3<br>x (2x + 3) = 90<br>\u21d2 2x\u00b2 + 3x \u2013 90 = 0<br>\u21d2 2x\u00b2 \u2013 12x + 15x \u2013 90 = 0<br>\u21d2 2x (x \u2013 6) + 15 (x \u2013 6) = 0<br>\u21d2 (x \u2013 6) (2x + 15) = 0<br>Either x \u2013 6 = 0, then x = 6<br>or 2x + 15 = 0 then 2x = -15 \u21d2 x =&nbsp;\u2212152&nbsp;which is not possible being negative<br>x = 6<br>Number of articles = 6<br>and price of each article = 2x + 3 = 2 x 6 + 3 = 12 + 3 = 15<\/p>\n<p>Question 12.<br>At t minutes past 2 pm, the time needed by the minute&#8217;s hand and a clock to show 3 pm was found to be 3 minutes less than t24&nbsp;minutes. Find t.<br>Solution:<br>We know that the time between 2 pm to 3 pm = 1 h = 60 minutes<br>Given that, at t minutes past 2 pm, the time needed by the min. hand of a clock to show 3 pm was found to be 3 min. less than&nbsp;t24&nbsp;min.<br>i.e., t = (t24&nbsp;\u2013 3) = 60<br>\u21d2 4t + t\u00b2 \u2013 12 = 240<br>\u21d2 t\u00b2 + 4t \u2013 252 = 0<br>\u21d2 t\u00b2 + 18t \u2013 14t \u2013 252 = 0 [by splitting the middle term]<br>\u21d2 t (t + 18) \u2013 14 (t + 18) = 0<br>\u21d2 (t + 18) (t \u2013 14) = 0 [since, time cannot be negative, so t \u2260 -18]<br>t = 14 min.<br>Hence, the required value of t is 14 minutes.<\/p>\n<p>We have provided complete details of RD Sharma Class 10 Solutions Chapter 8 Exercise 8.13. If you have any queries related to <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\"><strong>CBSE<\/strong><\/a>&nbsp;Class 10, feel free to ask us in the comment section below.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-class-10-solutions-chapter-8-exercise-813\"><\/span>FAQs on RD Sharma Class 10 Solutions Chapter 8 Exercise 8.13<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1631526028637\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-is-rd-sharma-class-10-solutions-chapter-8-exercise-813-helpful-for-board-exams\"><\/span>How is RD Sharma Class 10 Solutions Chapter 8 Exercise 8.13 helpful for board exams?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>For self-evaluation, RD Sharma Class 10 Solutions Chapter 8 Exercise 8.13 provides solutions with thorough descriptions as per term limits specified by the Board. Students will gain valuable experience solving these problems, allowing them to complete the assignment on time.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631526335890\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"is-the-rd-sharma-class-10-solutions-chapter-8-exercise-813-available-on-the-kopykitab-website\"><\/span>Is the RD Sharma Class 10 Solutions Chapter 8 Exercise 8.13 available on the Kopykitab website?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, the PDFs of RD Sharma Class 10 Solutions Chapter 8 Exercise 8.13 are available. These solutions are created in a unique method by Kopykitab\u2019s expert faculty.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631526528887\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"where-can-i-get-rd-sharma-class-10-solutions-chapter-8-exercise-813-free-pdf\"><\/span>Where can I get RD Sharma Class 10 Solutions Chapter 8 Exercise 8.13 Free PDF?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can get RD Sharma Class 10 Solutions Chapter 8 Exercise 8.13 Free PDF from the above article.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Class 10 Solutions Chapter 8 Exercise 8.13:&nbsp;The main focus of this exercise is on various applications of quadratic equations. The RD Sharma Class 10 Solutions&nbsp;created by Kopykitab experts is an excellent resource for students who want to clear up their doubts and focus on their weaker areas. Students can also refer to the &#8230; <a title=\"RD Sharma Class 10 Solutions Chapter 8 Quadratic Equations Exercise 8.13 (Updated for 2021-22)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-8-exercise-8-13\/\" aria-label=\"More on RD Sharma Class 10 Solutions Chapter 8 Quadratic Equations Exercise 8.13 (Updated for 2021-22)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":126891,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411,2985,73410],"tags":[3243,9206,73520,4388],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126850"}],"collection":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/users\/238"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=126850"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126850\/revisions"}],"predecessor-version":[{"id":127108,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126850\/revisions\/127108"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/126891"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=126850"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=126850"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=126850"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}