{"id":126855,"date":"2021-09-13T17:12:49","date_gmt":"2021-09-13T11:42:49","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=126855"},"modified":"2021-09-13T17:12:53","modified_gmt":"2021-09-13T11:42:53","slug":"rd-sharma-class-10-solutions-chapter-8-exercise-8-8","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-8-exercise-8-8\/","title":{"rendered":"RD Sharma Class 10 Solutions Chapter 8 Quadratic Equations Exercise 8.8 (Updated for 2021-22)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-126886\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-8-Exercise-8.8.jpg\" alt=\"RD Sharma Class 10 Solutions Chapter 8 Exercise 8.8\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-8-Exercise-8.8.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-8-Exercise-8.8-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 8 Exercise 8.8:&nbsp;<\/strong>Refer to the <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-for-maths\/\"><strong>RD Sharma Class 10 Solutions<\/strong><\/a> if you have any conceptual questions about these. The PDF below has more extensive solutions 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<\/strong><\/a> Exercise 8.8.<\/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-69d568374ce06\" 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\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #000000;color:#000000\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/label><input  type=\"checkbox\" id=\"item-69d568374ce06\"><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 eztoc-visibility-hide-by-default' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-8-exercise-8-8\/#download-rd-sharma-class-10-solutions-chapter-8-exercise-88-free-pdf\" title=\"Download RD Sharma Class 10 Solutions Chapter 8 Exercise 8.8 Free PDF\">Download RD Sharma Class 10 Solutions Chapter 8 Exercise 8.8 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-8-8\/#access-answers-to-rd-sharma-class-10-solutions-chapter-8-exercise-88-important-question-with-answers\" title=\"Access answers to RD Sharma Class 10 Solutions Chapter 8 Exercise 8.8- Important Question with Answers\">Access answers to RD Sharma Class 10 Solutions Chapter 8 Exercise 8.8- 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-8-8\/#faqs-on-rd-sharma-class-10-solutions-chapter-8-exercise-88\" title=\"FAQs on RD Sharma Class 10 Solutions Chapter 8 Exercise 8.8\">FAQs on RD Sharma Class 10 Solutions Chapter 8 Exercise 8.8<\/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-8-8\/#how-is-rd-sharma-class-10-solutions-chapter-8-exercise-88-helpful-for-board-exams\" title=\"How is RD Sharma Class 10 Solutions Chapter 8 Exercise 8.8 helpful for board exams?\">How is RD Sharma Class 10 Solutions Chapter 8 Exercise 8.8 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-8-8\/#is-the-rd-sharma-class-10-solutions-chapter-8-exercise-88-available-on-the-kopykitab-website\" title=\"Is the RD Sharma Class 10 Solutions Chapter 8 Exercise 8.8 available on the Kopykitab website?\">Is the RD Sharma Class 10 Solutions Chapter 8 Exercise 8.8 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-8-8\/#where-can-i-get-rd-sharma-class-10-solutions-chapter-8-exercise-88-free-pdf\" title=\"Where can I get RD Sharma Class 10 Solutions Chapter 8 Exercise 8.8 Free PDF?\">Where can I get RD Sharma Class 10 Solutions Chapter 8 Exercise 8.8 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-88-free-pdf\"><\/span>Download RD Sharma Class 10 Solutions Chapter 8 Exercise 8.8 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-8.8.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-8.8.pdf\">RD Sharma Class 10 Solutions Chapter 8 Exercise 8.8<\/a><\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-rd-sharma-class-10-solutions-chapter-8-exercise-88-important-question-with-answers\"><\/span>Access answers to RD Sharma Class 10 Solutions Chapter 8 Exercise 8.8- Important Question with Answers<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>RD Sharma Class 10 Solutions Chapter 8 Exercise 8.8 Question 1.<br>The speed of a boat in still water is 8 km\/hr. It can go 15 km upstream and 22 km downstream in 5 hours. Find the speed of the stream.<br>Solution:<br>Let the speed of stream = x km\/h<br>and speed of the boat in still water = 8 km\/h<br>Distance covered upstream = 15 km<br>and downstream = 22 km<br>Total time is taken = 5 hours<br>According to the conditions,<br><img src=\"https:\/\/farm2.staticflickr.com\/1757\/41860163124_cd5570e006_o.png\" alt=\"RD Sharma Class 10 Chapter 8 Quadratic Equations \" width=\"238\" height=\"251\"><br><img src=\"https:\/\/farm2.staticflickr.com\/1740\/42530378802_34fd3f1363_o.png\" alt=\"Quadratic Equations Class 10 RD Sharma \" width=\"349\" height=\"222\"><\/p>\n<p>RD Sharma Class 10 Solutions Chapter 8 Exercise 8.8 Question 2.<br>A train, traveling at a uniform speed of 360 km, would have taken 48 minutes less to travel the same distance if its speed were 5 km\/hr more. Find the original speed of the train. [NCERT Exemplar]<br>Solution:<br>Let the original speed of the train = x km\/h<br>Then, the increased speed of the train = (x + 5) km\/h [by given condition]<br>and distance = 360 km<br>According to the question,<br><img src=\"https:\/\/farm2.staticflickr.com\/1757\/41860163644_2c90c9e957_o.png\" alt=\"RD Sharma Class 10 Solutions Quadratic Equations \" width=\"353\" height=\"497\"><br><img src=\"https:\/\/farm2.staticflickr.com\/1757\/42530379212_867c76a5b0_o.png\" alt=\"RD Sharma Class 10 Solutions Chapter 8 Quadratic Equations \" width=\"354\" height=\"229\"><\/p>\n<p>RD Sharma Class 10 Solutions Chapter 8 Exercise 8.8 Question 3.<br>A fast train takes one hour less than a slow train for a journey of 200 km. If the speed of the slow train is 10 km\/hr less than that of the fast train, find the speed of the two trains.<br>Solution:<br>Total journey = 200 km<br>Let the speed of fast train = x km\/hr<br>Then speed of slow train = (x \u2013 10) km\/hr<br>According to the condition,<br><img src=\"https:\/\/farm2.staticflickr.com\/1726\/41860164614_433e19b0c3_o.png\" alt=\"RD Sharma Class 10 Pdf Chapter 8 Quadratic Equations \" width=\"348\" height=\"288\"><br>=&gt; x (x \u2013 50) + 40 (x \u2013 50) = 0<br>=&gt; (x \u2013 50) (x + 40) = 0<br>Either x \u2013 50 = 0, then x = 50<br>or x + 40 = 0, then x = -40 but it is not possible being negative<br>Speed of the fast train = 50 km\/hr<br>and speed of the slow train = 50 \u2013 40 = 10 km\/hr<\/p>\n<p>RD Sharma Class 10 Solutions Chapter 8 Exercise 8.8 Question 4.<br>A passenger train takes one hour less for a journey of 150 km if its speed is increased by 5 km\/hr from its usual speed. Find the usual speed of the train.<br>Solution:<br>Total journey = 150 km<br>Let the usual speed of the train = x km\/hr<br>According to the condition,<br><img src=\"https:\/\/farm2.staticflickr.com\/1738\/41860165424_3ee57029ba_o.png\" alt=\"RD Sharma Class 10 Solutions Quadratic Equations Ex 8.7 \" width=\"346\" height=\"274\"><br>=&gt; x (x + 30) \u2013 25 (x + 30) = 0<br>=&gt; (x + 30) (x \u2013 25) = 0<br>Either x + 30 = 0, then x = -30 but it is not possible being negative<br>or x \u2013 25 = 0, then x = 25<br>Usual speed of the train = 25 km\/hr<\/p>\n<p>Question 5.<br>The time taken by a person to cover 150 km was 2.5 hrs more than the time taken in the return journey. If he returned at a speed of 10 km\/hr more than the speed of going, what was the speed per hour in each direction?<br>Solution:<br>Distance = 150 km<br>Let the speed of the person while going = x km\/hr<br>Then the speed while returning = (x + 10) km\/hr<br>According to the condition,<br><img src=\"https:\/\/farm2.staticflickr.com\/1730\/41860166374_c73a077728_o.png\" alt=\"RD Sharma Solutions Class 10 Chapter 8 Quadratic Equations \" width=\"301\" height=\"174\"><br><img src=\"https:\/\/farm2.staticflickr.com\/1733\/41860165824_a988abe07f_o.png\" alt=\"Learncbse.In Class 10 Chapter 8 Quadratic Equations \" width=\"355\" height=\"137\"><br>=&gt; x (x + 30) \u2013 20 (x + 30) = 0<br>=&gt; (x + 30) (x \u2013 20) = 0<br>Either x + 30 = 0, then x = -30 which is not possible being negative<br>or (x \u2013 20) = 0 then x = 20<br>Usual speed of the man while going = 20 km\/hr<\/p>\n<p>Question 6.<br>A plane left 40 minutes late due to bad weather and to reach its destination, 1600 km away in time, it had to increase its speed by 400 km\/hr from its usual speed. Find the usual speed of the plane.<br>Solution:<br>Distance = 1600 km<br>Let usual speed of the plane = x km\/hr<br>Increased speed = (x + 400) km\/hr<br>According to the condition,<br><img src=\"https:\/\/farm2.staticflickr.com\/1735\/41860166944_d41310e1bd_o.png\" alt=\"Class 10 RD Sharma Solutions Chapter 8 Quadratic Equations \" width=\"366\" height=\"438\"><br>is not possible being negative or x \u2013 800 = 0, then x = 800<br>Usual speed of the plane = 800 km\/hr<\/p>\n<p>Question 7.<br>An airplane takes 1 hour less for a journey of 1200 km if its speed is increased by 100 km\/hr from its usual speed. Find its usual speed.<br>Solution:<br>Distance = 1200 km<br>Let usual speed of the aeroplane = x km\/hr<br>Increased speed = (x + 100) km\/hr<br>According to the condition,<br><img src=\"https:\/\/farm2.staticflickr.com\/1737\/40773981510_4e81ee3fcf_o.png\" alt=\"RD Sharma Class 10 Pdf Free Download Full Book Chapter 8 Quadratic Equations \" width=\"362\" height=\"304\"><br>=&gt; x (x + 400) \u2013 300 (x + 400) = 0<br>=&gt; (x + 400) (x \u2013 300) = 0<br>Either x \u2013 300 = 0, then x = 300<br>or x + 400 = 0, then x = -400 which is not possible being negative<br>Usual speed of the plane = 300 km\/hr<\/p>\n<p>Question 8.<br>A train travels at a certain average speed for a distance of 63 km and then travels a distance of 72 km at an average speed of 6 km\/hr more than the original speed. If it takes 3 hours to complete^total a journey, what is its original average speed? [NCERT Exemplar]<br>Solution:<br>Let its original average speed be x km\/h. Therefore<br>63x&nbsp;+&nbsp;72x+6&nbsp;= 3<br><img src=\"https:\/\/farm2.staticflickr.com\/1755\/41860167604_5dcb5d567b_o.png\" alt=\"RD Sharma Class 10 Solution Chapter 8 Quadratic Equations \" width=\"344\" height=\"316\"><\/p>\n<p>Question 9.<br>A train covers a distance of 90 km at a uniform speed. Had the speed been 15 km\/hr more, it would have taken 30 minutes less for the journey. Find the original speed of the train. [CBSE 2006C]<br>Solution:<br>Distance to be covered = 90 km<br>Let uniform-original speed = x km\/h<br>Increased speed = (x + 15) km\/hr<br>According to the condition,<br><img src=\"https:\/\/farm2.staticflickr.com\/1723\/41860168144_a3dfd2c471_o.png\" alt=\"RD Sharma Class 10 Pdf Ebook Chapter 8 Quadratic Equations \" width=\"354\" height=\"267\"><br>=&gt; x<sup>2<\/sup>&nbsp;+ 60x \u2013 45x \u2013 2700 = 0<br>=&gt; x (x + 60) \u2013 45 (x + 60) = 0<br>=&gt; (x + 60)(x \u2013 45) = 0<br>Either x + 60 = 0, then x = -60 which is not possible being negative<br>or x \u2013 45 = 0, then x = 45<br>Original speed of the train = 45 km\/hr<\/p>\n<p>Question 10.<br>A train travels 360 km at a uniform speed. If the speed had been 5 km\/hr more, it would have taken 1 hour less for the same journey. Find the speed of the train.<br>Solution:<br>Total distance = 360 km<br>Let uniform speed of the train = x km\/hr<br>Increased speed = (x + 5) km\/hr<br>According to the condition,<br><img src=\"https:\/\/farm2.staticflickr.com\/1760\/40773982100_bdeb028e67_o.png\" alt=\"RD Sharma Maths Class 10 Solutions Pdf Free Download Chapter 8 Quadratic Equations \" width=\"353\" height=\"280\"><br>=&gt; x (x + 45) \u2013 40 (x + 45) = 0<br>=&gt; (x + 45) (x \u2013 40) = 0<br>Either x + 45 = 0, then x = -45 but it is not possible being negative<br>or x \u2013 40 = 0, then x = 40<br>Speed of the train = 40 km\/hr<\/p>\n<p>Question 11.<br>An 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 11 km\/hr more than that of the passenger train, find the average speeds of the two trains.<br>Solution:<br>Distance between Mysore and Bangalore = 132 km<br>Let the speed of the passenger train=x km\/hr<br>Then speed of express train = x + 11<br>According to the condition,<br><img src=\"https:\/\/farm2.staticflickr.com\/1744\/40773982330_cb4c88e74b_o.png\" alt=\"RD Sharma Class 10 Book Pdf Free Download Chapter 8 Quadratic Equations \" width=\"286\" height=\"271\"><br>Either x + 44 = 0, then x = -44 but it is not possible being negative<br>or x \u2013 33 = 0, then x = 33<br>Speed of passenger train = 33 km\/hr<br>and speed of express train = 33 + 11 = 44 km\/hr<\/p>\n<p>Question 12.<br>An airplane left 50 minutes later than its scheduled time, and to reach the destination, 1250 km away, in time, it had to increase its speed by 250 km\/hr from its usual speed. Find its usual speed. (CBSE 2010)<br>Solution:<br>Distance = 1250 km<br>Usual speed = x km\/hr<br>Increased speed = (x + 250) km\/hr<br>According to the condition,<br><img src=\"https:\/\/farm2.staticflickr.com\/1751\/40773983100_0a5d5df0c9_o.png\" alt=\"Class 10 RD Sharma Chapter 8 Quadratic Equations \" width=\"326\" height=\"270\"><br><img src=\"https:\/\/farm2.staticflickr.com\/1749\/40773982790_1ec9d014ac_o.png\" alt=\"RD Sharma Maths Class 10 Solutions Chapter 8 Quadratic Equations \" width=\"353\" height=\"294\"><\/p>\n<p>Question 13.<br>While boarding an airplane, a passenger got hurt. The pilot showing promptness and concern, made arrangements to hospitalize the injured and so the plane started late by 30 minutes to reach the destination, 1500 km away, in time, the pilot increased the speed by 100 km\/hr. Find the original speed\/hour of the plane. [CBSE 2013]<br>Solution:<br>Distance = 1500 km<br>Let the original speed of the airplane = x km\/hr<br>Then increased speed = (x + 100) km\/hr<br>According to the condition,<br><img src=\"https:\/\/farm2.staticflickr.com\/1748\/40773983570_e4a554efcf_o.png\" alt=\"RD Sharma 10 Class Solutions Chapter 8 Quadratic Equations \" width=\"347\" height=\"308\"><br>=&gt; x<sup>2<\/sup>&nbsp;+ 100x = 300000<br>=&gt; x<sup>2<\/sup>+ 100x \u2013 300000 = 0<br>=&gt; x<sup>2<\/sup>&nbsp;+ 600x \u2013 500x \u2013 300000 = 0<br>=&gt; x (x + 600) \u2013 500(x + 600) = 0<br>=&gt; (x + 600) (x \u2013 500) = 0<br>Either x + 600 = 0, then x = -600 which is not possible being negative<br>or x \u2013 500 = 0, then x = 500<br>Original speed = 500 km\/hr<\/p>\n<p>Question 14.<br>A motorboat whose speed in still water is 18 km\/h, takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream. [CBSE 2014]<br>Solution:<br>Let the speed of the stream be x km\/hr,<br>Speed of the boat upstream = (18 \u2013 x) km\/hr<br>and Speed of the boat downstream = (18 + x) km\/hr<br>Distance = 24 km<br><img src=\"https:\/\/farm2.staticflickr.com\/1729\/40773983910_c65c1d17a2_o.png\" alt=\"RD Sharma Class 10 Textbook PDF Chapter 8 Quadratic Equations \" width=\"331\" height=\"217\"><br>48x = 324 \u2013 x<sup>2<\/sup><br>x<sup>2<\/sup>&nbsp;+ 48x \u2013 324 = 0<br>x<sup>2<\/sup>&nbsp;+ 54x \u2013 6x \u2013 324 = 0<br>x(x + 54) \u2013 6(x + 54) = 0<br>(x \u2013 6) (x + 54) = 0<br>x \u2013 6 = 0 or x + 54 = 0<br>x = 6 or x = \u2013 54<br>Since speed cannot be negative<br>Speed of stream, x = 6 km\/hr<\/p>\n<p>Question 15.<br>A car moves a distance of 2592 km with uniform speed. The number of hours taken for the journey is one-half the number representing the speed, in km\/ hour. Find the time taken to cover the distance. [CBSE 2017]<br>Solution:<br>Distance = 2592 km<br>Let the speed of the car = x km\/hr<br>and time taken =&nbsp;x2&nbsp;hour<br>We have, Distance = Speed x Time<br>2592 = x x&nbsp;x2<br>=&gt; 2592 =&nbsp;x22<br>=&gt; x<sup>2<\/sup>&nbsp;= 2592 x 2<br>=&gt; x = \u221a5184<br>=&gt; x = 72 km\/hr<br>and thus time taken =&nbsp;x2&nbsp;h =&nbsp;722&nbsp;= 36 hour<\/p>\n<p>We have provided complete details of RD Sharma Class 10 Solutions Chapter 8 Exercise 8.8. 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-88\"><\/span>FAQs on RD Sharma Class 10 Solutions Chapter 8 Exercise 8.8<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-1631525979328\" 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-88-helpful-for-board-exams\"><\/span>How is RD Sharma Class 10 Solutions Chapter 8 Exercise 8.8 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 8.8 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-1631526303328\" 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-88-available-on-the-kopykitab-website\"><\/span>Is the RD Sharma Class 10 Solutions Chapter 8 Exercise 8.8 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 8.8 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-1631526500114\" 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-88-free-pdf\"><\/span>Where can I get RD Sharma Class 10 Solutions Chapter 8 Exercise 8.8 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 8.8 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 8.8:&nbsp;Refer to the RD Sharma Class 10 Solutions if you have any conceptual questions about these. The PDF below has more extensive solutions to the RD Sharma Solutions for Class 10 Maths Chapter 8 Exercise 8.8. Download RD Sharma Class 10 Solutions Chapter 8 Exercise 8.8 Free &#8230; <a title=\"RD Sharma Class 10 Solutions Chapter 8 Quadratic Equations Exercise 8.8 (Updated for 2021-22)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-8-exercise-8-8\/\" aria-label=\"More on RD Sharma Class 10 Solutions Chapter 8 Quadratic Equations Exercise 8.8 (Updated for 2021-22)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":126886,"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\/126855"}],"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=126855"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126855\/revisions"}],"predecessor-version":[{"id":127103,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126855\/revisions\/127103"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/126886"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=126855"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=126855"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=126855"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}