{"id":125508,"date":"2023-09-08T19:27:00","date_gmt":"2023-09-08T13:57:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=125508"},"modified":"2023-11-08T10:07:21","modified_gmt":"2023-11-08T04:37:21","slug":"rd-sharma-class-10-solutions-chapter-3-exercise-3-6","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-3-exercise-3-6\/","title":{"rendered":"RD Sharma Class 10 Solutions Chapter 3 Exercise 3.6 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-125557\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-3-Exercise3.6.jpg\" alt=\"RD Sharma Class 10 Solutions Chapter 3 Exercise 3.6\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-3-Exercise3.6.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-3-Exercise3.6-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 3 Exercise 3.6:\u00a0<\/strong>The <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-for-maths\/\"><strong>RD Sharma Class 10 Solutions<\/strong><\/a> is an essential tool for students who want to improve their conceptual knowledge and exam confidence. For any assistance with this problem, students can refer to the <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-3-pair-of-linear-equations-in-two-variables\/\"><strong>RD Sharma Solutions for Class 10 Maths Chapter 3<\/strong><\/a> Pair of Linear Equations in Two Variables Exercise 3.6 PDF provided below.<\/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-69d7e14b4d4d3\" 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\" 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href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-3-exercise-3-6\/#download-rd-sharma-class-10-solutions-chapter-3-exercise-36-free-pdf\" title=\"Download RD Sharma Class 10 Solutions Chapter 3 Exercise 3.6 Free PDF\">Download RD Sharma Class 10 Solutions Chapter 3 Exercise 3.6 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-3-exercise-3-6\/#access-answers-to-rd-sharma-solutions-class-10-maths-chapter-3-exercise-36-important-question-with-answers\" title=\"Access answers to RD Sharma Solutions Class 10 Maths Chapter 3 Exercise 3.6- Important Question with Answers\">Access answers to RD Sharma Solutions Class 10 Maths Chapter 3 Exercise 3.6- 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\" 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href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-3-exercise-3-6\/#is-it-required-to-practice-all-of-the-questions-in-chapter-3-exercise-36-of-rd-sharma-solutions-for-class-10-maths\" title=\"Is it required to practice all of the questions in Chapter 3 Exercise 3.6 of RD Sharma Solutions for Class 10 Maths?\">Is it required to practice all of the questions in Chapter 3 Exercise 3.6 of RD Sharma Solutions for Class 10 Maths?<\/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-3-exercise-3-6\/#what-are-the-benefits-of-using-rd-sharma-class-10-solutions-chapter-3-exercise-36\" title=\"What are the benefits of using RD Sharma Class 10 Solutions Chapter 3 Exercise 3.6?\">What are the benefits of using RD Sharma Class 10 Solutions Chapter 3 Exercise 3.6?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-10-solutions-chapter-3-exercise-36-free-pdf\"><\/span>Download RD Sharma Class 10 Solutions Chapter 3 Exercise 3.6 Free PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/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-3-Exercise-3.6.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solution-Class-10-Chapter-3-Exercise-3.6.pdf\">RD Sharma Class 10 Solutions Chapter 3 Exercise 3.6<\/a><\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-rd-sharma-solutions-class-10-maths-chapter-3-exercise-36-important-question-with-answers\"><\/span>Access answers to RD Sharma Solutions Class 10 Maths Chapter 3 Exercise 3.6- Important Question with Answers<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>5 pens and 6 pencils together cost\u00a0<\/strong>\u20b9<strong>\u00a09, and 3 pens and 2 pencils cost\u00a0<\/strong>\u20b9<strong>\u00a05. Find the cost of 1 pen and 1 pencil.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s assume the cost of a pen and pencil is \u20b9 x and \u20b9 y, respectively.<\/p>\n<p>Then, forming equations according to the question,<\/p>\n<p>5x + 6y = 9 \u2026 (i)<\/p>\n<p>3x + 2y = 5 \u2026 (ii)<\/p>\n<p>On multiplying equation (i) by 2 and equation (ii) by 6, we get<\/p>\n<p>10x + 12y = 18 \u2026 (iii)<\/p>\n<p>18x + 12y = 30 \u2026 (iv)<\/p>\n<p>Now by subtracting equation (iii) from equation (iv), we get<\/p>\n<p>18x \u2013 10x + 12y \u2013 12y = 30 \u2013 18<\/p>\n<p>8x = 12<\/p>\n<p>x = 3\/2 = 1.5<\/p>\n<p>Putting x = 1.5 in equation (i), we find y.<\/p>\n<p>5(1.5) + 6y = 9<\/p>\n<p>6y = 9 \u2013 7.5<\/p>\n<p>y = (1.5)\/ 6 = 0.25<\/p>\n<p>Therefore, the cost of one pen = \u20b9 1.50, and so the cost of one pencil = \u20b9 0.25<\/p>\n<p><strong>2. 7 audio cassettes and 3 videocassettes cost\u00a0<\/strong>\u20b9<strong>\u00a01110, while 5 audio cassettes and 4 videocassettes cost\u00a0<\/strong>\u20b9\u00a0<strong>1350. Find the cost of audio cassettes and a video cassette.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s assume the cost of an audio cassette, and that of a video cassette be \u20b9 x and \u20b9 y, respectively. Then, forming equations according to the question, we have<\/p>\n<p>7x + 3y = 1110 \u2026 (i)<\/p>\n<p>5x + 4y = 1350 \u2026 (ii)<\/p>\n<p>On multiplying equation (i) by 4 and equation (ii) by 3,<\/p>\n<p>We get<\/p>\n<p>28x + 12y = 4440 \u2026 (iii)<\/p>\n<p>15x + 4y = 4050 \u2026 (iv)<\/p>\n<p>Subtracting equation (iv) from equation (iii),<\/p>\n<p>28x \u2013 13x + 12y \u2013 12y = 4440 \u2013 4050<\/p>\n<p>13x = 390<\/p>\n<p>\u21d2 x = 30<\/p>\n<p>On substituting x = 30 in equation (I),<\/p>\n<p>7(30) + 3y = 1110<\/p>\n<p>3y = 1110 \u2013 210<\/p>\n<p>y = 900\/ 3<\/p>\n<p>\u21d2 y = 300<\/p>\n<p>Therefore, it\u2019s found that the cost of one audio cassette = \u20b9 30<\/p>\n<p>And the cost of one video cassette = \u20b9 300<\/p>\n<p><strong>3. Reena has pens and pencils, which together are 40 in number. If she has 5 more pencils and 5 less pens, then the number of pencils would become 4 times the number of pens. Find the original number of pens and pencils.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s assume the number of pens and pencils is x and y, respectively.<\/p>\n<p>Forming equations according to the question, we have<\/p>\n<p>x + y = 40 \u2026 (i)<\/p>\n<p>(y+5) = 4(x-5)<\/p>\n<p>y + 5 = 4x \u2013 20<\/p>\n<p>5 + 20 = 4x \u2013 y<\/p>\n<p>4x \u2013 y = 25 \u2026 (ii)<\/p>\n<p>Adding equations (i) and (ii),<\/p>\n<p>We get<\/p>\n<p>x + 4x = 40 + 25<\/p>\n<p>5x = 65<\/p>\n<p>\u21d2 x = 13<\/p>\n<p>Putting x=13 in equation (i), we get<\/p>\n<p>13 + y = 40<\/p>\n<p>\u21d2 y = 40 \u2013 13 = 27<\/p>\n<p>Therefore, it\u2019s found that the number of pens Reena has is 13.<\/p>\n<p>And the number of pencils Reena has is 27.<\/p>\n<p><strong>4. 4 tables and 3 chairs, together, cost\u00a0<\/strong>\u20b9\u00a0<strong>2250, and 3 tables and 4 chairs cost\u00a0<\/strong>\u20b9<strong>\u00a01950. Find the cost of 2 chairs and 1 table.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s assume the cost of 1 table is \u20b9 x, and the cost of 1 chair is \u20b9 y.<\/p>\n<p>Then, according to the question,<\/p>\n<p>4x + 3y = 2250 \u2026 (i)<\/p>\n<p>3x + 4y = 1950 \u2026 (ii)<\/p>\n<p>On multiplying (i) with 3 and (ii) with 4,<\/p>\n<p>We get,<\/p>\n<p>12x + 9y = 6750 \u2026 (iii)<\/p>\n<p>12x + 16y = 7800 \u2026 (iv)<\/p>\n<p>Now, subtracting equation (iv) from (iii),<\/p>\n<p>We get,<\/p>\n<p>-7y = -1050<\/p>\n<p>y = 150<\/p>\n<p>Using y = 150 in (i), we find x<\/p>\n<p>4x + 3(150) = 2250<\/p>\n<p>4x = 2250 \u2013 450<\/p>\n<p>x = 1800\/ 4<\/p>\n<p>\u21d2 x = 450<\/p>\n<p>From the question, it\u2019s required to find the value of (x + 2y) \u21d2 450 + 2(150) = 750<\/p>\n<p>Therefore, the total cost of 2 chairs and 1 table is \u20b9 750.<\/p>\n<p><strong>5. 3 bags and 4 pens together cost\u00a0<\/strong>\u20b9<strong>\u00a0257, whereas 4 bags and 3 pens together cost\u00a0<\/strong>\u20b9<strong>324. Find the total cost of 1 bag and 10 pens.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let the cost of a bag and a pen be \u20b9 x and \u20b9 y, respectively.<\/p>\n<p>Then, according to the question,<\/p>\n<p>3x + 4y = 257 \u2026 (i)<\/p>\n<p>4x + 3y = 324 \u2026 (ii)<\/p>\n<p>On multiplying equation (i) by 3 and (ii) by 4,<\/p>\n<p>We get,<\/p>\n<p>9x + 12y = 770 \u2026 (iii)<\/p>\n<p>16x + 12y = 1296 \u2026 (iv)<\/p>\n<p>Subtracting equation (iii) from (iv), we get<\/p>\n<p>16x \u2013 9x = 1296 \u2013 771<\/p>\n<p>7x = 525<\/p>\n<p>x = 525\/7 = 75<\/p>\n<p>Hence, the cost of a bag = \u20b9 75<\/p>\n<p>Substituting x = 75 in equation (i),<\/p>\n<p>We get,<\/p>\n<p>3 x 75 + 4y = 257<\/p>\n<p>225 + 4y = 257<\/p>\n<p>4y = 257 \u2013 225<\/p>\n<p>4y = 32<\/p>\n<p>y = 32\/4 = 8<\/p>\n<p>Hence, the cost of a pen = \u20b9 8<\/p>\n<p>From the question, it\u2019s required to find the value of (x + 10y) \u21d2 75 +10(8) = 20<\/p>\n<p>Therefore, the total cost of 1 bag and 10 pens = 75 + 80 = \u20b9 155<\/p>\n<p><strong>6. 5 books and 7 pens together cost\u00a0<\/strong>\u20b9\u00a0<strong>79, whereas 7 books and 5 pens together cost\u00a0<\/strong>\u20b9\u00a0<strong>77. Find the total cost of 1 book and 2 pens.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s assume the cost of a book and a pen be \u20b9 x and \u20b9 y, respectively.<\/p>\n<p>Then, according to the question,<\/p>\n<p>5x + 7y = 79 \u2026 (i)<\/p>\n<p>7x + 5y = 77 \u2026 (ii)<\/p>\n<p>On multiplying equation (i) by 5 and (ii) by 7,<\/p>\n<p>We get,<\/p>\n<p>25x + 35y = 395 \u2026 (iii)<\/p>\n<p>49x + 35y = 539 \u2026 (iv)<\/p>\n<p>Subtracting equation (iii) from (iv),<\/p>\n<p>We have,<\/p>\n<p>49x \u2013 25x = 539 \u2013 395<\/p>\n<p>24x = 144<\/p>\n<p>x = 144\/24 = 6<\/p>\n<p>Hence, the cost of a book = \u20b9 6<\/p>\n<p>Substituting x= 6 in equation (i),<\/p>\n<p>We get<\/p>\n<p>5 (6) + 7y = 79<\/p>\n<p>30 + 7y = 79<\/p>\n<p>7y = 79 \u2013 30<\/p>\n<p>7y = 49<\/p>\n<p>y = 49\/ 7 = 7<\/p>\n<p>Hence, the cost of a pen = \u20b9 7<\/p>\n<p>From the question, it\u2019s required to find the value of (x + 2y) \u21d2 6 + 2(7) = 20<\/p>\n<p>Therefore, the total cost of 1 book and 2 pens = 6 + 14= \u20b9 20<\/p>\n<p><strong>7. Jamila sold a table and a chair for\u00a0<\/strong>\u20b9<strong>\u00a01050, thereby making a profit of 10% on the table and 25% on the chair. If she had taken a profit of 25% on the table and 10% on the chair, she would have got\u00a0<\/strong>\u20b9<strong>\u00a01065. Find the cost price of each.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let the cost price of one table and one chair be \u20b9 x and \u20b9 y, respectively.<\/p>\n<p>So,<\/p>\n<p>The selling price of the table, when it\u2019s sold at a profit of 10% = \u20b9 x + 10x\/100 = \u20b9 110x \/ 100<\/p>\n<p>The selling price of the chair, when it\u2019s sold at a profit of 25% = \u20b9 y + 25y\/100 = \u20b9 125y \/ 100<\/p>\n<p>Hence, according to the question,<\/p>\n<p>110x \/ 100 + 125y \/ 100 = 1050 \u2026 (i)<\/p>\n<p>Similarly,<\/p>\n<p>The selling price of the table, when it\u2019s sold at a profit of 25% = \u20b9 (x + 25x\/100) = \u20b9 125x\/ 100<\/p>\n<p>The selling price of the chair, when it\u2019s sold at a profit of 10% = \u20b9 (y + 10y\/100) = \u20b9 110y \/ 100<\/p>\n<p>Hence, again from the question<\/p>\n<p>125x \/ 100 + 110y \/ 100 = 1065 \u2026 (ii)<\/p>\n<p>Re-written (i) and (ii) with their simplest coefficients,<\/p>\n<p>11x\/10 + 5y\/4 = 1050\u2026\u2026.. (iii)<\/p>\n<p>5x\/4 + 11y\/10 = 1065\u2026\u2026.. (iv)<\/p>\n<p>Adding (iii) and (iv), we get<\/p>\n<p>(11\/ 10 + 5\/ 4)x + (5\/ 4 + 11\/ 10)y = 2115<\/p>\n<p>47\/ 20x + 47\/ 20y = 2115<\/p>\n<p>x + y = 2115(20\/ 47) = 900<\/p>\n<p>\u21d2 x = 900 \u2013 y \u2026\u2026. (v)<\/p>\n<p>Using (v) in (iii),<\/p>\n<p>11(900 \u2013 y)\/10 + 5y\/4 = 1050<\/p>\n<p>2(9900 -11y) +25y = 1050 x 20 [After taking LCM]<\/p>\n<p>19800 \u2013 22y + 25y = 21000<\/p>\n<p>3y = 1200<\/p>\n<p>\u21d2 y = 400<\/p>\n<p>Putting y = 400 in (v), we get<\/p>\n<p>x = 900 \u2013 400 = 500<\/p>\n<p>Therefore, the cost price of the table is \u20b9 500, and that of the chair is \u20b9 400.<\/p>\n<p>We have provided complete details of RD Sharma Class 10 Solutions Chapter 3 Exercise 3.6. If you have any queries related to <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\"><strong>CBSE<\/strong><\/a>\u00a0Class 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-3-exercise-36\"><\/span>FAQs on RD Sharma Class 10 Solutions Chapter 3 Exercise 3.6<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-1631107884468\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"where-can-i-download-rd-sharma-class-10-solutions-chapter-3-exercise-36-free-pdf\"><\/span>Where can I download RD Sharma Class 10 Solutions Chapter 3 Exercise 3.6 free PDF?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can download RD Sharma Class 10 Solutions Chapter 3 Exercise 3.6 free PDF from the above article.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631108091071\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"is-it-required-to-practice-all-of-the-questions-in-chapter-3-exercise-36-of-rd-sharma-solutions-for-class-10-maths\"><\/span>Is it required to practice all of the questions in Chapter 3 Exercise 3.6 of RD Sharma Solutions for Class 10 Maths?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, all of the questions in RD Sharma Class 10 Solutions Chapter 3 Exercise 3.6 must be learned. These questions may appear on both board exams and class tests. Students will be prepared for their board exams if they learn these questions.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631108643558\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"what-are-the-benefits-of-using-rd-sharma-class-10-solutions-chapter-3-exercise-36\"><\/span>What are the benefits of using RD Sharma Class 10 Solutions Chapter 3 Exercise 3.6?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>1. Correct answers according to the last CBSE guidelines and syllabus.<br \/>2. The RD Sharma Class 10 Solutions Chapter 3 Exercise 3.6 is written in simple language to assist students in their board examination, &amp; competitive examination preparation.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Class 10 Solutions Chapter 3 Exercise 3.6:\u00a0The RD Sharma Class 10 Solutions is an essential tool for students who want to improve their conceptual knowledge and exam confidence. For any assistance with this problem, students can refer to the RD Sharma Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in &#8230; <a title=\"RD Sharma Class 10 Solutions Chapter 3 Exercise 3.6 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-3-exercise-3-6\/\" aria-label=\"More on RD Sharma Class 10 Solutions Chapter 3 Exercise 3.6 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":125557,"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\/125508"}],"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=125508"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125508\/revisions"}],"predecessor-version":[{"id":504078,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125508\/revisions\/504078"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/125557"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=125508"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=125508"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=125508"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}