{"id":125501,"date":"2023-09-08T19:28:00","date_gmt":"2023-09-08T13:58:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=125501"},"modified":"2023-11-08T12:45:48","modified_gmt":"2023-11-08T07:15:48","slug":"rd-sharma-class-10-solutions-chapter-3-exercise-3-11","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-3-exercise-3-11\/","title":{"rendered":"RD Sharma Class 10 Solutions Chapter 3 Exercise 3.11 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-125562\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-3-Exercise3.11.jpg\" alt=\"RD Sharma Class 10 Solutions Chapter 3 Exercise 3.11\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-3-Exercise3.11.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-3-Exercise3.11-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 3 Exercise 3.11:\u00a0<\/strong>The main focus of this exercise is on various applications of linear equations. The <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-for-maths\/\"><strong>RD Sharma Class 10 Solutions<\/strong><\/a> created by Kopykitab experts is a powerful tool for students to clear their concerns and analyze their weak areas. Students can also 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 3.11 PDF, which is available 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-69d85627b0601\" 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\" 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href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-3-exercise-3-11\/#download-rd-sharma-class-10-solutions-chapter-3-exercise-311-free-pdf\" title=\"Download RD Sharma Class 10 Solutions Chapter 3 Exercise 3.11 Free PDF\">Download RD Sharma Class 10 Solutions Chapter 3 Exercise 3.11 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-3-11\/#access-answers-to-rd-sharma-solutions-class-10-maths-chapter-3-exercise-311-important-question-with-answers\" title=\"Access answers to RD Sharma Solutions Class 10 Maths Chapter 3 Exercise 3.11- Important Question with Answers\">Access answers to RD Sharma Solutions Class 10 Maths Chapter 3 Exercise 3.11- 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-3-11\/#is-it-required-to-practice-all-of-the-questions-in-chapter-3-exercise-311-of-rd-sharma-solutions-for-class-10-maths\" title=\"Is it required to practice all of the questions in Chapter 3 Exercise 3.11 of RD Sharma Solutions for Class 10 Maths?\">Is it required to practice all of the questions in Chapter 3 Exercise 3.11 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-3-11\/#what-are-the-benefits-of-using-rd-sharma-class-10-solutions-chapter-3-exercise-311\" title=\"What are the benefits of using RD Sharma Class 10 Solutions Chapter 3 Exercise 3.11?\">What are the benefits of using RD Sharma Class 10 Solutions Chapter 3 Exercise 3.11?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-10-solutions-chapter-3-exercise-311-free-pdf\"><\/span>Download RD Sharma Class 10 Solutions Chapter 3 Exercise 3.11 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-3.11.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-3.11.pdf\">RD Sharma Class 10 Solutions Chapter 3 Exercise 3.11<\/a><\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-rd-sharma-solutions-class-10-maths-chapter-3-exercise-311-important-question-with-answers\"><\/span>Access answers to RD Sharma Solutions Class 10 Maths Chapter 3 Exercise 3.11- Important Question with Answers<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>1. If in a rectangle, the length is increased and breadth reduced each by 2 units, the area is reduced by 28 square units. If, however the length is reduced by 1 unit and the breadth increased by 2 units, the area increases by 33 square units. Find the area of the rectangle.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s assume the length and breadth of the rectangle be x units and y units, respectively.<\/p>\n<p>Hence, the area of rectangle = xy sq. units<\/p>\n<p>From the question we have the following cases,<\/p>\n<p>Case 1:<\/p>\n<p>Length is increased by 2 units \u21d2 now, the new length is x+2 units<\/p>\n<p>Breadth is reduced by 2 units \u21d2 now, the new breadth is y-2 units<\/p>\n<p>And it\u2019s given that the area is reduced by 28 square units i.e., = xy \u2013 28<\/p>\n<p>So, the equation becomes<\/p>\n<p>(x+2)(y\u22122) = xy \u2212 28<\/p>\n<p>\u21d2 xy \u2212 2x + 2y \u2013 4 = xy \u2212 28<\/p>\n<p>\u21d2 \u22122x + 2y \u2013 4 + 28 = 0<\/p>\n<p>\u21d2 \u22122x + 2y + 24 = 0<\/p>\n<p>\u21d2 2x \u2212 2y \u2013 24 = 0 \u2026\u2026\u2026 (i)<\/p>\n<p>Case 2:<\/p>\n<p>Length is reduced by 1 unit \u21d2 now, the new length is x-1 units<\/p>\n<p>Breadth is increased by 2 units \u21d2 now, the new breadth is y+2 units<\/p>\n<p>And, it\u2019s given that the area is increased by 33 square units i.e. = i.e. = xy + 33<\/p>\n<p>So, the equation becomes<\/p>\n<p>(x\u22121)(y+2) = xy + 33<\/p>\n<p>\u21d2 xy + 2x \u2013 y \u2013 2 = x + 33<\/p>\n<p>\u21d2 2x \u2013 y \u2212 2 \u2212 33 = 0<\/p>\n<p>\u21d2 2x \u2013 y \u221235 = 0 \u2026\u2026\u2026.. (ii)<\/p>\n<p>Solving (i) and (ii),<\/p>\n<p>By using cross multiplication, we get<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables ex 3.11 - 1\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-3-50.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables ex 3.11 - 1\" \/><\/p>\n<p>x = 46\/2<\/p>\n<p>x = 23<\/p>\n<p>And,<\/p>\n<p>y = 22\/2<\/p>\n<p>y = 11<\/p>\n<p>Hence,<\/p>\n<p>The length of the rectangle is 23 units.<\/p>\n<p>The breadth of the rectangle is 11 units.<\/p>\n<p>So, the area of the actual rectangle = length x breadth,<\/p>\n<p>= x\u00d7y<\/p>\n<p>= 23 x 11<\/p>\n<p>= 253 sq. units<\/p>\n<p>Therefore, the area of rectangle is 253 sq. units.<\/p>\n<p><strong>2. The area of a rectangle remains the same if the length is increased by 7 metres and the breadth is decreased by 3 metres. The area remains unaffected if the length is decreased by 7 metres and the breadth is increased by 5 metres. Find the dimensions of the rectangle.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s assume the length and breadth of the rectangle be x units and y units, respectively.<\/p>\n<p>Hence, the area of rectangle = xy sq.units<\/p>\n<p>From the question we have the following cases,<\/p>\n<p>Case 1<\/p>\n<p>Length is increased by 7 metres \u21d2 now, the new length is x+7<\/p>\n<p>Breadth is decreased by 3 metres \u21d2 now, the new breadth is y-3<\/p>\n<p>And it\u2019s given, the area of the rectangle remains the same i.e. = xy.<\/p>\n<p>So, the equation becomes<\/p>\n<p>xy = (x+7)(y\u22123)<\/p>\n<p>xy = xy + 7y \u2212 3x \u2212 21<\/p>\n<p>3x \u2013 7y + 21 = 0 \u2026\u2026\u2026. (i)<\/p>\n<p>Case 2:<\/p>\n<p>Length is decreased by 7 metres \u21d2 now, the new length is x-7<\/p>\n<p>Breadth is increased by 5 metres \u21d2 now, the new breadth is y+5<\/p>\n<p>And it\u2019s given that, the area of the rectangle still remains the same i.e. = xy.<\/p>\n<p>So, the equation becomes<\/p>\n<p>xy = (x\u22127)(y+5)<\/p>\n<p>xy = xy \u2212 7y + 5x \u2212 35<\/p>\n<p>5x \u2013 7y \u2013 35 = 0 \u2026\u2026\u2026. (ii)<\/p>\n<p>Solving (i) and (ii),<\/p>\n<p>By using cross-multiplication, we get,<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables ex 3.11 - 2\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-3-51.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables ex 3.11 - 2\" \/><\/p>\n<p>x = 392\/14<\/p>\n<p>x = 28<\/p>\n<p>And,<\/p>\n<p>y = 210\/14<\/p>\n<p>y = 15<\/p>\n<p>Therefore, the length of the rectangle is 28 m. and the breadth of the actual rectangle is 15 m.<\/p>\n<p><span style=\"font-size: inherit; background-color: initial;\">We have provided complete details of RD Sharma Class 10 Solutions Chapter 3 Exercise 3.11. If you have any queries related to <\/span><a style=\"font-size: inherit; background-color: initial;\" href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\"><strong>CBSE<\/strong><\/a><span style=\"font-size: inherit; background-color: initial;\">\u00a0Class 10, feel free to ask us in the comment section below.<\/span><\/p>\n<div>\n<section>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-class-10-solutions-chapter-3-exercise-311\"><\/span>FAQs on RD Sharma Class 10 Solutions Chapter 3 Exercise 3.11<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<\/section>\n<\/div>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1631107924899\" 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-311-free-pdf\"><\/span>Where can I download RD Sharma Class 10 Solutions Chapter 3 Exercise 3.11 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 3.11 free PDF from the above article.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631108117157\" 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-311-of-rd-sharma-solutions-for-class-10-maths\"><\/span>Is it required to practice all of the questions in Chapter 3 Exercise 3.11 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 Solutions for Class 10 Maths Chapter 3 Exercise 3.11 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-1631108682872\" 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-311\"><\/span>What are the benefits of using RD Sharma Class 10 Solutions Chapter 3 Exercise 3.11?<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 3.11 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 3.11:\u00a0The main focus of this exercise is on various applications of linear equations. The RD Sharma Class 10 Solutions created by Kopykitab experts is a powerful tool for students to clear their concerns and analyze their weak areas. Students can also refer to the RD Sharma Solutions &#8230; <a title=\"RD Sharma Class 10 Solutions Chapter 3 Exercise 3.11 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-3-exercise-3-11\/\" aria-label=\"More on RD Sharma Class 10 Solutions Chapter 3 Exercise 3.11 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":125562,"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\/125501"}],"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=125501"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125501\/revisions"}],"predecessor-version":[{"id":504327,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125501\/revisions\/504327"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/125562"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=125501"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=125501"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=125501"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}