{"id":125505,"date":"2021-09-08T19:28:26","date_gmt":"2021-09-08T13:58:26","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=125505"},"modified":"2021-09-08T19:28:29","modified_gmt":"2021-09-08T13:58:29","slug":"rd-sharma-class-10-solutions-chapter-3-exercise-3-9","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-3-exercise-3-9\/","title":{"rendered":"RD Sharma Class 10 Solutions Chapter 3 Exercise 3.9 (Updated for 2021-22)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-125560\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-3-Exercise3.9.jpg\" alt=\"RD Sharma Class 10 Solutions Chapter 3 Exercise 3.9\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-3-Exercise3.9.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-3-Exercise3.9-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 3 Exercise 3.9:\u00a0<\/strong>Students have always been perplexed by age issues. These age-related problems can be easily solved using methods for solving linear pairs of equations in two variables. 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> to learn how to accomplish this correctly. For any reference relating to this exercise, students can use 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 3.9 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-69d56c0754ee6\" 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-3-9\/#is-it-required-to-practice-all-of-the-questions-in-chapter-3-exercise-39-of-rd-sharma-solutions-for-class-10-maths\" title=\"Is it required to practice all of the questions in Chapter 3 Exercise 3.9 of RD Sharma Solutions for Class 10 Maths?\">Is it required to practice all of the questions in Chapter 3 Exercise 3.9 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-3-9\/#what-are-the-benefits-of-using-rd-sharma-class-10-solutions-chapter-3-exercise-39\" title=\"What are the benefits of using RD Sharma Class 10 Solutions Chapter 3 Exercise 3.9?\">What are the benefits of using RD Sharma Class 10 Solutions Chapter 3 Exercise 3.9?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-10-solutions-chapter-3-exercise-39-free-pdf\"><\/span>Download RD Sharma Class 10 Solutions Chapter 3 Exercise 3.9 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-3.9.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-3.9.pdf\">RD Sharma Class 10 Solutions Chapter 3 Exercise 3.9<\/a><\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-rd-sharma-solutions-class-10-maths-chapter-3-exercise-39-important-question-with-answers\"><\/span>Access answers to RD Sharma Solutions Class 10 Maths Chapter 3 Exercise 3.9- Important Question with Answers<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br \/>A father is three times as old as his son. After twelve years, his age will be twice as that of his son then. Find their present ages. (C.B.S.E. 1992)<br \/>Solution:<br \/>Let present age of father = x<br \/>and that of son = y<br \/>According to the conditions,<br \/>x = 3y \u2026.(i)<br \/>After 12 years,<br \/>Age of father = x + 12<br \/>and age of son = y + 12<br \/>x + 12 = 2(y + 12)<br \/>x + 12 = 2y + 24<br \/>\u21d2 3y + 12 = 2y + 24 {From (i)}<br \/>\u21d2 3y \u2013 2y = 24 \u2013 12<br \/>\u21d2 y = 12<br \/>x = 3y = 3 x 12 = 36<br \/>Hence present age of father = 36 years and age of son = 12 years<\/p>\n<p>Question 2.<br \/>Ten years later, A will be twice as old as B and five years ago, A was three times as old as B. What are the present ages of A and B ? (C.B.S.E. 1992)<br \/>Solution:<br \/>Let present age of A = x years<br \/>and age of B = y years<br \/>10 years later<br \/>A\u2019s age will be = x + 10<br \/>and B\u2019s age will be = y + 10<br \/>x + 10 = 2(y + 10)<br \/>\u21d2 x + 10 = 2y + 20<br \/>\u21d2 x \u2013 2y = 20 \u2013 10<br \/>\u21d2 x \u2013 2y = 10 \u2026.(i)<br \/>5 years ago,<br \/>A\u2019s age was = x \u2013 5 years<br \/>and B\u2019s age was = y \u2013 5 years<br \/>x \u2013 5 = 3 (y \u2013 5)<br \/>\u21d2 x \u2013 5 = 3y \u2013 15<br \/>\u21d2 x \u2013 3y = 5 \u2013 15 = -10 \u2026.(ii)<br \/>Subtracting (ii) from (i) we get<br \/>y = 20<br \/>and x \u2013 2 x 20 = 10<br \/>\u21d2 x = 40 + 10 = 50<br \/>A\u2019s present age = 50 years<br \/>and B\u2019s present age = 20 years<\/p>\n<p>Question 3.<br \/>Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu ?<br \/>Solution:<br \/>Let present age of Nuri = x years<br \/>and age of Sonu = y years<br \/>5 years ago,<br \/>age of Nuri = (x \u2013 5) years<br \/>and age of Sonu = (y \u2013 5) years<br \/>x \u2013 5 = 3 (y \u2013 5) = 3y \u2013 15<br \/>\u21d2 x = 3y \u2013 15 + 5<br \/>\u21d2 x = 3y \u2013 10 \u2026.(i)<br \/>10 years later,<br \/>age of Nuri = x + 10<br \/>and age of Sonu = y + 10<br \/>x + 10 = 2 (y + 10) = 2y + 20<br \/>\u21d2 x = 2y + 20 \u2013 10 = 2y+ 10 \u2026.(ii)<br \/>From (i) and (ii)<br \/>3y \u2013 10 = 2y + 10 \u21d2 3y \u2013 2y = 10 + 10<br \/>\u21d2 y = 20<br \/>x = 3y \u2013 10 [from (i)]<br \/>x = 3 x 20 \u2013 10 = 60 \u2013 10 = 50 years<br \/>and age of Sonu = 20 years<\/p>\n<p>Question 4.<br \/>Six years hence a man\u2019s age will be three times the age of his son and three years ago, he was nine times as old as his son. Find their present ages. (C.B.S.E. 1994)<br \/>Solution:<br \/>Let present age of a man = x years<br \/>and age of his son = y years<br \/>6 years hence,<br \/>age of the man = x + 6<br \/>and age of his son = y + 6<br \/>x + 6 = 3 (y + 6)<br \/>\u21d2 x + 6 = 3y + 18<br \/>\u21d2 x \u2013 3y = 18 \u2013 6 = 12<br \/>\u21d2 x \u2013 3y = 12 \u2026.(i)<br \/>3 years ago,<br \/>the age of the man = x \u2013 3<br \/>and age of his son = y \u2013 3<br \/>x \u2013 3 = 9 (y \u2013 3)<br \/>\u21d2 x \u2013 3 = 9y \u2013 27<br \/>\u21d2 x \u2013 9y = -27 + 3<br \/>\u21d2 x \u2013 9y = -24 \u2026.(ii)<br \/>Subtracting (ii) from (i),<br \/>6y = 36<br \/>\u21d2 y = 6<br \/>From (i), x \u2013 3 x 6 = 12<br \/>\u21d2 x \u2013 18 = 12<br \/>\u21d2 x = 12 + 18 = 30<br \/>Present age of man = 30 years<br \/>and age of his son = 6 years<\/p>\n<div class=\"google-auto-placed ap_container\">\u00a0<\/div>\n<p>Question 5.<br \/>Ten years ago, a father was twelve times as old as his son and ten years hence, he will be twice as old as his son will be then. Find their present ages. (C.B.S.E. 1994)<br \/>Solution:<br \/>Let present age of father = x years<br \/>and age of his son = y years<br \/>10 years ago,<br \/>Father\u2019s age = x \u2013 10<br \/>and son\u2019s age = y \u2013 10<br \/>x \u2013 10 = 12(y \u2013 10)<br \/>\u21d2 x \u2013 10 = 12y \u2013 120<br \/>\u21d2 x \u2013 12y = -120 + 10 = -110<br \/>\u21d2 x \u2013 12y = -110 \u2026.(i)<br \/>10 years hence,<br \/>Father\u2019s age = x + 10<br \/>and his son\u2019s age = y + 10<br \/>10y = 120<br \/>y = 12<br \/>From (ii), x \u2013 2y = 10<br \/>x \u2013 2 x 10 = 10<br \/>\u21d2 x \u2013 24 = 10<br \/>\u21d2 x = 10 + 24<br \/>\u21d2 x = 34<br \/>Present age of father = 34 years<br \/>and age of his son = 12 years<\/p>\n<div class=\"google-auto-placed ap_container\">\u00a0<\/div>\n<p>Question 6.<br \/>The present age of a father is three years more than three times the age of the son. Three years hence father\u2019s age will be 10 years more than twice the age of the son. Determine their present ages. (C.B.S.E. 1994C)<br \/>Solution:<br \/>Let present age of father = x years<br \/>and age of his son = y years<br \/>x = 3y + 3 \u2026\u2026.(i)<br \/>3 years hence,<br \/>Father\u2019s age = (x + 3)<br \/>and his son\u2019s age = (y + 3)<br \/>x + 3 = 2 (y + 3) + 10 = 2y + 6 + 10<br \/>x + 3 = 2y + 16<br \/>\u21d2 x = 2y + 16 \u2013 3 = 2y + 13 \u2026.(ii)<br \/>From (i) and (ii)<br \/>3y + 3 = 2y + 13<br \/>\u21d2 3y \u2013 2y = 13 \u2013 3<br \/>\u21d2 y = 10<br \/>and x = 3y + 3 = 3 x 10 + 3 = 30 + 3 = 33<br \/>Present age of father = 33 years<br \/>and age of his son = 10 years<\/p>\n<div class=\"google-auto-placed ap_container\">\u00a0<\/div>\n<p>Question 7.<br \/>A father is three times as old\u2019as his son. In 12 years time, he will be twice as old as his son. Find the present ages of father and the son. (C.B.S.E. 1992, 1996)<br \/>Solution:<br \/>Let present age of father = x years<br \/>and age of son = y years<br \/>x = 3y \u2026\u2026\u2026(i)<br \/>12 years hence,<br \/>Father\u2019s age = x + 12<br \/>and son\u2019s age = y + 12<br \/>(x + 12) = 2 (y + 12)<br \/>\u21d2 x + 12 = 2y + 24<br \/>\u21d2 x = 2y + 24 \u2013 12 = 2y + 12 \u2026.(ii)<br \/>From (i) and (ii)<br \/>3y = 2y + 12<br \/>\u21d2 3y \u2013 2y = 12<br \/>\u21d2 y = 12<br \/>x = 3y = 3 x 12 = 36<br \/>Present age of father = 36 years and<br \/>age of son = 12 years<\/p>\n<div class=\"google-auto-placed ap_container\">\u00a0<\/div>\n<p>Question 8.<br \/>Father\u2019s age is three times the sum of ages of his two children. After 5 years his age will be twice the sum of ages of two children. Find the age of father. (C.B.S.E. 2003)<br \/>Solution:<br \/>Let father\u2019s present age = x years<br \/>and sum of ages of his two children = y<br \/>then x = 3y<br \/>\u21d2 y =\u00a0<span id=\"MathJax-Element-1-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-1\" class=\"math\"><span id=\"MathJax-Span-2\" class=\"mrow\"><span id=\"MathJax-Span-3\" class=\"mfrac\"><span id=\"MathJax-Span-4\" class=\"mn\">1<\/span><span id=\"MathJax-Span-5\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span>\u00a0x \u2026.(i)<br \/>After 5 years,<br \/>Age of father = x + 5<br \/>and sum of age of two children = y + 2 x 5 = y + 10<br \/>(x + 5) = 2(y + 10)<br \/>x + 5 = 2y + 20<br \/>\u21d2 x = 2y + 20 \u2013 5<br \/>x = 2y + 15 \u2026.(ii)<br \/>From (i)<br \/>x = 2 x\u00a0<span id=\"MathJax-Element-2-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-6\" class=\"math\"><span id=\"MathJax-Span-7\" class=\"mrow\"><span id=\"MathJax-Span-8\" class=\"mfrac\"><span id=\"MathJax-Span-9\" class=\"mn\">1<\/span><span id=\"MathJax-Span-10\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span>\u00a0x + 15<br \/>\u21d2 x =\u00a0<span id=\"MathJax-Element-3-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-11\" class=\"math\"><span id=\"MathJax-Span-12\" class=\"mrow\"><span id=\"MathJax-Span-13\" class=\"mfrac\"><span id=\"MathJax-Span-14\" class=\"mn\">2<\/span><span id=\"MathJax-Span-15\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span>\u00a0x + 15<br \/>\u21d2 x \u2013\u00a0<span id=\"MathJax-Element-4-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-16\" class=\"math\"><span id=\"MathJax-Span-17\" class=\"mrow\"><span id=\"MathJax-Span-18\" class=\"mfrac\"><span id=\"MathJax-Span-19\" class=\"mn\">1<\/span><span id=\"MathJax-Span-20\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span>\u00a0x = 15<br \/>\u21d2\u00a0<span id=\"MathJax-Element-5-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-21\" class=\"math\"><span id=\"MathJax-Span-22\" class=\"mrow\"><span id=\"MathJax-Span-23\" class=\"mfrac\"><span id=\"MathJax-Span-24\" class=\"mn\">1<\/span><span id=\"MathJax-Span-25\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span>\u00a0x = 15<br \/>\u21d2 x = 15 x 3 =45<br \/>Age of father = 45 years<\/p>\n<div class=\"google-auto-placed ap_container\">\u00a0<\/div>\n<p>Question 9.<br \/>Two years ago, a father was five times as old as his son. Two years later, his age will be 8 more than three times the age of the son. Find the present ages of father and son. (C.B.S.E. 2004)<br \/>Solution:<br \/>Let present age of father = x years<br \/>and age of his son = y years<br \/>2 years ago,<br \/>age of father = x \u2013 2<br \/>and age of son = y \u2013 2<br \/>x \u2013 2 = 5(y \u2013 2)<br \/>\u21d2 x \u2013 2 = 5y \u2013 10<br \/>\u21d2 x = 5y \u2013 10 + 2<br \/>\u21d2 x = 5y \u2013 8 \u2026\u2026\u2026(i)<br \/>2 years later,<br \/>age of father = x + 2<br \/>and age of son = y + 2<br \/>x + 2 = 3 (y + 2) + 8<br \/>\u21d2 x + 2 = 3y + 6 + 8<br \/>\u21d2 x = 3y + 14 \u2013 2 = 3y + 12 \u2026.(ii)<br \/>From,(i) and (ii)<br \/>5y \u2013 8 = 3y + 12<br \/>\u21d2 5y \u2013 3y = 12 + 8<br \/>\u21d2 2y = 20<br \/>\u21d2 y = 10<br \/>From (i)<br \/>x = 5y \u2013 8 = 5 x 10 \u2013 8 = 50 \u2013 8 = 42<br \/>Present age of father = 42 years<br \/>and age of son = 10 years<\/p>\n<p>Question 10.<br \/>A is elder to B by 2 years. A\u2019s father F is twice as old as A and B is twice as old as his sister S. If the ages of the father and sister differ by 40 years, find the age of A. (C.B.S.E. 1992C)<br \/>Solution:<br \/>Let age of A = x years<br \/>and age of B = y years<br \/>According to the conditions,<br \/>x = y + 2<br \/>\u21d2 y = x \u2013 2 \u2026.(i)<br \/>Age of A\u2019s\u2019 father = 2x<br \/>Age of B\u2019s sisters =\u00a0<span id=\"MathJax-Element-6-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-26\" class=\"math\"><span id=\"MathJax-Span-27\" class=\"mrow\"><span id=\"MathJax-Span-28\" class=\"mfrac\"><span id=\"MathJax-Span-29\" class=\"mi\">y<\/span><span id=\"MathJax-Span-30\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span><br \/>2x \u2013 2y = 40<br \/>4x \u2013 y = 80 \u2026.(ii)<br \/>4x \u2013 (x \u2013 2) = 80<br \/>\u21d2 4x \u2013 x + 2 = 80<br \/>3x = 80 \u2013 2 = 78<br \/>x = 26<br \/>A\u2019s age = 26 years<\/p>\n<p>Question 11.<br \/>The ages of two friends Ani and Biju differ by 3 years. Ani\u2019s father Dharam is twice as old as Ani and Biju are twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.<br \/>Solution:<br \/>Let age of Ani = x years<br \/>and age of Biju = y years<br \/>x \u2013 y = 3 \u2026.(i)<br \/>Ani\u2019s father Dharam\u2019s age = 2x<br \/>and Cathy\u2019s age =\u00a0<span id=\"MathJax-Element-7-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-31\" class=\"math\"><span id=\"MathJax-Span-32\" class=\"mrow\"><span id=\"MathJax-Span-33\" class=\"mfrac\"><span id=\"MathJax-Span-34\" class=\"mn\">1<\/span><span id=\"MathJax-Span-35\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0y<br \/>But 2x \u2013\u00a0<span id=\"MathJax-Element-8-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-36\" class=\"math\"><span id=\"MathJax-Span-37\" class=\"mrow\"><span id=\"MathJax-Span-38\" class=\"mfrac\"><span id=\"MathJax-Span-39\" class=\"mn\">1<\/span><span id=\"MathJax-Span-40\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0y = 30<br \/>\u21d2 4x \u2013 y = 60 \u2026.(ii)<br \/>Subtracting,<br \/>3x = 57<br \/>x = 19<br \/>and 4x \u2013 y = 60<br \/>\u21d2 4 x 19 \u2013 y = 60<br \/>\u21d2 76 \u2013 y = 60<br \/>\u21d2 76 \u2013 60 = y<br \/>\u21d2 y = 16<br \/>Anil\u2019s age = 19 years<br \/>and Biju\u2019s age = 16 years<\/p>\n<p>Question 12.<br \/>Two years ago, Salim was thrice as old as his daughter and six years later, he will be four years older than twice her age. How old are they now? [NCERT Exemplar]<br \/>Solution:<br \/>Let Salim and his daughter\u2019s age be x and y year respectively.<br \/>Now, by the first condition,<br \/>Two years ago, Salim was thrice as old as his daughter.<br \/>i. e., x \u2013 2 = 3(y \u2013 2)<br \/>\u21d2 x \u2013 2 = 3y \u2013 6<br \/>\u21d2 x \u2013 3y = -4 \u2026(i)<br \/>and by second condition,<br \/>six years later, Salim will be four years older than twice her age.<br \/>x + 6 = 2(y + 6) + 4<br \/>\u21d2 x + 6 = 2y + 12 + 4<br \/>\u21d2 x \u2013 2y = 16 \u2013 6<br \/>\u21d2x \u2013 2y = 10 \u2026(ii)<br \/>On subtracting Eq. (i) from Eq. (ii), we get<br \/>y = 14<br \/>Put the value of y in Eq. (ii), we get<br \/>x \u2013 2 x 14 = 10<br \/>\u21d2 x = 10 + 28<br \/>\u21d2 x = 38<br \/>Hence, Salim and his daughter\u2019s ages are 38 years and 14 years, respectively.<\/p>\n<p>Question 13.<br \/>The age of the father is twice the sum of the ages of his two children. After 20 years, his age will be equal to the sum of the ages of his children. Find the age of the father. [NCERT Exemplar]<br \/>Solution:<br \/>Let the present age (in year) of father and his two children be x, y and z year, respectively.<br \/>Now by given condition, x = 2(y + z) \u2026(i)<br \/>and after 20 years,<br \/>(x + 20) = (y + 20) + (z + 20)<br \/>\u21d2 y + z + 40 = x + 20<br \/>\u21d2 y + z = x \u2013 20<br \/>On putting the value of (y + z) in Eq. (i) and we get the present age of father<br \/>\u21d2 x = 2 (x \u2013 20)<br \/>x = 2x \u2013 40<br \/>\u21d2 x = 40<br \/>Hence, the father\u2019s age is 40 years.<\/p>\n<p>We have provided complete details of RD Sharma Class 10 Solutions Chapter 3 Exercise 3.9. 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-39\"><\/span>FAQs on RD Sharma Class 10 Solutions Chapter 3 Exercise 3.9<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-1631107908207\" 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-39-free-pdf\"><\/span>Where can I download RD Sharma Class 10 Solutions Chapter 3 Exercise 3.9 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 3.9 free PDF from the above article.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631108106982\" 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-39-of-rd-sharma-solutions-for-class-10-maths\"><\/span>Is it required to practice all of the questions in Chapter 3 Exercise 3.9 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 3.9 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-1631108665715\" 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-39\"><\/span>What are the benefits of using RD Sharma Class 10 Solutions Chapter 3 Exercise 3.9?<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 3.9 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 3.9:\u00a0Students have always been perplexed by age issues. These age-related problems can be easily solved using methods for solving linear pairs of equations in two variables. Refer to the RD Sharma Class 10 Solutions to learn how to accomplish this correctly. For any reference relating to this &#8230; <a title=\"RD Sharma Class 10 Solutions Chapter 3 Exercise 3.9 (Updated for 2021-22)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-3-exercise-3-9\/\" aria-label=\"More on RD Sharma Class 10 Solutions Chapter 3 Exercise 3.9 (Updated for 2021-22)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":125560,"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\/125505"}],"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=125505"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125505\/revisions"}],"predecessor-version":[{"id":125725,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125505\/revisions\/125725"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/125560"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=125505"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=125505"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=125505"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}