{"id":13750,"date":"2018-04-10T16:31:05","date_gmt":"2018-04-10T11:01:05","guid":{"rendered":"http:\/\/mycbseguide.com\/blog\/?p=13750"},"modified":"2025-10-08T11:41:19","modified_gmt":"2025-10-08T06:11:19","slug":"cbse-sample-papers-class-9-mathematics","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-9-mathematics\/","title":{"rendered":"Class 9 Maths Sample Paper 2024-25"},"content":{"rendered":"<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_76 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\" style=\"cursor:inherit\">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=\"Toggle Table of Content\"><span class=\"ez-toc-js-icon-con\"><span class=\"\"><span class=\"eztoc-hide\" style=\"display:none;\">Toggle<\/span><span class=\"ez-toc-icon-toggle-span\"><svg style=\"fill: #999;color:#999\" 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: #999;color:#999\" 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><\/span><\/span><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 eztoc-toggle-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:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-9-mathematics\/#Class_9_Maths_Sample_Paper_2025\" >Class 9 Maths Sample Paper 2025<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-9-mathematics\/#Sample_Paper_of_class_9_Math_%E2%80%93_in_PDF\" >Sample Paper of class 9 Math \u2013 in PDF<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-9-mathematics\/#The_Best_Model_Papers_for_Class_9_Maths_2024-25\" >The Best Model Papers for Class 9 Maths 2024-25<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-9-mathematics\/#CBSE_Sample_Papers_Class_9_Mathematics\" >CBSE Sample Papers Class 9 Mathematics<\/a><ul class='ez-toc-list-level-3' ><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-9-mathematics\/#Key_Points_about_CBSE_Class_9_Home_Exams\" >Key Points about CBSE Class 9 Home Exams:<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-9-mathematics\/#Key_Features_of_the_myCBSEguide_App\" >Key Features of the myCBSEguide App:<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"Class_9_Maths_Sample_Paper_2025\"><\/span>Class 9 Maths Sample Paper 2025<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>The <strong>CBSE <a href=\"https:\/\/mycbseguide.com\/cbse-sample-papers-class-9.html\">Class 9<\/a> Maths Sample Paper 2024-25<\/strong> includes both <strong>subjective<\/strong> and <strong>objective type questions<\/strong>, with a special focus on <strong>case study-based questions<\/strong> in the subjective section. Unlike Class 10, where students have the option to choose from multiple courses, all <strong>Class 9 Mathematics<\/strong> students study the same course content. he <strong>Class 9 Maths Sample Paper 2024-25<\/strong> is an indispensable tool for students looking to excel in their exams. It provides a comprehensive overview of the exam pattern, with well-structured questions designed according to the latest <strong>CBSE guidelines<\/strong>. Practicing with this <strong>Class 9 Maths Sample Paper 2024-25<\/strong> will help students understand the types of questions they can expect and assess their preparation level. Available on the <strong><a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\">myCBSEguide<\/a> App and <a href=\"https:\/\/mycbseguide.com\/\">myCBSEGuide<\/a> Website<\/strong>, these sample papers come with detailed solutions to help students grasp important concepts more effectively. By solving the sample paper, students can revise key topics and identify areas that need further attention, making it an ideal revision resource. Download the <strong>Class 9 Maths Sample Paper 2024-25<\/strong> today to ensure thorough and structured exam preparation. To help you prepare effectively, you can <strong>download the model question paper for Class 9 Maths<\/strong> for <strong>free<\/strong> from the <strong>myCBSEguide App<\/strong> or the student dashboard. These sample papers are designed to align with the updated exam pattern and provide a comprehensive understanding of the subject. <strong>Download the sample paper today<\/strong> and get ready to ace your <strong>Class 9 Maths exam<\/strong> with the best resources available.Teachers can also take advantage of the <strong><a href=\"https:\/\/play.google.com\/store\/apps\/details?id=com.examin8.exam\">Examin8<\/a> App and <a href=\"https:\/\/examin8.com\/\">Examin8<\/a> Website<\/strong>\u00a0to create custom exam papers with their name and logo.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-full aligncenter\" src=\"https:\/\/media-mycbseguide.s3.ap-south-1.amazonaws.com\/images\/blog\/09_maths_sample_paper.jpg\" alt=\"CBSE class 9 Maths Sample Paper 2022-23\" width=\"600\" height=\"300\" \/><\/p>\n<h2><span class=\"ez-toc-section\" id=\"Sample_Paper_of_class_9_Math_%E2%80%93_in_PDF\"><\/span>Sample Paper of class 9 Math \u2013 in PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><strong>CBSE Sample Question Papers (Solved) Class 9 Mathematics<\/strong> for the session 2023-24 are now available, following the updated <strong>marking scheme<\/strong> and <strong>blueprint<\/strong> issued by <strong>CBSE<\/strong>. These <strong>Class 9 Mathematics sample papers<\/strong> are designed to align with the latest exam pattern and will help students familiarize themselves with the types of questions asked in the exams. We are also offering <strong>guess papers<\/strong> for <strong>Class 9 Mathematics<\/strong> for the <strong>2025 annual exams<\/strong>. These sample question papers, available for <strong>free download<\/strong> in <strong>PDF format<\/strong> on the <strong><a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\">myCBSEguide<\/a> App and <a href=\"https:\/\/mycbseguide.com\/\">myCBSEGuide<\/a> Website<\/strong>, come with <strong>solutions<\/strong> to guide students through the exam preparation process. Practicing with these <strong>CBSE sample papers<\/strong> will not only boost confidence but also improve students&#8217; chances of scoring high marks in their <strong>Class 9 Mathematics exam<\/strong>. Teachers can also take advantage of the <strong><a href=\"https:\/\/play.google.com\/store\/apps\/details?id=com.examin8.exam\">Examin8<\/a> App and <a href=\"https:\/\/examin8.com\/\">Examin8<\/a> Website<\/strong>\u00a0to create custom exam papers with their name and logo.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"The_Best_Model_Papers_for_Class_9_Maths_2024-25\"><\/span>The Best Model Papers for Class 9 Maths 2024-25<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>For Class 9 students preparing for the CBSE exams, model papers are crucial for understanding the exam pattern and practicing effectively. While CBSE provides official model papers for board exams (Classes 10 and 12), Class 9 model papers are usually prepared by schools. This means the format and difficulty level can vary, making it important to have access to standardized, well-structured papers.<\/p>\n<p><strong>myCBSEguide<\/strong> offers model papers for Class 9 that strictly follow the latest CBSE guidelines. These papers are designed to provide a <strong>moderate difficulty level<\/strong>, matching the actual exam format. By practicing with these model papers, students can familiarize themselves with multiple-choice, short-answer, and long-answer questions.<\/p>\n<p>Key features of myCBSEguide model papers include:<\/p>\n<ul>\n<li><strong>Adherence to CBSE syllabus<\/strong>: Updated to reflect the latest curriculum.<\/li>\n<li><strong>Comprehensive coverage<\/strong>: Ensures all important topics are included.<\/li>\n<li><strong>Real exam experience<\/strong>: Simulates actual exam conditions.<\/li>\n<li><strong>Detailed solutions<\/strong>: Helps students understand concepts better.<\/li>\n<\/ul>\n<p>These model papers offer a reliable and consistent resource for students to practice and prepare confidently for their Class 9 exams. <strong>Download the myCBSEguide app or visit our website<\/strong> for free access to high-quality model papers and start your preparation today!<\/p>\n<p style=\"text-align: center;\"><strong>Sample Papers of Class 9 Maths 2025 with solution<\/strong><\/p>\n<p style=\"text-align: center;\"><a class=\"button\" href=\"https:\/\/mycbseguide.com\/dashboard\/category\/1234\/type\/2\"><strong>Download PDF\u00a0<\/strong><\/a><\/p>\n<h2><span class=\"ez-toc-section\" id=\"CBSE_Sample_Papers_Class_9_Mathematics\"><\/span><strong>CBSE Sample Papers Class 9 Mathematics<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>Unlike Class 10 and 12, CBSE does not conduct board exams for <strong>Class 9 students<\/strong>. Instead, these exams are home exams, where the responsibility of conducting the exam lies with the respective schools. However, schools must follow the guidelines issued by <strong>CBSE<\/strong> while preparing the question paper to ensure uniformity in terms of <strong>difficulty level<\/strong> and <strong>exam format<\/strong>.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"Key_Points_about_CBSE_Class_9_Home_Exams\"><\/span>Key Points about CBSE Class 9 Home Exams:<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<ul>\n<li><strong>No Board Exams for Class 9<\/strong>: The exams for Class 9 students are not board exams but are still an important part of the academic year.<\/li>\n<li><strong>Following CBSE Guidelines<\/strong>: While the exams are home-based, schools must strictly adhere to the guidelines provided by CBSE for the structure and content of the question paper. This ensures consistency and fairness across all schools. You can find detailed solutions to the <strong>Class 9 Maths Sample Paper 2024-25<\/strong> on the myCBSEguide website for better understanding.<\/li>\n<li><strong>Uniform Difficulty Level and Format<\/strong>: To maintain uniformity, CBSE provides model question papers and examination materials through its official portal. Schools are required to download the question papers directly from the <strong>CBSE website<\/strong>, ensuring that the format and difficulty level align with CBSE standards.<\/li>\n<\/ul>\n<p style=\"text-align: center;\"><strong>Class 09 &#8211; Mathematics<br \/>\nSample Paper &#8211; 01 (2024-25)<\/strong><\/p>\n<hr \/>\n<p><b>Maximum Marks: 80<br \/>\nTime Allowed: : 3 hours<\/b><\/p>\n<hr \/>\n<p><b>General Instructions:<\/b><\/p>\n<p>Read the following instructions carefully and follow them:<\/p>\n<ol>\n<li>This question paper contains 38 questions.<\/li>\n<li>This Question Paper is divided into 5 Sections A, B, C, D\u00a0and E.<\/li>\n<li>In Section A, Questions no. 1-18 are multiple choice questions (MCQs) and questions no. 19 and 20 are Assertion- Reason based questions of 1 mark each.<\/li>\n<li>In Section B, Questions no. 21-25 are very short answer (VSA) type questions, carrying 02 marks each.<\/li>\n<li>In Section C, Questions no. 26-31 are short answer (SA) type questions, carrying 03 marks each.<\/li>\n<li>In Section D, Questions no. 32-35 are long answer (LA) type questions, carrying 05 marks each.<\/li>\n<li>In Section E, Questions no. 36-38 are case study-based questions carrying 4 marks each with sub-parts of the values of 1,1 and 2 marks each respectively.<\/li>\n<li>All Questions are compulsory. However, an internal choice in 2 Questions of Section B, 2 Questions of Section C\u00a0and 2 Questions of Section D has been provided. An internal choice has been provided in all the 2 marks questions of Section E.<\/li>\n<li>Draw neat and clean figures wherever required.<\/li>\n<li>Take <span class=\"math-tex\">{tex}\\pi=22 \/ 7{\/tex}<\/span> wherever required if not stated.<\/li>\n<li>Use of calculators is not allowed.<\/li>\n<\/ol>\n<hr \/>\n<ol style=\"padding-left: 20px; list-style: decimal;\">\n<li style=\"text-align: center; clear: both; display: block;\"><b>Section A<\/b><\/li>\n<li>The equation of y-axis is:\n<div style=\"margin-left: 20px;\">\n<p>a)x = 0<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)y = x<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)y = 0<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)y <span class=\"math-tex\">{tex}\\ne{\/tex}<\/span>\u00a0x<\/p>\n<\/div>\n<\/li>\n<li>The side of a triangle is 12 cm, 16 cm, and 20 cm. Its area is\n<div style=\"margin-left: 20px;\">\n<p>a)100sq cm<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)90sq cm<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)120sq cm<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)96 sq cm<\/p>\n<\/div>\n<\/li>\n<li>In the figure, O is the centre of eh circle and <span class=\"math-tex\">{tex}\\angle AOB = {80^o}{\/tex}<\/span>. The value of x is :<br \/>\n<img decoding=\"async\" style=\"width: 130px; height: 110px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/testpaper\/1.01.09.10.67.Ques.JPG\" \/><\/p>\n<div style=\"margin-left: 20px;\">\n<p>a)<span class=\"math-tex\">{tex}{60^o}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)<span class=\"math-tex\">{tex}{30^o}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)<span class=\"math-tex\">{tex}{160^o}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)<span class=\"math-tex\">{tex}{40^o}{\/tex}<\/span><\/p>\n<\/div>\n<\/li>\n<li>In the given figure, AD is a median of <span class=\"math-tex\">{tex}\\triangle{\/tex}<\/span>ABC and E\u00a0is the midpoint of AD. If BE is joined and produced to meet AC in F then AF = ?<br \/>\n<img decoding=\"async\" style=\"width: 130px; height: 126px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/3yQ248f.png\" alt=\"\" data-imgur-src=\"3yQ248f.png\" \/><\/p>\n<div style=\"margin-left: 20px;\">\n<p>a)<span class=\"math-tex\">{tex}\\frac{1}{3} A C{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)<span class=\"math-tex\">{tex}\\frac{3}{4} AC{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)<span class=\"math-tex\">{tex}\\frac{2}{3} AC{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)<span class=\"math-tex\">{tex}\\frac{1}{2} A C{\/tex}<\/span><\/p>\n<\/div>\n<\/li>\n<li>The value of\u00a0<span class=\"math-tex\">{tex}\\left\\{8^{\\frac {-4} 3} \\div 2^{-2}\\right\\}^{\\frac 1 2}{\/tex}<\/span>, is\n<div style=\"margin-left: 20px;\">\n<p>a)4<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)2<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)<span class=\"math-tex\">{tex}\\frac{1}{2}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)<span class=\"math-tex\">{tex}\\frac{1}{4}{\/tex}<\/span><\/p>\n<\/div>\n<\/li>\n<li>In the adjoining fig, PQ = PR. If <span class=\"math-tex\">{tex}\\angle QPR{\/tex}<\/span> = 48<span class=\"math-tex\">{tex}^\\circ {\/tex}<\/span>, then value of x is:<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/quiz\/09\/mat\/q_20980.PNG\" alt=\"\" width=\"146\" height=\"110\" \/><\/p>\n<div style=\"margin-left: 20px;\">\n<p>a)132<sup>o<\/sup><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)114<span class=\"math-tex\">{tex}^\\circ {\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)104<sup>o<\/sup><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)96<sup>o<\/sup><\/p>\n<\/div>\n<\/li>\n<li>x = 2, y = 3 is a solution of the linear equation\n<div style=\"margin-left: 20px;\">\n<p>a)x + 2y = 8<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)x + y = 8<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)-x + y = 8<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)2x + y = 8<\/p>\n<\/div>\n<\/li>\n<li>If <span class=\"math-tex\">{tex}{x^2} + kx &#8211; 3 = (x &#8211; 3)(x + 1){\/tex}<\/span>, then the value of \u2018k\u2019 is\n<div style=\"margin-left: 20px;\">\n<p>a)-3<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)2<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)-2<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)3<\/p>\n<\/div>\n<\/li>\n<li>The value of <span class=\"math-tex\">{tex}\\left(\\frac{81}{16}\\right)^{\\frac{-3}{4}} \\times\\left\\{\\left(\\frac{25}{9}\\right)^{\\frac{-3}{2}} \\div\\left(\\frac{5}{2}\\right)^{-3}\\right\\}{\/tex}<\/span> is\n<div style=\"margin-left: 20px;\">\n<p>a)4<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)2<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)3<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)1<\/p>\n<\/div>\n<\/li>\n<li>In Triangle ABC which is right angled at B. Given that AB = 9cm, AC = 15cm and D, E are the mid-points of the sides AB and AC respectively. Find the length of BC?\n<div style=\"margin-left: 20px;\">\n<p>a)13cm<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)13.5cm<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)12cm<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)15cm<\/p>\n<\/div>\n<\/li>\n<li>The decimal representation of a rational number is\n<div style=\"margin-left: 20px;\">\n<p>a)always terminating<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)always non-terminating<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)either terminating or repeating<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)either terminating or non-repeating<\/p>\n<\/div>\n<\/li>\n<li>Which of the following points lie on the line y = 3x \u2013 4?\n<div style=\"margin-left: 20px;\">\n<p>a)(2, 2)<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)(4, 12)<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)(5, 15)<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)(3, 9)<\/p>\n<\/div>\n<\/li>\n<li>In a figure, if OP||RS, <span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>OPQ = 110\u00b0 and <span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>QRS = 130\u00b0, then <span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>PQR is equal to<br \/>\n<img decoding=\"async\" style=\"width: 150px; height: 63px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/WlweiAK.png\" alt=\"\" data-imgur-src=\"WlweiAK.png\" \/><\/p>\n<div style=\"margin-left: 20px;\">\n<p>a)40\u00b0<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)50\u00b0<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)70\u00b0<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)60\u00b0<\/p>\n<\/div>\n<\/li>\n<li>The product of any two irrational numbers is\n<div style=\"margin-left: 20px;\">\n<p>a)sometimes rational, sometimes irrational<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)always a rational number<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)always an integer<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)always an irrational number.<\/p>\n<\/div>\n<\/li>\n<li>In the given figure, O is the centre of a circle in which <span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>OAB = 20\u00b0 and <span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>OCB = 50\u00b0. Then, <span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>AOC = ?<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/GJ8slbu.png\" alt=\"\" width=\"121\" height=\"117\" data-imgur-src=\"GJ8slbu.png\" \/><\/p>\n<div style=\"margin-left: 20px;\">\n<p>a)20\u00b0<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)70\u00b0<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)60\u00b0<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)50\u00b0<\/p>\n<\/div>\n<\/li>\n<li>The perpendicular distance of the point P (4, 3) from x-axis is\n<div style=\"margin-left: 20px;\">\n<p>a)3<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)6<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)4<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)5<\/p>\n<\/div>\n<\/li>\n<li>The linear equation 3x &#8211;\u00a05y = 15 has\n<div style=\"margin-left: 20px;\">\n<p>a)no solution<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)infinitely many solutions<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)a unique solution<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)two solutions<\/p>\n<\/div>\n<\/li>\n<li>x<sup>3<\/sup> + y<sup>3<\/sup> + z<sup>3<\/sup> &#8211; 3xyz is\n<div style=\"margin-left: 20px;\">\n<p>a)(x + y + z)<sup>3<\/sup><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)(x + y + z)(x<sup>2<\/sup> + y<sup>2<\/sup> + z<sup>2<\/sup> &#8211; xy &#8211; yz &#8211; zx)<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)(x &#8211; y + z)<sup>3<\/sup><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)(x + y + z)<sup>3<\/sup> &#8211; 3xyz<\/p>\n<\/div>\n<\/li>\n<li><strong>Assertion (A):<\/strong> In <span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>ABC, E and F are the midpoints of AC and AB respectively. The altitude AP at BC intersects FE at Q. Then, AQ = QP.<br \/>\n<strong>Reason (R):<\/strong> Q is the midpoint of AP.<\/p>\n<div style=\"margin-left: 20px;\">\n<p>a)Both A and R are true and R is the correct explanation of A.<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)Both A and R are true but R is not the correct explanation of A.<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)A is true but R is false.<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)A is false but R is true.<\/p>\n<\/div>\n<\/li>\n<li><strong>Assertion (A):<\/strong>\u00a0<span class=\"math-tex\">{tex}\\sqrt{3}{\/tex}<\/span>\u00a0is an irrational number.<br \/>\n<strong>Reason (R):<\/strong> Square root of a positive integer which is not a perfect square is an irrational number.<\/p>\n<div style=\"margin-left: 20px;\">\n<p>a)Both A and R are true and R is the correct explanation of A.<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)Both A and R are true but R is not the correct explanation of A.<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)A is true but R is false.<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)A is false but R is true.<\/p>\n<p>To prepare effectively for your exams, download the <strong><a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\">myCBSEguide<\/a> App and <a href=\"https:\/\/mycbseguide.com\/\">myCBSEGuide<\/a> Website<\/strong>. This app offers a comprehensive range of study materials, including sample papers, solutions, and practice questions for <strong>CBSE<\/strong>, <strong>NCERT<\/strong>, <strong>JEE (Main)<\/strong>, <strong>NEET-UG<\/strong>, and <strong>NDA<\/strong> exams. It\u2019s designed to help students build a solid foundation and enhance their exam readiness.Teachers can also take advantage of the <strong><a href=\"https:\/\/play.google.com\/store\/apps\/details?id=com.examin8.exam\">Examin8<\/a> App and <a href=\"https:\/\/examin8.com\/\">Examin8<\/a> Website<\/strong>\u00a0to create custom exam papers with their name and logo. This powerful tool allows educators to generate personalized assessments, making it easier to track student progress and tailor their teaching approach.Whether you\u2019re a student or a teacher, <strong>myCBSEguide<\/strong> and <strong>Examin8<\/strong> are the perfect apps to streamline learning and teaching. Download them now to start preparing for success! Practicing the <strong>Class 9 Maths Sample Paper 2024-25<\/strong> is a great way to boost your confidence before the board exams.<\/p>\n<p style=\"text-align: center;\"><b>Section B<\/b><\/p>\n<\/div>\n<\/li>\n<li>Why is Axiom 5, in the list of Euclid&#8217;s axioms, considered a <strong>universal truth<\/strong>?<\/li>\n<li>In fig. AC = XD, C is the mid-point of AB and D is the mid-point of XY. Using a Euclid&#8217;s axiom, show that AB = XY.<br \/>\n<img decoding=\"async\" style=\"width: 184px; height: 113px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/deUBg1d.png\" alt=\"\" data-imgur-src=\"deUBg1d.png\" \/><\/li>\n<li>Name the quadrant in which the following points lie:\u00a0 (i) A(2, 9)\u00a0 \u00a0(ii) B(\u20133, 5)\u00a0 \u00a0 (iii) C(\u20134, \u20137)\u00a0 \u00a0(iv) D(3, \u20132)<\/li>\n<li>Examine,\u00a0whether\u00a0<span class=\"math-tex\">{tex}\\sqrt{5}-2{\/tex}<\/span>\u00a0is rational or irrational.\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Rationalise the denominator of:\u00a0<span class=\"math-tex\">{tex}\\frac{\\sqrt{5}}{2 \\sqrt{3}}{\/tex}<\/span>.<\/li>\n<li>A Joker&#8217;s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Find the ratio of the curved surface areas of two cones if their diameters of the bases are equal and slant heights are in the ratio 4 : 3.<\/li>\n<li style=\"text-align: center; clear: both; display: block;\"><b>Section C<\/b><\/li>\n<li>Give three rational numbers between <span class=\"math-tex\">{tex}\\frac {1} {3} {\/tex}<\/span> and <span class=\"math-tex\">{tex}\\frac {1} {2}{\/tex}<\/span>.<\/li>\n<li>The monthly profits (in Rs) of 100 shops are distributed as follows:<br \/>\n<table style=\"width: 100%;\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<tbody>\n<tr>\n<td width=\"200\">Profits per shop:<\/td>\n<td width=\"200\">0-50<\/td>\n<td width=\"200\">50-100<\/td>\n<td width=\"200\">100-50<\/td>\n<td width=\"200\">150-200<\/td>\n<td width=\"200\">200-250<\/td>\n<td width=\"200\">250-300<\/td>\n<\/tr>\n<tr>\n<td width=\"200\">No. of shops:<\/td>\n<td width=\"200\">12<\/td>\n<td width=\"200\">18<\/td>\n<td width=\"200\">27<\/td>\n<td width=\"200\">20<\/td>\n<td width=\"200\">17<\/td>\n<td width=\"200\">6<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Draw a histogram for the data and show the frequency polygon for it.<\/li>\n<li>In Figure\u00a0find the four angles A, B, C and D in the parallelogram ABCD.<br \/>\n<img decoding=\"async\" style=\"width: 140px; height: 98px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/sQgGKbJ.png\" alt=\"\" data-imgur-src=\"sQgGKbJ.png\" \/><\/li>\n<li>Find at least 3 solutions for the following linear equation in two variables: x + y \u2013 4 = 0<\/li>\n<li>The population of Delhi State in different census years is as given below:<br \/>\n<table style=\"width: 100%;\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<tbody>\n<tr>\n<td width=\"200\">Census year<\/td>\n<td width=\"200\">1961<\/td>\n<td width=\"200\">1971<\/td>\n<td width=\"200\">1981<\/td>\n<td width=\"200\">1991<\/td>\n<td width=\"200\">2001<\/td>\n<\/tr>\n<tr>\n<td width=\"200\">Population in Lakhs<\/td>\n<td width=\"200\">30<\/td>\n<td width=\"200\">55<\/td>\n<td width=\"200\">70<\/td>\n<td width=\"200\">110<\/td>\n<td width=\"200\">150<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Represent the above information with the help of a bar graph.<\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>A random survey of the number of children of various age groups playing football match in a park was found as follows<\/p>\n<table style=\"width: 100%;\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<tbody>\n<tr>\n<th scope=\"row\" width=\"200\">Age (in years)<\/th>\n<td width=\"200\">1-2<\/td>\n<td width=\"200\">2-3<\/td>\n<td width=\"200\">3-5<\/td>\n<td width=\"200\">5-7<\/td>\n<td width=\"200\">7-10<\/td>\n<td width=\"200\">10-15<\/td>\n<td width=\"200\">15-17<\/td>\n<\/tr>\n<tr>\n<th scope=\"row\" width=\"200\">Number of<br \/>\nchildren<\/th>\n<td width=\"200\">5<\/td>\n<td width=\"200\">4<\/td>\n<td width=\"200\">10<\/td>\n<td width=\"200\">12<\/td>\n<td width=\"200\">9<\/td>\n<td width=\"200\">10<\/td>\n<td width=\"200\">8<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Draw a histogram to represent the above data.<\/li>\n<li>Factorize: <span class=\"math-tex\">{tex}{x^3} + 13{x^2} + 32x + 20{\/tex}<\/span><\/li>\n<li style=\"text-align: center; clear: both; display: block;\"><b>Section D<\/b><\/li>\n<li>In each of the figures given below,\u00a0AB\u00a0<span class=\"math-tex\">{tex}\\parallel{\/tex}<\/span>\u00a0CD. Find the value of x<span class=\"math-tex\">{tex}^\\circ{\/tex}<\/span> in\u00a0each case.<br \/>\n<img decoding=\"async\" style=\"width: 110px; height: 102px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/v5tBKMZ.png\" alt=\"\" data-imgur-src=\"v5tBKMZ.png\" \/><\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Two lines AB and CD intersect at a point O such that\u00a0<span class=\"math-tex\">{tex}\\angle B O C+\\angle A O D=280^{\\circ}{\/tex}<\/span>,\u00a0as shown in the figure. Find all the four angles.<br \/>\n<img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/kK0BseF.png\" alt=\"\" data-imgur-src=\"kK0BseF.png\" \/><\/li>\n<li>What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm.\u00a0<span class=\"math-tex\">{tex}\\left( \\text{Use }\\pi \\text{ = 3}\\text{.14} \\right){\/tex}<\/span><\/li>\n<li>Two sides of a triangular field are 85 m and 154 m in length and its perimeter is 324 m. Find\u00a0the area of the field.\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Find the area of a triangular field whose sides are 91 m, 98 m and 105 m in length. Find the height corresponding to the longest side.<\/li>\n<li>If ax<sup>3<\/sup> + bx<sup>2 <\/sup>+ x &#8211; 6 has x + 2 as a factor and leaves a remainder 4 when divided by (x &#8211; 2), find the values of a and b.<\/li>\n<li style=\"text-align: center; clear: both; display: block;\"><b>Section E<\/b><\/li>\n<li><strong>Read the following text carefully and answer the questions that follow:<\/strong><br \/>\nAjay is writing a test which consists of \u2018True\u2019 or \u2018False\u2019 questions. One mark is awarded for every correct answer while \u00bc mark is deducted for every wrong answer. Ajay knew answers to some of the questions. Rest of the questions he attempted by guessing.<br \/>\n<img decoding=\"async\" style=\"height: 125px; width: 250px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/1665492082-p9utmh.jpg\" \/><\/p>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>If he answered 110 questions and got 80 marks and\u00a0answer to all questions,\u00a0he attempted by guessing were wrong, then how many questions did he answer correctly? (1)<\/li>\n<li>If he answered 110 questions and got 80 marks and\u00a0answer to all questions,\u00a0he attempted by guessing were wrong,\u00a0then how many questions did he guess? (1)<\/li>\n<li>If answer to all questions he attempted by guessing were wrong and answered 80 correctly, then how many marks he got? (2)<br \/>\n<strong>OR<\/strong><br \/>\nIf answer to all questions he attempted by guessing were wrong, then how many questions answered correctly to score 95 marks? (2)<\/li>\n<\/ol>\n<\/li>\n<li><strong>Read the following text carefully and answer the questions that follow:<\/strong><br \/>\nAs shown In the village of Surya there was a big pole PC. This pole was tied with a strong wire of 10 m length. Once there was a big spark on this pole, thus wires got damaged very badly. Any small fault was usually repaired with the help of a rope which normal board electricians were carrying on bicycles.<br \/>\nThis time electricians need a staircase of 10 m so that it can reach at point P on the pole and this should make 60<sup>o<\/sup> with line AC.<br \/>\n<img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1689679157-qdc26z.jpg\" \/><\/p>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>Show that\u00a0<span class=\"math-tex\">{tex}\\triangle{\/tex}<\/span>APC and\u00a0<span class=\"math-tex\">{tex}\\triangle{\/tex}<\/span>BPC are congruent. (1)<\/li>\n<li>Find the value of\u00a0<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>x. (1)<\/li>\n<li>Find the value of\u00a0<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>y. (2)<br \/>\n<strong>OR<\/strong><br \/>\nWhat is the value of\u00a0<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>PBC? (2)<\/li>\n<\/ol>\n<\/li>\n<li><strong>Read the following text carefully and answer the questions that follow:<\/strong><br \/>\nRohan draws a circle of radius 10 cm with the help of a compass and scale. He also draws two chords, AB and CD in such a way that the perpendicular distance from the center to AB and CD are 6 cm and 8 cm respectively. Now, he has some doubts that are given below.<br \/>\n<img decoding=\"async\" style=\"height: 132px; width: 150px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/1665491255-3854s2.jpg\" \/><\/p>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>Show that the perpendicular drawn from the Centre of a circle to a chord bisects the chord. (1)<\/li>\n<li>What is the length of CD? (1)<\/li>\n<li>What is the length of AB? (2)<br \/>\n<strong>OR<\/strong>How many circles can be drawn from given three noncollinear points? (2)<\/p>\n<p style=\"text-align: left;\">To boost your exam preparation and practice more questions, <strong>download the <a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\">myCBSEguide<\/a> App and <a href=\"https:\/\/mycbseguide.com\/\">myCBSEGuide<\/a> Website<\/strong>. The app offers <strong>comprehensive study material<\/strong> for CBSE, NCERT, JEE (Main), NEET-UG, and NDA exams, providing everything you need to excel. Whether you&#8217;re a student looking for practice papers, NCERT solutions, or detailed exam notes, myCBSEguide is the ultimate resource for effective study.<\/p>\n<h3 style=\"text-align: left;\"><span class=\"ez-toc-section\" id=\"Key_Features_of_the_myCBSEguide_App\"><\/span>Key Features of the myCBSEguide App:<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<ul>\n<li style=\"text-align: left;\"><strong>Complete Study Material<\/strong>: Access detailed notes, sample papers, model papers, and solutions for CBSE, JEE, NEET, and NDA.<\/li>\n<li style=\"text-align: left;\"><strong>Practice Papers<\/strong>: Get an extensive collection of practice questions and mock tests to prepare for exams.<\/li>\n<li style=\"text-align: left;\"><strong>NCERT Solutions<\/strong>: Find chapter-wise solutions for all subjects aligned with the CBSE curriculum.<\/li>\n<li style=\"text-align: left;\"><strong>User-Friendly Interface<\/strong>: Easily navigate through the app to find study resources tailored to your exam needs.<\/li>\n<\/ul>\n<p style=\"text-align: left;\">For <strong>teachers<\/strong>, the <strong><a href=\"https:\/\/play.google.com\/store\/apps\/details?id=com.examin8.exam\">Examin8<\/a> App and <a href=\"https:\/\/examin8.com\/\">Examin8<\/a> Website<\/strong>\u00a0offers the ability to create custom question papers with your own name and logo. This is a fantastic tool for schools and educators to generate personalized exam papers that adhere to the official guidelines and maintain exam integrity. The <strong>Class 9 Maths Sample Paper 2024-25<\/strong> includes all important topics and is designed according to the latest CBSE guidelines. Start your revision early with the <strong>Class 9 Maths Sample Paper 2024-25<\/strong> to avoid last-minute stress.<\/p>\n<p style=\"text-align: left;\"><strong>Download myCBSEguide<\/strong> for students and <strong>Examin8<\/strong> for teachers to make exam preparation more effective and personalized.<\/p>\n<p><strong>Class 09 &#8211; Mathematics<br \/>\nSample Paper &#8211; 01 (2024-25)<\/strong><\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<hr \/>\n<p style=\"text-align: center;\"><strong>Solution <\/strong><\/p>\n<ol style=\"padding-left: 20px;\">\n<li style=\"text-align: center; display: block;\"><b>Section A <\/b><\/li>\n<li>(a)\n<p style=\"display: inline;\">x = 0<\/p>\n<p><b>Explanation: <\/b>The value of abscissa or x-corrdinate is always zero at any point on y-axis.<\/p>\n<p>So, x = 0 is the equation of y-axis.<\/li>\n<li>(d)\n<p style=\"display: inline;\">96 sq cm<\/p>\n<p><b>Explanation: <\/b><span class=\"math-tex\">{tex}s = {{12 + 16 + 20} \\over 2} = {{48} \\over 2}{\/tex}<\/span>\u00a0= 24 cm<br \/>\nArea of triangle\u00a0 = <span class=\"math-tex\">{tex}\\sqrt {s\\left( {s &#8211; a} \\right)\\left( {s &#8211; b} \\right)\\left( {s &#8211; c} \\right)} {\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\sqrt {24\\left( {24 &#8211; 16} \\right)\\left( {24 &#8211; 12} \\right)\\left( {24 &#8211; 20} \\right)} {\/tex}<\/span><br \/>\n= <span class=\"math-tex\">{tex}\\sqrt {24 \\times 8 \\times 12 \\times 4} {\/tex}<\/span>\u00a0= 12 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> 8 = 96 sq cm<\/li>\n<li>(d)\n<p style=\"display: inline;\"><span class=\"math-tex\">{tex}{40^o}{\/tex}<\/span><\/p>\n<p><b>Explanation: <\/b>Angle made by a chord at the centre is twice the angle made by it on any point on the circumference.<br \/>\n<span class=\"math-tex\">{tex}x={\\angle AOB\\over2}={80^0\\over2}=40^0{\/tex}<\/span><\/li>\n<li>(a)\n<p style=\"display: inline;\"><span class=\"math-tex\">{tex}\\frac{1}{3} A C{\/tex}<\/span><\/p>\n<p><b>Explanation: <\/b>Let G be the mid-point of FC and join DG<br \/>\n<img decoding=\"async\" style=\"width: 120px; height: 116px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/OzIfcAh.png\" alt=\"\" data-imgur-src=\"OzIfcAh.png\" \/><br \/>\n\u200bIn\u00a0\u2206BCF,<br \/>\nG is the mid-point of FC and D is the mid-point of BC<br \/>\nThus, DG|| BF<br \/>\nDG || EF<br \/>\nNow, In \u2206 ADG,<br \/>\nE is the mid-point of AD and EF is parallel to DG.<br \/>\nThus, F is the mid-point of AG.<br \/>\nAF = FG = GC [G is the mid-point of FC]\nHence, AF =<span class=\"math-tex\">{tex} \\frac13{\/tex}<\/span>\u00a0AC<\/li>\n<li>(c)\n<p style=\"display: inline;\"><span class=\"math-tex\">{tex}\\frac{1}{2}{\/tex}<\/span><\/p>\n<p><b>Explanation: <\/b><span class=\"math-tex\">{tex}\\left\\{8^{\\frac {-4} 3} \\div 2^{-2}\\right\\}^{\\frac 1 2}{\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\left[\\left(2^{3}\\right)^{\\frac{-4}{3}} \\div 2^{-2}\\right]^{\\frac{1}{2}}\\\\{\/tex}<\/span><br \/>\n= <span class=\"math-tex\">{tex}\\left[2^{3 \\times \\frac{-4}{3}} \\div 2^{-2}\\right]^{\\frac{1}{2}}\\\\ {\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\left[2^{-4} \\div 2^{-2}\\right]^{\\frac{1}{2}}\\\\ {\/tex}<\/span><br \/>\n=<span class=\"math-tex\">{tex}\\left[2^{-4-(-2)} \\right]^{\\frac{1}{2}}\\\\ {\/tex}<\/span><br \/>\n=<span class=\"math-tex\">{tex}\\left[2^{-4+2} \\right]^{\\frac{1}{2}}\\\\ {\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\left[2^{-2} \\right]^{\\frac{1}{2}}\\\\ {\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex} \\left(\\frac{1}{2^{2}}\\right)^{\\frac{1}{2}}\\\\{\/tex}<\/span><br \/>\n=<span class=\"math-tex\">{tex} \\left(\\frac{1}{2}\\right)^{2\\times\\frac{1}{2}}\\\\{\/tex}<\/span><br \/>\n= <span class=\"math-tex\">{tex}\\frac{1}{2}{\/tex}<\/span><\/li>\n<li>(b)\n<p style=\"display: inline;\">114<span class=\"math-tex\">{tex}^\\circ {\/tex}<\/span><\/p>\n<p><b>Explanation: <\/b>It is an iscosceles triangle and hence angles opposite to equal sides are equal<br \/>\nAngle PQR and PRQ will be equal. Let suppose Angle PQR be Y<br \/>\nI.e Y+Y+48=180<br \/>\n= Y = 66<br \/>\nX = 180 &#8211; 66 = 114<\/li>\n<li>(a)\n<p style=\"display: inline;\">x + 2y = 8<\/p>\n<p><b>Explanation: <\/b>x + 2y = 8<br \/>\nWhen x=2 and y=3 then 2+ 2\u00d73=8<\/li>\n<li>(c)\n<p style=\"display: inline;\">-2<\/p>\n<p><b>Explanation: <\/b><span class=\"math-tex\">{tex}{x^2} + kx &#8211; 3 = (x &#8211; 3)(x + 1){\/tex}<\/span><\/p>\n<p>=&gt;\u00a0\u00a0\u00a0\u00a0 <span class=\"math-tex\">{tex}{x^2} + kx &#8211; 3 = {x^2} + \\left( { &#8211; 3 + 1} \\right)x + \\left( { &#8211; 3} \\right) \\times 1{\/tex}<\/span><\/p>\n<p>=&gt;\u00a0\u00a0\u00a0\u00a0 <span class=\"math-tex\">{tex}{x^2} + kx &#8211; 3 = {x^2} &#8211; 2x &#8211; 3{\/tex}<\/span><\/p>\n<p>On comparing the term, we get <span class=\"math-tex\">{tex}k=-2{\/tex}<\/span><\/li>\n<li>(d)\n<p style=\"display: inline;\">1<\/p>\n<p><b>Explanation: <\/b><span class=\"math-tex\">{tex}\\left(\\frac{81}{16}\\right)^{\\frac{-3}{4}} \\times\\left\\{\\left(\\frac{25}{9}\\right)^{\\frac{-3}{2}} \\div\\left(\\frac{5}{2}\\right)^{-3}\\right\\}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow\\left(\\frac{3}{2}\\right)^{4 \\times \\frac{-3}{4}} \\times\\left\\{\\left(\\frac{5}{3}\\right)^{2 \\times \\frac{-3}{2}} \\div\\left(\\frac{5}{2}\\right)^{-3}\\right\\}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow\\left(\\frac{3}{2}\\right)^{-3} \\times\\left\\{\\left(\\frac{5}{3}\\right)^{-3} \\div\\left(\\frac{5}{2}\\right)^{-3}\\right\\}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow\\left(\\frac{3}{2}\\right)^{-3} \\times\\left(\\frac{5}{3} \\times \\frac{2}{5}\\right)^{-3}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow\\left(\\frac{3}{2}\\right)^{-3} \\times\\left(\\frac{2}{3}\\right)^{-3}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow\\left(\\frac{3}{2} \\times \\frac{2}{3}\\right)^{-3}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0(1)<sup>-3<\/sup> = 1<\/li>\n<li>(c)\n<p style=\"display: inline;\">12cm<\/p>\n<p><b>Explanation: <\/b><\/p>\n<p><img decoding=\"async\" style=\"width: 200px; height: 144px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/iIYLRQ9.png\" alt=\"\" data-imgur-src=\"iIYLRQ9.png\" \/><br \/>\nApplying\u00a0pythagoras theorem in\u00a0<span class=\"math-tex\">{tex}\\triangle \\mathrm { ABC }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\mathrm { AC } ^ { 2 } = \\mathrm { AB } ^ { 2 } + \\mathrm { BC } ^ { 2 }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}15 ^ { 2 } = 9 ^ { 2 } + \\mathrm { BC } ^ { 2 }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}225 = 81 + \\mathrm { BC } ^ { 2 }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}225 &#8211; 81 = \\mathrm { BC } ^ { 2 }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\mathrm { BC } ^ { 2 } = 144{\/tex}<\/span><br \/>\nBC \u00a0= 12 cm<\/li>\n<li>(c)\n<p style=\"display: inline;\">either terminating or repeating<\/p>\n<p><b>Explanation: <\/b>Rational numbers can be represented in decimal forms rather than representing in fractions. They can easily be represented as decimals by just dividing numerator \u2018p\u2019 by denominator \u2018q\u2019 (as rational numbers is in the form of p\/q).<\/p>\n<p>A rational number can be expressed as a terminating or nonterminating, recurring decimal.<\/p>\n<p>For example:<\/p>\n<p>(i) <span class=\"math-tex\">{tex}5\\over 2{\/tex}<\/span>\u00a0= 2.5,<\/p>\n<p><span class=\"math-tex\">{tex}2\\over 8{\/tex}<\/span>= 0.25,<\/p>\n<p>7 = 7.0, etc., are rational numbers which are terminating decimals.<\/p>\n<p>(ii) <span class=\"math-tex\">{tex}5 \\over 9{\/tex}<\/span>= 0.555555555\u2026\u2026. = 0.5 \u0307,<\/p>\n<p><span class=\"math-tex\">{tex}1 \\over 6{\/tex}<\/span>\u00a0= 0.166666 \u2026.. = 0.16 \u0307<\/p>\n<p><span class=\"math-tex\">{tex}9 \\over 11{\/tex}<\/span>= 0.818181\u2026\u2026 = 0.81\u00a0 \u0307etc., are rational numbers which are nonterminating, recurring decimals.<\/li>\n<li>(a)\n<p style=\"display: inline;\">(2, 2)<\/p>\n<p><b>Explanation: <\/b><\/p>\n<p>When we put x=2 in the given equation,<br \/>\nThen, y = (3\u00d72) \u2013 4<br \/>\ny =6\u20134 = 2, so point is (2, 2)satisfied the given equation,<br \/>\nHence point (2, 2) will lie on the line y = 3x &#8211;\u00a04<\/li>\n<li>(d)\n<p style=\"display: inline;\">60\u00b0<\/p>\n<p><b>Explanation: <\/b>Produce OP to intersect RQ at point N.<br \/>\nNow, OP || RS and transversal RN intersects them at N and R respectively.<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\angle R N P=\\angle S R N{\/tex}<\/span>\u00a0(Alternate interior angles)<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>RNP = 130<sup>o<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\therefore \\angle P N Q=180^{\\circ}-130^{\\circ}=50^{\\circ}{\/tex}<\/span>\u00a0 (Linear pair)<br \/>\n<span class=\"math-tex\">{tex}\\angle O P Q=\\angle P N Q+\\angle P Q N{\/tex}<\/span>\u00a0(Exterior angle property)<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0110<sup>o<\/sup> = 50<sup>o<\/sup> +\u00a0<span class=\"math-tex\">{tex}\\angle P Q N{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\angle P Q N{\/tex}<\/span>\u00a0= 110<sup>o<\/sup> &#8211; 50<sup>o<\/sup> = 60<sup>o<\/sup> =\u00a0<span class=\"math-tex\">{tex}\\angle P Q R{\/tex}<\/span><\/li>\n<li>(a)\n<p style=\"display: inline;\">sometimes rational, sometimes irrational<\/p>\n<p><b>Explanation: <\/b>The product of any two irrational numbers is sometimes rational, sometimes irrational. It depends upon the numbers you choose.<br \/>\ne,g,\u00a0<span class=\"math-tex\">{tex}\\sqrt 2 \\times \\sqrt 2 = 2{\/tex}<\/span>\u00a0is rational whereas\u00a0<span class=\"math-tex\">{tex}\\sqrt 2 \\times \\sqrt3 = \\sqrt 6{\/tex}<\/span>\u00a0is irrational<\/li>\n<li>(c)\n<p style=\"display: inline;\">60\u00b0<\/p>\n<p><b>Explanation: <\/b>OA = OB <span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> <span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>OBA = <span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>OAB = 20\u00b0.<br \/>\nIn <span class=\"math-tex\">{tex}\\triangle{\/tex}<\/span>OAB,<br \/>\n<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>OAB + <span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>OBA + <span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>AOB = 180\u00b0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a020\u00b0 + 20\u00b0 + <span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>AOB = 180\u00b0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> <span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>AOB = 140\u00b0.<br \/>\nOB = OC <span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> <span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>OBC = <span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>OCB = 50\u00b0.<br \/>\nIn <span class=\"math-tex\">{tex}\\triangle{\/tex}<\/span>OCB,<br \/>\n<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>OCB + <span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>OBC + <span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>COB = 180\u00b0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> 50\u00b0 + 50\u00b0 + <span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>COB = 180\u00b0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> <span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>COB = 80\u00b0.<br \/>\n<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>AOB = 140\u00b0 <span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> <span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>AOC + <span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>COB = 140\u00b0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> <span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>AOC + 80\u00b0 = 140\u00b0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>AOC = 140\u00b0 &#8211;\u00a080\u00b0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> <span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>AOC = 60\u00b0.<\/li>\n<li>(a)\n<p style=\"display: inline;\">3<\/p>\n<p><b>Explanation: <\/b>The perpendicular distance of any point from x-axis is always equal to the value of ordinate.<\/li>\n<li>(b)\n<p style=\"display: inline;\">infinitely many solutions<\/p>\n<p><b>Explanation: <\/b>Given linear equation: 3x &#8211; 5y = 15 Or, x =\u00a0<span class=\"math-tex\">{tex}\\frac{5 y+15}{3}{\/tex}<\/span><br \/>\nWhen y = 0, x =\u00a0<span class=\"math-tex\">{tex}\\frac{15}{3}{\/tex}<\/span>\u00a0= 5<br \/>\nWhen y = 3, x =\u00a0<span class=\"math-tex\">{tex}\\frac{30}{3}{\/tex}<\/span>\u00a0= 10<br \/>\nWhen y = -3, x =\u00a0<span class=\"math-tex\">{tex}\\frac{0}{3}{\/tex}<\/span>\u00a0= 0<\/p>\n<table style=\"width: 100%;\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<tbody>\n<tr>\n<td>\u00a0 \u00a0 \u00a0xx<\/td>\n<td>\u00a0\u00a0\u00a0\u00a0\u00a0 5<\/td>\n<td>\u00a0 \u00a0 \u00a0 10<\/td>\n<td>\u00a0 \u00a0 0<\/td>\n<\/tr>\n<tr>\n<td>\u00a0 \u00a0 \u00a0yy<\/td>\n<td>\u00a0 \u00a0 \u00a0 0<\/td>\n<td>\u00a0 \u00a0 \u00a0 3<\/td>\n<td>\u00a0\u00a0\u00a0 -3<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Plot the points\u00a0A(5,0)\u00a0,\u00a0B(10,3) and\u00a0C(0,\u22123). Join the points and extend them in both the directions.<br \/>\nWe get infinite points that satisfy the given equation.<br \/>\nHence, the linear equation has infinitely many solutions.<br \/>\n<img decoding=\"async\" style=\"width: 300px; height: 300px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/Zc3i893.png\" alt=\"\" data-imgur-src=\"Zc3i893.png\" \/><\/li>\n<li>(b)\n<p style=\"display: inline;\">(x + y + z)(x<sup>2<\/sup> + y<sup>2<\/sup> + z<sup>2<\/sup> &#8211; xy &#8211; yz &#8211; zx)<\/p>\n<p><b>Explanation: <\/b><\/p>\n<p>x<sup>3<\/sup> + y<sup>3<\/sup> + z<sup>3\u00a0<\/sup>&#8211; 3xyz = x<sup>3<\/sup> + y<sup>3<\/sup> + 3x<sup>2<\/sup>y + 3xy<sup>2<\/sup> + z<sup>3\u00a0<\/sup>&#8211; 3xyz &#8211; 3x<sup>2<\/sup>y &#8211; 3xy<sup>2<\/sup><br \/>\n=\u00a0(x + y)<sup>3<\/sup> +z<sup>3<\/sup> &#8211; 3xy(x + y + z)<br \/>\n= (x + y + z) ((x + y)<sup>2<\/sup> + z<sup>2<\/sup> &#8211; (x + y)z) &#8211; 3xy (x + y + z)<br \/>\n= (x + y + z) (x<sup>2<\/sup> + 2x<sup>2<\/sup>y + y<sup>2<\/sup> + z2 &#8211; xy &#8211; xz &#8211; 3xy)<br \/>\n= (x + y + z) (x<sup>2<\/sup> + y<sup>2<\/sup> + z<sup>2<\/sup> &#8211; xy &#8211; yz &#8211; zx)<\/li>\n<li>(b)\n<p style=\"display: inline;\">Both A and R are true but R is not the correct explanation of A.<\/p>\n<p><b>Explanation: <\/b>In <span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>ABC, E and F are midpoint of the sides AC and AB respectively.<br \/>\nFE || BC [By mid-point theorem]\nNow, in <span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>ABP, F is mid-point of AB and FQ || BP. Q is mid-point of AP<br \/>\nAQ = QP<br \/>\n<img decoding=\"async\" style=\"width: 100px; height: 82px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/1657625583-u9gy5x.jpg\" alt=\"\" data-imgur-src=\"0nDSUnU.png\" \/><\/li>\n<li>(a)\n<p style=\"display: inline;\">Both A and R are true and R is the correct explanation of A.<\/p>\n<p><b>Explanation: <\/b>Both A and R are true and R is the correct explanation of A.<\/p>\n<p>To enhance your exam preparation and practice more questions, <strong>download the <a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\">myCBSEguide<\/a> app and <a href=\"https:\/\/mycbseguide.com\/\">myCBSEguide<\/a> Website<\/strong>. This app offers <strong>comprehensive study material<\/strong> for <strong>CBSE<\/strong>, <strong>NCERT<\/strong>, <strong>JEE (Main)<\/strong>, <strong>NEET-UG<\/strong>, and <strong>NDA exams<\/strong>, making it a one-stop resource for students aiming to excel. With <strong>sample papers<\/strong>, <strong>mock tests<\/strong>, <strong>chapter-wise solutions<\/strong>, and more, myCBSEguide ensures you have all the tools needed to succeed in your exams. Ensure comprehensive exam preparation with the <strong>Class 9 Maths Sample Paper 2024-25<\/strong>, available on myCBSEguide.\u00a0 The <strong>Class 9 Maths Sample Paper 2024-25<\/strong> offers a variety of question types, ensuring a well-rounded revision.<\/p>\n<p>For <strong>teachers<\/strong>, the <strong><a href=\"https:\/\/examin8.com\/\">Examin8<\/a> Website and <a href=\"https:\/\/play.google.com\/store\/apps\/details?id=com.examin8.exam\">Examin8<\/a> app<\/strong> is the ideal solution to create <strong>customized question papers<\/strong> with your school\u2019s name and logo. This app helps educators design unique and professional exam papers that adhere to official guidelines.<\/p>\n<p><strong>Download the myCBSEguide app<\/strong> for expert study material and exam prep resources, and for teachers, the <strong>Examin8 app<\/strong> to easily generate personalized question papers.<\/p>\n<p style=\"text-align: center;\"><b>Section B <\/b><\/p>\n<\/li>\n<li>Euclid&#8217;s Axiom 5 states that &#8220;The whole is greater than the part. Since this is true for anything in any part of the world. So, this is a universal truth.<\/li>\n<li>\u00a0In the above figure, we haveAB =\u00a0AC + BC = AC + AC = 2AC\u00a0 (Since, C is the mid-point of AB) ..(1)<br \/>\nXY =\u00a0XD + DY = XD + XD = 2XD (Since, D is the mid-point of XY) ..(2)<br \/>\nAlso, AC = XD (Given) ..(3)<br \/>\nFrom (1),(2)and(3), we get<br \/>\nAB = XY, According to Euclid,\u00a0things which are double of the same things are equal to one another.<\/li>\n<li>(i) I quadrant<br \/>\n(ii) II quadrant<br \/>\n(iii) III quadrant<br \/>\n(iv) IV quadrant<\/li>\n<li style=\"clear: both;\">Let x =\u00a0<span class=\"math-tex\">{tex}\\sqrt{5}-2{\/tex}<\/span>\u00a0be the rational number<br \/>\nSquaring on both sides, we get<br \/>\nx<sup>2<\/sup>\u00a0=\u00a0<span class=\"math-tex\">{tex}(\\sqrt{5}-2)^{2}{\/tex}<\/span><br \/>\nx<sup>2<\/sup>\u00a0= 5 + 4 &#8211;\u00a0<span class=\"math-tex\">{tex}4 \\sqrt{5}{\/tex}<\/span><br \/>\nx<sup>2<\/sup>\u00a0= 9 &#8211;\u00a0<span class=\"math-tex\">{tex}4 \\sqrt{5}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\sqrt{5}{\/tex}<\/span>\u00a0=\u00a0<span class=\"math-tex\">{tex}\\frac{x^{2}-9}{-4}{\/tex}<\/span>\u00a0= a rational number.<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\sqrt{5}{\/tex}<\/span>\u00a0is a rational number<br \/>\nSo, we arrive at a contradiction that\u00a0<span class=\"math-tex\">{tex} \\sqrt{5}{\/tex}<\/span>\u00a0is an irrational number.<br \/>\nSo\u00a0<span class=\"math-tex\">{tex}(\\sqrt{5}-2){\/tex}<\/span> is an irrational number.<\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p><span class=\"math-tex\">{tex}\\frac{\\sqrt{5}}{2 \\sqrt{3}}{\/tex}<\/span><br \/>\nMultiply both numerator and denominator by <span class=\"math-tex\">{tex}\\sqrt{3}{\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\frac{\\sqrt{5}}{2 \\sqrt{3}} \\times \\frac{\\sqrt{3}}{\\sqrt{3}}{\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\frac{\\sqrt{15}}{2 \\times 3}{\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\frac{\\sqrt{15}}{6}{\/tex}<\/span><\/li>\n<li style=\"clear: both;\">Radius of cap (r)\u00a0= 7 cm, Height of cap (h)\u00a0= 24 cm<br \/>\nSlant height of the cone (l) = <span class=\"math-tex\">{tex}\\sqrt{{{r}^{2}}+{{h}^{2}}}{\/tex}<\/span>\u00a0= <span class=\"math-tex\">{tex}\\sqrt{{{\\left( 7 \\right)}^{2}}+{{\\left( 24 \\right)}^{2}}}{\/tex}<\/span><br \/>\n= <span class=\"math-tex\">{tex}\\sqrt{49+576}{\/tex}<\/span><br \/>\n= <span class=\"math-tex\">{tex}\\sqrt{625}{\/tex}<\/span><br \/>\n= 25 cm<br \/>\nArea of sheet required to make a cap = CSA of cone = <span class=\"math-tex\">{tex}\\pi rl{\/tex}<\/span><br \/>\n= <span class=\"math-tex\">{tex}\\frac{22}{7}\\times 7\\times 25{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}=\\text{ }550\\text{ }c{{m}^{2}}{\/tex}<\/span>=550 cm<sup>2<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span> \u00a0Area of sheet required to make 10 caps =10\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span> 550 = 5500 cm<sup>2<\/sup><\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Since diameter of two cones are equal<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0Their radius are equal<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0r<sub>1<\/sub>\u00a0= r<sub>2<\/sub>\u00a0= r (say)<br \/>\nLet ratio be x,<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0Slant height &#8216;l<sub>1<\/sub>&#8216; of 1<sup>st<\/sup>\u00a0cone = 4x<br \/>\nSimilarly, slant height &#8216;l<sub>2<\/sub>&#8216; of 2<sup>nd<\/sup>\u00a0cone = 3x<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\frac { C . S . A _ { 1 } } { C . S . A _ { 2 } } = \\frac { \\pi r _ { 1 } l _ { 1 } } { \\pi r _ { 2 } l _ { 2 } } = \\frac { \\pi \\times r \\times 4 x } { \\pi \\times r \\times 3 x } = \\frac { 4 } { 3 }{\/tex}<\/span><\/li>\n<li style=\"text-align: center; display: block;\"><b>Section C <\/b><\/li>\n<li>Here\u00a0\u00a0\u00a0 a = <span class=\"math-tex\">{tex}\\frac {1} {3}{\/tex}<\/span>, b = <span class=\"math-tex\">{tex}\\frac {1} {2}{\/tex}<\/span>, n = 3<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\frac{b-a}{n+1}=\\frac{\\frac{1}{2}-\\frac{1}{3}}{3+1}{\/tex}<\/span><span class=\"math-tex\">{tex}=\\frac{\\frac{3-2}{6}}{4}{\/tex}<\/span><span class=\"math-tex\">{tex}=\\frac{\\frac{1}{6}}{4}=\\frac{1}{24}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0Three rational numbers between <span class=\"math-tex\">{tex}\\frac {1} {3}{\/tex}<\/span>\u00a0and <span class=\"math-tex\">{tex}\\frac {1} {2}{\/tex}<\/span>\u00a0are<br \/>\n<span class=\"math-tex\">{tex}\\frac{1}{3}+\\frac{1}{24}, \\frac{1}{3}+2\\left(\\frac{1}{24}\\right), \\frac{1}{3}+3\\left(\\frac{1}{24}\\right){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\frac{1}{3}+\\frac{1}{24}{\/tex}<\/span>,\u00a0<span class=\"math-tex\">{tex}\\frac{1}{3}+\\frac{1}{12}{\/tex}<\/span>,\u00a0<span class=\"math-tex\">{tex}\\frac{1}{3}+\\frac{1}{8}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\frac{3}{8}, \\frac{5}{12}, \\frac{11}{24}{\/tex}<\/span><\/li>\n<li>\u00a0Monthly profits (in Rs) of 100 shops<img decoding=\"async\" style=\"width: 250px; height: 207px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/dsGAaBk.png\" alt=\"\" data-imgur-src=\"dsGAaBk.png\" \/><\/li>\n<li>In\u00a0<span class=\"math-tex\">{tex}\\triangle{\/tex}<\/span>BCD, we have<br \/>\n<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>BDC +\u00a0<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>DCB +\u00a0<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>CBD = 180<sup>o<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a02a + 5a + 3a = 180<sup>o<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a010a = 180<sup>o<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0a = 18<sup>o<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>C = 5a = 5\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a018<sup>o<\/sup> = 90<sup>o<\/sup><br \/>\nSince opposite angles are equal in a parallelogram. Therefore,<br \/>\n<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>A =\u00a0<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>C\u00a0<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>A = 90<sup>o<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a02 (<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>A + <span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>B) = 360<sup>o <\/sup>[<span class=\"math-tex\">{tex}\\because{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>A =\u00a0<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>C and\u00a0<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>B =\u00a0<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>D]\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>A +\u00a0<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>B = 180<sup>o<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a090<sup>o<\/sup> +\u00a0<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>B = 180<sup>o\u00a0<\/sup>[<span class=\"math-tex\">{tex}\\because{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>A = 90<sup>o<\/sup>]\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>B = 90<sup>o<\/sup><br \/>\nHence,\u00a0<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>A = 90<sup>o<\/sup>,\u00a0\u00a0<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>B = 90<sup>o<\/sup>\u00a0<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>C = 90<sup>o<\/sup> and\u00a0<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>D = 90<sup>o<\/sup><\/li>\n<li>x + y \u2013 4 = 0<br \/>\n\u21d2 y = 4 \u2013 x<br \/>\nPut x = 0, then y = 4 \u2013 0 = 4<br \/>\nPut x = 1, then y = 4 \u2013 1 = 3<br \/>\nPut x = 2, then y = 4 \u2013 2 = 2<br \/>\nPut x = 3, then y = 4 \u2013 3 = 1<br \/>\n\u2234 (0, 4), (1, 3), (2, 2) and (3, 1) are the solutions of the equation x + y \u2013 4 = 0<\/li>\n<li style=\"clear: both;\">the population of Delhi State in different census years<img decoding=\"async\" style=\"width: 350px; height: 188px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/sNChKMP.png\" alt=\"\" data-imgur-src=\"sNChKMP.png\" \/>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Here, minimum class size = 2 &#8211; 1 = 1<br \/>\nAdjusted frequency of a class =\u00a0<span class=\"math-tex\">{tex}\\frac{Minimum \\ class \\ size}{Class \\ size \\ of \\ the \\ class}{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\times {\/tex}<\/span>\u00a0Frequency of the class<br \/>\n<strong>Frequency distribution after adjusting frequency<\/strong><\/p>\n<table style=\"width: 100%;\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<thead>\n<tr>\n<th scope=\"col\">Age<br \/>\n(in years)<\/th>\n<th scope=\"col\">Number of children<br \/>\n(Frequency)<\/th>\n<th scope=\"col\">Width of the<br \/>\nclass<\/th>\n<th scope=\"col\">Adjusted<br \/>\nfrequency<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td style=\"text-align: center;\">1-2<\/td>\n<td style=\"text-align: center;\">5<\/td>\n<td style=\"text-align: center;\">1<\/td>\n<td style=\"text-align: center;\"><span class=\"math-tex\">{tex}\\frac{1}{1}{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span> 5 = 5<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">2-3<\/td>\n<td style=\"text-align: center;\">4<\/td>\n<td style=\"text-align: center;\">1<\/td>\n<td style=\"text-align: center;\"><span class=\"math-tex\">{tex}\\frac{1}{1}{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a04 = 4<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">3-5<\/td>\n<td style=\"text-align: center;\">10<\/td>\n<td style=\"text-align: center;\">2<\/td>\n<td style=\"text-align: center;\"><span class=\"math-tex\">{tex}\\frac{1}{2}{\/tex}<\/span><span class=\"math-tex\">{tex}\\times{\/tex}<\/span>10 = 5<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">5-7<\/td>\n<td style=\"text-align: center;\">12<\/td>\n<td style=\"text-align: center;\">2<\/td>\n<td style=\"text-align: center;\"><span class=\"math-tex\">{tex}\\frac{1}{2}{\/tex}<\/span><span class=\"math-tex\">{tex}\\times{\/tex}<\/span>12 = 6<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">7-10<\/td>\n<td style=\"text-align: center;\">9<\/td>\n<td style=\"text-align: center;\">3<\/td>\n<td style=\"text-align: center;\"><span class=\"math-tex\">{tex}\\frac{1}{3}{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span> 9 = 3<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">10-15<\/td>\n<td style=\"text-align: center;\">10<\/td>\n<td style=\"text-align: center;\">5<\/td>\n<td style=\"text-align: center;\"><span class=\"math-tex\">{tex}\\frac{1}{5}{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span> 10 = 2<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">15-17<\/td>\n<td style=\"text-align: center;\">8<\/td>\n<td style=\"text-align: center;\">2<\/td>\n<td style=\"text-align: center;\"><span class=\"math-tex\">{tex}\\frac{1}{2}{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a08 = 4<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>The required histogram is as follows<br \/>\n<img decoding=\"async\" style=\"width: 250px; height: 140px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/Mb0TPHf.png\" alt=\"\" data-imgur-src=\"Mb0TPHf.png\" \/><\/li>\n<li><span class=\"math-tex\">{tex}{x^3} + 13{x^2} + 32x + 20{\/tex}<\/span><br \/>\nWe need to consider the factors of 20, which are <span class=\"math-tex\">{tex} \\pm 5, \\pm 4, \\pm 2, \\pm 1{\/tex}<\/span><br \/>\nLet us substitute -1 in the polynomial <span class=\"math-tex\">{tex}{x^3} + 13{x^2} + 32x + 20{\/tex}<\/span>, to get<br \/>\n<span class=\"math-tex\">{tex}{\\left( { &#8211; 1} \\right)^3} + 13{\\left( { &#8211; 1} \\right)^2} + 32\\left( { &#8211; 1} \\right) + 20 = &#8211; 1 + 13 &#8211; 32 + 20\\,\\, = &#8211; 20 + 20\\,\\, = 0{\/tex}<\/span><br \/>\nThus, according to factor theorem, we can conclude that <span class=\"math-tex\">{tex}\\left( {x + 1} \\right){\/tex}<\/span> is a factor of the polynomial <span class=\"math-tex\">{tex}{x^3} + 13{x^2} + 32x + 20{\/tex}<\/span><br \/>\nLet us divide the polynomial <span class=\"math-tex\">{tex}{x^3} + 13{x^2} + 32x + 20 by \\left( {x + 1} \\right){\/tex}<\/span>,to get<br \/>\n<img decoding=\"async\" style=\"width: 180px; height: 235px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/quesbank\/09\/maths\/ch-02\/img07.png\" \/><br \/>\n<span class=\"math-tex\">{tex}{x^3} + 13{x^2} + 32x + 20{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}= \\left( {x + 1} \\right)\\left( {{x^2} + 12x + 20} \\right){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}= \\left( {x + 1} \\right)\\left( {{x^2} + 2x + 10x + 20} \\right){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} = \\left( {x + 1} \\right)\\left[ {x\\left( {x + 2} \\right) + 10\\left( {x + 2} \\right)} \\right]{\/tex}<\/span><br \/>\nTherefore, we can conclude that on factorizing the polynomial <span class=\"math-tex\">{tex}{x^3} + 13{x^2} + 32x + 20{\/tex}<\/span>,\u00a0we get<br \/>\n<span class=\"math-tex\">{tex}\\left( {x + 1} \\right)\\left( {x &#8211; 10} \\right)\\left( {x + 2} \\right){\/tex}<\/span><\/li>\n<li style=\"text-align: center; display: block;\"><b>Section D <\/b><\/li>\n<li style=\"clear: both;\"><img decoding=\"async\" style=\"width: 110px; height: 86px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/kRfKEMo.png\" alt=\"\" data-imgur-src=\"kRfKEMo.png\" \/><br \/>\nDraw EF\u00a0<span class=\"math-tex\">{tex}\\parallel{\/tex}<\/span>\u00a0AB\u00a0<span class=\"math-tex\">{tex}\\parallel{\/tex}<\/span>\u00a0CD<br \/>\nNow, AB\u00a0<span class=\"math-tex\">{tex}\\parallel{\/tex}<\/span>\u00a0EF and BE is the transversal.<br \/>\nThen,<br \/>\n<span class=\"math-tex\">{tex}\\angle A B E=\\angle B E F{\/tex}<\/span>\u00a0[Alternate\u00a0Interior\u00a0Angles]\n<span class=\"math-tex\">{tex}\\Rightarrow \\angle B E F=35^{\\circ}{\/tex}<\/span><br \/>\nAgain, EF\u00a0<span class=\"math-tex\">{tex}\\parallel{\/tex}<\/span>\u00a0CD and DE is the transversal<br \/>\nThen,<br \/>\n<span class=\"math-tex\">{tex}\\angle D E F=\\angle F E D{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\angle F E D=65^{\\circ}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\therefore x^{\\circ}=\\angle B E F+\\angle F E D{\/tex}<\/span><br \/>\nx<span class=\"math-tex\">{tex}^\\circ{\/tex}<\/span>= 35<span class=\"math-tex\">{tex}^\\circ{\/tex}<\/span> + 65\u00b0<br \/>\nx<span class=\"math-tex\">{tex}^\\circ{\/tex}<\/span>= 100\u00b0<\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>We know that if two lines intersect, then the vertically-opposite angles are equal.<br \/>\nLet\u00a0<span class=\"math-tex\">{tex}\\angle B O C=\\angle A O D=x^{\\circ}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\angle B O C+\\angle A O D{\/tex}<\/span>\u00a0= 280<span class=\"math-tex\">{tex}^{\\circ}{\/tex}<\/span><br \/>\nx + x =\u00a0<span class=\"math-tex\">{tex}280^{\\circ}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a02x =\u00a0<span class=\"math-tex\">{tex}280^{\\circ}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0x =\u00a0\u00a0<span class=\"math-tex\">{tex}140^{\\circ}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\therefore \\angle B O C=\\angle A O D=140^{\\circ}{\/tex}<\/span><br \/>\nAlso, let\u00a0<span class=\"math-tex\">{tex}\\angle A O C=\\angle B O D=y^{\\circ}{\/tex}<\/span><br \/>\nWe know that the sum of all angles around a point is 360\u00b0<br \/>\n<span class=\"math-tex\">{tex}\\therefore \\angle A O C+\\angle B O C{\/tex}<\/span><span class=\"math-tex\">{tex}+\\angle B O D+\\angle A O D=360^{\\circ}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0y + 140 + y + 140 = 360\u00b0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a02y =\u00a0<span class=\"math-tex\">{tex}80^{\\circ}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0y = 40\u00b0<br \/>\nHence,\u00a0<span class=\"math-tex\">{tex}\\angle A O C=\\angle B O D=40^{\\circ}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\therefore \\angle B O C=\\angle A O D=140^{\\circ}{\/tex}<\/span>\u00a0and\u00a0<span class=\"math-tex\">{tex}\\angle A O C=\\angle B O D=40^{\\circ}{\/tex}<\/span><\/li>\n<li>Height of the conical tent <span class=\"math-tex\">{tex}\\left( h \\right){\/tex}<\/span>\u00a0= 8 m and Radius of the conical tent <span class=\"math-tex\">{tex}\\left( r \\right){\/tex}<\/span>\u00a0= 6 m<br \/>\nSlant height of the tent <span class=\"math-tex\">{tex}\\left( l \\right)=\\sqrt{{{r}^{2}}+{{h}^{2}}}{\/tex}<\/span><br \/>\n=<span class=\"math-tex\">{tex}\\sqrt{{{\\left( 6 \\right)}^{2}}+{{\\left( 8 \\right)}^{2}}}{\/tex}<\/span><br \/>\n=<span class=\"math-tex\">{tex}\\sqrt{36+64}{\/tex}<\/span><br \/>\n=<span class=\"math-tex\">{tex}\\sqrt{100}{\/tex}<\/span><br \/>\n= 10 m<br \/>\nArea of tarpaulin = Curved surface area of tent = <span class=\"math-tex\">{tex}\\pi rl{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}=3.14\\times 6\\times 10=188.4\\text{ }{{m}^{2}}{\/tex}<\/span><br \/>\nWidth of tarpaulin = 3 m<br \/>\nLet Length of tarpaulin = L<br \/>\n<span class=\"math-tex\">{tex}\\therefore {\/tex}<\/span>\u00a0Area of tarpaulin = <span class=\"math-tex\">{tex}Length\\times Breadth\\text{ }=\\text{ }L\\times 3{\/tex}<\/span>\u00a0= 3L<br \/>\nNow According to question, 3L = 188.4<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow {\/tex}<\/span>\u00a0L = 188.4\/3\u00a0= 62.8 m<br \/>\nThe extra length of the material required for stitching margins and cutting is 20 cm = 0.2 m.<br \/>\nSo the total length of tarpaulin bought is (62.8 + 0.2) m = 63 m<\/li>\n<li style=\"clear: both;\">Let:<br \/>\na = 85 m and b = 154 m<br \/>\nGiven that perimeter = 324 m<br \/>\nPerimeter= 2s = 324 m<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> s =<span class=\"math-tex\">{tex}\\frac{324}{2}{\/tex}<\/span>m<br \/>\nor, a + b + c = 324<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0c = 324 &#8211; 85 &#8211; 154 = 85 m<br \/>\nBy Herons&#8217;s formula, we have:<br \/>\nArea of triangle\u00a0<span class=\"math-tex\">{tex}=\\sqrt{s(s-a)(s-b)(s-c)}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}=\\sqrt{162(162-85)(162-154)(162-85)}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}=\\sqrt{162 \\times 77 \\times 8 \\times 77}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}=\\sqrt{1296 \\times 77 \\times 77}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}=\\sqrt{36 \\times 77 \\times 77 \\times 36}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}=36 \\times 77{\/tex}<\/span><br \/>\n= 2772 m<sup>2<\/sup><\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Let:<br \/>\na = 91 m, b = 98 m, and c = 105 m<br \/>\n<span class=\"math-tex\">{tex}\\therefore s=\\frac{a+b+c}{2}{\/tex}<\/span><span class=\"math-tex\">{tex}=\\frac{91+98+105}{2}{\/tex}<\/span>\u00a0= 147 m<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> s = 147 m<br \/>\nBy Heron&#8217;s formula, we have:<br \/>\nArea of triangle\u00a0<span class=\"math-tex\">{tex}=\\sqrt{s(s-a)(s-b)(s-c)}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}=\\sqrt{147(147-91)(147-98)(147-105)}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}=\\sqrt{147 \\times 56 \\times 49 \\times 42}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}=\\sqrt{7 \\times 3 \\times 7 \\times 2 \\times 2 \\times 2 \\times 7 \\times 7 \\times 7 \\times 7 \\times 3 \\times 2}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}=7 \\times 7 \\times 7 \\times 2 \\times 3 \\times 2{\/tex}<\/span><br \/>\n= 1446 m<sup>2<\/sup><br \/>\nWe know that the longest side is 105 m.<br \/>\nThus, we can find out the height of the triangle corresponding to 42 cm.<br \/>\nArea\u00a0of\u00a0triangle = 4116 m<sup>2<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\frac{1}{2} \\times{\/tex}<\/span>\u00a0Base\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a0Height = 4116\u00a0\u00a0<span class=\"math-tex\">{tex}\\Rightarrow \\frac{1}{2} \\times{\/tex}<\/span>(105)(Height) = 4116<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0Height\u00a0<span class=\"math-tex\">{tex}=\\frac{4116 \\times 2}{105}={\/tex}<\/span>\u00a078.4 m<\/li>\n<li>Let p(x) = ax<sup>3 <\/sup>+ bx<sup>2<\/sup> + x &#8211; 6 be the given polynomial.<br \/>\nNow,<br \/>\n(x + 2) is a factor of p(x)<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0p (-2) = 0 [<span class=\"math-tex\">{tex}\\because{\/tex}<\/span>\u00a0x + 2 = 0\u00a0<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0x = -2]\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0a(-2)<sup>3<\/sup> + b(-2)<sup>2<\/sup> + (-2) &#8211; 6 = 0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0-8a + 4b &#8211; 2 &#8211; 6 = 0\u00a0<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0-8a + 4b = 8\u00a0<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> -2a + b = 2 &#8230;(i)<br \/>\nIt is given that p(x) leaves the remainder 4 when it is divided by (x &#8211; 2).<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0p(2) = 4 [<span class=\"math-tex\">{tex}\\because{\/tex}<\/span>\u00a0x &#8211; 2 = 0\u00a0<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0x = 2]\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0a(2)<sup>3 <\/sup>+ b(2)<sup>2<\/sup> + 2 &#8211; 6 = 4<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a08a + 4b &#8211; 4 = 4\u00a0<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a08a + 4b = 0\u00a0<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a02a + b = 2 &#8230;(ii)<br \/>\nAdding (i) and (ii), we get<br \/>\n2b = 4\u00a0<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0b = 2<br \/>\nPutting b = 2 in (i), we get<br \/>\n&#8211; 2a + 2 = 2\u00a0<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0-2a = 0\u00a0<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0a = 0.<br \/>\nHence, a = 0 and b = 2<\/li>\n<li style=\"text-align: center; display: block;\"><b>Section E <\/b>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li style=\"text-align: left;\">Let the no of questions whose answer is known to Ajay be x and number questions attempted by guessing be y.<br \/>\nx + y = 110<br \/>\nx + 14y = 80\u00a0<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a04x + y = 320x + y = 110 &#8230;(1)<br \/>\n4x + y = 320\u00a0&#8230;(2)<br \/>\nSolving (1) and\u00a0(2)<br \/>\nx + y &#8211; 4x &#8211; y = 110 &#8211; 320 = -210<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0-3x = -210<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> x = 70<\/li>\n<li style=\"text-align: left;\">x + y = 110<br \/>\nx + 14y = 80 <span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a04x + y = 320<br \/>\nx + y = 110 &#8230;(1)<br \/>\n4x + y = 320 &#8230;(2)<br \/>\nSolving (1) and\u00a0(2)<br \/>\nx + y &#8211; 4x &#8211; y = 110 &#8211; 320 = &#8211; 210<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> &#8211; 3x = &#8211; 210<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0x = 70<br \/>\nPut x = 70 in (1)<br \/>\n70 + y = 110<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0y = 40<br \/>\n40 question he answered by guessing.<\/li>\n<li style=\"text-align: left;\"><span class=\"math-tex\">{tex}70 &#8211; 40 \\times \\frac{1}{4} = 70 &#8211; 10{\/tex}<\/span>\u00a0= 60 marks<br \/>\nHe scored 60 marks.x &#8211;\u00a0<span class=\"math-tex\">{tex}\\frac{1}{4}{\/tex}<\/span>(110 &#8211; x) = 95<br \/>\n<strong>OR<\/strong><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a04x &#8211; 110 + x = 380<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a05x = 380 + 110 = 490<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0x =\u00a0<span class=\"math-tex\">{tex}\\frac{490}{5}{\/tex}<\/span>\u00a0= 98<br \/>\nSo he answered 98 correct answers 12 by guessing.<\/li>\n<\/ol>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li style=\"text-align: left;\">In\u00a0<span class=\"math-tex\">{tex}\\triangle{\/tex}<\/span>APC and\u00a0<span class=\"math-tex\">{tex}\\triangle{\/tex}<\/span>BPC<br \/>\nAP = BP (Given)<br \/>\nCP = CP (common side)<br \/>\n<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>ACP =\u00a0<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>BCP = 90<sup>o<\/sup><br \/>\nBy RHS criteria\u00a0<span class=\"math-tex\">{tex}\\triangle{\/tex}<\/span>APC\u00a0<span class=\"math-tex\">{tex}\\cong{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\triangle{\/tex}<\/span>BPC<\/li>\n<li style=\"text-align: left;\">In\u00a0<span class=\"math-tex\">{tex}\\triangle{\/tex}<\/span>ACP<br \/>\n<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>APC +\u00a0<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>PAC +\u00a0<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>ACP = 180<sup>o<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0x + 60<sup>o<\/sup>\u00a0+ 90<sup>o<\/sup>\u00a0= 180<sup>o<\/sup>\u00a0(angle sum property of <span class=\"math-tex\">{tex}\\triangle{\/tex}<\/span>)<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>x = 180<sup>o<\/sup>\u00a0&#8211; 150<sup>o<\/sup>\u00a0= 30<sup>o<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>x = 30<sup>o<\/sup><\/li>\n<li style=\"text-align: left;\">In\u00a0<span class=\"math-tex\">{tex}\\triangle{\/tex}<\/span>APC and\u00a0<span class=\"math-tex\">{tex}\\triangle{\/tex}<\/span>BPC<br \/>\nCorresponding part of congruent triangle<br \/>\n<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>X =\u00a0<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>Y<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>Y = 30<sup>o<\/sup>\u00a0(given <span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>X = 30<sup>o<\/sup>)<br \/>\n<strong>OR<\/strong><br \/>\nIn\u00a0<span class=\"math-tex\">{tex}\\triangle{\/tex}<\/span>APC and\u00a0<span class=\"math-tex\">{tex}\\triangle{\/tex}<\/span>BPC<br \/>\nCorresponding part of congruent triangle<br \/>\n<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>PAC =\u00a0<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>PBC<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>PBC = 60<sup>o<\/sup>\u00a0(given <span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>PAC = 60<sup>o<\/sup>)<\/li>\n<\/ol>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li style=\"text-align: left;\">In\u00a0<span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>AOP and\u00a0<span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>BOP<br \/>\n<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>APO = <span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>BPO (Given)<br \/>\nOP = OP (Common)<br \/>\nAO = OB (radius of circle)<br \/>\n<span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>AOP\u00a0<span class=\"math-tex\">{tex}\\cong{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>BOP<br \/>\nAP = BP (CPCT)<\/li>\n<li style=\"text-align: left;\">In right\u00a0<span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>COQ<br \/>\nCO<sup>2<\/sup> = OQ<sup>2<\/sup> + CQ<sup>2<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a010<sup>2<\/sup> = 8<sup>2<\/sup> + CQ<sup>2<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0CQ<sup>2<\/sup> = 100 &#8211; 64 = 36<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0CQ = 6<br \/>\nCD = 2CQ<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0CD = 12 cm<!--EndFragment --><\/li>\n<li style=\"text-align: left;\">In right <span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>AOB<br \/>\nAO<sup>2<\/sup> = OP<sup>2<\/sup> + AP<sup>2<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a010<sup>2<\/sup> = 6<sup>2 <\/sup>+ AP<sup>2<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0AP<sup>2<\/sup> = 100 &#8211; 36 = 64<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0AP = 8<br \/>\nAB = 2AP<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0AB = 16 cm<br \/>\n<strong>OR<\/strong><br \/>\nThere is one and only one circle passing through three given non-collinear points.<br \/>\nTo improve your exam performance, <strong>download the <a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\">myCBSEguide<\/a> app and <a href=\"https:\/\/mycbseguide.com\/\">myCBSEguide<\/a> Website<\/strong>\u00a0for access to complete study materials for <strong>CBSE<\/strong>, <strong>NCERT<\/strong>, <strong>JEE (Main)<\/strong>, <strong>NEET-UG<\/strong>, and <strong>NDA exams<\/strong>. The app offers <strong>practice questions<\/strong>, <strong>sample papers<\/strong>, <strong>mock tests<\/strong>, and <strong>NCERT solutions<\/strong>, making it an essential resource for students aiming for top scores. Whether you need to revise concepts or test your knowledge, myCBSEguide provides everything to help you prepare effectively.For <strong>teachers<\/strong>, the <strong><a href=\"https:\/\/examin8.com\/\">Examin8<\/a> Website and <a href=\"https:\/\/play.google.com\/store\/apps\/details?id=com.examin8.exam\">Examin8<\/a> app<\/strong>\u00a0is the perfect tool to create <strong>personalized question papers<\/strong>. Teachers can design custom exams with their own name and logo, streamlining the paper creation process while maintaining exam integrity. Solving the <strong>Class 9 Maths Sample Paper 2024-25<\/strong> will help students familiarize themselves with the exam pattern.<strong>Download myCBSEguide now<\/strong> for comprehensive study material and exam resources, and <strong>Examin8 for teachers<\/strong> to create customized question papers easily.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<p><strong>Sample Papers for Class 9 all subjects 2024-25<\/strong><\/p>\n<ul>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-09-mathematics\/1234\/\"><strong>Mathematics<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-09-science\/1218\/\"><strong>Science<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-09-social-science\/1895\/\"><strong>Social Science<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-09-english-language-&amp;-literature\/1902\/\"><strong>English Language and Literature<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-09-%E0%A4%B9%E0%A4%BF%E0%A4%82%E0%A4%A6%E0%A5%80-%E0%A4%8F\/1904\/\"><strong>Hindi\u00a0 A<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-09-%E0%A4%B9%E0%A4%BF%E0%A4%82%E0%A4%A6%E0%A5%80-%E0%A4%AC\/1906\/\"><strong>Hindi B<\/strong><\/a><\/li>\n<\/ul>\n<p>To excel in your <strong>Class 9 exams<\/strong>, download <strong>sample papers<\/strong> for <strong>Science<\/strong>, <strong>Social Science<\/strong>, <strong>Mathematics<\/strong>, <strong>English Communicative<\/strong>, <strong>English Language and Literature<\/strong>, <strong>Hindi Course A<\/strong>, and <strong>Hindi Course B<\/strong> from the <strong> <a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\">myCBSEguide<\/a> app and <a href=\"https:\/\/mycbseguide.com\/\">myCBSEguide<\/a> Website<\/strong>. These sample papers are designed to help students practice effectively and boost their exam preparation. The <strong>Class 9 Maths Sample Paper 2024-25<\/strong> is an essential resource for students preparing for their final exams. Download the <strong>Class 9 Maths Sample Paper 2024-25<\/strong> from myCBSEguide for accurate exam preparation.<\/p>\n<p>myCBSEguide offers a wide range of resources, including:<\/p>\n<ul>\n<li><strong>Sample Papers with Solutions<\/strong>: Practice with real exam-like questions and detailed solutions.<\/li>\n<li><strong>Chapter-wise Test Papers<\/strong>: Ideal for focused practice on each subject.<\/li>\n<li><strong>NCERT Solutions<\/strong> &amp; <strong>NCERT Exemplar Solutions<\/strong>: Complete solutions for all NCERT exercises.<\/li>\n<li><strong>Quick Revision Notes<\/strong>: Ready-to-refer notes for last-minute revision.<\/li>\n<li><strong>CBSE Guess Papers<\/strong> &amp; <strong>Important Question Papers<\/strong>: Focus on high-weightage topics.<\/li>\n<\/ul>\n<p><strong>Download the myCBSEguide app<\/strong> today for easy access to all these resources and get the best exam preparation materials in one place!<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Class 9 Maths Sample Paper 2025 The CBSE Class 9 Maths Sample Paper 2024-25 includes both subjective and objective type questions, with a special focus on case study-based questions in the subjective section. Unlike Class 10, where students have the option to choose from multiple courses, all Class 9 Mathematics students study the same course &#8230; <a title=\"Class 9 Maths Sample Paper 2024-25\" class=\"read-more\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-9-mathematics\/\" aria-label=\"More on Class 9 Maths Sample Paper 2024-25\">Read more<\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1339,1987,1376,1989],"tags":[1855,1527,12,1341,1933,1959,1967,1344,1049,321,1340],"class_list":["post-13750","post","type-post","status-publish","format-standard","hentry","category-cbse-sample-papers","category-class-9-sample-papers","category-mathematics-cbse-class-09","category-maths-sample-papers","tag-cbse-class-9","tag-cbse-question-paper","tag-cbse-sample-papers","tag-cbse-sample-papers-2018-19","tag-cbse-sample-papers-2023","tag-cbse-sample-papers-2024","tag-cbse-sample-papers-2025","tag-class-9","tag-class-9-mathematics","tag-mathematics","tag-model-question-papers"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.0 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Class 9 Maths Sample Paper 2024-25 | myCBSEguide<\/title>\n<meta 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