{"id":30244,"date":"2023-08-03T14:36:16","date_gmt":"2023-08-03T09:06:16","guid":{"rendered":"https:\/\/mycbseguide.com\/blog\/?p=30244"},"modified":"2025-10-08T17:18:47","modified_gmt":"2025-10-08T11:48:47","slug":"class-10-maths-sample-paper","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/class-10-maths-sample-paper\/","title":{"rendered":"Class 10 Basic Maths Sample Paper 2025"},"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\/class-10-maths-sample-paper\/#Class_10_Basic_Maths_Sample_Paper_2025\" >Class 10 Basic 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\/class-10-maths-sample-paper\/#Class_10_%E2%80%93_Basic_Mathematics_Sample_Paper_%E2%80%93_01_2024-25\" >Class 10 &#8211; Basic Mathematics \nSample Paper &#8211; 01 (2024-25)<\/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\/class-10-maths-sample-paper\/#Solutions\" >Solutions<\/a><ul class='ez-toc-list-level-3' ><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/mycbseguide.com\/blog\/class-10-maths-sample-paper\/#myCBSEguide_App_for_Board_Students\" >myCBSEguide App for Board Students<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/mycbseguide.com\/blog\/class-10-maths-sample-paper\/#To_excel_in_the_Class_10_Basic_Maths_exam_2025_practicing_with_high-quality_sample_papers_is_essential_Our_sample_paper_for_Class_10_Basic_Maths_2025_offers_a_comprehensive_set_of_questions_that_follow_the_CBSE_exam_pattern\" >To excel in the Class 10 Basic Maths exam 2025, practicing with high-quality sample papers is essential. Our sample paper for Class 10 Basic Maths 2025 offers a comprehensive set of questions that follow the CBSE exam pattern.<\/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\/class-10-maths-sample-paper\/#Why_Choose_myCBSEguide_for_Your_Exam_Preparation\" >Why Choose myCBSEguide for Your Exam Preparation?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"Class_10_Basic_Maths_Sample_Paper_2025\"><\/span>Class 10 Basic Maths Sample Paper 2025<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>Prepare for your exams with the <strong>Class 10 Basic Maths Sample Paper 2025<\/strong> available on <strong>myCBSEguide<\/strong>.\u00a0By solving the <strong>Class 10 Basic Maths Sample Paper 2025<\/strong>, you can get familiar with key topics such as <strong>arithmetic<\/strong>, <strong>algebra<\/strong>, and <strong>geometry<\/strong>. Regular practice will boost your problem-solving skills and improve your confidence. Access the latest <strong>Model papers<\/strong> and study.\u00a0Access Mock Papers through the <a href=\"https:\/\/play.google.com\/store\/search?q=EXAMIN8&amp;c=apps\"><strong>myCBSEguide app<\/strong><\/a>, which provides detailed solutions and expert guidance. The <strong>Class 10 Basic Maths Sample Paper 2025<\/strong> is designed to mirror the actual exam format, making it an ideal resource for mock test practice. Stay one step ahead in your <strong><a href=\"https:\/\/mycbseguide.com\/cbse-sample-papers-class-10.html\">Class 10<\/a> Basic Maths<\/strong> preparation by using the <strong>Class 10 Basic Maths Sample Paper 2025. <\/strong>The <strong><a href=\"https:\/\/play.google.com\/store\/search?q=EXAMIN8&amp;c=apps\">Examin8 app<\/a><\/strong> allows educators to design high-quality practice tests and exams that cater to their specific teaching requirements, ensuring a streamlined and efficient exam preparation process for students.<\/p>\n<p><strong>Basic Maths<\/strong> has been newly introduced as subject for <strong>Class 10<\/strong> students under the <strong>CBSE<\/strong> curriculum. This subject is designed for students who prefer a less advanced mathematical approach, focusing on practical topics. <strong>Basic Maths<\/strong> covers essential concepts such as <strong>arithmetic<\/strong>, <strong>algebra<\/strong>, <strong>geometry<\/strong>, and <strong>statistics<\/strong>, without the complexity of <strong>Standard Maths<\/strong>. Download the <strong>myCBSEguide app<\/strong> for <strong>Basic Maths sample papers<\/strong> and study resources. And for more Model Papers visit <strong><a href=\"https:\/\/mycbseguide.com\/\">myCBSEGuide<\/a><\/strong> Website.<\/p>\n<p style=\"text-align: left;\"><strong>Maths Standard Sample Paper 2025 with Solution<\/strong> <a class=\"button\" style=\"font-weight: bold;\" href=\"https:\/\/mycbseguide.com\/dashboard\/category\/1202\/type\/2\">Download as PDF<\/a><\/p>\n<p style=\"text-align: left;\"><strong>Maths Basic Sample Paper 2025 with Solution<\/strong> <a class=\"button\" style=\"font-weight: bold;\" href=\"https:\/\/mycbseguide.com\/dashboard\/category\/1202\/type\/2\">Download as PDF<\/a><\/p>\n<p style=\"text-align: left;\"><strong>Class 11 Sample Papers<\/strong> <a class=\"button\" style=\"font-weight: bold;\" href=\"https:\/\/mycbseguide.com\/cbse-sample-papers-class-10.html\">Download as PDF<\/a><\/p>\n<h2 style=\"text-align: center;\"><span class=\"ez-toc-section\" id=\"Class_10_%E2%80%93_Basic_Mathematics_Sample_Paper_%E2%80%93_01_2024-25\"><\/span><strong>Class 10 &#8211; Basic Mathematics<br \/>\nSample Paper &#8211; 01 (2024-25)<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\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<p>To prepare effectively for the upcoming exams, practicing with the <strong>Class 10 Basic Maths Sample Paper 2025<\/strong> is essential. These sample papers are designed to help you understand the format of the exam and get a feel for the types of questions that might appear. By solving the <strong>2025<\/strong> <strong>Class 10 Basic Maths Sample Paper<\/strong>, you can identify areas that need improvement and focus on mastering them before the exams.<\/p>\n<p style=\"text-align: center;\"><b>Section A<\/b><\/p>\n<ol style=\"padding-left: 20px; list-style: decimal;\">\n<li>The product of a rational number and an irrational number is\n<div style=\"margin-left: 20px;\">\n<p>a)both rational and irrational number<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)an integer<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)an irrational number only<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)a rational number only<\/p>\n<\/div>\n<\/li>\n<li>The LCM of the smallest prime number and the smallest odd composite number is:\n<div style=\"margin-left: 20px;\">\n<p>a)10<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)9<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)6<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)18<\/p>\n<\/div>\n<\/li>\n<li>4x<sup>2<\/sup> &#8211; 20x + 25 = 0 have\n<div style=\"margin-left: 20px;\">\n<p>a)Real roots<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)No Real roots<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)Real and Equal roots<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)Real and Distinct roots<\/p>\n<\/div>\n<\/li>\n<li>The value of <strong>p<\/strong> if (-2, p) lies on the line represented by the equation 2x &#8211; 3y + 7 = 0, is\n<div style=\"margin-left: 20px;\">\n<p>a)-1<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)<span class=\"math-tex\">{tex}\\frac{13}{2}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)<span class=\"math-tex\">{tex}-\\frac{13}{2}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)1<\/p>\n<\/div>\n<\/li>\n<li>The roots of the quadratic equation 4x<sup>2<\/sup> &#8211; 5x + 4 = 0 are\n<div style=\"margin-left: 20px;\">\n<p>a)irrational<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)rational and equal<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)rational and distinct<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)not real<\/p>\n<\/div>\n<\/li>\n<li>If A = (-1, 2), B = (2, -1)\u00a0and C = (3, 1)\u00a0are any three vertices of a parallelogram, then find D (a, b)\n<div style=\"margin-left: 20px;\">\n<p>a)a = -2, b = 0<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)a = 2, b = 0<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)a = -2, b = 6<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)a = 0, b = 4<\/p>\n<\/div>\n<\/li>\n<li>ABCD is a trapezium such that BC || AD and AB = 4 cm. If the diagonals AC and BD intersect at O such that <span class=\"math-tex\">{tex} \\frac{AO}{OC}{\/tex}<\/span> =<span class=\"math-tex\">{tex}\\frac{DO}{OB}{\/tex}<\/span> =\u00a0<span class=\"math-tex\">{tex}\\frac{1}{2}{\/tex}<\/span>\u00a0then, Find DC.\n<div style=\"margin-left: 20px;\">\n<p>a)8 cm<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)6 cm<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)9 cm<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)7 cm<\/p>\n<\/div>\n<\/li>\n<li>In a \u25b3ABC, it is given that AD is the internal bisector of \u2220A. If AB = 10 cm, AC = 14 cm and BC = 6 cm, the CD = ?<img decoding=\"async\" style=\"width: 286px; height: 248px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/xOLAh4y.png\" alt=\"\" data-imgur-src=\"xOLAh4y.png\" \/>\n<div style=\"margin-left: 20px;\">\n<p>a)3.5 cm<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)7 cm<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)4.8 cm<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)10.5 cm<\/p>\n<\/div>\n<\/li>\n<li>In the given figure, AB is a tangent to the circle centered at O. If OA = 6 cm and <span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>OAB = 30<sup>o<\/sup>, then the radius of the circle is:<br \/>\n<img decoding=\"async\" style=\"height: 87px; width: 120px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1689233226-qqa7v3.jpg\" \/><\/p>\n<div style=\"margin-left: 20px;\">\n<p>a)<span class=\"math-tex\">{tex}3\\sqrt 3{\/tex}<\/span>\u00a0cm<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)2 cm<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)<span class=\"math-tex\">{tex}\\sqrt 3{\/tex}<\/span> cm<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)3 cm<\/p>\n<\/div>\n<\/li>\n<li>If 5 tan A = 3, then the value of cot A is:\n<div style=\"margin-left: 20px;\">\n<p>a)<span class=\"math-tex\">{tex} \\frac{3}{4}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)<span class=\"math-tex\">{tex} \\frac{5}{3}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)<span class=\"math-tex\">{tex} \\frac{3}{5}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)<span class=\"math-tex\">{tex} \\frac{4}{5}{\/tex}<\/span><\/p>\n<\/div>\n<\/li>\n<li>A vertical pole 10 m long casts a shadow of length 5 m on the ground. At the same time, a tower casts a shadow of length 12.5 m on the ground. The height of the tower is:\n<div style=\"margin-left: 20px;\">\n<p>a)22 m<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)25 m<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)24 m<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)20 m<\/p>\n<\/div>\n<\/li>\n<li>If <span class=\"math-tex\">{tex}\\sin A = \\frac{{12}}{{13}}{\/tex}<\/span>, then tan A =\n<div style=\"margin-left: 20px;\">\n<p>a)<span class=\"math-tex\">{tex}\\frac{{13}}{{12}}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)<span class=\"math-tex\">{tex}\\frac{{12}}{5}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)<span class=\"math-tex\">{tex}\\frac{{13}}{5}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)<span class=\"math-tex\">{tex}\\frac{5}{{12}}{\/tex}<\/span><\/p>\n<\/div>\n<\/li>\n<li>The angle described by the hour hand in 2 hours is\n<div style=\"margin-left: 20px;\">\n<p>a)360<sup>o<\/sup><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)60<sup>o<\/sup><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)90<sup>o<\/sup><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)30<sup>o<\/sup><\/p>\n<\/div>\n<\/li>\n<li>Area of a sector of angle p (in degrees) of a circle with radius R is\n<div style=\"margin-left: 20px;\">\n<p>a)<span class=\"math-tex\">{tex}\\frac { p } { 360 } \\times 2 \\pi R{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)<span class=\"math-tex\">{tex}\\frac { \\mathrm { p } } { 180 } \\times \\pi \\mathrm { R } ^ { 2 }{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)<span class=\"math-tex\">{tex}\\frac { p } { 180 } \\times 2 \\pi \\mathrm { R }{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)<span class=\"math-tex\">{tex}\\frac { p } { 720 } \\times 2 \\pi R ^ { 2 }{\/tex}<\/span><\/p>\n<\/div>\n<\/li>\n<li>If a digit is chosen at random from the digits 1, 2, 3, 4, 5, 6, 7, 8, 9, then the probability that it is odd and is a multiple of 3 is\n<div style=\"margin-left: 20px;\">\n<p>a)<span class=\"math-tex\">{tex}\\frac{1}{9}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)<span class=\"math-tex\">{tex}\\frac{2}{9}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)<span class=\"math-tex\">{tex}\\frac{2}{3}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)<span class=\"math-tex\">{tex}\\frac{1}{3}{\/tex}<\/span><\/p>\n<\/div>\n<\/li>\n<li>The median and mode of distribution are 20 and 18, then the mean is\n<div style=\"margin-left: 20px;\">\n<p>a)21<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)22<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)20<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)18<\/p>\n<\/div>\n<\/li>\n<li>A sphere is placed inside a right circular cylinder so as to touch the top, base and lateral surface of the cylinder. If the radius of the sphere is r, then the volume of the cylinder is\n<div style=\"margin-left: 20px;\">\n<p>a)2<span class=\"math-tex\">{tex}\\pi{\/tex}<\/span>r<sup>3<\/sup><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)8<span class=\"math-tex\">{tex}\\pi{\/tex}<\/span>r<sup>3<\/sup><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)<span class=\"math-tex\">{tex}\\frac{8}{3} \\pi r^{3}{\/tex}<\/span><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)4<span class=\"math-tex\">{tex}\\pi{\/tex}<\/span>r<sup>3<\/sup><\/p>\n<\/div>\n<\/li>\n<li>The wickets taken by a bowler in 10 cricket matches are 2, 6, 4, 5, 0, 3, 1, 3, 2, 3. The mode of the data is\n<div style=\"margin-left: 20px;\">\n<p>a)1<\/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)4<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)3<\/p>\n<\/div>\n<\/li>\n<li><strong>Assertion (A):<\/strong> The points (k, 2 &#8211; 2k), (-k +1, 2k) and (-4 &#8211; k, 6 &#8211; 2k) are collinear if k = <span class=\"math-tex\">{tex}\\frac{1}{2}{\/tex}<\/span><br \/>\n<strong>Reason (R):<\/strong> Three points A, B and C are collinear in the same straight line if AB + BC = AC<\/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> H.C.F. of smallest prime and smallest composite is 2.<br \/>\n<strong>Reason (R):<\/strong> Smallest prime is 2 and smallest composite is 4 so their H.C.F. is 2.<\/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>Maximize your exam preparation by downloading the <a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\"><strong>myCBSEguide app<\/strong><\/a>. This all-in-one learning platform provides comprehensive <strong>study material for CBSE<\/strong>, <strong>NCERT<\/strong>, <strong>JEE (Main)<\/strong>, <strong>NEET-UG<\/strong>, and <strong>NDA exams<\/strong>. With access to <strong>practice questions<\/strong>, <strong>sample papers<\/strong>, <strong>previous year papers<\/strong>, and <strong>detailed solutions<\/strong>, the app helps you study smarter and achieve better results. Get ahead of the competition and prepare efficiently with <a href=\"https:\/\/mycbseguide.com\/\"><strong>myCBSEguide<\/strong><\/a>.<\/p>\n<p><strong>Examin8 App: Empower Teachers to Create Custom Exam Papers<\/strong><br \/>\nFor teachers looking to create personalized assessments, the <a href=\"https:\/\/play.google.com\/store\/search?q=EXAMIN8&amp;c=apps\"><strong>Examin8 app<\/strong><\/a> offers a simple and effective solution. Teachers can easily generate <strong>customized question papers<\/strong> with their own name, logo, and branding. The <strong>Examin8 app<\/strong> allows educators to design high-quality practice tests and exams that cater to their specific teaching requirements, ensuring a streamlined and efficient exam preparation process for students.<\/p>\n<p>Familiarity with the exam format is key to performing well, and the <strong>Class 10 Basic Maths Sample Paper 2025<\/strong> is the perfect tool for this. It provides an accurate representation of the structure, question types, and marking scheme of the<strong> Basic Maths<\/strong> <strong>CBSE Class 10 exam 2025<\/strong>. By practicing with the <strong> Sample Paper <\/strong><strong>Class 10 Basic Maths 2025<\/strong>, you will feel more confident and prepared when sitting for your final exams.<\/p>\n<\/div>\n<\/li>\n<li style=\"text-align: center; clear: both; display: block;\"><b>Section B<\/b><\/li>\n<li>Solve the following pair of equations by substitution method:<br \/>\n7x \u2013 15y = 2 &#8230;(1)<br \/>\nx + 2y = 3 &#8230;(2)<\/li>\n<li>In the given figure,<span class=\"math-tex\">{tex}\\frac{{AO}}{{OC}} = \\frac{{BO}}{{OD}} = \\frac{1}{2}{\/tex}<\/span> and AB = 5cm, Find the value of DC.<br \/>\n<img decoding=\"async\" style=\"height: 89px; width: 93px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/quesbank\/10\/math\/ch06\/image729.png\" \/><\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Prove that, In a right triangle, the square on the hypotenuse is equal to the sum of the squares of the other two sides.<\/li>\n<li>The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.<\/li>\n<li>If tan (A + B) = <span class=\"math-tex\">{tex}\\sqrt3{\/tex}<\/span>\u00a0and\u00a0tan (A &#8211; B) =\u00a0<span class=\"math-tex\">{tex}\\frac{1}{\\sqrt3}{\/tex}<\/span>; 0\u00b0 &lt; A + B <span class=\"math-tex\">{tex}\\leq{\/tex}<\/span>\u00a090\u00b0; A &gt; B, then find A and B.<\/li>\n<li>Find the area of a sector of a circle with radius 6 cm, if the angle of the sector is 60<sup>o<\/sup>.\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>An arc of length 15 cm subtends an angle of 45\u00b0 at the centre of a circle. Find in terms of <span class=\"math-tex\">{tex}\\pi{\/tex}<\/span> , the radius of the circle.<\/li>\n<li style=\"text-align: center; clear: both; display: block;\"><b>Section C<\/b><\/li>\n<li>Show that <span class=\"math-tex\">{tex}5-\\sqrt{3}{\/tex}<\/span>\u00a0is irrational.<\/li>\n<li>Find the zeroes of the given quadratic polynomial and verify the relationship between the zeroes and their coefficients for\u00a0<span class=\"math-tex\">{tex}x^2+x-12{\/tex}<\/span><\/li>\n<li>The ratio of incomes of two persons is 9 : 7 and the ratio of their expenditures is 4 : 3. If each of them manages to save \u20b9 2000 per month, then find their monthly incomes.\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Solve the system of the equation:<br \/>\n99x + 101y = 499<br \/>\n101x + 99y = 501<\/li>\n<li>The radii of two concentric circles are 13 cm and 8 cm.AB is a diameter of the bigger circle.BD is a tangent to the smaller circle touching it at D. Find the length of AD.<\/li>\n<li>In <span class=\"math-tex\">{tex}\\triangle A B C{\/tex}<\/span>,\u00a0right angled at B, if\u00a0<span class=\"math-tex\">{tex}\\tan A = \\frac { 1 } { \\sqrt { 3 } }{\/tex}<\/span>.\u00a0Find the value of cos A cos C &#8211; sin A sin C\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>If cosec<span class=\"math-tex\">{tex}\\theta{\/tex}<\/span>\u00a0= 2, show that\u00a0<span class=\"math-tex\">{tex}\\left( \\cot \\theta + \\frac { \\sin \\theta } { 1 + \\cos \\theta } \\right){\/tex}<\/span> = 2.<\/li>\n<li>A die is thrown once. What is the probability that it shows\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>a &#8216;3&#8217;,<\/li>\n<li>a &#8216;5&#8217;,<\/li>\n<li>an odd number<\/li>\n<li>a number greater than 4?<\/li>\n<\/ol>\n<\/li>\n<li style=\"text-align: center; clear: both; display: block;\"><b>Section D<\/b><\/li>\n<li>The difference of squares of two numbers is 204. The square of the smaller number is 4 less than 10 times the larger number. Find the two numbers.\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>A dealer\u00a0sells a toy for \u20b9\u00a024 and gains as much per cent as the cost price of the toy. Then find the cost price of the toy.38319<\/li>\n<li>If a perpendicular is drawn from the vertex containing the right angle to the right angle triangle to the hypotenuse, then prove that the triangle on each side of the perpendicular is similar to each other and to the original triangle. Also, prove that the square of the perpendicular is equal to the product of the lengths of the two parts of the hypotenuse.<\/li>\n<li>The interior of a building is in the form of cylinder of diameter 4.3 m and height 3.8 m, surmounted by a cone whose vertical angle is a right angle. Find the area of the surface and the volume of the building. (Use <span class=\"math-tex\">{tex} \\pi {\/tex}<\/span>\u00a0= 3.14).\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>The boilers are used in thermal power plants to store water and then used to produce steam. One such boiler consists of a cylindrical part in middle and two hemispherical parts at its both ends. Length of the cylindrical part is 7m and radius of cylindrical part is <span class=\"math-tex\">{tex}\\frac {7} {2}{\/tex}<\/span>m.<br \/>\nFind the total surface area and the volume of the boiler. Also, find the ratio of the volume of cylindrical part to the volume of one hemispherical part.<br \/>\n<img decoding=\"async\" style=\"height: 280px; width: 140px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1689139358-t2ns3a.jpg\" \/><\/li>\n<li>The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:<br \/>\n<table style=\"width: 100%;\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<tbody>\n<tr>\n<td style=\"text-align: center;\"><strong>Expenditure (in \u20b9)<\/strong><\/td>\n<td style=\"text-align: center;\"><strong>Frequency<\/strong><\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">1000-1500<\/td>\n<td style=\"text-align: center;\">24<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">1500-2000<\/td>\n<td style=\"text-align: center;\">40<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">2000-2500<\/td>\n<td style=\"text-align: center;\">33<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">2500-3000<\/td>\n<td style=\"text-align: center;\">28<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">3000-3500<\/td>\n<td style=\"text-align: center;\">30<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">3500-4000<\/td>\n<td style=\"text-align: center;\">22<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">4000-4500<\/td>\n<td style=\"text-align: center;\">16<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">4500-5000<\/td>\n<td style=\"text-align: center;\">7<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/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 \/>\nElpis Technology is a TV manufacturer company. It produces\u00a0smart TV sets not only for the Indian market but also exports them to many foreign countries. Their TV sets have been in demand every time but due to the Covid-19 pandemic, they are not getting sufficient spare parts, especially chips to accelerate the production.\u00a0They have to work in a limited capacity due to the\u00a0lack of raw materials.<br \/>\n<img decoding=\"async\" style=\"height: 90px; width: 300px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1642242777-tpuyrz.jpg\" alt=\"\" \/><\/p>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>They produced 600 sets in the third year and 700 sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find an increase in the production of TV\u00a0every year. (1)<\/li>\n<li>They produced 600 sets in the third year and 700 sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find in which year production of TV is 1000. (1)<\/li>\n<li>They produced 600 sets in the third year and 700 sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find\u00a0the production in the 10th\u00a0year. (2)<br \/>\n<strong>OR<\/strong><br \/>\nThey produced 600 sets in the third year and 700 sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find the total production in first 7 years. (2)<\/li>\n<\/ol>\n<\/li>\n<li><strong>Read the following text carefully and answer the questions that follow:<\/strong><br \/>\nThere are two routes to travel from source A to destination B by bus. First bus reaches at B via point C and second bus reaches from A to B directly. The position of A, B and C are represented in the following graph:<br \/>\nBased on the above information, answer the following questions.<br \/>\n<img decoding=\"async\" style=\"height: 59px; width: 250px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/1629864532-awcwet.jpg\" alt=\"\" \/><br \/>\n<strong>Scale: <\/strong>x-axis : 1 unit = 1 km<br \/>\ny-axis: 1 unit = 1 km<br \/>\n<img decoding=\"async\" style=\"height: 269px; width: 300px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1681983468-kbepbv.jpg\" \/><\/p>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>If the fare for the second bus is \u20b915\/km, then what will be the fare to reach to the destination by this bus? (1)<\/li>\n<li>What is the distance between A and B? (1)<\/li>\n<li>What is the distance between A and C? (2)<br \/>\n<strong>OR<\/strong><br \/>\nIf it is assumed that both buses have same speed, then by which bus do you want to travel from A to B? (2)<\/li>\n<\/ol>\n<\/li>\n<li><strong>Read the following text carefully and answer the questions that follow:<\/strong><br \/>\nIn a society, there are many multistory buildings. The RWA of the society wants to install a tower and a water tank so that all the households can get water without using water pumps.<br \/>\nFor this they have measured the height of the tallest building in the society and now they want to install a tower that will be taller than that so that the level of water must be higher than the tallest building in their society. Here is one solution they have found and now they want to check if it will work or not.<br \/>\nFrom a point on the ground 40 m away from the foot of a tower, the angle of elevation of the top of the tower is 300. the angle of elevation of the top of the water tank is 45<sup>o<\/sup>.<br \/>\n<img decoding=\"async\" style=\"height: 154px; width: 250px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/1663861466-md72t9.jpg\" \/><\/p>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>What is the height of the tower? (1)<\/li>\n<li>What is the height of the water tank? (1)<\/li>\n<li>At what distance from the bottom of the tower\u00a0the angle of elevation of the top of the tower is 45<sup>o<\/sup>. (2)<br \/>\n<strong>OR<\/strong><br \/>\nWhat will be the angle of elevation of the top of the water tank from the place at\u00a0<span class=\"math-tex\">{tex}\\frac{40}{\\sqrt{3}}{\/tex}<\/span>m from the bottom of the tower. (2)<br \/>\n<strong>Download the <a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\">myCBSEguide App<\/a> for Comprehensive Exam Preparation<\/strong><br \/>\nEnhance your exam preparation by downloading the <strong>myCBSEguide app<\/strong>. It offers complete study resources for <strong>CBSE<\/strong>, <strong>NCERT<\/strong>, <strong>JEE (Main)<\/strong>, <strong>NEET-UG<\/strong>, and <strong>NDA exams<\/strong>, including practice questions, sample papers, and detailed solutions. Stay ahead in your studies with curated materials for every subject. Visit our <strong>website<a href=\"https:\/\/mycbseguide.com\/\"> myCBSEGuide<\/a><\/strong><strong>Examin8 App for Teachers: Create Custom Question Papers<\/strong><br \/>\nFor educators, the <a href=\"https:\/\/examin8.com\/blog\/free-mobile-app\/\"><strong>Examin8 app<\/strong><\/a> is the perfect tool to design personalized test papers with custom branding, including your own name and logo. Easily create tailored assessments for students, ensuring that each paper meets specific learning objectives. Get started with our <strong>website<\/strong> <a href=\"https:\/\/examin8.com\/\"><strong>Examin8<\/strong><\/a> today to streamline your teaching process.Use the <strong>Class 10 Basic Maths Sample Paper 2025<\/strong> as a tool to assess your readiness for the final exam. By practicing with these sample papers, you can evaluate your strengths and weaknesses and get a clear idea of how well you are prepared for the actual test. The<strong> 2025<\/strong> <strong>Class 10 Basic Maths Sample Paper <\/strong>\u00a0is designed to mirror the actual exam format, making it an ideal resource for mock test practice.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<p style=\"text-align: center; page-break-before: always;\"><strong>Class 10 &#8211; Mathematics<br \/>\nBasic Sample Paper &#8211; 01 (2024-25)<\/strong><\/p>\n<hr \/>\n<h2 style=\"text-align: center;\"><span class=\"ez-toc-section\" id=\"Solutions\"><\/span><strong>Solutions<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\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;\">both rational and irrational number<\/p>\n<p><b>Explanation: <\/b><\/p>\n<p>The product of a rational number and an irrational number\u00a0can be either a rational number or an irrational number.<br \/>\ne.g\u00a0<span class=\"math-tex\">{tex} \\sqrt5{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span> <span class=\"math-tex\">{tex}\\sqrt2{\/tex}<\/span>\u00a0=\u00a0<span class=\"math-tex\">{tex} \\sqrt10{\/tex}<\/span>\u00a0which is\u00a0 irrational<br \/>\nbut \u00a0<span class=\"math-tex\">{tex} \\sqrt8{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span> <span class=\"math-tex\">{tex} \\sqrt2{\/tex}<\/span>\u00a0=\u00a0<span class=\"math-tex\">{tex} \\sqrt16{\/tex}<\/span>\u00a0\u00a0= 4 which is a rational number<br \/>\nThus, the product of two irrational numbers can be either rational or irrational<br \/>\nsimilarly, the product of rational and irrational numbers can be either rational or irrational<br \/>\n5\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span> <span class=\"math-tex\">{tex} \\sqrt2{\/tex}<\/span>\u00a0= 5<span class=\"math-tex\">{tex} \\sqrt2{\/tex}<\/span>\u00a0 which is irrational.<br \/>\nbut 0\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span> <span class=\"math-tex\">{tex} \\sqrt3{\/tex}<\/span> \u00a0= 0\u00a0which is rational.<\/li>\n<li>(d)\n<p style=\"display: inline;\">18<\/p>\n<p><b>Explanation: <\/b>Smallest prime no. = 2<br \/>\nSmallest composite no. = 9<br \/>\nLCM [2, 9] = 2\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a09 = 18<\/li>\n<li>(c)\n<p style=\"display: inline;\">Real and Equal roots<\/p>\n<p><b>Explanation: <\/b>D = b<sup>2<\/sup> &#8211; 4ac<br \/>\nD = (-20)<sup>2<\/sup> &#8211; 4\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a04\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span> 25<br \/>\nD = 400 &#8211; 400<br \/>\nD = 0. Hence Real and equal roots.<\/li>\n<li>(d)\n<p style=\"display: inline;\">1<\/p>\n<p><b>Explanation: <\/b>2x &#8211; 3y + 7 = 0<br \/>\n2(-2) -3p + 7 = 0<br \/>\n3p = 3<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> p =\u00a01<\/li>\n<li>(d)\n<p style=\"display: inline;\">not real<\/p>\n<p><b>Explanation: <\/b>not real<\/li>\n<li>(d)\n<p style=\"display: inline;\">a = 0, b = 4<\/p>\n<p><b>Explanation: <\/b>In parallelogram ABCD, diagonals AC and AD bisect each other at O.<br \/>\nO is mid-point of AC.<br \/>\n<img decoding=\"async\" style=\"width: 180px; height: 120px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/GPsGxOh.png\" alt=\"\" data-imgur-src=\"GPsGxOh.png\" \/><br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0Co-ordinates of O will be<br \/>\n<span class=\"math-tex\">{tex}\\left(\\frac{-1+3}{2}, \\frac{2+1}{2}\\right) \\text { or }\\left(\\frac{2}{2}, \\frac{3}{2}\\right) \\text { or }\\left(1, \\frac{3}{2}\\right){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\because{\/tex}<\/span>\u00a0O is mid-point of BD<br \/>\n<span class=\"math-tex\">{tex}\\therefore \\frac{2+a}{2}=1 \\text { and } \\frac{-1+b}{2}=\\frac{3}{2} \\Rightarrow 2+a=2{\/tex}<\/span><br \/>\nand -1 + b = 3\u00a0<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0b = 3 + 1 = 4<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0a = 0, b = 4<\/li>\n<li>(a)\n<p style=\"display: inline;\">8 cm<\/p>\n<p><b>Explanation: <\/b><\/p>\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/Picture1_1410873774952.jpg\" alt=\"\" data-imgur-src=\"Picture1_1410873774952.jpg\" \/><\/p>\n<p>In \u0394AOB and \u0394COD,<br \/>\n\u2220 AOB = \u2220COD\u00a0 \u00a0 \u00a0(Vertically opposite angles)<br \/>\n<span class=\"math-tex\">{tex}\\frac {AO} {OC}{\/tex}<\/span>= <span class=\"math-tex\">{tex}\\frac {DO} {OB}{\/tex}<\/span> \u00a0 (Given)<br \/>\nTherefore according to SAS similarity criterion, we have<br \/>\n\u0394AOB <span class=\"math-tex\">{tex}\\sim{\/tex}<\/span> \u0394COD<br \/>\n<span class=\"math-tex\">{tex}\\frac {AO}{OC}{\/tex}<\/span>= <span class=\"math-tex\">{tex}\\frac {BO}{OD}{\/tex}<\/span> = <span class=\"math-tex\">{tex}\\frac {AB}{DC}{\/tex}<\/span>[corresponding sides of similar triangles are proportional]\n\u21d2 <span class=\"math-tex\">{tex}\\frac 12{\/tex}<\/span>= <span class=\"math-tex\">{tex}\\frac 4{DC}{\/tex}<\/span><br \/>\n\u21d2 DC = 8 cm<\/li>\n<li>(a)\n<p style=\"display: inline;\">3.5 cm<\/p>\n<p><b>Explanation: <\/b>By using angle bisector theore in\u00a0\u25b3ABC, we have<br \/>\n<span class=\"math-tex\">{tex}\\frac{\\mathrm{AB}}{\\mathrm{AC}}=\\frac{\\mathrm{BD}}{\\mathrm{DC}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\frac{10}{14}=\\frac{6-x}{x}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow 10 x=84-14 x{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow 24 x=84 {\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow x=3.5{\/tex}<\/span><br \/>\nHence, the correct answer is 3.5.<\/li>\n<li>(d)\n<p style=\"display: inline;\">3 cm<\/p>\n<p><b>Explanation: <\/b>sin 30<sup>o<\/sup>\u00a0=\u00a0<span class=\"math-tex\">{tex}\\frac{OB}{OA}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\frac12{\/tex}<\/span>\u00a0=\u00a0<span class=\"math-tex\">{tex}\\frac r 6{\/tex}<\/span><br \/>\nr = 3 cm<\/li>\n<li>(b)\n<p style=\"display: inline;\"><span class=\"math-tex\">{tex} \\frac{5}{3}{\/tex}<\/span><\/p>\n<p><b>Explanation: <\/b>5 tan A = 3<br \/>\ntan A =\u00a0<span class=\"math-tex\">{tex}\\frac{3}{5}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\cot A=\\frac{1}{\\operatorname{tan} A}=\\frac{5}{3}{\/tex}<\/span><\/li>\n<li>(b)\n<p style=\"display: inline;\">25 m<\/p>\n<p><b>Explanation: <\/b>25 m<\/li>\n<li>(b)\n<p style=\"display: inline;\"><span class=\"math-tex\">{tex}\\frac{{12}}{5}{\/tex}<\/span><\/p>\n<p><b>Explanation: <\/b>Let BC = 12cm and AB = 13cm<br \/>\n<img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/quiz\/10\/math\/49092_e1.jpg\" \/><br \/>\n<span class=\"math-tex\">{tex}\\therefore {\/tex}<\/span> AC = <span class=\"math-tex\">{tex}\\sqrt {{{\\left( {13k} \\right)}^2} + {{\\left( {12k} \\right)}^2}} {\/tex}<\/span> = <span class=\"math-tex\">{tex}\\sqrt {169{k^2} &#8211; 144{k^2}} {\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow {\/tex}<\/span> AC = <span class=\"math-tex\">{tex}\\sqrt {25{k^2}} {\/tex}<\/span> = <span class=\"math-tex\">{tex}5k{\/tex}<\/span> 5k<br \/>\n<span class=\"math-tex\">{tex}\\therefore {\/tex}<\/span><span class=\"math-tex\">{tex}\\tan {\\text{A}} = \\frac{{12k}}{{5k}} = \\frac{{12}}{5}{\/tex}<\/span><\/li>\n<li>(b)\n<p style=\"display: inline;\">60<sup>o<\/sup><\/p>\n<p><b>Explanation: <\/b><\/p>\n<p><span class=\"math-tex\">{tex}\\because {\/tex}<\/span> Angle described by the hour hand in 12 hours = 360<sup>o<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\therefore {\/tex}<\/span> Angle described by the hour hand in 2 hours = <span class=\"math-tex\">{tex}\\frac{{{{360}^ \\circ }}}{{12}}{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a02\u00a0= 60<sup>o<\/sup><\/li>\n<li>(d)\n<p style=\"display: inline;\"><span class=\"math-tex\">{tex}\\frac { p } { 720 } \\times 2 \\pi R ^ { 2 }{\/tex}<\/span><\/p>\n<p><b>Explanation: <\/b><\/p>\n<p>Area of the\u00a0sector of angle p of a circle with radius R<\/p>\n<p><span class=\"math-tex\">{tex}= {\\theta \\over 360} \\times \\pi r ^2 = {p \\over 360} \\times \\pi R ^2{\/tex}<\/span><\/p>\n<p><span class=\"math-tex\">{tex}= {p \\over 2(360)} \\times 2 \\pi R ^2 = {p \\over 720} \\times 2\\pi R ^2{\/tex}<\/span><\/li>\n<li>(b)\n<p style=\"display: inline;\"><span class=\"math-tex\">{tex}\\frac{2}{9}{\/tex}<\/span><\/p>\n<p><b>Explanation: <\/b><\/p>\n<p>Total numbers of digits for 1 to 9(n) = 9<br \/>\nNumber divisible by 3(m) = 3, 6, 9<br \/>\nOdd numbers out of 3, 6, 9 = 3, 9<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0Probability <span class=\"math-tex\">{tex}=\\frac{m}{n}=\\frac{2}{9}{\/tex}<\/span><\/li>\n<li>(a)\n<p style=\"display: inline;\">21<\/p>\n<p><b>Explanation: <\/b>3 Median = 2<span class=\"math-tex\">{tex}\\times{\/tex}<\/span> Mean + Mode<br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow {\/tex}<\/span> 3\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span> 20 = 2<span class=\"math-tex\">{tex}\\times{\/tex}<\/span> Mean + 18<br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow {\/tex}<\/span> 2 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span>Mean = 60 &#8211; 18 = 42<br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow {\/tex}<\/span> Mean = 21<\/li>\n<li>(a)\n<p style=\"display: inline;\">2<span class=\"math-tex\">{tex}\\pi{\/tex}<\/span>r<sup>3<\/sup><\/p>\n<p><b>Explanation: <\/b>Volume of a sphere = (4\/3)<span class=\"math-tex\">{tex}\\pi{\/tex}<\/span>r<sup>3<\/sup><br \/>\nVolume of a cylinder = <span class=\"math-tex\">{tex}\\pi{\/tex}<\/span>r<sup>2<\/sup>h<br \/>\nGiven, sphere is placed inside a right circular cylinder so as to touch the top, base and lateral surface of the cylinder and the radius of the sphere is\u00a0<i>r.<\/i><br \/>\nThus, height of the cylinder = diameter = 2r and base radius = r<br \/>\nVolume of the cylinder = <span class=\"math-tex\">{tex}\\pi{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a0r<sup>2<\/sup>\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a02r = 2<span class=\"math-tex\">{tex}\\pi{\/tex}<\/span>r<sup>3<\/sup><\/li>\n<li>(d)\n<p style=\"display: inline;\">3<\/p>\n<p><b>Explanation: <\/b>In the given data, the frequency of 3 is more than those other wickets taken by a bowler.<br \/>\nTherefore, Mode of given data is 3.<\/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><\/p>\n<p>Both A and R are true and R is the correct explanation of A.<\/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>Smallest prime is 2 and the smallest composite is 4 so H.C.F. of 2 and 4 is 4.<\/p>\n<p><strong>Download the myCBSEguide App for Comprehensive Exam Preparation<\/strong><br \/>\nFor effective exam preparation, download the <a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\"><strong>myCBSEguide app<\/strong><\/a>, which offers complete <strong>study material for CBSE<\/strong>, <strong>NCERT<\/strong>, <strong>JEE (Main)<\/strong>, <strong>NEET-UG<\/strong>, and <strong>NDA exams<\/strong>. With a vast collection of practice questions, sample papers, and chapter-wise solutions, this app is your one-stop resource to study smarter and achieve better results.<\/p>\n<p><strong>Examin8 App: Empower Teachers to Create Custom Papers<\/strong><br \/>\nTeachers can enhance their teaching and assessment process using the <a href=\"https:\/\/play.google.com\/store\/search?q=EXAMIN8&amp;c=apps\"><strong>Examin8 app<\/strong><\/a>. This powerful tool allows educators to design <strong>customized exam papers<\/strong> with their name, logo, and branding. Whether for practice tests or final assessments, <a href=\"https:\/\/examin8.com\/\"><strong>Examin8<\/strong><\/a> Website provides an easy solution for creating professional, personalized question papers.<\/p>\n<p>An effective exam strategy can be developed by practicing with the <strong><a href=\"https:\/\/mycbseguide.com\/cbse-sample-papers-class-10.html\">Class 10<\/a> Basic Maths Sample Paper 2025<\/strong>. Solving sample papers not only helps you learn the material but also improves your time-management skills. The <strong>Class 10 Basic Maths Sample Paper 2025<\/strong> will help you understand how to allocate time for each section, ensuring that you can complete your exam within the given time frame.<\/li>\n<li style=\"text-align: center; display: block;\"><b>Section B <\/b><\/li>\n<li><strong>Step 1:<\/strong>\u00a0By substitution method,\u00a0we pick either of the equations and write one variable in terms of the other.<br \/>\n7x \u2013 15y = 2 &#8230;(1)<br \/>\nand\u00a0x + 2y = 3 &#8230;(2)<br \/>\nLet us consider the Equation (2):<br \/>\nx + 2y = 3<br \/>\nand write it as x = 3 \u2013 2y &#8230;(3)<br \/>\n<strong>Step 2:<\/strong>\u00a0Now substitute the value of x in Equation (1)<br \/>\nWe get 7(3 \u2013 2y) \u2013 15y = 2<br \/>\ni.e., 21 \u2013 14y \u2013 15y = 2<br \/>\ni.e., \u2013 29y = \u201319<br \/>\nTherefore\u00a0<span class=\"math-tex\">{tex}\\mathrm{y}=\\frac{19}{29}{\/tex}<\/span><br \/>\n<strong>Step 3: <\/strong>Substituting this value of y in Equation (3), we get<br \/>\n<span class=\"math-tex\">{tex}\\mathrm{x}=3-2\\left(\\frac{19}{29}\\right)=\\frac{49}{29}{\/tex}<\/span><br \/>\nTherefore, the solution is x =\u00a0<span class=\"math-tex\">{tex}\\frac{49}{29}, \\mathrm{y}=\\frac{19}{29}{\/tex}<\/span><\/li>\n<li style=\"clear: both;\">In \u200b<span class=\"math-tex\">{tex}\\triangle {\/tex}<\/span>AOB and \u200b<span class=\"math-tex\">{tex}\\triangle {\/tex}<\/span>\u200bCOD<br \/>\n\u200b<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>\u200b AOB = \u200b<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>\u200b COD [Vertically opposite angles]\n<span class=\"math-tex\">{tex}\\frac{{AO}}{{OC}} = \\frac{{BO}}{{OD}} \\Rightarrow \\frac{{AO}}{{OB}} = \\frac{{OC}}{{OD}}{\/tex}<\/span>[Given]\n\u200b<span class=\"math-tex\">{tex}\\therefore {\/tex}<\/span>\u200b \u200b<span class=\"math-tex\">{tex}\\triangle {\/tex}<\/span>\u200bAOB \u200b<span class=\"math-tex\">{tex} \\sim {\/tex}<\/span>\u200b <span class=\"math-tex\">{tex}\\triangle {\/tex}<\/span>COD [By SAS similarity]\n\u200b<span class=\"math-tex\">{tex}\\therefore {\/tex}<\/span>\u200b<span class=\"math-tex\">{tex}\\frac{{AO}}{{OC}} = \\frac{{BO}}{{OD}} = \\frac{{AB}}{{CD}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\frac{1}{2} = \\frac{{AB}}{{DC}}\\left[ {\\frac{{AO}}{{OC}} = \\frac{{BO}}{{OD}} = \\frac{1}{2}} \\right]{\/tex}<\/span>is given<br \/>\n\u200b<span class=\"math-tex\">{tex}\\Rightarrow {\/tex}<\/span>\u200b <span class=\"math-tex\">{tex}\\frac{1}{2} = \\frac{5}{{DC}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow {\/tex}<\/span> DC = 10 cm<\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Given :\u00a0<span class=\"math-tex\">{tex}\\triangle \\mathrm{ABC}{\/tex}<\/span>\u00a0is right angle at B<br \/>\nTo prove:\u00a0\u00a0<span class=\"math-tex\">{tex}A C^{2}=A B^{2}+B C^{2}{\/tex}<\/span><br \/>\n<img decoding=\"async\" style=\"height: 200px; width: 200px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/u1wwzZM.png\" alt=\"\" \/><\/p>\n<p>Construction: Draw\u00a0<span class=\"math-tex\">{tex}B D \\perp A C{\/tex}<\/span><\/p>\n<p>Proof:In\u00a0<span class=\"math-tex\">{tex}\\triangle{\/tex}<\/span>ABC and\u00a0<span class=\"math-tex\">{tex}\\triangle{\/tex}<\/span>ABD<\/p>\n<p><span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>ABC=<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>ADB=90<\/p>\n<p><span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>A is common<\/p>\n<p><span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>ABC<span class=\"math-tex\">{tex}\\sim{\/tex}<\/span><span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>ADB<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\frac{A D}{A B}=\\frac{A B}{A C}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0AD. AC = AB<sup>2<\/sup>\u00a0&#8230;(i)<br \/>\nSimilarly\u00a0<span class=\"math-tex\">{tex}\\Delta \\mathrm{BDC} \\sim \\Delta \\mathrm{ABC}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\frac{C D}{B C}=\\frac{B C}{A C}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0CD.AC = BC<sup>2<\/sup>\u00a0&#8230;(ii)<br \/>\nAdding (i) and (ii)<br \/>\nAD.AC + CD. AC= AB<sup>2<\/sup> + BC<sup>2<\/sup><br \/>\nAC(AD + CD) = AB<sup>2<\/sup> + BC<sup>2<\/sup><br \/>\nAC\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a0AC = AB<sup>2<\/sup> + BC<sup>2<\/sup><br \/>\nAC<sup>2<\/sup> = AB<sup>2<\/sup> + BC<sup>2<\/sup><br \/>\nHence Proved<\/li>\n<li><img decoding=\"async\" style=\"height: 96px; width: 150px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/quesbank\/10\/math\/ch10\/image180.jpg\" \/><br \/>\nWe know that the tangent at any point of a circle is\u00a0<span class=\"math-tex\">{tex}\\perp{\/tex}<\/span> to the radius through the point of contact.<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>OPA = 90<sup>o<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0OA<sup>2<\/sup> = OP<sup>2<\/sup> + AP<sup>2<\/sup> [By Pythagoras theorem]\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0 (5)<sup>2<\/sup> = (OP)<sup>2<\/sup> + (4)<sup>2<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0 25 = (OP)<sup>2<\/sup> + 16<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0 OP<sup>2 <\/sup>= 9<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0 OP = 3 cm<\/li>\n<li>We have,tan (A + B) = <span class=\"math-tex\">{tex}\\sqrt3{\/tex}<\/span><span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>tan(A+B) = tan 60\u00b0<br \/>\nA + B = 60\u00b0&#8230;&#8230;&#8230;.(i)<br \/>\nAgain, tan (A &#8211; B ) =<span class=\"math-tex\">{tex}\\frac{1}{\\sqrt3}{\/tex}<\/span><span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>tan(A-B) = tan 30\u00b0<br \/>\nA &#8211; B = 30\u00b0&#8230;&#8230;&#8230;.(ii)<br \/>\nAdding,\u00a0(i) and (ii)<br \/>\n2A = 90\u00b0<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0A = <span class=\"math-tex\">{tex}\\frac{90^\\circ}{2}{\/tex}<\/span>\u00a0=45\u00b0<br \/>\nPutting\u00a0A=45<span class=\"math-tex\">{tex}^o{\/tex}<\/span>\u00a0in equation (i),<br \/>\nB = 60\u00b0 &#8211; A = 60\u00b0 &#8211; 45\u00b0 = 15\u00b0<br \/>\nTherefore,A = 45\u00b0 and B = 15\u00b0.<\/li>\n<li style=\"clear: both;\">\u00a0 <img decoding=\"async\" style=\"height: 98px; width: 100px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/quesbank\/10\/math\/ch12\/image483.jpg\" \/><br \/>\nWe know that Area of sector =<span class=\"math-tex\">{tex}\\frac { \\theta } { 360 } \\pi r ^ { 2 }{\/tex}<\/span><br \/>\nHere,\u00a0<span class=\"math-tex\">{tex} \\theta = 60 , r = 6{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span> Required area <span class=\"math-tex\">{tex}= \\frac { 60 } { 360 } \\times \\frac { 22 } { 7 } \\times ( 6 ) ^ { 2 }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}= \\frac { 1 } { 6 } \\times \\frac { 22 } { 7 } \\times 36{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}= \\frac { 22 \\times 6 } { 7 }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}= \\frac { 132 } { 7 } = 18 \\frac { 6 } { 7 } c m ^ { 2 }{\/tex}<\/span><\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Length of arc = 15cm<br \/>\n<span class=\"math-tex\">{tex}\\theta = 45^\\circ {\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\therefore {\/tex}<\/span>\u00a0length of arc =\u00a0<span class=\"math-tex\">{tex}\\frac{\\theta }{{360^\\circ }} \\times 2\\pi r{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow 15 = \\frac{{45^\\circ }}{{360^\\circ }} \\times 2 \\times \\pi \\times r{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow r = \\frac{{15 \\times 360^\\circ }}{{45^\\circ \\times 2 \\times \\pi }}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow r = \\frac{{60}}{\\pi }cm{\/tex}<\/span><\/li>\n<li style=\"text-align: center; display: block;\"><b>Section C <\/b><\/li>\n<li>Let us assume, to the contrary, that\u00a0<span class=\"math-tex\">{tex}5-\\sqrt{3}{\/tex}<\/span>\u00a0is rational.<br \/>\nThat is, we can find coprime numbers a and b (b \u2260 0) such that\u00a0<span class=\"math-tex\">{tex}5-\\sqrt{3}=\\frac{a}{b}{\/tex}<\/span><br \/>\nTherefore,\u00a0<span class=\"math-tex\">{tex}5-\\frac{a}{b}=\\sqrt{3}{\/tex}<\/span><br \/>\nRearranging this equation, we get\u00a0<span class=\"math-tex\">{tex}\\sqrt{3}=5-\\frac{a}{b}=\\frac{5 b-a}{b}{\/tex}<\/span><br \/>\nSince a and b are integers, we get\u00a0<span class=\"math-tex\">{tex}5-\\frac{a}{b}{\/tex}<\/span>\u00a0is rational, and so\u00a0<span class=\"math-tex\">{tex}\\sqrt{3}{\/tex}<\/span>\u00a0is rational.<br \/>\nBut this contradicts the fact that\u00a0<span class=\"math-tex\">{tex}\\sqrt{3}{\/tex}<\/span>\u00a0is irrational<br \/>\nThis contradiction has arisen because of our incorrect assumption that\u00a0<span class=\"math-tex\">{tex}5-\\sqrt{3}{\/tex}<\/span>\u00a0is rational.<br \/>\nSo, we conclude that\u00a0<span class=\"math-tex\">{tex}5-\\sqrt{3}{\/tex}<\/span>\u00a0is irrational.<\/li>\n<li>x<sup>2<\/sup> + x &#8211; 12<br \/>\nLet p(x) = x<sup>2<\/sup> + x &#8211; 12<br \/>\nFor zeroes of p(x), p(x) = 0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow x ^ { 2 } + x &#8211; 12 = 0{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow x ^ { 2 } + 4 x &#8211; 3 x &#8211; 12 = 0{\/tex}<\/span><br \/>\nBy the method of splitting the middle term<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow x ( x + 4 ) &#8211; 3 ( x + 4 ) = 0{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow ( x + 4 ) ( x &#8211; 3 ) = 0{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow x + 4 = 0 \\text { or } x &#8211; 3 = 0{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow x = &#8211; 4 \\text { or } x = 3{\/tex}<\/span><br \/>\nx = -4, 3<br \/>\nSo, the zeroes of x<sup>2<\/sup> + x &#8211; 12 are -4 and 3.<br \/>\nSum of zeroes<br \/>\n<span class=\"math-tex\">{tex}= ( &#8211; 4 ) + 3 = &#8211; 1 = \\frac { &#8211; 1 } { 1 } = \\frac { &#8211; \\text { Coefficient of } \\mathrm { x } } { \\text { Coefficient of } \\mathrm { x } ^ { 2 } }{\/tex}<\/span><br \/>\nProduct of its zeroes<br \/>\n<span class=\"math-tex\">{tex}= ( &#8211; 4 ) \\times ( 3 ) = &#8211; 12 = \\frac { &#8211; 12 } { 1 } = \\frac { \\text { Constant term } } { \\text { Coefficient of } \\mathrm { x } ^ { 2 } }{\/tex}<\/span>Hence the relation between zeroes and coefficient is verified.<\/li>\n<li style=\"clear: both;\">Let us denote the incomes of the two-person by \u20b9\u00a09x and \u20b9\u00a07x and their expenditures by \u20b9\u00a04y and \u20b9\u00a03y respectively.<br \/>\nThen the equations formed in the situation is given by :<br \/>\n9x \u2013 4y = 2000 &#8230;(i)<br \/>\nand 7x \u2013 3y = 2000 &#8230;(2)<br \/>\n<strong>Step 1: <\/strong>Multiply Equation (1) by 3 and Equation (2) by 4 to make the coefficients of y equal. Then, we get the equations:<br \/>\n27x \u2013 12y = 6000 &#8230;(3)<br \/>\n28x \u2013 12y = 8000 &#8230;(4)<br \/>\n<strong>Step 2:<\/strong> Subtract Equation (3) from Equation (4) to eliminate y, because the coefficients of y are the same. So, we get<br \/>\n(28x \u2013 27x) \u2013 (12y \u2013 12y) = 8000 \u2013 6000<br \/>\ni.e., x = 2000<br \/>\n<strong>Step 3:<\/strong> Substituting this value of x in (1), we get<br \/>\n9(2000) \u2013 4y = 2000<br \/>\ni.e., y = 4000<br \/>\nSo, the solution of the equations is x = 2000, y = 4000. Therefore, the monthly incomes of the persons are \u20b918,000 and \u20b914,000\u00a0respectively.<\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Given, 99x + 101y = 499 &#8230;.(i)<br \/>\n101x + 99y = 501&#8230; (ii)<br \/>\nAdding eqn. (i) and (ii),<br \/>\n( 99x + 101y ) + (101x + 99y ) = 499 + 501<br \/>\n99x + 101y + 101x + 99y = 1000<br \/>\n200x + 200y = 1000<br \/>\nx + y = 5 &#8230;(iii)<br \/>\nSubtracting eqn. (ii) from eqn. (i), we get<br \/>\n( 99x + 101y ) &#8211; (101x + 99y ) = 499 &#8211; 501<br \/>\n99x + 101y &#8211; 101x &#8211; 99y = -2<br \/>\n-2x + 2y = &#8211; 2<br \/>\nor, x &#8211; y= 1 &#8230;&#8230;.. (iv)<br \/>\nAdding equations (iii) and (iv)<br \/>\nx + y + x &#8211; y = 5 + 1<br \/>\n2x = 6<br \/>\n<span class=\"math-tex\">{tex}\\therefore {\/tex}<\/span>\u00a0x = 3<br \/>\nSubstituting the value of x in eqn. (iii), we get<br \/>\n3 + y = 5<br \/>\ny = 2<br \/>\nHence the value of x and y of given equation are 3 and 2\u00a0respectively.<\/li>\n<li>Produce BD to meet the bigger circle at E. Join AE.<br \/>\n<img decoding=\"async\" style=\"width: 141px; height: 133px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/w89DYGC.png\" alt=\"\" data-imgur-src=\"w89DYGC.png\" \/><br \/>\nThen,\u00a0<span class=\"math-tex\">{tex}\\angle AEB=90^o{\/tex}<\/span>[ angle in semi-circle]\nClearly,\u00a0<span class=\"math-tex\">{tex}OD\\bot BE{\/tex}<\/span><br \/>\nNow, in\u00a0<span class=\"math-tex\">{tex}\\triangle AEB{\/tex}<\/span>, O and D are the mid-points of AB and BE, respectively.<br \/>\nTherefore,by mid-point theorem, we have<br \/>\n<span class=\"math-tex\">{tex}OD=\\frac{1}{2}AE{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow AE=2\\times 8=16\\;cm{\/tex}<\/span>\u00a0[OD = radius of smaller circle = 8 cm ]\nIn right-angled <span class=\"math-tex\">{tex}\\triangle ODB{\/tex}<\/span>,<br \/>\n<span class=\"math-tex\">{tex}OB^2=OD^2+BD^2{\/tex}<\/span>\u00a0[By pythagoras theorem]\n<span class=\"math-tex\">{tex}\\Rightarrow BD^2=169-64=105{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow BC=\\sqrt{105}\\;cm=DE{\/tex}<\/span>\u00a0[BD = DE]\nNow, in right-angled\u00a0<span class=\"math-tex\">{tex}\\triangle AED,{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}AD^2=AE^2+ED^2{\/tex}<\/span>\u00a0[by pythagoras theorem]\n<span class=\"math-tex\">{tex}\\Rightarrow AD=\\sqrt{(16)^2+(\\sqrt{105})^2}=19\\;cm{\/tex}<\/span><\/li>\n<li style=\"clear: both;\"><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/xed1QWA.png\" alt=\"\" data-imgur-src=\"xed1QWA.png\" \/><br \/>\nwe have,<br \/>\n<span class=\"math-tex\">{tex}\\tan A = \\frac { 1 } { \\sqrt { 3 } }{\/tex}<\/span><span class=\"math-tex\">{tex}= tan30^\\circ{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\therefore A = 30^\\circ{\/tex}<\/span><br \/>\nIn\u00a0<span class=\"math-tex\">{tex}\\triangle{\/tex}<\/span>ABC, we have<br \/>\n<strong><strong><span class=\"math-tex\">{tex}\\angle A+\\angle B+\\angle C=180^\\circ{\/tex}<\/span><\/strong><\/strong><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow30^\\circ+90^\\circ+\\angle C=180^\\circ{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow120^\\circ+\\angle C=180^\\circ{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow\\angle C=180^\\circ-120^\\circ=60^\\circ{\/tex}<\/span><br \/>\nSo,<br \/>\ncos A.cos C &#8211; sin A.sin C<br \/>\n<span class=\"math-tex\">{tex}=\\cos30^\\circ.\\cos60^\\circ-\\sin30^\\circ.\\sin60^\\circ{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}=\\frac{\\sqrt3}2\\cdot\\frac12-\\frac12\\cdot\\frac{\\sqrt3}2=0{\/tex}<\/span><\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Let us draw a triangle ABC in which\u00a0<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>B =\u00a090\u00b0.<br \/>\n<img decoding=\"async\" style=\"width: 150px; height: 144px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/a4okXzn.png\" alt=\"\" data-imgur-src=\"a4okXzn.png\" \/><br \/>\nLet\u00a0<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>A =\u00a0<span class=\"math-tex\">{tex}\\theta{\/tex}<\/span>\u00b0.<br \/>\nThen,\u00a0<span class=\"math-tex\">{tex}\\text{cosec} \\theta = \\frac { A C } { B C } = \\frac { 2 } { 1 }{\/tex}<\/span><br \/>\nLet BC = 1k and AC = 2k, where k is positive.<br \/>\nBy Pythagoras&#8217; theorem, we have<br \/>\nAC<sup>2<\/sup>\u00a0= AB<sup>2 <\/sup>+ BC<sup>2<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>AB<sup>2<\/sup>\u00a0= AC<sup>2<\/sup>\u00a0&#8211; BC<sup>2<\/sup><br \/>\n= (2k)<sup>2<\/sup>\u00a0&#8211; (1k)<sup>2<\/sup>\u00a0= 4k<sup>2<\/sup>\u00a0&#8211; 1k<sup>2<\/sup>\u00a0= 3k<sup>2<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\quad B C = \\sqrt 3k{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\sin \\theta = \\frac { B C } { A C } = \\frac { 1 k } { 2 k } = \\frac { 1 } { 2 }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\cos \\theta = \\frac { A B } { A C } = \\frac { \\sqrt { 3 k } } { 2 k } = \\frac { \\sqrt { 3 } } { 2 }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\cot \\theta = \\frac { \\cos \\theta } { \\sin \\theta } = \\left( \\frac { \\sqrt { 3 } } { 2 } \\times \\frac { 2 } { 1 } \\right) = \\sqrt { 3 }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\left[ \\cot \\theta + \\frac { \\sin \\theta } { 1 + \\cos \\theta } \\right] = \\left[ \\sqrt { 3 } + \\frac { \\frac { 1 } { 2 } } { 1 + \\frac { \\sqrt { 3 } } { 2 } } \\right]{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}= \\left( \\sqrt { 3 } + \\frac { 1 } { 2 + \\sqrt { 3 } } \\right) = \\left( \\frac { 2 \\sqrt { 3 } + 3 + 1 } { 2 + \\sqrt { 3 } } \\right){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}= \\left( \\frac { 2 \\sqrt { 3 } + 4 } { 2 + \\sqrt { 3 } } \\right) = 2 \\left( \\frac { \\sqrt { 3 } + 2 } { 2 + \\sqrt { 3 } } \\right) = 2{\/tex}<\/span><br \/>\nHence,\u00a0<span class=\"math-tex\">{tex}\\left[ \\cot \\theta + \\frac { \\sin \\theta } { 1 + \\cos \\theta } \\right] ={\/tex}<\/span>2<\/li>\n<li>When we throw a\u00a0die, possible outcomes are 1,2,3,4,5 and 6.\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>Let E<sub>1<\/sub>\u00a0be the event of getting a 3.<br \/>\nThen, the number of favourable outcomes = 1.<br \/>\nThe favourable outcome\u00a0is\u00a03.<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0P(getting a 3) = P(E<sub>1<\/sub>)\u00a0<span class=\"math-tex\">{tex}= \\frac { 1 } { 6 }{\/tex}<\/span>.<\/li>\n<li>Let E<sub>2<\/sub> be the event of getting a 5<br \/>\nThen, the number of favourable outcomes = 1.<br \/>\nThe favourable outcome\u00a0is 5.<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0 P(getting a 5) = P(E<sub>2<\/sub>) <span class=\"math-tex\">{tex}= \\frac { 1 } { 6 }{\/tex}<\/span>.<\/li>\n<li>Let E<sub>3<\/sub> be the event of getting an odd number.<br \/>\nNumber of favourable outcomes = 3.<br \/>\nThe favourable outcomes are 1,3,5.<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0P(getting an odd number)=<span class=\"math-tex\">{tex}P \\left( E _ { 3 } \\right) = \\frac { 3 } { 6 } = \\frac { 1 } { 2 }{\/tex}<\/span>.<\/li>\n<li>Let E<sub>4<\/sub> be the event of getting a number greater than 4.<br \/>\nNumber of favourable outcomes = 2.<br \/>\nThen, the favourable outcomes are 5,6.<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0P(getting a number greater than 4) = P(E<sub>4<\/sub>)<span class=\"math-tex\">{tex}= \\frac { 2 } { 6 } = \\frac { 1 } { 3 }{\/tex}<\/span>.<\/li>\n<\/ol>\n<\/li>\n<li style=\"text-align: center; display: block;\"><b>Section D <\/b><\/li>\n<li style=\"clear: both;\">Let the numbers are x and (x &gt; y)<br \/>\nx<sup>2<\/sup> &#8211; y<sup>2<\/sup> = 204 \u2026(i)<br \/>\ny<sup>2<\/sup> = 10x &#8211; 4 \u2026(ii)<br \/>\nBy (i) and (ii)<br \/>\nx<sup>2<\/sup> &#8211; 10x + 4 &#8211; 204 = 0<br \/>\nx<sup>2<\/sup> &#8211; 10x + 200 = 0<br \/>\n(x &#8211; 20) (x + 10) = 0<br \/>\nx = 20, x = -10 (rejected)<br \/>\ny = 14<\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>According to question a dealer sells a toy for Rs. 24 and gains as much per cent as the cost price of the toy.<br \/>\nLet the cost price of the toy be Rs.\u00a0x.<br \/>\nThen, Gain = x%<br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow{\/tex}<\/span>\u00a0Gain = Rs.\u00a0<span class=\"math-tex\">{tex} \\left( x \\times \\frac { x } { 100 } \\right){\/tex}<\/span>\u00a0= Rs.\u00a0<span class=\"math-tex\">{tex} \\frac { x ^ { 2 } } { 100 }{\/tex}<\/span>.<br \/>\nTherefore,\u00a0S.P. = C.P. + Gain =\u00a0<span class=\"math-tex\">{tex} x + \\frac { x ^ { 2 } } { 100 }{\/tex}<\/span>.<br \/>\nBut, S.P. =Rs.\u00a024.<br \/>\n<span class=\"math-tex\">{tex} \\therefore \\quad x + \\frac { x ^ { 2 } } { 100 } = 24{\/tex}<\/span>\u00a0[Given]\n<span class=\"math-tex\">{tex} \\Rightarrow{\/tex}<\/span>\u00a0100x + x<sup>2<\/sup>\u00a0= 2400<br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow{\/tex}<\/span>\u00a0x<sup>2<\/sup>\u00a0+ 100x &#8211; 2400 = 0<br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow{\/tex}<\/span>\u00a0x<sup>2<\/sup>\u00a0+ 120x -20x -2400 = 0<br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow \\quad x ( x + 120 ) &#8211; 20 ( x + 120 ) = 0{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow{\/tex}<\/span>\u00a0(x + 120) (x &#8211; 20) = 0<br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow{\/tex}<\/span>\u00a0x = 20, -120<br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow{\/tex}<\/span>\u00a0x = 20 [because x &gt; 0 ]\nHence, the cost price of the toy is Rs.\u00a020.<\/li>\n<li><strong>GIVEN <\/strong>A right triangle ABC right angled at B i.e.\u00a0<span class=\"math-tex\">{tex}B D \\perp A C.{\/tex}<\/span><br \/>\n<strong>PROOF <\/strong>We have,<br \/>\n<img decoding=\"async\" style=\"width: 180px; height: 150px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/k2BUeQP.png\" alt=\"\" data-imgur-src=\"k2BUeQP.png\" \/><br \/>\n<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>ABD + <span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>DBC = 90\u00b0<br \/>\nAlso,\u00a0<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>C + <span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>DBC + <span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>BDC = 180\u00b0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>C + <span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>DBC + 90\u00b0 = 180\u00b0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>C +\u00a0<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>DBC = 90\u00b0<br \/>\nBut,\u00a0<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>ABD + <span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>DBC = 90\u00b0<br \/>\n<span class=\"math-tex\">{tex}\\therefore \\quad \\angle A B D + \\angle D B C = \\angle C + \\angle D B C{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\quad \\angle A B D = \\angle C{\/tex}<\/span>\u00a0&#8230;(i)<br \/>\nThus, in <span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>ADB and <span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>BDC, we have<br \/>\n<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>ABD =\u00a0<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>C [From (i)]\nand, <span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>ADB = <span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>BDC [Each equal to 90\u00b0]\nSo, by AA-similarity criterion, we obtain\u00a0<strong><span class=\"math-tex\">{tex}\\Delta A D B \\sim \\Delta B D C.{\/tex}<\/span>\u00a0 &#8230;.(ii)<\/strong><br \/>\nIn <span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>ADB and <span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>ABC, we have<br \/>\n<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>ADB = <span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>ABC [Each equal to 90\u00b0]\nand, <span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>A =\u00a0 <span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>A [Common]\nSo, by AA-similarity criterion, we obtain <span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>ADB ~ <span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>ABC &#8230;..(iii)<br \/>\nIn <span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>BDC and <span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>ABC, we have<br \/>\n<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>BDC = <span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>ABC [Each equal to 90\u00b0]\n<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>C = <span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>C [Common]\nSo, by AA &#8211; similarity criterion, we obtain <span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>BDC ~ <span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>ABC &#8230;&#8230;(iv)<br \/>\nFrom (ii), we have<br \/>\n<span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>ADB ~ <span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>BDC<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\quad \\frac { A D } { B D } = \\frac { B D } { D C }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\quad B D ^ { 2 } = A D \\times D C{\/tex}<\/span><\/li>\n<li style=\"clear: both;\">r<sub>1<\/sub> = Radius of the base of the cylinder = <span class=\"math-tex\">{tex}\\frac{4.3}{2}{\/tex}<\/span> m = 2.15 m<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0r<sub>2<\/sub> = Radius of the base of the cone = 2.15 m, h<sub>1<\/sub> = Height of the cylinder = 3.8 m<br \/>\n<img decoding=\"async\" style=\"width: 200px; height: 222px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/bhr7VQm.png\" alt=\"\" data-imgur-src=\"bhr7VQm.png\" \/><br \/>\nIn <span class=\"math-tex\">{tex}\\triangle{\/tex}<\/span>VOA, we have<br \/>\n<span class=\"math-tex\">{tex}\\sin 45^{\\circ}=\\frac{O A}{V A} \\Rightarrow \\frac{1}{\\sqrt{2}}=\\frac{2.15}{V A} \\Rightarrow V A=(\\sqrt{2} \\times 2.15) \\mathrm{m}=(1.414 \\times 2.15) \\mathrm{m}=3.04 \\mathrm{m}{\/tex}<\/span><br \/>\nClearly, <span class=\"math-tex\">{tex}\\triangle{\/tex}<\/span>VOA is an isosceles triangle. Therefore, VO = OA = 2.15 m<br \/>\nThus, we have<br \/>\nh<sub>2<\/sub> = Height of the cone = VO = 2.15 m, l<sub>2<\/sub> = Slant height of the cone = VA = 3.04 m<br \/>\nLet S be the Surface area of the building. Then,<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0S = Surface area of the cylinder + Surface area of cone<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0S = (2<span class=\"math-tex\">{tex}\\pi{\/tex}<\/span>r<sub>1<\/sub>h<sub>1<\/sub> + <span class=\"math-tex\">{tex}\\pi{\/tex}<\/span>r<sub>2<\/sub>l<sub>2<\/sub>) m<sup>2<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0S = (2<span class=\"math-tex\">{tex}\\pi{\/tex}<\/span>r<sub>1<\/sub>h<sub>1<\/sub> + <span class=\"math-tex\">{tex}\\pi{\/tex}<\/span>r<sub>1<\/sub>l<sub>2<\/sub>) m<sup>2<\/sup>\u00a0[<span class=\"math-tex\">{tex}\\because{\/tex}<\/span>\u00a0r<sub>1<\/sub>\u00a0=\u00a0r<sub>2<\/sub> &#8211; 2.15 m]\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0S = <span class=\"math-tex\">{tex}\\pi{\/tex}<\/span>r<sub>1<\/sub>(2h<sub>1<\/sub>\u00a0+ l<sub>2<\/sub>) m<sup>2<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0S = 3.14 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> 2.15 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a0(2 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a03.8 + 3.04) m<sup>2<\/sup> = 3.14 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a02.15 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a010.64 m<sup>2<\/sup> = 71.83 m<sup>2<\/sup><br \/>\nLet U be the volume of the building. Then,<br \/>\nV = Volume of the cylinder + Volume of the cone<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\quad V=\\left(\\pi r_{1}^{2} h_{1}+\\frac{1}{3} \\pi r_{2}^{2} h_{2}\\right) \\mathrm{m}^{3}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\quad V=\\left(\\pi r_{1}^{2} h_{1}+\\frac{1}{3} \\pi r_{1}^{2} h_{2}\\right) \\mathrm{m}^{3} \\quad\\left[\\because r_{2}=r_{1}\\right]{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\quad V=\\pi r_{1}^{2}\\left(h_{1}+\\frac{1}{3} h_{2}\\right) \\mathrm{m}^{3}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\quad V=3.14 \\times 2.15 \\times 2.15 \\times\\left(3.8+\\frac{2.15}{3}\\right) \\mathrm{m}^{3}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\quad V=[3.14 \\times 2.15 \\times 2.15 \\times(3.8+0.7166)] \\mathrm{m}^{3}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\quad V=(3.14 \\times 2.15 \\times 2.15 \\times 4.5166) \\mathrm{m}^{3}=65.55 \\mathrm{m}^{3}{\/tex}<\/span><\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Given that,<br \/>\n<img decoding=\"async\" style=\"height: 280px; width: 140px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1689139358-t2ns3a.jpg\" \/><br \/>\nLength of cylindrical part = 7 m<br \/>\nRadius of cylindrical part = <span class=\"math-tex\">{tex}\\frac{7}{2}{\/tex}<\/span> m<br \/>\nTotal surface area of figure = 2<span class=\"math-tex\">{tex}\\pi{\/tex}<\/span>rh + 2(2<span class=\"math-tex\">{tex}\\pi{\/tex}<\/span>r<sup>2<\/sup>)<br \/>\n= <span class=\"math-tex\">{tex}2 \\pi\\left[\\frac{7}{2} \\times 7+2 \\times\\left(\\frac{7}{2}\\right)^2\\right]{\/tex}<\/span><br \/>\n= 308 m<sup>2<\/sup><br \/>\nVolume of boiler = Volume of cylindrical part + Volume of two hemispherical parts<br \/>\n= <span class=\"math-tex\">{tex}\\pi{\/tex}<\/span>r<sup>2<\/sup>h <span class=\"math-tex\">{tex}+\\left(\\frac{4}{3}\\right) \\pi r^3{\/tex}<\/span><br \/>\n= <span class=\"math-tex\">{tex}\\pi\\left(\\frac{7}{2}\\right)^2 \\times(7)+\\left(\\frac{4}{3}\\right) \\pi\\left(\\frac{7}{2}\\right)^3{\/tex}<\/span><br \/>\n= 269.5 + 179.66<br \/>\n= 449.167 m<sup>3<\/sup><br \/>\nRequired ratio = <span class=\"math-tex\">{tex}\\frac{\\text { Volume of cylindrical part }}{\\text { Volume of one hemispherical part }}{\/tex}<\/span><br \/>\n= <span class=\"math-tex\">{tex}\\frac{269.5}{89.83}{\/tex}<\/span><br \/>\n= 3<\/li>\n<li>We may observe from the given data that maximum class frequency is 40 belonging to 1500 &#8211; 2000 interval.<br \/>\nClass size (h) = 500<br \/>\nMode\u00a0= l +\u00a0<span class=\"math-tex\">{tex} \\frac { f &#8211; f _ { 1 } } { 2 f &#8211; f _ { 1 } &#8211; f _ { 2 } } \\times{\/tex}<\/span>\u00a0h<br \/>\nLower limit <span class=\"math-tex\">{tex}(l) {\/tex}<\/span>of modal class = 1500<br \/>\nFrequency (f) of modal class = 40<br \/>\nFrequency (f<sub>1<\/sub>) of class preceding modal class = 24<br \/>\nFrequency (f<sub>2<\/sub>) of class succeeding modal class = 33<br \/>\nmode\u00a0= 1500 +\u00a0<span class=\"math-tex\">{tex}\\frac { 40 &#8211; 24 } { 2 \\times 40 &#8211; 24 &#8211; 33 } \\times{\/tex}<\/span>\u00a0500<br \/>\n= 1500 +\u00a0<span class=\"math-tex\">{tex} \\frac { 16 } { 80 &#8211; 57 } \\times{\/tex}<\/span>\u00a0500<br \/>\n= 1500 + 347.826<br \/>\n= 1847.826 \u2248 1847.83<\/p>\n<table style=\"width: 100%;\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<tbody>\n<tr>\n<td style=\"text-align: center;\"><span style=\"display: none;\">\u00a0<\/span><span style=\"display: none;\">\u00a0<\/span><strong>Expenditure (in \u20b9.)<\/strong><\/td>\n<td style=\"text-align: center;\"><strong>Number of families f<sub>i<\/sub><\/strong><\/td>\n<td style=\"text-align: center;\"><strong>x<sub>i<\/sub><\/strong><\/td>\n<td style=\"text-align: center;\"><strong>d<sub>i<\/sub> = x<sub>i<\/sub> &#8211; 2750<\/strong><\/td>\n<td style=\"text-align: center;\"><b>u<sub>i<\/sub><\/b><\/td>\n<td style=\"text-align: center;\"><span style=\"font-size: 10.8333px;\"><b>u<sub>i<\/sub>f<sub>i<\/sub><\/b><\/span><\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">1000-1500<\/td>\n<td style=\"text-align: center;\">24<\/td>\n<td style=\"text-align: center;\">1250<\/td>\n<td style=\"text-align: center;\">-1500<\/td>\n<td style=\"text-align: center;\">-3<\/td>\n<td style=\"text-align: center;\">-72<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">1500-2000<\/td>\n<td style=\"text-align: center;\">40<\/td>\n<td style=\"text-align: center;\">1750<\/td>\n<td style=\"text-align: center;\">-1000<\/td>\n<td style=\"text-align: center;\">-2<\/td>\n<td style=\"text-align: center;\">-80<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">2000-2500<\/td>\n<td style=\"text-align: center;\">33<\/td>\n<td style=\"text-align: center;\">2250<\/td>\n<td style=\"text-align: center;\">-500<\/td>\n<td style=\"text-align: center;\">-1<\/td>\n<td style=\"text-align: center;\">-33<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">2500-3000<\/td>\n<td style=\"text-align: center;\">28<\/td>\n<td style=\"text-align: center;\">2750=a<\/td>\n<td style=\"text-align: center;\">0<\/td>\n<td style=\"text-align: center;\">0<\/td>\n<td style=\"text-align: center;\">0<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">3000-3500<\/td>\n<td style=\"text-align: center;\">30<\/td>\n<td style=\"text-align: center;\">3250<\/td>\n<td style=\"text-align: center;\">500<\/td>\n<td style=\"text-align: center;\">1<\/td>\n<td style=\"text-align: center;\">30<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">3500-4000<\/td>\n<td style=\"text-align: center;\">22<\/td>\n<td style=\"text-align: center;\">3750<\/td>\n<td style=\"text-align: center;\">1000<\/td>\n<td style=\"text-align: center;\">2<\/td>\n<td style=\"text-align: center;\">44<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">4000-4500<\/td>\n<td style=\"text-align: center;\">16<\/td>\n<td style=\"text-align: center;\">4250<\/td>\n<td style=\"text-align: center;\">1500<\/td>\n<td style=\"text-align: center;\">3<\/td>\n<td style=\"text-align: center;\">48<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">4500-5000<\/td>\n<td style=\"text-align: center;\">7<\/td>\n<td style=\"text-align: center;\">4750<\/td>\n<td style=\"text-align: center;\">2000<\/td>\n<td style=\"text-align: center;\">4<\/td>\n<td style=\"text-align: center;\">28<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\"><\/td>\n<td style=\"text-align: center;\"><span class=\"math-tex\">{tex}\\Sigma f _ { i }{\/tex}<\/span>\u00a0\u00a0= 200<\/td>\n<td style=\"text-align: center;\"><\/td>\n<td style=\"text-align: center;\"><\/td>\n<td style=\"text-align: center;\"><\/td>\n<td style=\"text-align: center;\"><span class=\"math-tex\">{tex}\\Sigma f _ { i } d _ { i }{\/tex}<\/span>\u00a0\u00a0= &#8211; 35<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Mean\u00a0<span class=\"math-tex\">{tex}\\overline { x } = a + \\frac { \\Sigma f _ { i } d _ { i } } { \\Sigma f _ { i } } \\times{\/tex}<\/span>\u00a0h<br \/>\n<span class=\"math-tex\">{tex}\\overline { x }{\/tex}<\/span>\u00a0= 2750 + <span class=\"math-tex\">{tex}\\frac { &#8211; 35 } { 200 } \\times{\/tex}<\/span> 500<br \/>\n<span class=\"math-tex\">{tex}\\overline { x }{\/tex}<\/span>\u00a0\u00a0= 2750 &#8211; 87.5<br \/>\n<span class=\"math-tex\">{tex}\\overline { x } {\/tex}<\/span>\u00a0= 2662.5<\/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;\">Since the production increases uniformly by a fixed number every year. Therefore, the sequence formed by the production in different years is an A.P. Let a be the first term and d be the common difference of the A.P. formed i.e., &#8216;a&#8217; denotes the production in the first year and d denotes the number of units by which the production increases every year.<br \/>\nWe have, a<sub>3<\/sub> = 600 and<br \/>\na<sub>3\u00a0<\/sub>= 600<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> 600 = a + 2d<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0a = 600 &#8211; 2d &#8230;(i)<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0a<sub>7 <\/sub>= 700<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0a<sub>7\u00a0<\/sub>= 700<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> 700 = a + 6d<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0a = 700 &#8211; 6d &#8230;(ii)<br \/>\nFrom (i) and (ii)<br \/>\n600 &#8211; 2d = 700 &#8211; 6d<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a04d = 100<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0d = 25<\/li>\n<li style=\"text-align: left;\">Since the production increases uniformly by a fixed number every year. Therefore, the sequence formed by the production in different years is an A.P. Let a be the first term and d be the common difference of the A.P. formed i.e., &#8216;a&#8217; denotes the production in the first year and d denotes the number of units by which the production increases every year.<br \/>\nWe know that first term = a = 550 and common difference = d = 25<br \/>\na<sub>n\u00a0<\/sub>= 1000<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> 1000 = a + (n &#8211; 1)d<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a01000 = 550 + 25n &#8211; 25<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a01000 &#8211; 550 + 25 = 25n<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0475 = 25n<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> n = <span class=\"math-tex\">{tex}\\frac{475}{25}{\/tex}<\/span>\u00a0= 19<\/li>\n<li style=\"text-align: left;\">Since the production increases uniformly by a fixed number every year. Therefore, the sequence formed by the production in different years is an A.P. Let a be the first term and d be the common difference of the A.P. formed i.e., &#8216;a&#8217; denotes the production in the first year and d denotes the number of units by which the production increases every year.<br \/>\nThe production in the 10th term is given by a<sub>10.<\/sub>\u00a0Therefore,\u00a0production in the 10th year = a<sub>10<\/sub> = a + 9d = 550 + 9 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a025 = 775. So, production in 10th year is of\u00a0775 TV sets.<br \/>\n<strong>OR<\/strong><br \/>\nSince the production increases uniformly by a fixed number every year. Therefore, the sequence formed by the production in different years is an A.P. Let a be the first term and d be the common difference of the A.P. formed i.e., &#8216;a&#8217; denotes the production in the first year and d denotes the number of units by which the production increases every year.<br \/>\nTotal production in 7 years =\u00a0Sum of 7 terms of the A.P. with first term a (= 550) and d (= 25).<br \/>\n<span class=\"math-tex\">{tex}S_n=\\frac{n}{2}[2 a+(n-1) d]{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow S_7=\\frac{7}{2}[2 \\times 550+(7-1) 25]{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow S_7=\\frac{7}{2}[2 \\times 550+(6) \\times 25]{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow S_7=\\frac{7}{2}[1100+150]{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> S<sub>7\u00a0<\/sub>= 4375<\/li>\n<\/ol>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li style=\"text-align: left;\">Distance travelled by second bus = 7.2 km<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span> Total fare = 7.2 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> 15 = \u20b908<\/li>\n<li style=\"text-align: left;\">Required distance = <span class=\"math-tex\">{tex}\\sqrt{(2+2)^{2}+(3+3)^{2}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}= \\sqrt{4^{2}+6^{2}}=\\sqrt{16+36}{\/tex}<\/span> <span class=\"math-tex\">{tex}= 2 \\sqrt{13}{\/tex}<\/span> km <span class=\"math-tex\">{tex}\\approx{\/tex}<\/span> 7.2 km<\/li>\n<li style=\"text-align: left;\">Required distance = <span class=\"math-tex\">{tex}\\sqrt{(3+2)^{2}+(2+3)^{2}}{\/tex}<\/span><br \/>\n= <span class=\"math-tex\">{tex}\\sqrt{5^{2}+5^{2}}=5 \\sqrt{2}{\/tex}<\/span> km<br \/>\n<strong>OR<\/strong><br \/>\nDistance between B and C<br \/>\n= <span class=\"math-tex\">{tex}\\sqrt{(3-2)^{2}+(2-3)^{2}}{\/tex}<\/span> = <span class=\"math-tex\">{tex}\\sqrt{1+1}=\\sqrt{2}{\/tex}<\/span> km<br \/>\nThus, distance travelled by first bus to reach to B<br \/>\n= AC + CB = <span class=\"math-tex\">{tex}5 \\sqrt{2}+\\sqrt{2}=6 \\sqrt{2}{\/tex}<\/span> km <span class=\"math-tex\">{tex}\\approx{\/tex}<\/span> 8.48 km<br \/>\nand distance travelled by second bus to reach to B<br \/>\n= AB = <span class=\"math-tex\">{tex}2 \\sqrt{13}{\/tex}<\/span> km = 7.2 km<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span> Distance of first bus is greater than distance of the cond bus, therefore second bus should be chosen.<\/li>\n<li style=\"text-align: left;\"><img decoding=\"async\" style=\"height: 145px; width: 200px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1663762240-28x7ug.jpg\" \/><br \/>\nLet BC be the tower of height h \u00a0and CD be the water tank of height h<sub>1<\/sub><br \/>\nIn\u00a0<span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>ABD, we have<br \/>\n<span class=\"math-tex\">{tex}\\tan 45 ^ { \\circ } = \\frac { B D } { A B }{\/tex}<\/span><br \/>\n<strong><span class=\"math-tex\">{tex}\\Rightarrow 1 = \\frac { h + h _ { 1 } } { 40 }{\/tex}<\/span><\/strong><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow h + h _ { 1 } = 40{\/tex}<\/span>\u00a0&#8230;(1)<br \/>\nIn\u00a0<span class=\"math-tex\">{tex}\\Delta A B C,{\/tex}<\/span>\u00a0we have<br \/>\n<span class=\"math-tex\">{tex}\\tan 30 ^ { \\circ } = \\frac { B C } { A B }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\quad \\frac { 1 } { \\sqrt { 3 } } = \\frac { h } { 40 }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow h = \\frac { 40 } { \\sqrt { 3 } } = \\frac { 40 \\sqrt { 3 } } { 3 } {\/tex}<\/span>\u00a0= 23.1\u00a0m<br \/>\nThus height of the tower is 23.1 m.<\/li>\n<li style=\"text-align: left;\"><img decoding=\"async\" class=\"alignleft\" style=\"height: 145px; width: 200px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1663762240-28x7ug.jpg\" \/><br \/>\nLet BC be the tower of height h \u00a0and CD be the water tank of height h<sub>1<\/sub><br \/>\nIn\u00a0<span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>ABD, we have<br \/>\n<span class=\"math-tex\">{tex}\\tan 45 ^ { \\circ } = \\frac { B D } { A B }{\/tex}<\/span><br \/>\n<strong><span class=\"math-tex\">{tex}\\Rightarrow 1 = \\frac { h + h _ { 1 } } { 40 }{\/tex}<\/span><\/strong><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow h + h _ { 1 } = 40{\/tex}<\/span>\u00a0&#8230;(1)<br \/>\nIn\u00a0<span class=\"math-tex\">{tex}\\Delta A B C,{\/tex}<\/span>\u00a0we have<br \/>\n<span class=\"math-tex\">{tex}\\tan 30 ^ { \\circ } = \\frac { B C } { A B }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\quad \\frac { 1 } { \\sqrt { 3 } } = \\frac { h } { 40 }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow h = \\frac { 40 } { \\sqrt { 3 } } = \\frac { 40 \\sqrt { 3 } } { 3 } {\/tex}<\/span>\u00a0= 23.1\u00a0m<br \/>\nThus height of the tower is 23.1 m.<br \/>\nSubstituting the value of h in (1), we have<br \/>\n<span class=\"math-tex\">{tex}23.1 + h_1 = 40{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0h<sub>1<\/sub> = 40 &#8211; 23.1<br \/>\n= 6.9 m<br \/>\nThus height of the tank is 6.9 m.<\/li>\n<li style=\"text-align: left;\"><strong><img decoding=\"async\" style=\"height: 78px; width: 200px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1663861762-4mmg54.jpg\" \/><\/strong><br \/>\nIn the<strong>\u00a0<span class=\"math-tex\">{tex}\\triangle ABC {\/tex}<\/span>\u00a0<\/strong>if<strong>\u00a0<span class=\"math-tex\">{tex}\\angle CAB =45^\\circ{\/tex}<\/span>\u00a0<\/strong>then<br \/>\n<strong><span class=\"math-tex\">{tex}\\cot 45^\\circ =\\frac y{23.1}=1{\/tex}<\/span><\/strong><br \/>\ny = 23.1 m<br \/>\nThus the angle of elevation will be 45<sup>o<\/sup>\u00a0at 23.1 m.<br \/>\n<strong>OR<\/strong><br \/>\n<img decoding=\"async\" style=\"height: 148px; width: 200px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1723444721-nurbe9.jpg\" \/><br \/>\nGiven that<strong> <span class=\"math-tex\">{tex}AB=\\dfrac{40}{\\sqrt3}{\/tex}<\/span><br \/>\n<img decoding=\"async\" style=\"height: 131px; width: 200px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1663862064-qygbkx.jpg\" \/><\/strong><br \/>\nIn the\u00a0<span class=\"math-tex\">{tex}\\triangle ABD{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\cot \\theta=\\frac{AB}{BD} =\\frac{ \\frac{40 }{\\sqrt3} }{40}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\cot \\theta=\\frac1{\\sqrt3}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\theta=60^\\circ{\/tex}<\/span><br \/>\nHence the angle of elevation would be\u00a060<sup>o<\/sup>.<br \/>\n<strong>Download the myCBSEguide App for Comprehensive Exam Preparation<\/strong><br \/>\nPrepare effectively for your exams with the <a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\"><strong>myCBSEguide app<\/strong><\/a>, offering complete study material for <strong>CBSE<\/strong>, <strong>NCERT<\/strong>, <strong>JEE (Main)<\/strong>, <strong>NEET-UG<\/strong>, and <strong>NDA exams<\/strong>. This app provides access to a wide range of practice questions, sample papers, and detailed solutions, helping you study smarter and achieve your academic goals. Download the <strong>myCBSEguide app<\/strong> today and get everything you need to excel in your exams.<strong>Examin8 App for Teachers: Create Custom Exam Papers Easily<\/strong><br \/>\nTeachers can enhance their teaching experience with the <a href=\"https:\/\/play.google.com\/store\/search?q=EXAMIN8&amp;c=apps\"><strong>Examin8 app<\/strong><\/a>. This powerful tool allows educators to create <strong>customized question papers<\/strong> featuring their name and logo.\u00a0Visit our <strong><a href=\"https:\/\/examin8.com\/\">Examin8<\/a> Website. <\/strong>The <strong>Class 10 Basic Maths Sample Paper 2025<\/strong> is an excellent resource to help you solidify your understanding of key mathematical concepts. These sample papers cover all major topics in the syllabus, ensuring that you get ample practice for topics like algebra, geometry, and statistics. Regular practice with the <strong>Class 10 Basic Maths Sample Paper 2025<\/strong> can enhance your problem-solving skills and build confidence for the exam.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<h3><span class=\"ez-toc-section\" id=\"myCBSEguide_App_for_Board_Students\"><\/span>myCBSEguide App for Board Students<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p style=\"text-align: left;\"><a class=\"button\" style=\"font-weight: bold;\" href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\">Download myCBSEguide app<\/a><\/p>\n<h3><span class=\"ez-toc-section\" id=\"To_excel_in_the_Class_10_Basic_Maths_exam_2025_practicing_with_high-quality_sample_papers_is_essential_Our_sample_paper_for_Class_10_Basic_Maths_2025_offers_a_comprehensive_set_of_questions_that_follow_the_CBSE_exam_pattern\"><\/span>To excel in the <strong>Class 10 Basic Maths exam 2025<\/strong>, practicing with high-quality <strong>sample papers<\/strong> is essential. Our <strong>sample paper for Class 10 Basic Maths 2025<\/strong> offers a comprehensive set of questions that follow the <strong>CBSE exam pattern<\/strong>.<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"Why_Choose_myCBSEguide_for_Your_Exam_Preparation\"><\/span><strong>Why Choose myCBSEguide for Your Exam Preparation?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>1. Comprehensive Study Resources<\/strong><br \/>\n<a href=\"https:\/\/mycbseguide.com\/\"><strong>myCBSEguide<\/strong><\/a> offers a complete range of study materials for <strong>CBSE<\/strong>, <strong>NCERT<\/strong>, <strong>JEE (Main)<\/strong>, <strong>NEET-UG<\/strong>, and <strong>NDA exams<\/strong>. From textbooks to sample papers and previous years\u2019 questions, all resources are curated to match the latest <strong>CBSE syllabus<\/strong> and <strong>exam patterns<\/strong>, ensuring that you are fully prepared for your exams.<\/p>\n<p><strong>2. Expert-Curated Content<\/strong><br \/>\nOur study materials are developed by experienced educators who understand the nuances of the <strong>CBSE exam<\/strong>. Whether you are looking for <strong>concepts explanations<\/strong>, <strong>practice questions<\/strong>, or <strong>detailed solutions<\/strong>, <strong>myCBSEguide<\/strong> ensures the content is accurate, reliable, and aligned with the <strong>latest marking schemes<\/strong>.<\/p>\n<p><strong>3. 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You\u2019ll always have access to the most relevant and current study resources.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Class 10 Basic Maths Sample Paper 2025 Prepare for your exams with the Class 10 Basic Maths Sample Paper 2025 available on myCBSEguide.\u00a0By solving the Class 10 Basic Maths Sample Paper 2025, you can get familiar with key topics such as arithmetic, algebra, and geometry. Regular practice will boost your problem-solving skills and improve your &#8230; <a title=\"Class 10 Basic Maths Sample Paper 2025\" class=\"read-more\" href=\"https:\/\/mycbseguide.com\/blog\/class-10-maths-sample-paper\/\" aria-label=\"More on Class 10 Basic Maths Sample Paper 2025\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1346,46,1998,1347,2004],"tags":[1042,12,1959,1967],"class_list":["post-30244","post","type-post","status-publish","format-standard","hentry","category-cbse","category-cbse-class-10","category-class-10-sample-papers","category-mathematics","category-maths-sample-papers-class-10-sample-papers","tag-cbse-class-10-mathematics","tag-cbse-sample-papers","tag-cbse-sample-papers-2024","tag-cbse-sample-papers-2025"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.0 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Class 10 Basic Maths Sample Paper 2025 | myCBSEguide<\/title>\n<meta name=\"description\" content=\"CBSE class 10 Maths Basic Sample Paper 2025 with solutions in PDF for free download. 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