{"id":13717,"date":"2023-01-10T12:47:59","date_gmt":"2023-01-10T07:17:59","guid":{"rendered":"http:\/\/mycbseguide.com\/blog\/?p=13717"},"modified":"2025-10-08T17:48:08","modified_gmt":"2025-10-08T12:18:08","slug":"cbse-sample-papers-class-10-mathematics","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-10-mathematics\/","title":{"rendered":"Class 10 Maths Sample Papers 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\/cbse-sample-papers-class-10-mathematics\/#Class_10_Maths_Standard_Sample_Papers_2025\" >Class 10 Maths Standard Sample Papers 2025\u00a0<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-10-mathematics\/#Sample_Paper_of_Maths_Class_10_%E2%80%93_PDF_with_Solution\" >Sample Paper of Maths Class 10 &#8211;\u00a0 PDF with Solution<\/a><ul class='ez-toc-list-level-3' ><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-10-mathematics\/#Maths_Standard_Sample_Paper_2025_with_Solution_Download_as_PDF\" >Maths Standard Sample Paper 2025 with Solution Download as PDF<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-10-mathematics\/#Maths_Basic_Sample_Paper_2025_with_Solution_Download_as_PDF\" >Maths Basic Sample Paper 2025 with Solution Download as PDF<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-10-mathematics\/#CBSE_Class_10_Maths_Sample_Papers_2024-25\" >CBSE Class 10 Maths Sample Papers 2024-25<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-10-mathematics\/#Model_Papers_Class_10_Maths_Standard_2025\" >Model Papers Class 10 Maths Standard 2025<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-7\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-10-mathematics\/#Solution\" >Solution<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-8\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-10-mathematics\/#Sample_Papers_for_Class_10_2024-25_in_PDF\" >Sample Papers for Class 10 2024-25 in PDF<\/a><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"Class_10_Maths_Standard_Sample_Papers_2025\"><\/span><strong>Class 10 Maths Standard Sample Papers 2025\u00a0<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>CBSE offers two distinct courses in <strong><a href=\"https:\/\/mycbseguide.com\/cbse-sample-papers-class-10.html\">Class 10<\/a> Maths<\/strong>: <strong>Maths Basic<\/strong> and <strong>Maths Standard<\/strong>. This page provides <strong>Class 10 Maths Standard Sample Papers 2025<\/strong> and their solution specifically tailored for students enrolled in the <strong>Maths Standard<\/strong> course. <strong>Sample Model Test Papers<\/strong> for <strong>Class 10 Maths<\/strong> with detailed solutions are available on the <a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\"><strong>myCBSEguide app<\/strong><\/a>. Teachers can use the <a href=\"https:\/\/play.google.com\/store\/apps\/details?id=com.examin8.exam\"><strong>Examin8<\/strong><\/a>\u00a0app to create <strong>customized exam papers<\/strong> with their <strong>own branding<\/strong>, including <strong>name<\/strong> and <strong>logo<\/strong>. For students, the <strong><a href=\"https:\/\/mycbseguide.com\/\">myCBSEguide<\/a> website<\/strong> offers a range of study resources, sample papers, and solutions to enhance exam preparation. These sample papers are designed by the latest <strong>marking scheme<\/strong> and <strong>blueprint<\/strong> issued by CBSE for the <strong>2024-25 academic session<\/strong>.<\/p>\n<p>By practicing the <strong>Class 10 Maths Standard Sample Papers 2025<\/strong>, students can get a comprehensive understanding of the <strong>exam format<\/strong>, <strong>question types<\/strong>, and <strong>mark distribution<\/strong>. This practice will help improve <strong>problem-solving skills<\/strong> and enhance time management, ensuring students are thoroughly prepared for the <strong>Class 10 board exams<\/strong>.<\/p>\n<h2 class=\"sub-title\"><span class=\"ez-toc-section\" id=\"Sample_Paper_of_Maths_Class_10_%E2%80%93_PDF_with_Solution\"><\/span>Sample Paper of Maths Class 10 &#8211;\u00a0 PDF with Solution<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><strong>Class 10 Maths Sample Papers 2025 \u2013 Practice with the Latest Marking Scheme and Blueprint<\/strong><\/p>\n<p>We are offering <strong>Class 10 Maths Sample Papers 2025<\/strong> designed according to the latest <strong>marking scheme<\/strong> and <strong>blueprint<\/strong> for both <strong>Maths Basic<\/strong> and <strong>Maths Standard<\/strong> courses. <strong>Maths<\/strong> is a subject that requires consistent practice, but it is also one of the most scoring subjects in the <strong>CBSE board exams<\/strong>. Start practicing today with the <strong>Class 10 Maths Sample Papers 2025<\/strong> and ensure you are fully prepared for your <strong>board exams<\/strong><\/p>\n<h3 style=\"text-align: left;\"><span class=\"ez-toc-section\" id=\"Maths_Standard_Sample_Paper_2025_with_Solution_Download_as_PDF\"><\/span>Maths Standard Sample Paper 2025 with Solution <a class=\"button\" style=\"font-weight: bold;\" href=\"https:\/\/mycbseguide.com\/dashboard\/category\/1202\/type\/2\">Download as PDF<\/a><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3 style=\"text-align: left;\"><span class=\"ez-toc-section\" id=\"Maths_Basic_Sample_Paper_2025_with_Solution_Download_as_PDF\"><\/span>Maths Basic Sample Paper 2025 with Solution <a class=\"button\" style=\"font-weight: bold;\" href=\"https:\/\/mycbseguide.com\/dashboard\/category\/1202\/type\/2\">Download as PDF<\/a><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignright\" src=\"https:\/\/media-mycbseguide.s3.ap-south-1.amazonaws.com\/images\/blog\/10%20Maths%20Book.jpg\" alt=\"CBSE Sample Papers Class 10 Mathematics 2019\" width=\"100\" height=\"123\" \/><\/p>\n<h2><span class=\"ez-toc-section\" id=\"CBSE_Class_10_Maths_Sample_Papers_2024-25\"><\/span>CBSE Class 10 Maths Sample Papers 2024-25<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><strong>Download Class 10 Maths Sample Papers 2025 with Solutions \u2013 Free PDF on myCBSEguide<\/strong><\/p>\n<p><a href=\"https:\/\/mycbseguide.com\/\"><strong>myCBSEguide<\/strong><\/a> <strong>Website<\/strong> offers <strong>Class 10 Maths Sample Papers 2025<\/strong> with solutions in <strong>PDF format<\/strong> for <strong>free download<\/strong>. These sample papers are designed to help students understand the updated <strong>exam pattern<\/strong> and <strong>question types<\/strong>. You may notice that the format of the <strong>Class 10 Maths Sample Papers 2025<\/strong> is slightly different from previous years, with a reduced number of straightforward questions and an increased focus on <strong>real-life application-based problems<\/strong>.<\/p>\n<p>Many questions in the <strong>2025 Maths sample papers(SQP)<\/strong> aim to assess your <strong>competency<\/strong> in applying concepts to practical situations.Start practicing today by downloading the <strong>Class 10 Maths Sample Papers 2025<\/strong> with solutions on <a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\"><strong>myCBSEguide<\/strong><\/a> to boost your confidence and score well in your <strong>board exams<\/strong>.<\/p>\n<h2 style=\"text-align: center;\"><span class=\"ez-toc-section\" id=\"Model_Papers_Class_10_Maths_Standard_2025\"><\/span><strong>Model Papers Class 10 Maths Standard 2025<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full\" src=\"https:\/\/media-mycbseguide.s3.ap-south-1.amazonaws.com\/images\/blog\/10_maths_sample_paper.jpg\" alt=\"CBSE class 10 Maths Sample Papers\" width=\"600\" height=\"300\" \/><\/p>\n<p style=\"text-align: center;\"><strong>Class 10 &#8211; Mathematics<br \/>\nMaths Standard SP &#8211; 01 (2024-25)<\/strong><\/p>\n<hr \/>\n<p><b>Maximum Marks: 80<br \/>\nTime Allowed: : 3 hours<\/b><\/p>\n<hr \/>\n<p><b>General Instructions:<\/b><\/p>\n<ol>\n<li>This Question Paper has 5 Sections A, B, C, D and E.<\/li>\n<li>Section A has 20 MCQs carrying 1 mark each<\/li>\n<li>Section B has 5 questions carrying 02 marks each.<\/li>\n<li>Section C has 6 questions carrying 03 marks each.<\/li>\n<li>Section D has 4 questions carrying 05 marks each.<\/li>\n<li>Section E has 3 case based integrated units of assessment carrying 04 marks each.<\/li>\n<li>All Questions are compulsory. However, an internal choice in 2 Qs of 5 marks, 2 Qs of 3 marks and 2 Questions of 2 marks has been provided. An internal choice has been provided in the 2 marks questions of Section E<\/li>\n<li>Draw neat figures wherever required. Take <span class=\"math-tex\">{tex}\\pi=\\frac{22}{7}{\/tex}<\/span>\u00a0wherever required if not stated.<\/li>\n<\/ol>\n<hr \/>\n<ol style=\"padding-left: 20px; list-style: decimal;\">\n<li style=\"text-align: center; clear: both; display: block;\"><b>Section A<\/b><\/li>\n<li>The HCF of the smallest 2-digit number and the smallest composite number is\n<div style=\"margin-left: 20px;\">a) 4<\/div>\n<div style=\"margin-left: 20px;\">b) 10<\/div>\n<div style=\"margin-left: 20px;\">c) 20<\/div>\n<div style=\"margin-left: 20px;\">d) 2<\/div>\n<\/li>\n<li>In the given figure, graph of a polynomial p(x) is given. Number of zeroes of p(x) is:<br \/>\n<img decoding=\"async\" style=\"height: 122px; width: 168px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1708082561-uc6sxn.jpg\" \/><\/p>\n<div style=\"margin-left: 20px;\">a) 2<\/div>\n<div style=\"margin-left: 20px;\">b) 4<\/div>\n<div style=\"margin-left: 20px;\">c) 5<\/div>\n<div style=\"margin-left: 20px;\">d) 3<\/div>\n<\/li>\n<li>The pair of linear equations y = 0 and y = &#8211; 6 has:<br \/>\n<img decoding=\"async\" style=\"width: 250px; height: 126px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1681814375-9r2bdk.jpg\" alt=\"\" \/><\/p>\n<div style=\"margin-left: 20px;\">a) no solution<\/div>\n<div style=\"margin-left: 20px;\">b) only solution (0, 0)<\/div>\n<div style=\"margin-left: 20px;\">c) infinitely many solutions<\/div>\n<div style=\"margin-left: 20px;\">d) a unique solution<\/div>\n<\/li>\n<li>The roots of the quadratic equation 2x<sup>2<\/sup> &#8211; x &#8211; 6 = 0 are\n<div style=\"margin-left: 20px;\">a) <span class=\"math-tex\">{tex}2, \\frac{-3}{2}{\/tex}<\/span><\/div>\n<div style=\"margin-left: 20px;\">b) <span class=\"math-tex\">{tex}2, \\frac{3}{2}{\/tex}<\/span><\/div>\n<div style=\"margin-left: 20px;\">c) <span class=\"math-tex\">{tex}-2, \\frac{-3}{2}{\/tex}<\/span><\/div>\n<div style=\"margin-left: 20px;\">d) <span class=\"math-tex\">{tex}-2, \\frac{3}{2}{\/tex}<\/span><\/div>\n<\/li>\n<li>The 5th term of an AP is -3 and its common difference is -4. The sum of its first 10 terms is\n<div style=\"margin-left: 20px;\">a) 30<\/div>\n<div style=\"margin-left: 20px;\">b) -50<\/div>\n<div style=\"margin-left: 20px;\">c) -30<\/div>\n<div style=\"margin-left: 20px;\">d) 50<\/div>\n<\/li>\n<li>The point on the x-axis which is equidistant from points (-1, 0)\u00a0and (5, 0)\u00a0is\n<div style=\"margin-left: 20px;\">a) (0, 3)<\/div>\n<div style=\"margin-left: 20px;\">b) (2, 0)<\/div>\n<div style=\"margin-left: 20px;\">c) (3, 0)<\/div>\n<div style=\"margin-left: 20px;\">d) (0, 2)<\/div>\n<\/li>\n<li>The centre of the circle having end points of its one diameter as (-4, 2) and (4, -3) is\n<div style=\"margin-left: 20px;\">a) (2, -1)<\/div>\n<div style=\"margin-left: 20px;\">b) (0, -1)<\/div>\n<div style=\"margin-left: 20px;\">c) (1, -1)<\/div>\n<div style=\"margin-left: 20px;\">d) \u00a0(0,<span class=\"math-tex\">{tex}\\frac{-1}{2}{\/tex}<\/span>)<\/div>\n<\/li>\n<li>In the given figure, DE || BC. If AD = 2 units, DB = AE = 3 units and EC = x units, then the value of x is:<br \/>\n<img decoding=\"async\" style=\"height: 131px; width: 120px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1688982744-hh3wyu.jpg\" \/><\/p>\n<div style=\"margin-left: 20px;\">a) 2<\/div>\n<div style=\"margin-left: 20px;\">b) 3<\/div>\n<div style=\"margin-left: 20px;\">c) <span class=\"math-tex\">{tex}\\frac{9}{2}{\/tex}<\/span><\/div>\n<div style=\"margin-left: 20px;\">d) 5<\/div>\n<\/li>\n<li>In Figure, TP and TQ are tangents drawn to the circle with centre at O. If <span class=\"math-tex\">{tex}\\angle \\text { POQ}{\/tex}<\/span> = 115<sup>o<\/sup> then <span class=\"math-tex\">{tex}\\angle \\text { PTQ}{\/tex}<\/span> is<br \/>\n<img decoding=\"async\" style=\"height: 112px; width: 150px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1710830366-zyh92h.jpg\" \/><\/p>\n<div style=\"margin-left: 20px;\">a) 55<sup>o<\/sup><\/div>\n<div style=\"margin-left: 20px;\">b) 115<sup>o<\/sup><\/div>\n<div style=\"margin-left: 20px;\">c) 65<sup>o<\/sup><\/div>\n<div style=\"margin-left: 20px;\">d) 57.5<sup>o<\/sup><\/div>\n<\/li>\n<li>Quadrilateral ABCD is circumscribed to a circle. If AB = 6 cm, BC = 7 cm and CD = 4 cm then the length of AD is\n<div style=\"margin-left: 20px;\">a) 6 cm<\/div>\n<div style=\"margin-left: 20px;\">b) 4 cm<\/div>\n<div style=\"margin-left: 20px;\">c) 7 cm<\/div>\n<div style=\"margin-left: 20px;\">d) 3 cm<\/div>\n<\/li>\n<li>8 (cos<sup>2<\/sup> A + sin<sup>2<\/sup>\u00a0 A) is equal to:\n<div style=\"margin-left: 20px;\">a) 8<\/div>\n<div style=\"margin-left: 20px;\">b) 0<\/div>\n<div style=\"margin-left: 20px;\">c) 9<\/div>\n<div style=\"margin-left: 20px;\">d) 1<\/div>\n<\/li>\n<li>If 2 tan A = 3, then the value of\u00a0<span class=\"math-tex\">{tex}\\frac{4 \\sin A+3 \\cos A}{4 \\sin A-3 \\cos A}{\/tex}<\/span>\u00a0is\n<div style=\"margin-left: 20px;\">a) 3<\/div>\n<div style=\"margin-left: 20px;\">b) does not exist<\/div>\n<div style=\"margin-left: 20px;\">c) <span class=\"math-tex\">{tex}\\frac{7}{\\sqrt{13}}{\/tex}<\/span><\/div>\n<div style=\"margin-left: 20px;\">d) <span class=\"math-tex\">{tex}\\frac{1}{\\sqrt{13}}{\/tex}<\/span><\/div>\n<\/li>\n<li>A kite is flying at a height of 30 m from the ground. The length of string from the kite to the ground is 60 m. Assuming that there is no slack in the string, the angle of elevation of the kite at the ground is\n<div style=\"margin-left: 20px;\">a) 30\u00b0<\/div>\n<div style=\"margin-left: 20px;\">b) 45\u00b0<\/div>\n<div style=\"margin-left: 20px;\">c) 90\u00b0<\/div>\n<div style=\"margin-left: 20px;\">d) 60\u00b0<\/div>\n<\/li>\n<li>The area of the sector of a circle with radius 6 cm which subtends an angle of 60<sup>o<\/sup> at the centre of the circle is:\n<div style=\"margin-left: 20px;\">a) <span class=\"math-tex\">{tex}\\frac {152}{7}{\/tex}<\/span>cm<sup>2<\/sup><\/div>\n<div style=\"margin-left: 20px;\">b) <span class=\"math-tex\">{tex}\\frac {132}{7}{\/tex}<\/span>cm<sup>2<\/sup><\/div>\n<div style=\"margin-left: 20px;\">c) <span class=\"math-tex\">{tex}\\frac {142}{7}{\/tex}<\/span>cm<sup>2<\/sup><\/div>\n<div style=\"margin-left: 20px;\">d) <span class=\"math-tex\">{tex}\\frac {122}{7}{\/tex}<\/span>cm<sup>2<\/sup><\/div>\n<\/li>\n<li>The length of an arc that subtends an angle of 24<sup>o<\/sup> at the centre of a circle with 5 cm radius is\n<div style=\"margin-left: 20px;\">a) <span class=\"math-tex\">{tex}\\frac{3 \\pi}{2} {\/tex}<\/span> cm<\/div>\n<div style=\"margin-left: 20px;\">b) <span class=\"math-tex\">{tex}\\frac{5 \\pi}{3} {\/tex}<\/span> cm<\/div>\n<div style=\"margin-left: 20px;\">c) <span class=\"math-tex\">{tex}\\frac{\\pi}{3} {\/tex}<\/span> cm<\/div>\n<div style=\"margin-left: 20px;\">d) <span class=\"math-tex\">{tex}\\frac{2 \\pi}{3} {\/tex}<\/span> cm<\/div>\n<\/li>\n<li>The probablity that a number selected at random from the numbers 1, 2, 3, &#8230;, 15\u00a0is a multiple of 4, is\n<div style=\"margin-left: 20px;\">a) <span class=\"math-tex\">{tex}\\frac{2}{15}{\/tex}<\/span><\/div>\n<div style=\"margin-left: 20px;\">b) <span class=\"math-tex\">{tex}\\frac{1}{5}{\/tex}<\/span><\/div>\n<div style=\"margin-left: 20px;\">c) <span class=\"math-tex\">{tex}\\frac{4}{15}{\/tex}<\/span><\/div>\n<div style=\"margin-left: 20px;\">d) <span class=\"math-tex\">{tex}\\frac{1}{15}{\/tex}<\/span><\/div>\n<\/li>\n<li>A box contains 90 discs, numbered from 1 to 90. If one disc\u00a0is drawn at random from the box, the probability that it bears a prime number less than 23,\u00a0is\n<div style=\"margin-left: 20px;\">a) <span class=\"math-tex\">{tex}\\frac{10}{90}{\/tex}<\/span><\/div>\n<div style=\"margin-left: 20px;\">b) <span class=\"math-tex\">{tex}\\frac{7}{90}{\/tex}<\/span><\/div>\n<div style=\"margin-left: 20px;\">c) <span class=\"math-tex\">{tex}\\frac{9}{89}{\/tex}<\/span><\/div>\n<div style=\"margin-left: 20px;\">d) <span class=\"math-tex\">{tex}\\frac{4}{45}{\/tex}<\/span><\/div>\n<\/li>\n<li>The cumulative frequency table is useful in determining\n<div style=\"margin-left: 20px;\">a) All of these<\/div>\n<div style=\"margin-left: 20px;\">b) Mean<\/div>\n<div style=\"margin-left: 20px;\">c) Median<\/div>\n<div style=\"margin-left: 20px;\">d) Mode<\/div>\n<\/li>\n<li><strong>Assertion (A):<\/strong> In a solid hemisphere of radius 10 cm, a right cone of same radius is removed out. The surface area of the remaining solid is 570.74\u00a0cm<sup>2<\/sup>\u00a0[Take <span class=\"math-tex\">{tex}\\pi{\/tex}<\/span>\u00a0= 3.14 and <span class=\"math-tex\">{tex}\\sqrt 2{\/tex}<\/span>\u00a0= 1.4]\nReason (R):\u00a0<strong>Reason (R):<\/strong>\u00a0Expression used here to calculate Surface area of remaining solid = Curved surface area of hemisphere +\u00a0Curved surface area of cone<\/p>\n<div style=\"margin-left: 20px;\">a) Both A and R are true and R is the correct explanation of A.<\/div>\n<div style=\"margin-left: 20px;\">b) Both A and R are true but R is not the correct explanation of A.<\/div>\n<div style=\"margin-left: 20px;\">c) A is true but R is false.<\/div>\n<div style=\"margin-left: 20px;\">d) A is false but R is true.<\/div>\n<\/li>\n<li><strong>Assertion (A):<\/strong> If n<sup>th<\/sup> term of an A.P. is 7 &#8211; 4n, then its common difference\u00a0is -4.<br \/>\n<strong>Reason (R):<\/strong> Common difference of an A.P. is given by d = a<sub>n &#8211; 1<\/sub> &#8211; a<sub>n<\/sub><\/p>\n<div style=\"margin-left: 20px;\">a) Both A and R are true and R is the correct explanation of A.<\/div>\n<div style=\"margin-left: 20px;\">b) Both A and R are true but R is not the correct explanation of A.<\/div>\n<div style=\"margin-left: 20px;\">c) A is true but R is false.<\/div>\n<div style=\"margin-left: 20px;\">d) A is false but R is true.<br \/>\n<strong>Download myCBSEguide App for Comprehensive Exam Preparation<\/strong>For effective exam preparation, download the <a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\"><strong>myCBSEguide.com<\/strong><\/a> app, which offers complete study material for <strong>CBSE<\/strong>, <strong>NCERT<\/strong>, <strong>JEE (Main)<\/strong>, <strong>NEET-UG<\/strong>, and <strong>NDA<\/strong> exams. With a wide range of practice questions, sample papers, and detailed solutions, <strong>myCBSEguide<\/strong> helps students enhance their learning and perform better in exams.Teachers can also utilize the <a href=\"https:\/\/play.google.com\/store\/apps\/details?id=com.examin8.exam\"><strong>Examin8.com<\/strong><\/a> app to create customized exam papers with their own branding, including name and logo, making it easy to assess and engage students. Start preparing today with the <strong><a href=\"https:\/\/mycbseguide.com\/\">myCBSEguide<\/a> Website<\/strong> and <a href=\"https:\/\/examin8.com\/\"><strong>Examin8<\/strong><\/a> <strong>Website<\/strong> for a more streamlined and effective study experience.<\/div>\n<\/li>\n<li style=\"text-align: center; clear: both; display: block;\"><b>Section B<\/b><\/li>\n<li>An army contingent of 612 members is to march behind an army band of 48 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?<\/li>\n<li><span class=\"math-tex\">{tex}\\triangle ABC{\/tex}<\/span>\u00a0is a right triangle right-angled at A and\u00a0<span class=\"math-tex\">{tex}A D \\perp B C{\/tex}<\/span>.\u00a0If BC = 13 cm and AC = 5 cm, find\u00a0the ratio of the areas of\u00a0<span class=\"math-tex\">{tex}\\triangle ABC{\/tex}<\/span>\u00a0and\u00a0<span class=\"math-tex\">{tex}\\triangle ADC{\/tex}<\/span>.<br \/>\n<img decoding=\"async\" style=\"width: 180px; height: 133px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/UzyIbO4.png\" alt=\"\" data-imgur-src=\"UzyIbO4.png\" \/><\/li>\n<li>If PA and PB are two tangents drawn from a point P to a circle with centre O touching it at A and B, prove that OP is perpendicular bisector of AB.<br \/>\n<img decoding=\"async\" style=\"width: 150px; height: 121px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/RTlle3A.png\" alt=\"\" data-imgur-src=\"RTlle3A.png\" \/><\/li>\n<li>Prove that:<br \/>\n<span class=\"math-tex\">{tex}\\frac{1+\\tan \\text A}{2 \\sin \\text A}+\\frac{1+\\cot\\text A}{2 \\cos \\text A}{\/tex}<\/span> = cosec A + sec A<\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>If sin<span class=\"math-tex\">{tex}\\theta{\/tex}<\/span>\u00a0+ cos <span class=\"math-tex\">{tex}\\theta{\/tex}<\/span>\u00a0= <span class=\"math-tex\">{tex}\\sqrt{3}{\/tex}<\/span> , then prove that tan <span class=\"math-tex\">{tex}\\theta{\/tex}<\/span>\u00a0+ cot <span class=\"math-tex\">{tex}\\theta{\/tex}<\/span>\u00a0= 1<\/li>\n<li>Find the area of the sector of a circle of radius 7 cm and of central angle 90<sup>o<\/sup>. Also, find the area of corresponding major sector.\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>A horse is tethered to one corner of a field which is in the shape of an equilateral triangle of side 12 m. If the length of the rope is 7 m, find the area of the field which the horse cannot graze. Take <span class=\"math-tex\">{tex}\\sqrt3{\/tex}<\/span>\u00a0= 1.732. Write the answer correct to 2 places of decimal.<\/li>\n<li style=\"text-align: center; clear: both; display: block;\"><b>Section C<\/b><\/li>\n<li>Prove that <span class=\"math-tex\">{tex}( 5 + 3 \\sqrt { 2 } ){\/tex}<\/span> is irrational.<\/li>\n<li>Find the zeroes of the quadratic polynomial 6x<sup>2\u00a0<\/sup>&#8211; 3 &#8211; 7x and verify the relationship between the zeroes and the coefficients of the polynomial.<\/li>\n<li>How many terms of the AP 9,17,25,&#8230; must be taken so that their sum is 636?\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>If the 8th term of an AP is 37 and the 15th term is 15 more than the 12th term. Find the AP. Hence find the sum of first 15 terms of the AP.<\/li>\n<li>Two concentric circles are of diameters 30 cm and 18 cm. Find the length of the chord of the larger circle which touches the smaller circle.\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>In figure,\u00a0O\u00a0is the centre of a circle. PT and PQ are tangents to the circle from an external point P. If <span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>TPQ = 70\u00b0, find <span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>TRQ.<br \/>\n<img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/bXrGgti.png\" alt=\"\" data-imgur-src=\"bXrGgti.png\" \/><\/li>\n<li>If sec<span class=\"math-tex\">{tex}\\theta{\/tex}<\/span>\u00a0+ tan<span class=\"math-tex\">{tex}\\theta{\/tex}<\/span>\u00a0= p, show that\u00a0<span class=\"math-tex\">{tex}\\frac { p ^ { 2 } &#8211; 1 } { p ^ { 2 } + 1 } = \\sin \\theta{\/tex}<\/span><\/li>\n<li>The frequency distribution given below shows the weight of 40 students of a class. Find the median weight of the students.<br \/>\n<table style=\"width: 100%;\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<tbody>\n<tr>\n<td style=\"text-align: center;\">Weight (in kg)<\/td>\n<td style=\"text-align: center;\">Number of Students<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">40 &#8211; 45<\/td>\n<td style=\"text-align: center;\">9<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">45 &#8211; 50<\/td>\n<td style=\"text-align: center;\">5<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">50 &#8211; 55<\/td>\n<td style=\"text-align: center;\">8<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">55 &#8211; 60<\/td>\n<td style=\"text-align: center;\">9<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">60 &#8211; 65<\/td>\n<td style=\"text-align: center;\">6<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">65 &#8211; 70<\/td>\n<td style=\"text-align: center;\">3<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n<li style=\"text-align: center; clear: both; display: block;\"><b>Section D<\/b><\/li>\n<li>Two water taps together can fill a tank in 6 hours. The tap of larger diameter takes 9 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>A train covered a certain distance at a uniform speed. If the train would have been 6 km\/h faster, it would have taken 4 hours less than the scheduled time. And, if the train were slower by 6 km\/h, it would have taken 6 hours more than the scheduled time. Find the length of the journey.<\/li>\n<li>From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60<sup>o<\/sup> and the angle of depression of its foot is 30<sup>o<\/sup>. Determine the height of the tower.<\/li>\n<li>A toy is in the form of a cylinder of diameter <span class=\"math-tex\">{tex}2 \\sqrt { 2 } m{\/tex}<\/span> and height 3.5 m surmounted by a cone whose vertical angle is 90\u00b0. Find total surface area of the toy.\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>In Figure, from a solid cube of side 7 cm, a cylinder of radius 2.1 cm and height 7 cm is scooped out. Find the total surface area of the remaining solid.<br \/>\n<img decoding=\"async\" style=\"width: 100px; height: 110px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1679118169-t2897p.jpg\" \/><\/li>\n<li>A survey conducted on 20 families in a locality by a group of students resulted in the following frequency table for the number of family members in a family.<br \/>\n<table style=\"width: 100%;\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<tbody>\n<tr>\n<td><strong>Family size<\/strong><\/td>\n<td>\n<p style=\"text-align: center;\">1 &#8211; 3<\/p>\n<\/td>\n<td>\n<p style=\"text-align: center;\">3 &#8211; 5<\/p>\n<\/td>\n<td>\n<p style=\"text-align: center;\">5 &#8211; 7<\/p>\n<\/td>\n<td>\n<p style=\"text-align: center;\">7 &#8211; 9<\/p>\n<\/td>\n<td>\n<p style=\"text-align: center;\">9 &#8211; 11<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td><strong>Number\u00a0of families<\/strong><\/td>\n<td>\n<p style=\"text-align: center;\">7<\/p>\n<\/td>\n<td>\n<p style=\"text-align: center;\">8<\/p>\n<\/td>\n<td>\n<p style=\"text-align: center;\">2<\/p>\n<\/td>\n<td>\n<p style=\"text-align: center;\">4<\/p>\n<\/td>\n<td>\n<p style=\"text-align: center;\">1<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Determine the mean and mode of the above data.<\/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 \/>\n<img decoding=\"async\" style=\"height: 165px; width: 250px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1691560192-g2ttbr.jpg\" \/><br \/>\nLokesh, a production manager in Mumbai, hires a taxi everyday to go to his office. The taxi charges in Mumbai consists of a fixed charges together with the charges for the distance covered. His office is at a distance of 10 km from his home. For a distance of 10 km to his office, Lokesh paid \u20b9 105. While coming back home, he took another route. He covered a distance of 15 km and the charges paid by him were \u20b9 155.<\/p>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>What are the fixed charges? (1)<\/li>\n<li>What are the charges per km? (1)<\/li>\n<li>If fixed charges are \u20b9 20 and charges per km are \u20b9 10, then how much Lokesh have to pay for travelling a distance of 10 km? (2)<br \/>\n<strong>OR<\/strong><br \/>\nFind the total amount paid by Lokesh for travelling 10 km from home to office and 25 km from office to home. [Fixed charges and charges per km are as in (i) &amp; (ii). (2)<\/li>\n<\/ol>\n<\/li>\n<li><strong>Read the following text carefully and answer the questions that follow:<\/strong><br \/>\nTwo poles, 30 feet and 50 feet tall, are 40 feet apart and perpendicular to the ground. The poles are supported by wires attached from the top of each pole to the bottom of the other, as in the figure. A coupling is placed at C where the two wires cross.<br \/>\n<img decoding=\"async\" style=\"height: 202px; width: 250px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1671162084-fhumkm.jpg\" \/><\/p>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>What is the horizontal distance from C to the taller pole? (1)<\/li>\n<li>How high above the ground is the coupling? (1)<\/li>\n<li>How far down the wire from the smaller pole is the coupling? (2)<br \/>\n<strong>OR<\/strong><br \/>\nFind the length of line joining the top of the two poles. (2)<\/li>\n<\/ol>\n<\/li>\n<li><strong>Read the following text carefully and answer the questions that follow:<\/strong><br \/>\nReena has a 10 m <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> 10 m kitchen garden attached to her kitchen. She divides it into a 10 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> 10 grid and wants to grow some vegetables and herbs used in the kitchen. She puts some soil and manure in that and sow a green chilly plant at A, a coriander plant at B and a tomato plant at C. Her friend Kavita visited the garden and praised the plants grown there. She pointed out that they seem to be in a straight line. See the below diagram carefully:<br \/>\n<img decoding=\"async\" style=\"height: 197px; width: 250px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1671173488-k66cp3.jpg\" \/><\/p>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>Find the distance between A and B? (1)<\/li>\n<li>Find the mid-point of the distance AB? (1)<\/li>\n<li>Find the distance between B and C? (2)<br \/>\n<strong>OR<\/strong><br \/>\nFind the mid point of BC. (2)<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<p><strong>Enhance Your Exam Preparation with myCBSEguide App<\/strong><\/p>\n<p>Looking for the best tools to boost your exam preparation? Download the <a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\"><strong>myCBSEguide.com<\/strong><\/a> app, your one-stop solution for comprehensive study material for <strong>CBSE<\/strong>, <strong>NCERT<\/strong>, <strong>JEE (Main)<\/strong>, <strong>NEET-UG<\/strong>, and <strong>NDA<\/strong> exams. The app offers a vast collection of practice questions, sample papers, NCERT solutions, and more to help you master every topic and perform better in your exams.<\/p>\n<p>For educators, the <a href=\"https:\/\/play.google.com\/store\/apps\/details?id=com.examin8.exam\"><strong>Examin8.com<\/strong><\/a> app is an ideal platform to create customized question papers with personal branding, including name and logo. Also <a href=\"https:\/\/mycbseguide.com\/\"><strong>myCBSEguide<\/strong><\/a> and <a href=\"https:\/\/examin8.com\/\"><strong>Examin8<\/strong><\/a> apps provide all the resources you need to succeed, whether you&#8217;re a student or a teacher.<\/p>\n<p style=\"text-align: center; page-break-before: always;\"><strong>Class 10 &#8211; Mathematics<br \/>\nMaths Standard SP &#8211; 01 (2024-25)<\/strong><\/p>\n<hr \/>\n<h2 style=\"text-align: center;\"><span class=\"ez-toc-section\" id=\"Solution\"><\/span><strong>Solution <\/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>(d) 2<br \/>\n<b>Explanation: <\/b> Smallest two digit number is 10\u00a0and smallest composite number is 4 .<br \/>\nClearly, 2 is the\u00a0greatest\u00a0factor of 4 and 10,\u00a0so their H.C.F. is 2.<\/li>\n<li>(b) 4<br \/>\n<b>Explanation: <\/b> Number of zeros= number of times the graph touches x-axis.<br \/>\nHere the graph touches x-axis 4 times.<\/li>\n<li>(a) no solution<br \/>\n<b>Explanation: <\/b> Since, we have y = 0 and y = -6 are two parallel lines.<br \/>\ntherefore, no solution exists.<\/li>\n<li>(a) <span class=\"math-tex\">{tex}2, \\frac{-3}{2}{\/tex}<\/span><br \/>\n<b>Explanation: <\/b> The given quadratic equation is\u00a02x<sup>2<\/sup> &#8211; x &#8211; 6 = 0<br \/>\n2x<sup>2<\/sup>\u00a0&#8211; x &#8211; 6 = 0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a02x<sup>2<sub>\u00a0<\/sub><\/sup>&#8211; 4x + 3x &#8211; 6 = 0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a02x (x-2) + 3 (x-2) = 0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0x &#8211; 2 = 0 or 2x + 3 = 0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0x = 2 or x = -3\/2<br \/>\nThus, the roots of the given equation are 2 and -3\/2<\/li>\n<li>(b) -50<br \/>\n<b>Explanation: <\/b> a + 4d = -3 and d = -4<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0a = -3 + 16<br \/>\nSo, a = 13.<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}S_{10}=\\frac{10}{2}[2 a+9 d]=5[2 \\times 13+9 \\times(-4)]=-50{\/tex}<\/span>.<\/li>\n<li>(b) (2, 0)<br \/>\n<b>Explanation: <\/b> Let the required point be P(x, 0). Then,<br \/>\n<span class=\"math-tex\">{tex}P A^{2}=P B^{2} \\Rightarrow(x+1)^{2}=(x-5)^{2}\\\\ \\Rightarrow x^{2}+2 x+1=x^{2}-10 x+25\\\\ \\Rightarrow 12 x=24 \\Rightarrow x=2{\/tex}<\/span><br \/>\nSo, the required point is P(2, 0).<\/li>\n<li>(d) \u00a0(0,<span class=\"math-tex\">{tex}\\frac{-1}{2}{\/tex}<\/span>)<br \/>\n<b>Explanation: <\/b> Centare of circle is equal to mid point of diameter.<br \/>\n<img decoding=\"async\" style=\"height: 80px; width: 150px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1716194444-7x4ymz.jpg\" \/><br \/>\nMidpoint (O) = \u00a0(0,<span class=\"math-tex\">{tex}\\frac{-1}{2}{\/tex}<\/span>)<br \/>\nCoordinate of O =\u00a0\u00a0(0,<span class=\"math-tex\">{tex}\\frac{-1}{2}{\/tex}<\/span>)<\/li>\n<li>(c) <span class=\"math-tex\">{tex}\\frac{9}{2}{\/tex}<\/span><br \/>\n<b>Explanation: <\/b> According to Basic Proportionality theorem.<br \/>\n<span class=\"math-tex\">{tex}\\frac {\\text {AD}}{\\text {DB}}{\/tex}<\/span>\u00a0=\u00a0<span class=\"math-tex\">{tex}\\frac {\\text {AE}}{\\text {EC}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\frac 23{\/tex}<\/span>\u00a0=\u00a0<span class=\"math-tex\">{tex}\\frac 3x{\/tex}<\/span><br \/>\nx =\u00a0<span class=\"math-tex\">{tex}\\frac 92{\/tex}<\/span><\/li>\n<li>(c) 65<sup>o<\/sup><br \/>\n<b>Explanation: <\/b> OPTQ is a quadrilateral<br \/>\n<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>\u00a0O+<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span> P + <span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>T +\u00a0 <span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>Q = 360<sup>o<\/sup><br \/>\n115<sup>o<\/sup> + 90<sup>o<\/sup> + <span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>PTQ + 90<sup>o<\/sup> = 360<sup>o<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>\u00a0PTQ = 360<sup>o<\/sup> &#8211; 295<sup>o<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>\u00a0PTQ = 65<sup>o<\/sup><\/li>\n<li>(d) 3 cm<br \/>\n<b>Explanation: <\/b> A quadrilateral ABCD is circumscribed to a circle with centre O.<br \/>\nAB = 6 cm, BC = 7 cm, CD = 4 cm, AD = 7 cm<br \/>\nABCD circumscribed to a circle.<br \/>\nAB + CD = BC + AD<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> 6 + 4 = 7 + AD<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a010 = 7 + AD<br \/>\nAD = 10 &#8211;\u00a07 = 3 cm<\/li>\n<li>(a) 8<br \/>\n<b>Explanation: <\/b> 8(cos<sup>2<\/sup> A + sin<sup>2<\/sup> A) = 8(1) = 8<\/li>\n<li>(a) 3<br \/>\n<b>Explanation: <\/b> 2tan A =\u00a03<br \/>\ntan A =\u00a0<span class=\"math-tex\">{tex}\\frac32{\/tex}<\/span>\u00a0=<span class=\"math-tex\">{tex}\\frac Pb{\/tex}<\/span><br \/>\nh =\u00a0<span class=\"math-tex\">{tex}\\sqrt{P^2+b^2}{\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\sqrt{3^2+2^2}{\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\sqrt{13}{\/tex}<\/span><br \/>\nNow, Sin A =\u00a0<span class=\"math-tex\">{tex}\\frac Ph{\/tex}<\/span>\u00a0=\u00a0<span class=\"math-tex\">{tex}\\frac{3}{\\sqrt{13}}{\/tex}<\/span><br \/>\nCos A =\u00a0<span class=\"math-tex\">{tex}\\frac bh{\/tex}<\/span>\u00a0=\u00a0<span class=\"math-tex\">{tex}\\frac{2}{\\sqrt{13}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\frac{4Sin\\ A + 3Cos\\ A}{4Sin\\ A-3cos\\ A}{\/tex}<\/span>=\u00a0<span class=\"math-tex\">{tex}\\frac {4(\\frac {3}{\\sqrt {13}}) + 3(\\frac {2}{\\sqrt {13}})}{4(\\frac{3}{\\sqrt{13}}) &#8211; 3(\\frac{2}{\\sqrt{13}})}{\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\frac{\\frac{12}{\\sqrt{13}}+\\frac{6}{\\sqrt{13}}}{\\frac{12}{\\sqrt{13}} &#8211; {\\frac{6}{\\sqrt{13}}}}{\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\frac{18}{6}{\/tex}<\/span>\u00a0= 3<\/li>\n<li>(a) 30\u00b0<br \/>\n<b>Explanation: <\/b> Let AB be the tower and B be the kite.<br \/>\nLet AC be the horizontal and let BC\u00a0<span class=\"math-tex\">{tex}\\perp{\/tex}<\/span>\u00a0AC.<br \/>\nLet\u00a0<span class=\"math-tex\">{tex}\\angle{\/tex}<\/span>CAB =\u00a0<span class=\"math-tex\">{tex}\\theta{\/tex}<\/span>.<br \/>\nBC = 30 m and AB = 60 m. Then,<br \/>\n<span class=\"math-tex\">{tex}\\frac{B C}{A B}{\/tex}<\/span>\u00a0= sin\u00a0<span class=\"math-tex\">{tex}\\theta{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0sin\u00a0<span class=\"math-tex\">{tex}\\theta{\/tex}<\/span>\u00a0=\u00a0<span class=\"math-tex\">{tex}\\frac{30}{60}=\\frac{1}{2} \\Rightarrow{\/tex}<\/span>\u00a0sin\u00a0<span class=\"math-tex\">{tex}\\theta{\/tex}<\/span>\u00a0= sin 30<sup>o<\/sup>\u00a0<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\theta{\/tex}<\/span>\u00a0= 30<sup>o<\/sup>.<br \/>\n<img decoding=\"async\" style=\"width: 130px; height: 74px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/fW7h7eJ.png\" alt=\"\" data-imgur-src=\"fW7h7eJ.png\" \/><\/li>\n<li>(b) <span class=\"math-tex\">{tex}\\frac {132}{7}{\/tex}<\/span>cm<sup>2<\/sup><br \/>\n<b>Explanation: <\/b> Angle of the sector is 60\u00b0<br \/>\nArea of sector = (<span class=\"math-tex\">{tex}\\frac {\\theta}{360^o}{\/tex}<\/span>) <span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\pi{\/tex}<\/span>r<sup>2<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0Area of the sector with angle 60<sup>o<\/sup>\u00a0= (<span class=\"math-tex\">{tex}\\frac {60^o}{360^o}{\/tex}<\/span>) <span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\pi{\/tex}<\/span>r<sup>2<\/sup>\u00a0cm<sup>2<\/sup><br \/>\n= (<span class=\"math-tex\">{tex}\\frac {36}{6}{\/tex}<\/span>)<span class=\"math-tex\">{tex}\\pi{\/tex}<\/span>\u00a0cm<sup>2<\/sup><br \/>\n= 6 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a0(<span class=\"math-tex\">{tex}\\frac {22}{7}{\/tex}<\/span>) cm<sup>2<\/sup><br \/>\n= <span class=\"math-tex\">{tex}\\frac {132}{7}{\/tex}<\/span>\u00a0cm<sup>2<\/sup><\/li>\n<li>(d) <span class=\"math-tex\">{tex}\\frac{2 \\pi}{3} {\/tex}<\/span> cm<br \/>\n<b>Explanation: <\/b> Given that, <span class=\"math-tex\">{tex}\\theta{\/tex}<\/span> = 30<sup>o<\/sup> and r = 4 cm<br \/>\nLength of an arc <span class=\"math-tex\">{tex}=\\frac{\\theta}{360^{\\circ}}(2 \\pi r){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}=\\frac{30^{\\circ}}{360^{\\circ}} \\times 2 \\pi \\times 4{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}=\\frac{2 \\pi \\times 4}{12}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span> Length of an arc <span class=\"math-tex\">{tex}=\\frac{2 \\pi}{3}{\/tex}<\/span><\/li>\n<li>(b) <span class=\"math-tex\">{tex}\\frac{1}{5}{\/tex}<\/span><br \/>\n<b>Explanation: <\/b> Total outcomes = 15<br \/>\n(<span class=\"math-tex\">{tex}\\because{\/tex}<\/span> 15 numbers are given)<br \/>\nFavourable outcomes for a multiple of 4 = 3\u00a0(i.e. 4, 8, 12)<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0Probability of selecting a number which is a multiple of 4 = <span class=\"math-tex\">{tex}\\frac{3}{15}=\\frac{1}{5}{\/tex}<\/span><\/li>\n<li>(d) <span class=\"math-tex\">{tex}\\frac{4}{45}{\/tex}<\/span><br \/>\n<b>Explanation: <\/b> Number of disc\u00a0in a box = 90<br \/>\nNumbered on it are 1 to 90 Prime numbers less than 23 are = 2, 3, 5, 7, 11, 13, 17, 19 = 8<br \/>\nProbability of a number being a prime less than\u00a0<span class=\"math-tex\">{tex}23=\\frac{8}{90}=\\frac{4}{45}{\/tex}<\/span><\/li>\n<li>(c) Median<br \/>\n<b>Explanation: <\/b> Median<\/li>\n<li>(d) A is false but R is true.<br \/>\n<b>Explanation: <\/b> A is false but R is true.<\/li>\n<li>(a) Both A and R are true and R is the correct explanation of A.<br \/>\n<b>Explanation: <\/b> Both are correct. Reason is the correct explanation.<br \/>\nAssertion,<br \/>\na<sub>n<\/sub> = 7 &#8211; 4n<br \/>\nd = a<sub>n &#8211; 1<\/sub>\u00a0&#8211; a<sub>n<\/sub><br \/>\n= 7 &#8211; 4(n + 1) &#8211; (7 &#8211; 4n)<br \/>\n= 7 &#8211; 4n &#8211; 4 &#8211; 7 + 4n = -4<br \/>\n<strong>Download myCBSEguide App for Comprehensive Exam Preparation<\/strong>To enhance your exam preparation, download the <a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\"><strong>myCBSEguide<\/strong><\/a>\u00a0app today! It offers complete study materials for <strong>CBSE<\/strong>, <strong>NCERT<\/strong>, <strong>JEE (Main)<\/strong>, <strong>NEET-UG<\/strong>, and <strong>NDA<\/strong> exams. With an extensive collection of practice questions, sample papers, NCERT solutions, and detailed explanations, the app is designed to help you study smarter and perform better. By regularly practicing <strong>Class 10 Maths Sample Papers 2025<\/strong>, students can gain a clear understanding of the exam format and question types.For teachers, the <a href=\"https:\/\/examin8.com\/\"><strong>Examin8<\/strong><\/a>\u00a0app provides an excellent platform to create personalized question papers with your own name, logo, and branding. Whether you&#8217;re a student or an educator, <strong>myCBSEguide<\/strong> and <strong>Examin8<\/strong> are the perfect apps to simplify your exam preparation and assessment process and for more visit <a href=\"https:\/\/mycbseguide.com\/\"><strong>myCBSEGuide<\/strong><\/a> Website and <a href=\"https:\/\/examin8.com\/\"><strong>Examin8<\/strong><\/a> Website.<\/li>\n<li style=\"text-align: center; display: block;\"><b>Section B <\/b><\/li>\n<li>Clearly, the maximum number of columns = HCF (612, 48).<br \/>\n<img decoding=\"async\" style=\"width: 135px; height: 100px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/d6Uaw6m.png\" alt=\"\" data-imgur-src=\"d6Uaw6m.png\" \/><br \/>\nNow,\u00a0<span class=\"math-tex\">{tex} 612 {\/tex}<\/span> =<span class=\"math-tex\">{tex}2 \\times 2 \\times 3 \\times 3 \\times 17{\/tex}<\/span> <span class=\"math-tex\">{tex}= \\left( 2 ^ { 2 } \\times 3 ^ { 2 } \\times 17 \\right){\/tex}<\/span><br \/>\nand\u00a0<span class=\"math-tex\">{tex} 48= 2 \\times 2 \\times 2 \\times 2 \\times 3 = \\left( 2 ^ { 4 } \\times 3 \\right){\/tex}<\/span>.<br \/>\n<span class=\"math-tex\">{tex} \\therefore{\/tex}<\/span>\u00a0HCF (612, 48)\u00a0<span class=\"math-tex\">{tex} = \\left( 2 ^ { 2 } \\times 3 \\right) = ( 4 \\times 3 ) {\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\therefore{\/tex}<\/span>\u00a0HCF (612, 48)= 12.<br \/>\n<span class=\"math-tex\">{tex} \\therefore{\/tex}<\/span>\u00a0 Maximum number of columns in which they can march = 12.<\/li>\n<li>Given,\u00a0BC = 13 cm and AC = 5 cm and\u00a0<span class=\"math-tex\">{tex}A D \\perp B C{\/tex}<\/span><br \/>\n<img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/UzyIbO4.png\" data-imgur-src=\"UzyIbO4.png\" \/><br \/>\nIn\u00a0<span class=\"math-tex\">{tex}\\triangle BAC{\/tex}<\/span>\u00a0and\u00a0<span class=\"math-tex\">{tex}\\triangle A D C {\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\angle A = \\angle D =90^\\circ{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\angle A C B = \\angle D C A{\/tex}<\/span>\u00a0(common)<br \/>\n<span class=\"math-tex\">{tex}\\triangle \\mathrm { BAC } \\sim \\triangle \\mathrm { ADC }{\/tex}<\/span>[by AA similarity]\n<span class=\"math-tex\">{tex}\\therefore \\frac {AC}{AD} = \\frac {BC}{AC} = \\frac {AB}{DC}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\therefore \\frac { \\operatorname { ar } ( \\triangle A B C ) } { \\operatorname { ar } ( \\triangle A D C ) }{\/tex}<\/span><span class=\"math-tex\">{tex}=\\frac {\\frac{1}{2} AB . AC}{\\frac {1}{2}DC. AD} = \\frac { B C ^ { 2 } } { A C ^ { 2 } }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}= \\frac { 13 ^ { 2 } } { 5 ^ { 2 } } = \\frac { 169 } { 25 }{\/tex}<\/span><br \/>\nTherefore, the ratio of the areas of\u00a0<span class=\"math-tex\">{tex}\\triangle ABC{\/tex}<\/span>\u00a0and\u00a0<span class=\"math-tex\">{tex}\\triangle ADC{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}= 169 : 25.{\/tex}<\/span><\/li>\n<li>Let OP intersect AB at a point C,<br \/>\n<img decoding=\"async\" style=\"width: 150px; height: 121px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/RTlle3A.png\" alt=\"\" data-imgur-src=\"RTlle3A.png\" \/><br \/>\nClearly, <span class=\"math-tex\">{tex}\\angle APO=\\angle BPO{\/tex}<\/span>&#8230;(i) [<span class=\"math-tex\">{tex}\\because{\/tex}<\/span> O is bisector of <span class=\"math-tex\">{tex}\\angle APB{\/tex}<\/span>]\nNow, in <span class=\"math-tex\">{tex}\\triangle ACP{\/tex}<\/span> and <span class=\"math-tex\">{tex}\\triangle BCP{\/tex}<\/span>.<br \/>\n<span class=\"math-tex\">{tex}AP\\;=\\;BP{\/tex}<\/span> [<span class=\"math-tex\">{tex}\\because{\/tex}<\/span> length of tangents drawn from an external point to a circle are equal ]\n<span class=\"math-tex\">{tex}PC\\;=\\;PC{\/tex}<\/span> [<span class=\"math-tex\">{tex}\\because{\/tex}<\/span> common sides]\nand <span class=\"math-tex\">{tex}\\angle APO=\\angle BPO{\/tex}<\/span> [From Eq.(i)]\n<span class=\"math-tex\">{tex}\\therefore \\triangle ACP \\cong \\triangle BCP{\/tex}<\/span> [by SAS congruence rule]\nThen, <span class=\"math-tex\">{tex}AC\\;=\\;BC{\/tex}<\/span> [by congruence side of congruence triangle]\nand <span class=\"math-tex\">{tex}\\angle ACP=\\angle BCP{\/tex}<\/span> [by congruence angle of congruence triangle]\n<span class=\"math-tex\">{tex}\\angle ACP=\\angle BCP{\/tex}<\/span><span class=\"math-tex\">{tex}=\\frac{1}{2}\\times 180^o=90^o{\/tex}<\/span> [AB is a straight line]\nHence, OP is a perpendicular bisector of AB.<\/li>\n<li style=\"clear: both;\">LHS = <span class=\"math-tex\">{tex}\\frac{1+\\tan A}{2 \\sin A}+\\frac{1+\\cot A}{2 \\cos A}{\/tex}<\/span><br \/>\n= <span class=\"math-tex\">{tex}\\frac{\\cos A+\\sin A}{2 \\sin A \\cos A}+\\frac{\\sin A+\\cos A}{2 \\cos A \\sin A}{\/tex}<\/span><br \/>\n= <span class=\"math-tex\">{tex}\\frac{2(\\cos A+\\sin A)}{2 \\sin A \\cos A}{\/tex}<\/span><br \/>\n= cosec A + sec A<br \/>\n= RHS<\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Given that: sin <span class=\"math-tex\">{tex}\\theta{\/tex}<\/span> + cos <span class=\"math-tex\">{tex}\\theta{\/tex}<\/span>\u00a0=<span class=\"math-tex\">{tex}\\sqrt{3}{\/tex}<\/span><br \/>\nor (sin <span class=\"math-tex\">{tex}\\theta{\/tex}<\/span>\u00a0+ cos <span class=\"math-tex\">{tex}\\theta{\/tex}<\/span>)<sup>2<\/sup> = 3<br \/>\nor sin<sup>2<\/sup> <span class=\"math-tex\">{tex}\\theta{\/tex}<\/span>\u00a0+ cos<sup>2<\/sup> <span class=\"math-tex\">{tex}\\theta{\/tex}<\/span>\u00a0+ 2sin<span class=\"math-tex\">{tex}\\theta{\/tex}<\/span>\u00a0cos<span class=\"math-tex\">{tex}\\theta{\/tex}<\/span>\u00a0= 3<br \/>\n2sin<span class=\"math-tex\">{tex}\\theta{\/tex}<\/span>\u00a0cos<span class=\"math-tex\">{tex}\\theta{\/tex}<\/span>\u00a0= 2\u00a0 \u00a0 \u00a0[As sin<sup>2<\/sup> <span class=\"math-tex\">{tex}\\theta{\/tex}<\/span>\u00a0+ cos<sup>2<\/sup> <span class=\"math-tex\">{tex}\\theta{\/tex}<\/span>\u00a0= 1]\nor sin <span class=\"math-tex\">{tex}\\theta{\/tex}<\/span>\u00a0cos <span class=\"math-tex\">{tex}\\theta{\/tex}<\/span>\u00a0= 1 = sin<sup>2<\/sup> <span class=\"math-tex\">{tex}\\theta{\/tex}<\/span>\u00a0+ cos<sup>2 <\/sup><span class=\"math-tex\">{tex}\\theta{\/tex}<\/span><br \/>\nLHS=1<br \/>\nRHS=<span class=\"math-tex\">{tex}\\frac{Sin \\theta}{Cos \\theta} + \\frac{Cos \\theta}{Sin \\theta}{\/tex}<\/span><br \/>\nTaking LCM<br \/>\nSubstitute the value of LHS in RHS we get,<br \/>\ntan<span class=\"math-tex\">{tex}\\theta{\/tex}<\/span>\u00a0+ cot<span class=\"math-tex\">{tex}\\theta{\/tex}<\/span>\u00a0= 1<\/li>\n<li style=\"clear: both;\">Given that,<br \/>\nRadius of circle = 7 cm<br \/>\nCentral angle = 90<sup>o<\/sup><br \/>\nNow, area of minor sector of circle<br \/>\n= <span class=\"math-tex\">{tex}\\frac{\\pi r^2 \\theta}{360^{\\circ}}{\/tex}<\/span><br \/>\n= <span class=\"math-tex\">{tex}\\frac{\\pi(7)^2}{4}{\/tex}<\/span><br \/>\n= <span class=\"math-tex\">{tex}\\frac{22 \\times 7 \\times 7}{7 \\times 4}{\/tex}<\/span><br \/>\n= 38.5 cm<sup>2<\/sup><br \/>\nArea of complete circle<br \/>\n= <span class=\"math-tex\">{tex}\\pi{\/tex}<\/span>r<sup>2<\/sup><br \/>\n= <span class=\"math-tex\">{tex}\\pi{\/tex}<\/span>(7)<sup>2<\/sup><br \/>\n= 154 cm<sup>2<\/sup><br \/>\nNow, area of major sector<br \/>\n= Area of complete circle &#8211; Area of minor sector<br \/>\n= 154 &#8211; 38.5<br \/>\n= 115.5 cm<sup>2<\/sup><\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p><img decoding=\"async\" style=\"width: 190px; height: 177px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/eMKIT1L.png\" alt=\"\" data-imgur-src=\"eMKIT1L.png\" \/><br \/>\nArea which cannot be grazed = (area of equilateral\u00a0<span class=\"math-tex\">{tex}\\triangle ABC{\/tex}<\/span> &#8211; (area of the sector with r = 7m,<span class=\"math-tex\">{tex}\\theta = 60 ^ { \\circ }{\/tex}<\/span>)<br \/>\n<span class=\"math-tex\">{tex}= \\left[ \\frac { \\sqrt { 3 } } { 4 } \\times ( 12 ) ^ { 2 } &#8211; \\frac { 22 } { 7 } \\times ( 7 ) ^ { 2 } \\times \\frac { 60 } { 360 } \\right] \\mathrm { m } ^ { 2 }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}= \\left[ ( \\sqrt { 3 } \\times 12 \\times 3 ) &#8211; \\frac { ( 22 \\times 7 ) } { 6 } \\right]{\/tex}<\/span><br \/>\n= 62.35 &#8211; 25.66 m<sup>2<\/sup><br \/>\n= 36.68 m<sup>2<\/sup><\/li>\n<li style=\"text-align: center; display: block;\"><b>Section C <\/b><\/li>\n<li>Let <span class=\"math-tex\">{tex}5 + 3 \\sqrt { 2 }{\/tex}<\/span> is rational. It can be written in the form <span class=\"math-tex\">{tex} \\frac pq{\/tex}<\/span>.<br \/>\n<span class=\"math-tex\">{tex}(5 + 3 \\sqrt { 2 }) = \\frac pq{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} 3 \\sqrt { 2 } = \\frac pq &#8211; 5{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} 3 \\sqrt { 2 } = \\frac{ p &#8211; 5q}q {\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\sqrt { 2 } = \\frac{ p &#8211; 5q}{3q} {\/tex}<\/span><br \/>\nAs p &#8211; 5q and 3q are integers .<br \/>\nSo, <span class=\"math-tex\">{tex}\\frac{ p &#8211; 5q}{3q} {\/tex}<\/span> is rational number .<br \/>\nBut <span class=\"math-tex\">{tex} \\sqrt { 2 }{\/tex}<\/span> is not rational number .<br \/>\nSince a rational number cannot be equal to an irrational number. Our assumption that\u00a0<span class=\"math-tex\">{tex}5 + 3 \\sqrt { 2 }{\/tex}<\/span> is rational wrong.<br \/>\nHence,\u00a0<span class=\"math-tex\">{tex}5 + 3 \\sqrt { 2 }{\/tex}<\/span> is irrational.<\/li>\n<li>The given polynomial is<br \/>\np(x) = 6x<sup>2\u00a0<\/sup>-7x -3<br \/>\nFactorize the above quadratic polynomial, we have<br \/>\n6x<sup>2\u00a0<\/sup>-7x -3 =\u00a0 6x<sup>2\u00a0<\/sup>-9x + 2x &#8211; 3<br \/>\n= 3x(2x &#8211; 3) + 1(2x &#8211; 3)<br \/>\n= (3x + 1)(2x &#8211; 3)<br \/>\nFor p(x) = 0, either 3x + 1 = 0 or 2x &#8211; 3 = 0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\quad x = \\frac { &#8211; 1 } { 3 } \\text { or } x = \\frac { 3 } { 2 }{\/tex}<\/span><br \/>\n<strong><em><u>Verification<\/u><\/em><\/strong>:we have a = 6, b = -7, c = -3<br \/>\nSum of zeroes =\u00a0<span class=\"math-tex\">{tex}\\frac { &#8211; 1 } { 3 } + \\frac { 3 } { 2 } = \\frac { 7 } { 6 }{\/tex}<\/span><br \/>\nAlso,\u00a0<span class=\"math-tex\">{tex}\\frac { &#8211; b } { a } = \\frac { &#8211; ( &#8211; 7 ) } { 6 } = \\frac { 7 } { 6 }{\/tex}<\/span><br \/>\nNow, product of zeroes=\u00a0<span class=\"math-tex\">{tex}\\left( &#8211; \\frac { 1 } { 3 } \\right) \\times \\frac { 3 } { 2 } = \\frac { &#8211; 1 } { 2 }{\/tex}<\/span><br \/>\nAlso,\u00a0<span class=\"math-tex\">{tex}\\frac { c } { a } = \\frac { &#8211; 3 } { 6 } = \\frac { &#8211; 1 } { 2 } {\/tex}<\/span><\/li>\n<li style=\"clear: both;\">According to the question,we have,<br \/>\na=9. Therefore, common difference d =17-9=8<br \/>\nlet the required number of terms be n.<br \/>\nTherefore, S<sub>n<\/sub>=636<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span><span class=\"math-tex\">{tex}\\frac{n}{2}{\/tex}<\/span>[2a+(n-1)d]=636<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span><span class=\"math-tex\">{tex}\\frac{n}{2}{\/tex}<\/span>[2(9)+(n-1)8]=636<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>n[18+8n-8]=1272<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>8n<sup>2<\/sup>+10n-1272=0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>4n<sup>2<\/sup>+5n-636=0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>4n<sup>2<\/sup>+53n-48n-636=0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>n(4n+53)-12(4n+53)=0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>(4n+53)(n-12)=0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>4n+53=0 or n-12=0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>n=<span class=\"math-tex\">{tex}\\frac{{ &#8211; 53}}{4}{\/tex}<\/span>\u00a0or\u00a0 n=12<br \/>\nSince number of terms cannot neither be negative nor fraction, n=12<br \/>\nhence, the required number of terms is 12.<\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Let 1st term of the AP be a and common difference be d.<br \/>\nNow a<sub>8<\/sub>\u00a0= 37<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0a + 7d = 37<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0a = 37 &#8211; 7d ..(i)<br \/>\nAlso, a<sub>15<\/sub>\u00a0= a<sub>12\u00a0<\/sub>+ 15<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0a + 14d = a + 11d + 15<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a03d = 15\u00a0<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0d = 5<br \/>\nSubstituting in eq. (i), we get<br \/>\na = 37 &#8211; 7(5) = 2<br \/>\nAP. is 2, 7, 12, 17, &#8230;<br \/>\nS<sub>15<\/sub>\u00a0=\u00a0<span class=\"math-tex\">{tex}\\frac{15}{2}[2 \\times 2+14 \\times 5]{\/tex}<\/span><br \/>\n= 15(37) = 555<\/li>\n<li style=\"clear: both;\">Let <span class=\"math-tex\">{tex}OA = R = \\frac{{30}}{2} = 15cm{\/tex}<\/span>, <span class=\"math-tex\">{tex}OB = r = \\frac{{18}}{2} = 9cm{\/tex}<\/span><br \/>\n<img decoding=\"async\" style=\"width: 150px; height: 149px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/cbse\/10\/mathematics\/rd\/ch08\/ch08_33.PNG\" alt=\"\" \/><br \/>\nSince AC is the tangent to the circle with radius 9 cm, we have OB\u22a5AC.<br \/>\nHence, by applying the Pythagoras Theorem, we have,<br \/>\nOA<sup>2<\/sup> = OB<sup>2<\/sup> + AB<sup>2<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> 15<sup>2<\/sup> = 9<sup>2<\/sup> + AB<sup>2<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> AB<sup>2<\/sup> = 15<sup>2<\/sup> &#8211; 9<sup>2<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> AB<sup>2<\/sup> = 225 &#8211; 81 = 144<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span> AB = 12 cm<br \/>\nWe know that the perpendicular drawn from the centre of a circle to any chord of the circle bisects the chord.<br \/>\nHere, OB is the perpendicular and AC is the length of the chord of the circle with radius 15 cm.<br \/>\nSo,<br \/>\nAC = 2 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a0AB = 2\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a012 = 24 cm<br \/>\nLength of the chord of the larger circle which touches the smaller circle = 24 cm.<\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p><img decoding=\"async\" style=\"width: 200px; height: 124px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/TvLSJRZ.png\" alt=\"\" data-imgur-src=\"TvLSJRZ.png\" \/><br \/>\nwe know that ,\u00a0angle subtended by an arc at centre of the circle is twice the angle subtended by it in alternate segment.<br \/>\n<span class=\"math-tex\">{tex}\\angle \\mathrm { TOQ } + \\angle \\mathrm { TPQ } = 180 ^ { \\circ }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span><span class=\"math-tex\">{tex}\\angle \\mathrm { TOQ } = 110 ^ { \\circ }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\angle \\mathrm { TOQ } = 2 \\angle \\mathrm { TRQ }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}110 ^ { \\circ } = 2 \\angle \\mathrm { TRQ } \\Rightarrow \\angle \\mathrm { TRQ } = 55 ^ { \\circ }{\/tex}<\/span><\/li>\n<li>We have,<br \/>\n<span class=\"math-tex\">{tex}\\mathrm { LHS } = \\frac { p ^ { 2 } &#8211; 1 } { p ^ { 2 } + 1 } = \\frac { ( \\sec \\theta + \\tan \\theta ) ^ { 2 } &#8211; 1 } { ( \\sec \\theta + \\tan \\theta ) ^ { 2 } + 1 }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\sec ^ { 2 } \\theta + \\tan ^ { 2 } \\theta + 2 \\sec \\theta \\tan \\theta &#8211; 1 } { \\sec ^ { 2 } \\theta + \\tan ^ { 2 } \\theta + 2 \\sec \\theta \\tan \\theta + 1 }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\left( \\sec ^ { 2 } \\theta &#8211; 1 \\right) + \\tan ^ { 2 } \\theta + 2 \\sec \\theta \\tan \\theta } { \\sec ^ { 2 } \\theta + 2 \\sec \\theta \\tan \\theta + \\left( 1 + \\tan ^ { 2 } \\theta \\right) }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 2 } \\theta + \\tan ^ { 2 } \\theta + 2 \\sec \\theta \\tan \\theta } { \\sec ^ { 2 } \\theta + 2 \\sec \\theta \\tan \\theta + \\sec ^ { 2 } \\theta }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { 2 \\tan ^ { 2 } \\theta + 2 \\tan \\theta \\sec \\theta } { 2 \\sec ^ { 2 } \\theta + 2 \\sec \\theta \\tan \\theta } = \\frac { 2 \\tan \\theta ( \\tan \\theta + \\sec \\theta ) } { 2 \\sec \\theta ( \\sec \\theta + \\tan \\theta ) } = \\frac { \\tan \\theta } { \\sec \\theta } = \\frac { \\sin \\theta } { \\cos \\theta \\cdot \\sec \\theta } = \\sin \\theta{\/tex}<\/span>=RHS<\/li>\n<li>The frequency distribution table for calculations of median is as follow:<br \/>\n<table style=\"width: 100%;\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<thead>\n<tr>\n<th scope=\"col\">Class Interval<\/th>\n<th scope=\"col\">Frequency (f)<\/th>\n<th scope=\"col\">Cumulative frequency<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td style=\"text-align: center;\">40-45<\/td>\n<td style=\"text-align: center;\">9<\/td>\n<td style=\"text-align: center;\">9<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">45-50<\/td>\n<td style=\"text-align: center;\">5<\/td>\n<td style=\"text-align: center;\">14<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">50-55<\/td>\n<td style=\"text-align: center;\">8<\/td>\n<td style=\"text-align: center;\">22<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">55-60<\/td>\n<td style=\"text-align: center;\">9<\/td>\n<td style=\"text-align: center;\">31<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">60-65<\/td>\n<td style=\"text-align: center;\">6<\/td>\n<td style=\"text-align: center;\">37<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">65-70<\/td>\n<td style=\"text-align: center;\">3<\/td>\n<td style=\"text-align: center;\">40<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>We know,<br \/>\nMedian =\u00a0<span class=\"math-tex\">{tex}l+\\left\\{h \\times \\frac{\\frac{N}{2}-c f}{f}\\right\\}{\/tex}<\/span><br \/>\nHere,<br \/>\nl denotes lower limit of median class<br \/>\nh denotes width of median class<br \/>\nf denotes frequency of median class<br \/>\ncf denotes cumulative frequency of the class preceding the median class<br \/>\nN denotes sum of frequency<br \/>\n<span class=\"math-tex\">{tex}\\frac {N}{2}{\/tex}<\/span>\u00a0=\u00a020<br \/>\ncumulative frequency just greater than 20 is 22<br \/>\nTherefore, median class is 50-55<br \/>\nMedian class is 50 &#8211; 55<br \/>\nSo,<br \/>\nl = 50,<br \/>\nh = 5,<br \/>\nf = 8,<br \/>\ncf = cf of preceding class = 14<br \/>\n<span class=\"math-tex\">{tex}\\frac {N}{2}{\/tex}<\/span>\u00a0= 20<br \/>\nO substituting all the above values in the formula, we get<br \/>\nMedian =\u00a0<span class=\"math-tex\">{tex}l+\\left\\{\\mathrm{h} \\times \\frac{\\frac{\\mathrm{N}}{2}-c f}{f}\\right\\}{\/tex}<\/span><br \/>\nMedian =\u00a0<span class=\"math-tex\">{tex}50+\\left\\{5 \\times \\frac{20-14}{8}\\right\\}{\/tex}<\/span><br \/>\nMedian =\u00a0<span class=\"math-tex\">{tex}50+\\left\\{5 \\times \\frac{6}{8}\\right\\}{\/tex}<\/span><br \/>\nMedian =\u00a0<span class=\"math-tex\">{tex}50+\\left\\{5 \\times \\frac{3}{4}\\right\\}{\/tex}<\/span><br \/>\nMedian =\u00a0<span class=\"math-tex\">{tex}50+\\left\\{\\frac{15}{4}\\right\\}{\/tex}<\/span><br \/>\nMedian = 50 + 3.75<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0Median = 53.75<\/li>\n<li style=\"text-align: center; display: block;\"><b>Section D <\/b><\/li>\n<li style=\"clear: both;\">Let the time taken by the smaller pipe to fill the tank be x hr.<br \/>\nThe time is taken by the larger pipe = (x &#8211; 9)hr.<br \/>\nPart of the tank filled by a smaller pipe in 1 hour =\u00a0<span class=\"math-tex\">{tex}\\frac { 1 } { x }{\/tex}<\/span><br \/>\nPart\u00a0of the tank filled by a larger pipe in 1 hour =\u00a0<span class=\"math-tex\">{tex}\\frac { 1 } { x &#8211; 9 }{\/tex}<\/span><br \/>\nAccording to question,<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\frac { 1 } { x } + \\frac { 1 } { x &#8211; 9 } = \\frac { 1 } { 6 }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\frac { x &#8211; 9 + x } { x ^ { 2 } &#8211; 9 x } = \\frac { 1 } { 6 }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a06(2x &#8211; 9) = x<sup>2<\/sup>\u00a0&#8211; 9x<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a012x &#8211; 54 = x<sup>2\u00a0<\/sup>&#8211; 9x<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0x<sup>2<\/sup>\u00a0&#8211; 21x + 54 = 0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0x<sup>2<\/sup>\u00a0&#8211; 18x &#8211; 3x + 54 = 0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0x(x &#8211; 18) &#8211; 3(x &#8211; 18) = 0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0x &#8211; 18 = 0 or x &#8211; 3 = 0<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0x = 18 or x = 3<br \/>\nAs time can&#8217;t be less than 9. Then,\u00a0x\u00a0<span class=\"math-tex\">{tex}\\neq{\/tex}<\/span>\u00a03.<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0x = 18<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0x &#8211; 9 = 18 &#8211; 9 = 9<br \/>\nTherefore, the smaller pipe and larger pipe take 18 hours and 9 hours to fill the tank respectively.<\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Suppose the actual speed of the train be x km\/hr and the actual time taken be y hours. Then,<br \/>\nDistance covered =(xy) km &#8230;&#8230;&#8230;..(i) [<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0Distance = Speed\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>Time]\nIf the speed is increased by <span class=\"math-tex\">{tex}6km\/hr{\/tex}<\/span>, then time of journey is reduced by 4 hours i.e., when speed is <span class=\"math-tex\">{tex}(x + 6)km\/hr{\/tex}<\/span>, time of journey is <span class=\"math-tex\">{tex}(y-4)hours.{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0Distance covered = <span class=\"math-tex\">{tex}(x + 6) (y &#8211; 4){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span><span class=\"math-tex\">{tex}xy = (x + 6) (y &#8211; 4){\/tex}<\/span> [Using (i)]\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span><span class=\"math-tex\">{tex}-4x + 6y -24=0{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span><span class=\"math-tex\">{tex}-2x + 3y -12=0{\/tex}<\/span> &#8230;&#8230;&#8230;.(ii)<br \/>\nwhen the speed is reduced by 6 km\/hr, then the time of journey is increased by 6 hours i.e.,<br \/>\nwhen speed is <span class=\"math-tex\">{tex}(x-6)km\/hr{\/tex}<\/span>, time of journey is <span class=\"math-tex\">{tex}(y &#8211; 6)hours.{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0Distance covered =<span class=\"math-tex\">{tex}(x -6) (y + 6){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}xy = (x &#8211; 6) (y + 6){\/tex}<\/span> [Using (i)]\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}6x &#8211; 6y -36 =0{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}x &#8211; y -6=0 {\/tex}<\/span>&#8230;&#8230;&#8230;..(iii)<br \/>\nThus, we obtain the following system of equations:<br \/>\n<span class=\"math-tex\">{tex}-2x + 3y -12 =0{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}x &#8211; y &#8211; 6=0{\/tex}<\/span><br \/>\nBy using cross-multiplication, we have<br \/>\n<span class=\"math-tex\">{tex}\\frac { x } { 3 \\times &#8211; 6 &#8211; ( &#8211; 1 ) \\times &#8211; 12 } = \\frac { &#8211; y } { &#8211; 2 \\times &#8211; 6 &#8211; 1 \\times &#8211; 12 } = \\frac { 1 } { &#8211; 2 \\times &#8211; 1 &#8211; 1 \\times 3 }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\quad \\frac { x } { &#8211; 30 } = \\frac { &#8211; y } { 24 } = \\frac { 1 } { &#8211; 1 }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}x =30\\ and\\ y=24.{\/tex}<\/span><br \/>\nPutting the values of x and y in equation (i), we obtain<br \/>\nDistance= <span class=\"math-tex\">{tex}(30\\times24)km =720km.{\/tex}<\/span><br \/>\nHence, the length of the journey is <span class=\"math-tex\">{tex}720 km{\/tex}<\/span>.<\/li>\n<li>Given,<br \/>\n<img decoding=\"async\" style=\"width: 150px; height: 254px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/TgB4l84.png\" alt=\"\" data-imgur-src=\"TgB4l84.png\" \/><br \/>\nAB be the building.<br \/>\nEC be the tower.<br \/>\nA is the point from where elevation of tower is 60\u00b0 and the angle of depression of its foot is 45<sup>o<\/sup><br \/>\n<span class=\"math-tex\">{tex}EC = DE\u00a0+ CD{\/tex}<\/span><br \/>\nalso,\u00a0<span class=\"math-tex\">{tex}CD = AB = 7 m.{\/tex}<\/span><br \/>\nand\u00a0<span class=\"math-tex\">{tex} BC = AD{\/tex}<\/span><br \/>\nIn right <span class=\"math-tex\">{tex}\\triangle{\/tex}<\/span>ABC,<br \/>\ntan45<sup>o<\/sup> =<span class=\"math-tex\">{tex}\\frac{AB}{BC}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a01 =\u00a0<span class=\"math-tex\">{tex}\\frac{7}{BC}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span><span class=\"math-tex\">{tex}\u00a0BC = 7 m = AD{\/tex}<\/span><br \/>\nIn\u00a0\u00a0right <span class=\"math-tex\">{tex}\\triangle{\/tex}<\/span>ADE,<br \/>\n<span class=\"math-tex\">{tex}\\tan 60^{\\circ}{\/tex}<\/span>\u00a0=\u00a0<span class=\"math-tex\">{tex}\\frac{DE}{AD}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\sqrt 3{\/tex}<\/span>\u00a0=\u00a0<span class=\"math-tex\">{tex}\\frac{DE}{7}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0DE = 7<span class=\"math-tex\">{tex}\\sqrt 3{\/tex}<\/span><br \/>\nHeight of the tower<span class=\"math-tex\">{tex} = EC = DE + CD = {\/tex}<\/span>(7<span class=\"math-tex\">{tex}\\sqrt 3{\/tex}<\/span>\u00a0+ 7) m = 7(<span class=\"math-tex\">{tex}\\sqrt 3{\/tex}<\/span>\u00a0+ 1) m.<\/li>\n<li style=\"clear: both;\"><img decoding=\"async\" style=\"width: 100px; height: 145px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/KbB2nZ6.png\" alt=\"\" data-imgur-src=\"KbB2nZ6.png\" \/><br \/>\nAccording to question,\u00a0<span class=\"math-tex\">{tex}\\angle C = 90 ^ { \\circ }{\/tex}<\/span><br \/>\nLet, AC = BC = x<br \/>\nAccording to pythagoras theorem,<br \/>\n<span class=\"math-tex\">{tex}\\therefore A B ^ { 2 } = A C ^ { 2 } + B C ^ { 2 }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}A B ^ { 2 } = x ^ { 2 } + x ^ { 2 }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\therefore 2 x ^ { 2 } = ( 2 \\sqrt { 2 } ) ^ { 2 }{\/tex}<\/span><br \/>\nor, x = 2m<br \/>\n<span class=\"math-tex\">{tex}r = \\sqrt { 2 } \\mathrm { m }{\/tex}<\/span>\u00a0(given)<br \/>\nRadius of the cylinder = Radius of the cone<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0Slant height of conical portion,\u00a0x = 2 m<br \/>\nTotal surface area of toy = curved surface area of cylinder + Area of base of cylinder + Curved surface area of cone<br \/>\n<span class=\"math-tex\">{tex}= 2 \\pi r h + \\pi r ^ { 2 } + \\pi r l{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}= \\pi r( 2h + r + l){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}= \\pi r [ 7 + \\sqrt { 2 } + 2 ] m ^ { 2 }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}= \\pi \\sqrt { 2 } [ 9 + \\sqrt { 2 } ] m ^ { 2 }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}= \\pi [ 2 + 9 \\sqrt { 2 } ] \\mathrm { m } ^ { 2 }{\/tex}<\/span><br \/>\nThe total surface area of the toy\u00a0<span class=\"math-tex\">{tex}= \\pi [ 2 + 9 \\sqrt { 2 } ] \\mathrm { m } ^ { 2 }{\/tex}<\/span><\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>We have;<br \/>\nA Cube,<br \/>\nCube&#8217;s\u00a0<span class=\"math-tex\">{tex}\\frac {length}{Edge}{\/tex}<\/span>, a = 7 cm<br \/>\nA Cylinder:<br \/>\nCylinder&#8217;s Radius, r = 2.1 cm or r =\u00a0<span class=\"math-tex\">{tex}\\frac {21}{10}{\/tex}<\/span>cm<br \/>\nCylinder&#8217;s Height, h = 7 cm<br \/>\n<span class=\"math-tex\">{tex}\\because{\/tex}<\/span>\u00a0A cylinder is scooped out from a cube,<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0TSA of the resulting cuboid:<br \/>\n=\u00a0TSA of whole Cube &#8211; 2 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a0(Area of upper circle or Area of lower circle) + CSA of the scooped out Cylinder<br \/>\n= 6a<sup>2<\/sup> + 2<span class=\"math-tex\">{tex}\\pi{\/tex}<\/span>rh &#8211; 2\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a0(<span class=\"math-tex\">{tex}\\pi{\/tex}<\/span>r<sup>2<\/sup>)<br \/>\n= 6\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a0(7)<sup>2<\/sup> + 2\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a0(22 <span class=\"math-tex\">{tex}\\div{\/tex}<\/span>\u00a07\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a02.1\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a07) &#8211; 2\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a0[22 <span class=\"math-tex\">{tex}\\div{\/tex}<\/span>\u00a07\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a0(2.1)<sup>2<\/sup>]\n= 6\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a049 + (44 <span class=\"math-tex\">{tex}\\div{\/tex}<\/span>\u00a07 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a014.7) &#8211; (44 <span class=\"math-tex\">{tex}\\div{\/tex}<\/span>\u00a07 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a04.41)<br \/>\n= 294 + 92.4 &#8211; 27.72<br \/>\n= 294 + 64.68<br \/>\n= 358.68 cm<sup>2<\/sup><br \/>\nHence, the total surface area of the remaining solid is 358.68 cm<sup>2<\/sup><\/li>\n<li>\n<table style=\"width: 100%;\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<tbody>\n<tr>\n<td style=\"text-align: center;\"><strong>Family size<\/strong><\/td>\n<td style=\"text-align: center;\"><strong>X<sub>i<\/sub><\/strong><\/td>\n<td style=\"text-align: center;\"><strong>F<sub>i<\/sub><\/strong><\/td>\n<td style=\"text-align: center;\"><strong>f<sub>i<\/sub>x<sub>i<\/sub><\/strong><\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">1 &#8211; 3<\/td>\n<td style=\"text-align: center;\">2<\/td>\n<td style=\"text-align: center;\">7<\/td>\n<td style=\"text-align: center;\">14<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">3 &#8211; 5<\/td>\n<td style=\"text-align: center;\">4<\/td>\n<td style=\"text-align: center;\">8<\/td>\n<td style=\"text-align: center;\">32<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">5 &#8211; 7<\/td>\n<td style=\"text-align: center;\">6<\/td>\n<td style=\"text-align: center;\">2<\/td>\n<td style=\"text-align: center;\">12<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">7 &#8211; 9<\/td>\n<td style=\"text-align: center;\">8<\/td>\n<td style=\"text-align: center;\">2<\/td>\n<td style=\"text-align: center;\">16<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">9 &#8211; 11<\/td>\n<td style=\"text-align: center;\">10<\/td>\n<td style=\"text-align: center;\"><u>1<\/u><\/td>\n<td style=\"text-align: center;\"><u>10<\/u><\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\"><\/td>\n<td style=\"text-align: center;\"><\/td>\n<td style=\"text-align: center;\"><u>20<\/u><\/td>\n<td style=\"text-align: center;\"><u>84<\/u><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Mean = <span class=\"math-tex\">{tex}\\frac{\\sum \\mathrm{f}_{\\mathrm{i}} \\mathrm{x}_{\\mathrm{i}}}{\\sum \\mathrm{f}_{\\mathrm{i}}}{\/tex}<\/span>\u00a0= <span class=\"math-tex\">{tex}\\frac{84}{20}{\/tex}<\/span>\u00a0= 4.2<br \/>\nMode : Modal Class = 3-5<br \/>\nI = 3, f1 = 8, f0 = 7, f2 = 2, h = 2<br \/>\nMode = l\u00a0+ <span class=\"math-tex\">{tex}\\left(\\frac{\\mathrm{f}_1-\\mathrm{f}_0}{2 \\mathrm{f}_1-\\mathrm{f}_0-\\mathrm{f}_2}\\right){\/tex}<\/span> <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> h<br \/>\n= 3 + <span class=\"math-tex\">{tex}\\left(\\frac{8-7}{16-7-2}\\right){\/tex}<\/span> <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> 2<br \/>\n= <span class=\"math-tex\">{tex}\\frac{23}{7}{\/tex}<\/span> or\u00a03.287<\/li>\n<li style=\"text-align: center; display: block;\"><b>Section E <\/b>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li style=\"text-align: left;\">Let the fixed charge be \u20b9 x and per kilometer charge be \u20b9 y<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span> x + 10y = 105 &#8230;(i)<br \/>\nx + 15y = 155 &#8230;(ii)<br \/>\nFrom (i) and (ii)<br \/>\n5y = 50<br \/>\ny = <span class=\"math-tex\">{tex}\\frac{50}{5}{\/tex}<\/span> = 10<br \/>\nFrom equation (i)<br \/>\nx + 100 = 105<br \/>\nx = 105 &#8211; 100 = 5<br \/>\nFixed charges = \u20b9 5<\/li>\n<li style=\"text-align: left;\">Let the fixed charge be \u20b9 x and per kilometer charge be \u20b9 y<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span> x + 10y = 105 &#8230;(1)<br \/>\nx + 15y = 155 &#8230;(2)<br \/>\nFrom (1) and (2)<br \/>\n5y = 50<br \/>\ny = <span class=\"math-tex\">{tex}\\frac{50}{5}{\/tex}<\/span> = 10<br \/>\nFrom equation (1)<br \/>\nx + 100 = 105<br \/>\nx = 105 &#8211; 100 = 5<br \/>\nPer km charges = \u20b9 10<\/li>\n<li style=\"text-align: left;\">Let the fixed charge be \u20b9 a and per kilometer charge be \u20b9 b<br \/>\na + 10b<br \/>\n20 + 10 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> 10 = \u20b9 120<br \/>\n<strong>OR<\/strong><br \/>\nTotal amount = x + 10y + x + 25y<br \/>\n= 2x + 35y<br \/>\n= 2 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> 5 + 35 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> 10<br \/>\n= 10 + 350<br \/>\n= \u20b9 360<\/li>\n<\/ol>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li style=\"text-align: left;\"><img decoding=\"async\" style=\"height: 157px; width: 150px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1673246700-qksmzw.jpg\" \/><br \/>\n<span class=\"math-tex\">{tex}\\triangle{\/tex}<\/span>ABD <span class=\"math-tex\">{tex}\\sim{\/tex}<\/span> <span class=\"math-tex\">{tex}\\triangle{\/tex}<\/span>CED (by AA criteria)<br \/>\n<span class=\"math-tex\">{tex}\\frac{30}{a}=\\frac{40}{x}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\frac{x}{a}=\\frac{40}{30}{\/tex}<\/span><br \/>\na = <span class=\"math-tex\">{tex}\\frac{30}{40}x{\/tex}<\/span><br \/>\nAgain<br \/>\n<img decoding=\"async\" style=\"height: 158px; width: 150px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1672308298-e4r5sx.jpg\" \/><br \/>\n<span class=\"math-tex\">{tex}\\frac{40-x}{40}=\\frac{a}{50}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\frac{40-x}{40}=\\frac{30\\times x}{40\\times50}{\/tex}<\/span><br \/>\n8000 &#8211; 200x = 120x<br \/>\n8000 = 320x<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span> x = 25 feet<\/li>\n<li style=\"text-align: left;\"><img decoding=\"async\" style=\"height: 157px; width: 150px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1673246700-qksmzw.jpg\" \/><br \/>\n<span class=\"math-tex\">{tex}\\frac{x}{40}=\\frac{a}{30}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\frac{25}{40}=\\frac{a}{30}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\frac{25\\times30}{40}=a{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\frac{75}{4}{\/tex}<\/span> = a<br \/>\na = 18.75 feet<\/li>\n<li style=\"text-align: left;\"><img decoding=\"async\" style=\"height: 173px; width: 150px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1672309301-9u6akg.jpg\" \/><br \/>\nAD =\u00a0<span class=\"math-tex\">{tex}\\sqrt{{30^2+40}^2}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}= \\sqrt{900+1600}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}=\\sqrt{2500}{\/tex}<\/span><br \/>\nAD = 50 feet<br \/>\nIn\u00a0<span class=\"math-tex\">{tex}\\triangle{\/tex}<\/span>CED<br \/>\nCD =\u00a0<span class=\"math-tex\">{tex}\\sqrt{18.75^2+25^2}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}= \\sqrt{976.5625}{\/tex}<\/span><br \/>\n= 31.25 feet<br \/>\nAC = AD &#8211; CD<br \/>\n= 50 &#8211; 31.25<br \/>\n= 18.75 feet<br \/>\n<strong>OR<\/strong><br \/>\n<span class=\"math-tex\">{tex}\\sqrt{40^2+20^2}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}= \\sqrt{1600+400}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}=\\sqrt{2000}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}= 20\\sqrt{5}{\/tex}<\/span> feet<\/li>\n<\/ol>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li style=\"text-align: left;\"><img decoding=\"async\" style=\"height: 19px; width: 200px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1672224661-t547cy.jpg\" \/><br \/>\nAB = <span class=\"math-tex\">{tex}\\sqrt{{(5-2)}^2+{(4-2)^2}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}= \\sqrt{9+4}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}= \\sqrt{13}{\/tex}<\/span><\/li>\n<li style=\"text-align: left;\">Middle point AB = <span class=\"math-tex\">{tex}\\left(\\frac{2+5}2,\\frac{2+4}2\\right){\/tex}<\/span><br \/>\n= (3.5, 3)<\/li>\n<li style=\"text-align: left;\"><img decoding=\"async\" style=\"height: 24px; width: 200px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1672224762-xnjktw.jpg\" \/><br \/>\nBC = <span class=\"math-tex\">{tex}\\sqrt{{(7-5)^2}+{(6-4)^2}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}= \\sqrt{4+4}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}= 2 \\sqrt{2}{\/tex}<\/span><br \/>\n<strong>OR<\/strong><br \/>\n<img decoding=\"async\" style=\"height: 24px; width: 200px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1672224905-hfrd7g.jpg\" \/><br \/>\nMiddle point of BC = <span class=\"math-tex\">{tex}\\left(\\frac{5+7}2,\\frac{4+6}2\\right){\/tex}<\/span><br \/>\n= (6, 5)<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<p><strong>Boost Your Exam Preparation with myCBSEguide App<\/strong><\/p>\n<p>For thorough exam practice and effective preparation, download the <a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\"><strong>myCBSEguide<\/strong><\/a>\u00a0app. It offers complete <strong>study material<\/strong> for <strong>CBSE<\/strong>, <strong>NCERT<\/strong>, <strong>JEE (Main)<\/strong>, <strong>NEET-UG<\/strong>, and <strong>NDA exams<\/strong>. The app provides access to an extensive range of <strong>practice questions<\/strong>, <strong>sample papers<\/strong>, and <strong>NCERT solutions<\/strong> to help you prepare comprehensively and improve your exam performance. The <strong>Class 10 Maths Sample Papers 2025<\/strong> are an essential resource for students preparing for their upcoming exams.<\/p>\n<p>Teachers can also take advantage of the <a href=\"https:\/\/play.google.com\/store\/apps\/details?id=com.examin8.exam\"><strong>Examin8<\/strong><\/a>\u00a0app, which allows them to create customized <strong>exam papers<\/strong> featuring their own name and logo, making it easier to conduct personalized assessments. Whether you&#8217;re a student aiming to excel or a teacher designing tailored exams,<a href=\"https:\/\/mycbseguide.com\/\"> <strong>myCBSEguide<\/strong><\/a> and <a href=\"https:\/\/examin8.com\/\"><strong>Examin8<\/strong><\/a> Website are your go-to tools for academic success.<\/p>\n<p style=\"text-align: center;\"><strong>\u00a0Login into our <a href=\"https:\/\/mycbseguide.com\/dashboard\/\">student dashboard<\/a>\u00a0for more.<\/strong><\/p>\n<p style=\"text-align: center;\"><b><strong><a class=\"button\" href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide&amp;referrer=utm_source%3Dmycbse_bottom%26utm_medium%3Dtext%26utm_campaign%3Dmycbseads\">Download myCBSEguide App<\/a><\/strong><\/b><\/p>\n<h2><span class=\"ez-toc-section\" id=\"Sample_Papers_for_Class_10_2024-25_in_PDF\"><\/span><strong>Sample Papers for Class 10 2024-25 in PDF<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<ul>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-10-mathematics\/1202\/\"><strong>Mathematics<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-10-science\/1176\/\"><strong>Science<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-10-social-science\/1896\/\"><strong>Social Science<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-10-english-language-and-literature\/1903\/\"><strong>English Language and Literature<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-10-%E0%A4%B9%E0%A4%BF%E0%A4%82%E0%A4%A6%E0%A5%80-%E0%A4%8F\/1905\/\"><strong>Hindi Course A<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-10-%E0%A4%B9%E0%A4%BF%E0%A4%82%E0%A4%A6%E0%A5%80-%E0%A4%AC\/1907\/\"><strong>Hindi Course B<\/strong><\/a><\/li>\n<\/ul>\n<p><strong>Download Class 10 Sample Papers 2025 for Mathematics, Science, English, and More on <a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\">myCBSEguide<\/a><\/strong><\/p>\n<p>To practice effectively and excel in your exams, download the <strong>Sample Papers for Class 10 Mathematics 2025<\/strong>, <strong>Science<\/strong>, <strong>Social Science<\/strong>, <strong>English Communicative<\/strong>, <strong>English Language and Literature<\/strong>, <strong>Hindi Course A<\/strong>, and <strong>Hindi Course B<\/strong> from the <strong>myCBSEguide<\/strong> app or <a href=\"https:\/\/mycbseguide.com\/\"><strong>myCBSEguide <\/strong><\/a><strong>website<\/strong>. <strong>myCBSEguide<\/strong> offers <strong>sample papers with solutions<\/strong>, <strong>chapter-wise test papers<\/strong>, <strong>NCERT solutions<\/strong>, and <strong>NCERT Exemplar solutions<\/strong> to help students improve their understanding and problem-solving skills.<\/p>\n<p>Additionally, the app provides <strong>quick revision notes<\/strong>, <strong>CBSE guess papers<\/strong>, and <strong>important question papers<\/strong>, ensuring you are well-prepared for the board exams. Available through the <strong>myCBSEguide app<\/strong> and website, these resources are perfect for <strong>CBSE students<\/strong> who want to perform at their best in the upcoming exams.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Class 10 Maths Standard Sample Papers 2025\u00a0 CBSE offers two distinct courses in Class 10 Maths: Maths Basic and Maths Standard. This page provides Class 10 Maths Standard Sample Papers 2025 and their solution specifically tailored for students enrolled in the Maths Standard course. Sample Model Test Papers for Class 10 Maths with detailed solutions &#8230; <a title=\"Class 10 Maths Sample Papers 2025\" class=\"read-more\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-10-mathematics\/\" aria-label=\"More on Class 10 Maths Sample Papers 2025\">Read more<\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1339,1998,1347,2004],"tags":[448,1042,12,1341,1933,1959,1967,321,1340],"class_list":["post-13717","post","type-post","status-publish","format-standard","hentry","category-cbse-sample-papers","category-class-10-sample-papers","category-mathematics","category-maths-sample-papers-class-10-sample-papers","tag-cbse-class-10","tag-cbse-class-10-mathematics","tag-cbse-sample-papers","tag-cbse-sample-papers-2018-19","tag-cbse-sample-papers-2023","tag-cbse-sample-papers-2024","tag-cbse-sample-papers-2025","tag-mathematics","tag-model-question-papers"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.0 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Class 10 Maths Sample Papers 2025<\/title>\n<meta name=\"description\" content=\"Class 10 Maths Sample Papers 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