{"id":8723,"date":"2017-12-20T12:50:24","date_gmt":"2017-12-20T07:20:24","guid":{"rendered":"http:\/\/mycbseguide.com\/blog\/?p=8723"},"modified":"2025-10-08T17:36:48","modified_gmt":"2025-10-08T12:06:48","slug":"sample-paper-for-class-10-maths","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/sample-paper-for-class-10-maths\/","title":{"rendered":"Sample paper for class 10 Maths 2018-19"},"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\/sample-paper-for-class-10-maths\/#Class_10_Mathematics_Sample_papers\" >Class 10 Mathematics Sample papers<\/a><ul class='ez-toc-list-level-3' ><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/mycbseguide.com\/blog\/sample-paper-for-class-10-maths\/#Section_A_Question_numbers_1_to_6_carry_1_mark_each\" >Section A \nQuestion numbers 1 to 6 carry 1 mark each<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/mycbseguide.com\/blog\/sample-paper-for-class-10-maths\/#Section_B_Question_numbers_7_to_12_carry_2_marks_each\" >Section B \nQuestion numbers 7 to 12 carry 2 marks each.<\/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\/sample-paper-for-class-10-maths\/#Section_C_Question_numbers_13_to_22_carry_3_marks_each\" >Section C \nQuestion numbers 13 to 22 carry 3 marks each.<\/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\/sample-paper-for-class-10-maths\/#Section_D_Question_numbers_23_to_30_carry_4_marks_each\" >Section D \nQuestion numbers 23 to 30 carry 4 marks each.<\/a><\/li><\/ul><\/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\/sample-paper-for-class-10-maths\/#Sample_paper_for_class_10_Maths\" >Sample paper for class 10 Maths<\/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\/sample-paper-for-class-10-maths\/#Sample_paper_for_class_10_all_subjects\" >Sample paper for class 10 all subjects<\/a><\/li><\/ul><\/nav><\/div>\n<p><strong>Download CBSE sample paper for class 10 Maths<\/strong> for board exams available for download in myCBSEguide mobile app. The best app for CBSE students now provides class 10 maths sample paper includes all questions from Mathematics &#8211; Textbook for class X &#8211; NCERT Publication, Guidelines for Mathematics Laboratory in Schools, class X &#8211; CBSE Publication, Laboratory Manual &#8211; Mathematics, secondary stage &#8211; NCERT Publication, Mathematics exemplar problems for class X, NCERT Publication. In the session 2018-19, CBSE will conduct board exam for all CBSE class 10 students which will cover the whole book.<\/p>\n<p style=\"text-align: center;\"><strong>Download Updated sample paper for class 10 maths 2018-19\u00a0<a class=\"button\" href=\"https:\/\/mycbseguide.com\/downloads\/cbse-class-10-mathematics\/1202\/cbse-sample-papers\/2\/\">Click Here<\/a><\/strong><\/p>\n<p>For study on the go download <a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide&amp;referrer=utm_source%3Dmycbse_bottom%26utm_medium%3Dtext%26utm_campaign%3Dmycbseads\"><strong>myCBSEguide app<\/strong><\/a> for android phones. Sample paper for class 10 Maths and other subjects are available for download as PDF in app too.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"Class_10_Mathematics_Sample_papers\"><\/span><strong>Class 10 Mathematics Sample papers<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><strong>General Instructions:<\/strong><\/p>\n<ol>\n<li>All questions are compulsory.<\/li>\n<li>The question paper consists of 30 questions divided into four sections A, B, C and D.<\/li>\n<li>Section A contains 6 questions of 1 mark each. Section B contains 6 questions of 2 marks each. Section C contains 10 questions of 3 marks each. Section D contains 8 questions of 4 marks each.<\/li>\n<li>There is no overall choice. However, an internal choice has been provided in four questions of 3 marks each and three questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.<\/li>\n<li>Use of calculators is not permitted.<\/li>\n<\/ol>\n<hr \/>\n<h3 style=\"text-align: center;\"><span class=\"ez-toc-section\" id=\"Section_A_Question_numbers_1_to_6_carry_1_mark_each\"><\/span><strong>Section A<br \/>\nQuestion numbers 1 to 6 carry 1 mark each<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>1. Write whether the rational number <\/strong><span class=\"math-tex\">{tex}\\frac{7}{{75}}{\/tex}<\/span><strong> will have a terminating decimal expansion or a nor-terminating repeating decimal expansion.<\/strong><br \/>\n<strong>Ans.<\/strong> Non terminating repeating decimal expansion.<\/p>\n<p><strong>2. Find the value(s) of k, if the quadratic equation <\/strong><span class=\"math-tex\">{tex}3{x^2} &#8211; k\\sqrt 3 x + 4 = 0{\/tex}<\/span><strong> has equal roots.<\/strong><br \/>\n<strong>Ans. <\/strong><span class=\"math-tex\">{tex}k = \\pm 4{\/tex}<\/span><\/p>\n<p><strong>3. Find the eleventh term from the last term of the AP:<\/strong><br \/>\n<strong>27, 23, 19, &#8230;, \u201365.<\/strong><br \/>\n<strong>Ans. <\/strong><span class=\"math-tex\">{tex}{a_{11}} = &#8211; 25{\/tex}<\/span><br \/>\n<strong>4. Find the coordinates of the point on y-axis which is nearest to the point (\u20132, 5).<\/strong><br \/>\n<strong>Ans. <\/strong><span class=\"math-tex\">{tex}\\left( {0,\\,5} \\right){\/tex}<\/span><br \/>\n<strong>5. In given figure, <\/strong><span class=\"math-tex\">{tex}ST||RQ,{\\text{ }}PS = 3\\,cm{\/tex}<\/span><strong> and <\/strong><span class=\"math-tex\">{tex}SR = 4{\\text{ }}cm.{\/tex}<\/span><strong> Find the ratio of the area of <\/strong><span class=\"math-tex\">{tex}\\Delta PST{\/tex}<\/span><strong> to the area of <\/strong><span class=\"math-tex\">{tex}\\Delta PRQ.{\/tex}<\/span><br \/>\n<strong><img loading=\"lazy\" decoding=\"async\" id=\"Picture 1\" class=\"alignnone\" style=\"height: 111px; width: 144px;\" src=\"http:\/\/media-mycbseguide.s3.amazonaws.com\/images\/cbse\/10\/mathematics\/sp18\/image001.jpg\" alt=\"Sample paper for class 10 Maths\" width=\"144\" height=\"111\" \/><\/strong><br \/>\n<strong>Ans. <\/strong><span class=\"math-tex\">{tex}9:\\,49{\/tex}<\/span><br \/>\n<strong>6. If <\/strong><span class=\"math-tex\">{tex}cos{\\text{ }}A = \\frac{2}{5},{\/tex}<\/span><strong> find the value of <\/strong><span class=\"math-tex\">{tex}4 + 4\\,ta{n^2}A{\/tex}<\/span><br \/>\n<strong>Ans. <\/strong>25<\/p>\n<h3 style=\"text-align: center;\"><span class=\"ez-toc-section\" id=\"Section_B_Question_numbers_7_to_12_carry_2_marks_each\"><\/span><strong>Section B<br \/>\nQuestion numbers 7 to 12 carry 2 marks each.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>7. If two positive integers p and q are written as <\/strong><span class=\"math-tex\">{tex}p = {a^2}{b^3}{\\text{ }}and{\\text{ }}q = {a^3}{b};{\\text{ }}a,\\;b{\/tex}<\/span><strong> are prime numbers, then verify:<\/strong><br \/>\n<span class=\"math-tex\">{tex}LCM\\left( {p,q} \\right) \\times HCF\\left( {p,q} \\right) = pq{\/tex}<\/span><\/p>\n<p><strong>Ans. <\/strong><span class=\"math-tex\">{tex}LCM(p,\\,q) = {a^3}{b^3}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}HCF(p,\\,q) = {a^2}b{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}LCM\\left( {p,q} \\right) \\times HCF\\left( {p,q} \\right) {\/tex}<\/span><span class=\"math-tex\">{tex} = {a^5}{b^4} = \\left( {{a^2}{b^3}} \\right)\\left( {{a^3}b} \\right) = pq{\/tex}<\/span><br \/>\n<strong>8. The sum of first n terms of an AP is given by <\/strong><span class=\"math-tex\">{tex}{S_n} = 2{n^2} + 3n.{\/tex}<\/span><strong> Find the sixteenth term of the AP.<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Ans.<\/strong> S<sub>n <\/sub>= 2n<sup>2<\/sup> + 3n<\/p>\n<p style=\"text-align: justify;\">S<sub>1<\/sub> = 5 = a<sub>1<\/sub><br \/>\nS<sub>2 <\/sub>= a<sub>1<\/sub> + a<sub>2<\/sub> = 14 <span class=\"math-tex\">{tex} \\Rightarrow {\/tex}<\/span> a<sub>2<\/sub> = 9<br \/>\nd = a<sub>2<\/sub> \u2013 a<sub>1<\/sub> = 4<br \/>\na<sub>16<\/sub> = a<sub>1<\/sub> + 15d = 5 + 15(4) = 65<br \/>\n<strong>9. Find the value(s) of k for which the pair of linear equations <\/strong><span class=\"math-tex\">{tex}kx + y = {k^2}{\\text{ }}and{\\text{ }}x + ky = 1{\/tex}<\/span><strong> have infinitely many solutions.<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Ans.<\/strong> For pair of equations kx + 1y = k<sup>2<\/sup> and 1x + ky = 1<br \/>\nWe have: <span class=\"math-tex\">{tex}\\frac{{{a_1}}}{{{a_2}}} = \\frac{k}{1},\\,\\frac{{{b_1}}}{{{b_2}}} = \\frac{1}{k},\\,\\frac{{{c_1}}}{{{c_2}}} = \\frac{{{k^2}}}{1}{\/tex}<\/span><br \/>\nFor infinitely many solutions, <span class=\"math-tex\">{tex}\\frac{{{a_1}}}{{{a_2}}} = \\frac{{{b_1}}}{{{b_2}}} = \\frac{{{c_1}}}{{{c_2}}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\therefore \\frac{k}{1} = \\frac{1}{k} \\Rightarrow {k^2} = 1 \\Rightarrow k = 1,\\, &#8211; 1\\,&#8230;(i){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}and\\,\\frac{1}{k} = \\frac{{{k^2}}}{1} \\Rightarrow {k^3} = 1 \\Rightarrow k = 1\\,&#8230;(ii){\/tex}<\/span><br \/>\nFrom (i) and (ii), k = 1<br \/>\n<strong>10. If <\/strong><span class=\"math-tex\">{tex}\\left( {1,\\,\\frac{p}{3}} \\right){\/tex}<\/span><strong> is the mid-point of the line segment joining the points (2, 0) and <\/strong><span class=\"math-tex\">{tex}\\left( {0,\\,\\frac{2}{9}} \\right),{\/tex}<\/span><strong> then show that the line <\/strong><span class=\"math-tex\">{tex}5x + 3y + 2 = 0{\/tex}<\/span><strong> passes through the point <\/strong><span class=\"math-tex\">{tex}\\left( { &#8211; 1,{\\text{ }}3p} \\right).{\/tex}<\/span><\/p>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>Since <span class=\"math-tex\">{tex}\\left( {1,\\,\\frac{p}{3}} \\right){\/tex}<\/span> is the mid-point of the line segment joining the points <span class=\"math-tex\">{tex}\\left( {2,\\,0} \\right)\\,and\\,\\left( {0,\\,\\frac{2}{9}} \\right){\/tex}<\/span> therefore, <span class=\"math-tex\">{tex}\\frac{p}{3} = \\frac{{0 + \\frac{2}{9}}}{2} \\Rightarrow p = \\frac{1}{3}{\/tex}<\/span><br \/>\nThe line <span class=\"math-tex\">{tex}5x + 3y + 2 = 0{\/tex}<\/span> passes through the point <span class=\"math-tex\">{tex}\\left( {-1,{\\text{ }}1} \\right){\\text{ }}as{\\text{ }}5\\left( {-1} \\right) + 3\\left( 1 \\right) + 2 = 0{\/tex}<\/span><br \/>\n<strong>11. A box contains cards numbered 11 to 123. A card is drawn at random from the box. Find the probability that the number on the drawn card is<\/strong><br \/>\n<strong>(i) a square number<\/strong><br \/>\n<strong>(ii) a multiple of 7<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>(i) P(square number)<span class=\"math-tex\">{tex} = \\frac{8}{{113}}{\/tex}<\/span><br \/>\n(ii) P(multiple of 7)<span class=\"math-tex\">{tex} = \\frac{{16}}{{113}}{\/tex}<\/span><br \/>\n<strong>12. A box contains 12 balls of which some are red in colour. If 6 more red balls are put in the box and a ball is drawn at random, the probability of drawing a red ball doubles than what it was before. Find the number of red balls in the bag.<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Ans.<\/strong> Let number of red balls be = x<br \/>\n<span class=\"math-tex\">{tex}\\therefore P(red\\,ball) = \\frac{x}{{12}}{\/tex}<\/span><br \/>\nIf 6 more red balls are added:<br \/>\nThe number of red balls = x + 6<br \/>\n<span class=\"math-tex\">{tex}P(red\\,ball) = \\frac{{x + 6}}{{18}}{\/tex}<\/span><br \/>\nSince, <span class=\"math-tex\">{tex}\\frac{{x + 6}}{{18}} = 2\\left( {\\frac{x}{{12}}} \\right) \\Rightarrow x = 3{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\therefore {\/tex}<\/span> There are 3 red balls in the bag.<\/p>\n<h3 style=\"text-align: center;\"><span class=\"ez-toc-section\" id=\"Section_C_Question_numbers_13_to_22_carry_3_marks_each\"><\/span><strong>Section C<br \/>\nQuestion numbers 13 to 22 carry 3 marks each.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>13. Show that exactly one of the numbers <\/strong><span class=\"math-tex\">{tex}n,{\\text{ }}n + 2{\\text{ }}or{\\text{ }}n + 4{\/tex}<\/span><strong> is divisible by 3.<\/strong><\/p>\n<p><strong>Ans. <\/strong>Let n = 3k, 3k + 1 or 3k + 2.<br \/>\n(i) When n = 3k:<br \/>\nn is divisible by 3.<br \/>\nn + 2 = 3k + 2 <span class=\"math-tex\">{tex} \\Rightarrow {\/tex}<\/span> n + 2 is not divisible by 3.<br \/>\nn + 4 = 3k + 4 = 3(k + 1) + 1 <span class=\"math-tex\">{tex} \\Rightarrow {\/tex}<\/span> n + 4 is not divisible by 3.<\/p>\n<p>(ii) When n = 3k + 1:<br \/>\nn is not divisible by 3.<br \/>\nn + 2 = (3k + 1) + 2 = 3k + 3 = 3(k + 1) <span class=\"math-tex\">{tex} \\Rightarrow {\/tex}<\/span> n + 2 is divisible by 3.<br \/>\nn + 4 = (3k + 1) + 4 = 3k + 5 = 3(k + 1) + 2 <span class=\"math-tex\">{tex} \\Rightarrow {\/tex}<\/span> n + 4 is not divisible by 3.<\/p>\n<p>(iii) When n = 3k + 2:<br \/>\nn is not divisible by 3.<br \/>\nn + 2 = (3k + 2) + 2 = 3k + 4 = 3(k + 1) + 1 <span class=\"math-tex\">{tex} \\Rightarrow {\/tex}<\/span> n + 2 is not divisible by 3.<br \/>\nn + 4 = (3k + 2) + 4 = 3k + 6 = 3(k + 2) <span class=\"math-tex\">{tex} \\Rightarrow {\/tex}<\/span> n + 4 is divisible by 3.<br \/>\nHence exactly one of the numbers n, n + 2 or n + 4 is divisible by 3.<br \/>\n<strong>14. Find all the zeroes of the polynomial <\/strong><span class=\"math-tex\">{tex}3{x^4}{\\text{ + }}6{x^3} &#8211; 2{x^2} &#8211; 10x &#8211; 5{\/tex}<\/span><strong> if two of its zeroes are <\/strong><span class=\"math-tex\">{tex}\\sqrt {\\frac{5}{3}} \\,and\\, &#8211; \\sqrt {\\frac{5}{3}} .{\/tex}<\/span><\/p>\n<p><strong>Ans. <\/strong>Since <span class=\"math-tex\">{tex}\\sqrt {\\frac{5}{3}} \\,and\\, &#8211; \\sqrt {\\frac{5}{3}} {\/tex}<\/span> are the two zeroes therefore, <span class=\"math-tex\">{tex}\\left( {x &#8211; \\sqrt {\\frac{5}{3}} } \\right)\\left( {x + \\sqrt {\\frac{5}{3}} } \\right) = \\frac{1}{3}\\left( {3{x^2} &#8211; 5} \\right){\/tex}<\/span> is a factor of given polynomial.<\/p>\n<p>We divide the given polynomial by <span class=\"math-tex\">{tex}3{x^2} &#8211; 5.{\/tex}<\/span><\/p>\n<p><img decoding=\"async\" id=\"Picture_x0020_16\" style=\"width: 200px; height: 151px;\" src=\"http:\/\/media-mycbseguide.s3.amazonaws.com\/images\/cbse\/10\/mathematics\/sp18\/image015.jpg\" \/><\/p>\n<p>For other zeroes, <span class=\"math-tex\">{tex}{x^2} + 2x + 1 = 0 \\Rightarrow {(x + 1)^2} = 0,{\/tex}<\/span> <span class=\"math-tex\">{tex}\\,x = &#8211; 1, &#8211; 1{\/tex}<\/span><\/p>\n<p><span class=\"math-tex\">{tex}\\because {\/tex}<\/span> Zeroes of the given polynomial are <span class=\"math-tex\">{tex}\\sqrt {\\frac{5}{3}} , &#8211; \\sqrt {\\frac{5}{3}} , &#8211; 1\\,and\\, &#8211; 1.{\/tex}<\/span><\/p>\n<p><strong>15. Seven times a two digit number is equal to four times the number obtained by reversing the order of its digits. If the difference of the digits is 3, determine the number.<\/strong><\/p>\n<p><strong>Ans. <\/strong>Let the ten\u2019s and the units digit be y and x respectively.<br \/>\nSo, the number is 10y + x.<br \/>\nThe number when digits are reversed is 10x + y.<br \/>\nNow, 7(10y + x) = 4(10x + y) <span class=\"math-tex\">{tex} \\Rightarrow {\/tex}<\/span> 2y = x \u2026(i)<br \/>\nAlso x \u2013 y = 3 \u2026(ii)<br \/>\nSolving (1) and (2), we get y = 3 and x = 6.<br \/>\nHence the number is 36<\/p>\n<p><strong>16. In what ratio does the x-axis divide the line segment joining the points <\/strong><span class=\"math-tex\">{tex}\\left( {-4,\\;-6} \\right){\\text{ }}and{\\text{ }}\\left( {-1,{\\text{ }}7} \\right)?{\/tex}<\/span><strong> Find the co-ordinates of the point of division.<\/strong><br \/>\n<strong>OR<\/strong><br \/>\n<strong>The points <\/strong><span class=\"math-tex\">{tex}A\\left( {4,-2} \\right),B\\left( {7,2} \\right),C\\left( {0,9} \\right){\\text{ }}and\\,D\\left( {-3,5} \\right){\/tex}<\/span><strong> form a parallelogram. Find the length of the altitude of the parallelogram on the base AB.<\/strong><\/p>\n<p><strong>Ans.<\/strong> Let x-axis divides the line segment joining <span class=\"math-tex\">{tex}\\left( {-4,-6} \\right){\\text{ }}and{\\text{ }}\\left( {-1,7} \\right){\/tex}<\/span> at the point P in the ratio 1 : k.<br \/>\nNow, coordinates of point of division <span class=\"math-tex\">{tex}P\\left( {\\frac{{ &#8211; 1 &#8211; 4k}}{{k + 1}},\\frac{{7 &#8211; 6k}}{{k + 1}}} \\right){\/tex}<\/span><br \/>\nSince P lies on x-axis, therefore <span class=\"math-tex\">{tex}\\frac{{7 &#8211; 6k}}{{k + 1}} = 0{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow 7 &#8211; 6k = 0{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow k = \\frac{7}{6}{\/tex}<\/span><br \/>\nHence the ratio is <span class=\"math-tex\">{tex}1:\\,\\frac{7}{6} = 6:7{\/tex}<\/span><br \/>\nNow, the coordinates of P are <span class=\"math-tex\">{tex}\\left( {\\frac{{ &#8211; 34}}{{13}},0} \\right).{\/tex}<\/span><br \/>\nOR<br \/>\nLet the height of parallelogram taking AB as base be h.<\/p>\n<p>Now AB <span class=\"math-tex\">{tex} = \\sqrt {{{\\left( {7 &#8211; 4} \\right)}^2} + {{\\left( {2 + 2} \\right)}^2}} = \\sqrt {{3^2} + {4^2}} = 5\\,units.{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}Area\\left( {\\Delta ABC} \\right) = \\frac{1}{2}{\/tex}<\/span><span class=\"math-tex\">{tex}\\left[ {4\\left( {2 &#8211; 9} \\right) + 7\\left( {9 + 2} \\right) + 0\\left( { &#8211; 2 &#8211; 2} \\right)} \\right]{\/tex}<\/span><span class=\"math-tex\">{tex} = \\frac{{49}}{2}sq\\,units.{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}Now,\\,\\frac{1}{2} \\times AB \\times h = \\frac{{49}}{2}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow \\frac{1}{2} \\times 5 \\times h = \\frac{{49}}{2}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow h = \\frac{{49}}{5} = 9.8\\,units.{\/tex}<\/span><\/p>\n<p><strong>17. In given figure <\/strong><span class=\"math-tex\">{tex}\\angle 1 = \\angle 2\\,and\\,\\Delta NSQ \\cong \\Delta MTR,{\/tex}<\/span><strong> then prove that <\/strong><span class=\"math-tex\">{tex}\\Delta PTS \\sim \\Delta PRQ{\/tex}<\/span><br \/>\n<strong><img decoding=\"async\" id=\"Picture 3\" style=\"height: 122px; width: 157px;\" src=\"http:\/\/media-mycbseguide.s3.amazonaws.com\/images\/cbse\/10\/mathematics\/sp18\/image002.jpg\" \/><\/strong><br \/>\n<strong>OR<\/strong><br \/>\n<strong>In an equilateral triangle ABC, D is a point on the side BC such that <\/strong><span class=\"math-tex\">{tex}BD = \\frac{1}{3}BC.{\/tex}<\/span><strong> Prove that <\/strong><span class=\"math-tex\">{tex}9A{D^2} = 7A{B^2}.{\/tex}<\/span><br \/>\n<strong><img decoding=\"async\" id=\"Picture 2\" style=\"height: 115px; width: 106px;\" src=\"http:\/\/media-mycbseguide.s3.amazonaws.com\/images\/cbse\/10\/mathematics\/sp18\/image003.jpg\" \/><\/strong><\/p>\n<p><strong>Ans. <\/strong><span class=\"math-tex\">{tex}\\angle {\\text{SQN = }}\\angle {\\text{TRM}}{\/tex}<\/span><span class=\"math-tex\">{tex}{\\text{(CPCT as }}\\Delta {\\text{NSQ}} \\cong \\Delta {\\text{MTR)}}{\/tex}<\/span><br \/>\n<img decoding=\"async\" id=\"Picture 7\" style=\"height: 125px; width: 165px;\" src=\"http:\/\/media-mycbseguide.s3.amazonaws.com\/images\/cbse\/10\/mathematics\/sp18\/image004.png\" \/><br \/>\nSince, <span class=\"math-tex\">{tex}\\angle P + \\angle 1 + \\angle 2 = \\angle P + \\angle PQR + \\angle PRQ{\/tex}<\/span><span class=\"math-tex\">{tex}\\left( {Angle\\,sum\\,property} \\right){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow \\angle 1 + \\angle 2 = \\angle PQR + \\angle PRQ{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow 2\\angle 1 = 2\\angle PQR{\/tex}<\/span><span class=\"math-tex\">{tex}(as\\,\\angle 1 = \\angle 2\\,and\\,\\angle PQR = \\angle PRQ){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\angle 1 = \\angle PQR{\/tex}<\/span><br \/>\nAlso <span class=\"math-tex\">{tex}\\angle 2 = \\angle PRQ{\/tex}<\/span><br \/>\nAnd <span class=\"math-tex\">{tex}\\angle SPT = \\angle QPR(common){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Delta PTS \\sim \\Delta PRQ{\/tex}<\/span><span class=\"math-tex\">{tex}(By\\,AAA\\,similarity\\,criterion){\/tex}<\/span><\/p>\n<p>OR<br \/>\n<img decoding=\"async\" id=\"Picture 8\" style=\"height: 128px; width: 153px;\" src=\"http:\/\/media-mycbseguide.s3.amazonaws.com\/images\/cbse\/10\/mathematics\/sp18\/image005.png\" \/><br \/>\nConstruction: Draw <span class=\"math-tex\">{tex}AP \\bot BC{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}In\\,\\Delta ADP,\\,A{D^2} = A{P^2} + D{P^2}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}A{D^2} = A{P^2} + {\\left( {BP &#8211; BD} \\right)^2}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}A{D^2} = A{P^2} + B{P^2} + B{D^2} &#8211; 2\\left( {BP} \\right)\\left( {BD} \\right){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}A{D^2} = A{B^2} + {\\left( {\\frac{1}{3}BC} \\right)^2} &#8211; 2\\left( {\\frac{{BC}}{2}} \\right)\\left( {\\frac{{BC}}{3}} \\right){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}A{D^2} = \\frac{7}{9}A{B^2}\\left( {\\because BC = AB} \\right){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}9A{D^2} = 7A{B^2}{\/tex}<\/span><\/p>\n<p><strong>18. In given figure <\/strong><span class=\"math-tex\">{tex}XY{\\text{ }}and\\;X&#8217;Y'{\/tex}<\/span><strong> are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting <\/strong><span class=\"math-tex\">{tex}XY{\/tex}<\/span><strong> at A and <\/strong><span class=\"math-tex\">{tex}X&#8217;Y'{\/tex}<\/span><strong> at B. Prove that <\/strong><span class=\"math-tex\">{tex}\\angle AOB = 90^\\circ .{\/tex}<\/span><br \/>\n<strong><img decoding=\"async\" id=\"Picture 4\" style=\"height: 124px; width: 192px;\" src=\"http:\/\/media-mycbseguide.s3.amazonaws.com\/images\/cbse\/10\/mathematics\/sp18\/image006.jpg\" \/><\/strong><\/p>\n<p><strong>Ans. <\/strong>Join OC<br \/>\nIn <span class=\"math-tex\">{tex}\\Delta OPA{\\text{ }}and{\\text{ }}\\Delta OCA{\/tex}<\/span><br \/>\nOP = OC (radii of same circle)<br \/>\nPA = CA (length of two tangents)<br \/>\n<img decoding=\"async\" id=\"Picture 9\" style=\"height: 139px; width: 216px;\" src=\"http:\/\/media-mycbseguide.s3.amazonaws.com\/images\/cbse\/10\/mathematics\/sp18\/image007.jpg\" \/><br \/>\nAO = AO (Common)<br \/>\n<span class=\"math-tex\">{tex}\\therefore \\,\\Delta OPA \\cong \\Delta OCA{\/tex}<\/span> (By SSS congruency criterion)<\/p>\n<p>Hence, <span class=\"math-tex\">{tex}\\angle 1 = \\angle 2\\,(CPCT){\/tex}<\/span><br \/>\nSimilarly <span class=\"math-tex\">{tex}\\angle 3 = \\angle 4{\/tex}<\/span><br \/>\nNow, <span class=\"math-tex\">{tex}\\angle PAB + \\angle QBA = 180^\\circ {\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow 2\\angle 2 + 2\\angle 4 = 180^\\circ {\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow \\angle 2 + \\angle 4 = 90^\\circ {\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow \\angle AOB = 90^\\circ \\,(Angle\\,sum\\,property){\/tex}<\/span><\/p>\n<p><strong>19. Evaluate: <\/strong><span class=\"math-tex\">{tex}\\frac{{\\cos e{c^2}63^\\circ + {{\\tan }^2}24^\\circ }}{{{{\\cot }^2}66^\\circ + {{\\sec }^2}27^\\circ }} + {\/tex}<\/span><span class=\"math-tex\">{tex}\\frac{{{{\\sin }^2}63^\\circ + \\cos 63^\\circ \\sin 27^\\circ + \\sin 27^\\circ \\sec 63^\\circ }}{{2\\left( {\\cos e{c^2}65^\\circ &#8211; {{\\tan }^2}25^\\circ } \\right)}}{\/tex}<\/span><br \/>\n<strong>OR<\/strong><br \/>\n<strong>If <\/strong><span class=\"math-tex\">{tex}sin\\theta + cos\\theta = \\sqrt 2 ,{\/tex}<\/span><strong> then evaluate: <\/strong><span class=\"math-tex\">{tex}tan\\theta + cot\\theta {\/tex}<\/span><\/p>\n<p><strong>Ans. <\/strong><span class=\"math-tex\">{tex}\\frac{{\\cos e{c^2}63^\\circ + {{\\tan }^2}24^\\circ }}{{{{\\cot }^2}66^\\circ + {{\\sec }^2}27^\\circ }} + {\/tex}<\/span><span class=\"math-tex\">{tex}\\frac{{{{\\sin }^2}63^\\circ + \\cos 63^\\circ \\sin 27^\\circ + \\sin 27^\\circ \\sec 63^\\circ }}{{2(\\cos e{c^2}65^\\circ &#8211; {{\\tan }^2}25^\\circ )}}{\/tex}<\/span><br \/>\n<img decoding=\"async\" id=\"Picture 10\" style=\"height: 89px; width: 476px;\" src=\"http:\/\/media-mycbseguide.s3.amazonaws.com\/images\/cbse\/10\/mathematics\/sp18\/image008.png\" \/><br \/>\n<span class=\"math-tex\">{tex} = 1 + \\frac{{1 + 1}}{{2(1)}} = 2{\/tex}<\/span><\/p>\n<p>OR<br \/>\n<span class=\"math-tex\">{tex}\\sin \\theta + \\cos \\theta = \\sqrt 2 {\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow {\\left( {\\sin \\theta + \\cos \\theta } \\right)^2} = {\\left( {\\sqrt 2 } \\right)^2}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow {\\sin ^2}\\theta + {\\cos ^2}\\theta + 2\\sin \\theta \\cos \\theta = 2{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow 1 + 2\\sin \\theta \\cos \\theta = 2{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow \\sin \\theta \\cos \\theta = \\frac{1}{2}\\,&#8230;(i){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}we\\,know,\\,{\\sin ^2}\\theta + {\\cos ^2}\\theta = 1\\,&#8230;(ii){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}Dividing\\,(ii)\\,by\\,(i)\\,we\\,get{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\frac{{{{\\sin }^2}\\theta + {{\\cos }^2}\\theta }}{{\\sin \\theta \\cos \\theta }} = \\frac{1}{{1\/2}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow \\tan \\theta + \\cot \\theta = 2{\/tex}<\/span><\/p>\n<p><strong>20. In given figure ABPC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.<\/strong><br \/>\n<strong><img decoding=\"async\" id=\"Picture 5\" style=\"height: 145px; width: 154px;\" src=\"http:\/\/media-mycbseguide.s3.amazonaws.com\/images\/cbse\/10\/mathematics\/sp18\/image009.jpg\" \/><\/strong><\/p>\n<p><strong>Ans. <\/strong>We know, AC = r<br \/>\n<span class=\"math-tex\">{tex}In\\,\\Delta ACB,\\,B{C^2} = A{C^2} + A{B^2}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow BC = AC\\sqrt 2 \\left( {\\because AB = AC} \\right){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow BC = r\\sqrt 2 {\/tex}<\/span><br \/>\nRequired area = <span class=\"math-tex\">{tex}ar\\left( {\\Delta ACB} \\right){\/tex}<\/span> + ar(semicircle on BC as diameter) \u2013ar(quadrant<\/p>\n<p>ABPC<br \/>\n<span class=\"math-tex\">{tex} = \\frac{1}{2} \\times r \\times r + \\frac{1}{2} \\times \\pi \\times {\\left( {\\frac{{r\\sqrt 2 }}{2}} \\right)^2} &#8211; \\frac{1}{4}\\pi {r^2}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} = \\frac{{{r^2}}}{2} + \\frac{{\\pi {r^2}}}{4} &#8211; \\frac{{\\pi {r^2}}}{4}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} = \\frac{{{r^2}}}{2} = \\frac{{196}}{2}c{m^2} = 98\\,c{m^2}{\/tex}<\/span><\/p>\n<p><strong>21. Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km\/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?<\/strong><br \/>\n<strong>OR<\/strong><br \/>\n<strong>A cone of maximum size is carved out from a cube of edge 14 cm. Find the surface area of the remaining solid after the cone is carved out.<\/strong><\/p>\n<p><strong>Ans.<\/strong> Let the area that can be irrigated in 30 minute be A <span class=\"math-tex\">{tex}{m^2}.{\/tex}<\/span><br \/>\nWater flowing in canal in 30 minutes <span class=\"math-tex\">{tex} = \\left( {10,000 \\times \\frac{1}{2}} \\right)m = 5000\\,m{\/tex}<\/span><br \/>\nVolume of water flowing out in 30 minutes <span class=\"math-tex\">{tex} = \\left( {5000 \\times 6 \\times 1.5} \\right)\\,{m^3} = 45000\\,{m^3}\\,&#8230;(i){\/tex}<\/span><br \/>\nVolume of water required to irrigate the field <span class=\"math-tex\">{tex} = A \\times \\frac{8}{{100}}\\,{m^3}\\,&#8230;(ii){\/tex}<\/span><br \/>\nEquating (i) and (ii), we get<br \/>\n<span class=\"math-tex\">{tex}A \\times \\frac{8}{{100}} = 45000{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}A = 562500\\,{m^2}{\/tex}<\/span><\/p>\n<p>Or<br \/>\n<span class=\"math-tex\">{tex}l = \\sqrt {{7^2} + {{14}^2}} = 7\\sqrt 5 {\/tex}<\/span><br \/>\nSurface area of remaining solid<span class=\"math-tex\">{tex} = 6{l^2} &#8211; \\pi {r^2} + \\pi rl,{\/tex}<\/span> where r and l are the radius and slant height of the cone.<br \/>\n<img decoding=\"async\" id=\"Picture 11\" style=\"height: 145px; width: 170px;\" src=\"http:\/\/media-mycbseguide.s3.amazonaws.com\/images\/cbse\/10\/mathematics\/sp18\/image010.png\" \/><br \/>\n<span class=\"math-tex\">{tex} = 6 \\times 14 \\times 4 &#8211; \\frac{{22}}{7} \\times 7 \\times 7 + \\frac{{22}}{7} \\times 7 \\times 7\\sqrt 5 {\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} = 1176 &#8211; 154 + 154\\sqrt 5 {\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} = \\left( {1022 + 154\\sqrt 5 } \\right)c{m^2}{\/tex}<\/span><br \/>\n<strong>22. Find the mode of the following distribution of marks obtained by the students in an examination:<\/strong><\/p>\n<table class=\"mobile\" border=\"1\" cellspacing=\"0\" cellpadding=\"0\">\n<tbody>\n<tr>\n<td><strong>Marks obtained<\/strong><\/td>\n<td><strong>0-20<\/strong><\/td>\n<td><strong>20-40<\/strong><\/td>\n<td><strong>40-60<\/strong><\/td>\n<td><strong>60-80<\/strong><\/td>\n<td><strong>80-100<\/strong><\/td>\n<\/tr>\n<tr>\n<td><strong>Number of students<\/strong><\/td>\n<td><strong>15<\/strong><\/td>\n<td><strong>18<\/strong><\/td>\n<td><strong>21<\/strong><\/td>\n<td><strong>29<\/strong><\/td>\n<td><strong>17<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong>Given the mean of the above distribution is 53, using empirical relationship estimate the value of its median.<\/strong><\/p>\n<p><strong>Ans. <\/strong><span class=\"math-tex\">{tex}Mode = \\ell + \\left( {\\frac{{{f_2} &#8211; {f_0}}}{{2{f_2} &#8211; {f_0} &#8211; {f_2}}}} \\right) \\times h{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} = 60 + \\left( {\\frac{{29 &#8211; 21}}{{58 &#8211; 21 &#8211; 17}}} \\right) \\times 20 = 68{\/tex}<\/span><br \/>\nSo, the mode marks is 68.<br \/>\nEmpirical relationship between the three measures of central tendencies is:<br \/>\n<span class=\"math-tex\">{tex}3\\,Median = Mode + 2\\,Mean{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}3\\,Median = 68 + 2 \\times 53{\/tex}<\/span><span class=\"math-tex\">{tex}Median = 58\\,marks{\/tex}<\/span><\/p>\n<h3 style=\"text-align: center;\"><span class=\"ez-toc-section\" id=\"Section_D_Question_numbers_23_to_30_carry_4_marks_each\"><\/span><strong>Section D<br \/>\nQuestion numbers 23 to 30 carry 4 marks each.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>23. A train travelling at a uniform speed for 360 km would have taken 48 minutes less to travel the same distance if its speed were 5 km\/hour more. Find the original speed of the train.<\/strong><br \/>\n<strong>OR<\/strong><br \/>\n<strong>Check whether the equation <\/strong><span class=\"math-tex\">{tex}5{x^2}-6x-\\;2 = 0{\/tex}<\/span><strong> has real roots and if it has, find them by the method of completing the square. Also verify that roots obtained satisfy the given equation.<\/strong><\/p>\n<p><strong>Ans. <\/strong>Let original speed of the train be x km\/h.<br \/>\nTime taken at original speed <span class=\"math-tex\">{tex} = \\frac{{360}}{x}\\,hours{\/tex}<\/span><br \/>\nTime taken at increased speed<span class=\"math-tex\">{tex} = \\frac{{360}}{{x + 5}}\\,hours{\/tex}<\/span><br \/>\nNow, <span class=\"math-tex\">{tex}\\frac{{360}}{x} &#8211; \\frac{{360}}{{x + 5}} = \\frac{{48}}{{60}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow 360\\left[ {\\frac{1}{x} &#8211; \\frac{1}{{x + 5}}} \\right] = \\frac{4}{5}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow {x^2} + 5x &#8211; 2250 = 0{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow x = 45\\,or\\, &#8211; 50{\/tex}<\/span><span class=\"math-tex\">{tex}(as\\,speed\\,cannot\\,be\\,negative){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow x = 45\\,km\/h{\/tex}<\/span><br \/>\nOR<br \/>\nDiscriminant <span class=\"math-tex\">{tex} = {b^2} &#8211; 4ac = 36 &#8211; 4 \\times 5 \\times ( &#8211; 2) = 76 &gt; 0{\/tex}<\/span><br \/>\nSo, the given equation has two distinct real roots<br \/>\n<span class=\"math-tex\">{tex}5{x^2} &#8211; 6x &#8211; 2 = 0{\/tex}<\/span><\/p>\n<p>Multiplying both sides by 5.<br \/>\n<span class=\"math-tex\">{tex}{(5x)^2} &#8211; 2 \\times (5x) \\times 3 = 10{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow {(5x)^2} &#8211; 2 \\times (5x) \\times 3 + {3^2} = 10 + {3^2}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow {(5x &#8211; 3)^2} = 19{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow 5x &#8211; 3 = \\pm \\sqrt {19} {\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow x = \\frac{{3 \\pm \\sqrt {19} }}{5}{\/tex}<\/span><br \/>\nVerification: <span class=\"math-tex\">{tex}5{\\left( {\\frac{{3 + \\sqrt {19} }}{5}} \\right)^2} &#8211; 6\\left( {\\frac{{3 + \\sqrt {19} }}{5}} \\right) &#8211; 2{\/tex}<\/span><span class=\"math-tex\">{tex} = \\frac{{9 + 6\\sqrt {19} + 19}}{5} &#8211; \\frac{{18 + 6\\sqrt {19} }}{5} &#8211; \\frac{{10}}{5} = 0{\/tex}<\/span><br \/>\nSimilarly, <span class=\"math-tex\">{tex}5{\\left( {\\frac{{3 &#8211; \\sqrt {19} }}{5}} \\right)^2} &#8211; 6\\left( {\\frac{{3 &#8211; \\sqrt {19} }}{5}} \\right) &#8211; 2 = 0{\/tex}<\/span><br \/>\n<strong>24. An AP consists of 37 terms. The sum of the three middle most terms is 225 and the sum of the last three terms is 429. Find the AP.<\/strong><\/p>\n<p><strong>Ans. <\/strong>Let the three middle most terms of the AP be a \u2013 d, a, a + d.<br \/>\nWe have, (a \u2013 d) + a + (a + d) = 225<br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow 3a = 225 \\Rightarrow a = 75{\/tex}<\/span><br \/>\nNow, the AP is<br \/>\na \u2013 18d,\u2026,a \u2013 2d, a \u2013 d, a, a + d, a + 2d,\u2026, a + 18d<\/p>\n<p>Sum of last three terms:<br \/>\n<span class=\"math-tex\">{tex}(a + 18d) + (a + 17d) + (a + 16d) = 429{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow 3a + 51d = 429 \\Rightarrow a + 17d = 143{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow 75 + 17d = 143 \\Rightarrow d = 4{\/tex}<\/span><br \/>\nNow, first term <span class=\"math-tex\">{tex} = a &#8211; 18d = 75 &#8211; 18(4) = 3{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\therefore {\/tex}<\/span>The AP is 3, 7, 11, \u2026, 147.<\/p>\n<p><strong>25. Show that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.<\/strong><br \/>\n<strong>OR<\/strong><br \/>\n<strong>Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.<\/strong><\/p>\n<p><strong>Ans. <\/strong>Given: A right triangle ABC right angled at B.<br \/>\nTo prove: <span class=\"math-tex\">{tex}A{C^2} = A{B^2} + B{C^2}{\/tex}<\/span><br \/>\nConstruction: Draw <span class=\"math-tex\">{tex}BD \\bot AC{\/tex}<\/span><br \/>\nProof: In <span class=\"math-tex\">{tex}\\Delta ADB\\,and\\,\\Delta ABC{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\angle ADB = \\angle ABC(each\\,90^\\circ ){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\angle BAD = \\angle CAB(common){\/tex}<\/span><br \/>\n<img decoding=\"async\" id=\"Picture 12\" style=\"height: 100px; width: 217px;\" src=\"http:\/\/media-mycbseguide.s3.amazonaws.com\/images\/cbse\/10\/mathematics\/sp18\/image011.png\" \/><br \/>\n<span class=\"math-tex\">{tex}\\Delta ADB \\sim \\Delta ABC{\/tex}<\/span><span class=\"math-tex\">{tex}(By\\,AA\\,similarity\\,criterion){\/tex}<\/span><br \/>\nNow, <span class=\"math-tex\">{tex}\\frac{{AD}}{{AB}} = \\frac{{AB}}{{AC}}{\/tex}<\/span>(corresponding sides are proportional)<br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow A{B^2} = AD \\times AC\\,&#8230;(i){\/tex}<\/span><br \/>\nSimilarly <span class=\"math-tex\">{tex}\\Delta BDC \\sim \\Delta ABC{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow B{C^2} = CD \\times AC\\,&#8230;(ii){\/tex}<\/span><\/p>\n<p>Adding (1) and (2)<br \/>\n<span class=\"math-tex\">{tex}A{B^2} + B{C^2} = AD \\times AC + CD \\times AC{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow A{B^2} + B{C^2} = AC \\times (AD + CD){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow A{B^2} + B{C^2} = A{C^2},\\,Hence\\,proved.{\/tex}<\/span><br \/>\nOR<br \/>\nGiven: <span class=\"math-tex\">{tex}\\Delta ABC \\sim \\Delta PQR{\/tex}<\/span><br \/>\nTo prove: <span class=\"math-tex\">{tex}\\frac{{ar(\\Delta ABC)}}{{ar(\\Delta PQR)}} = {\\left( {\\frac{{AB}}{{PQ}}} \\right)^2} = {\\left( {\\frac{{BC}}{{QR}}} \\right)^2} = {\\left( {\\frac{{CA}}{{RP}}} \\right)^2}{\/tex}<\/span><br \/>\nConstruction: Draw <span class=\"math-tex\">{tex}AM \\bot BC,\\,PN \\bot QR{\/tex}<\/span><br \/>\n<img decoding=\"async\" id=\"Picture 13\" style=\"height: 115px; width: 308px;\" src=\"http:\/\/media-mycbseguide.s3.amazonaws.com\/images\/cbse\/10\/mathematics\/sp18\/image012.png\" \/><br \/>\n<span class=\"math-tex\">{tex}\\frac{{ar(\\Delta ABC)}}{{ar(\\Delta PQR)}} = \\frac{{\\frac{1}{2} \\times BC \\times AM}}{{\\frac{1}{2} \\times QR \\times PN}}{\/tex}<\/span><span class=\"math-tex\">{tex} = \\frac{{BC}}{{QR}} \\times \\frac{{AM}}{{PN}}\\,&#8230;(i){\/tex}<\/span><br \/>\nIn <span class=\"math-tex\">{tex}\\Delta ABM\\,and\\,\\Delta PQN{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\angle B = \\angle Q(\\because \\Delta ABC \\sim \\Delta PQR){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\angle M = \\angle N(each\\,90^\\circ ){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Delta ABM \\sim \\Delta PQN{\/tex}<\/span> <span class=\"math-tex\">{tex}(AA\\,similarity\\,criterion){\/tex}<\/span><br \/>\nTherefore, <span class=\"math-tex\">{tex}\\frac{{AM}}{{PN}} = \\frac{{AB}}{{PQ}}\\,&#8230;(ii){\/tex}<\/span><br \/>\nBut <span class=\"math-tex\">{tex}\\frac{{AB}}{{PQ}} = \\frac{{BC}}{{QR}} = \\frac{{CA}}{{RP}}\\left( {\\Delta ABC \\sim \\Delta PQR} \\right)\\,&#8230;(iii){\/tex}<\/span><\/p>\n<p>Hence, <span class=\"math-tex\">{tex}\\frac{{ar(\\Delta ABC)}}{{ar(\\Delta PQR)}} = \\frac{{BC}}{{QR}} \\times \\frac{{AM}}{{PN}}\\,from\\,(i){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} = \\frac{{AB}}{{PQ}} \\times \\frac{{AB}}{{PQ}}[from\\,(ii)\\,and\\,(iii)]{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} = {\\left( {\\frac{{AB}}{{PQ}}} \\right)^2}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\frac{{ar(\\Delta ABC)}}{{ar(\\Delta PQR)}} = {\\left( {\\frac{{AB}}{{PQ}}} \\right)^2} = {\\left( {\\frac{{BC}}{{QR}}} \\right)^2}{\/tex}<\/span><span class=\"math-tex\">{tex} = {\\left( {\\frac{{CA}}{{RP}}} \\right)^2}{\\text{Using}}\\,(iii){\/tex}<\/span><br \/>\n<strong>26. Draw a triangle ABC with side <\/strong><span class=\"math-tex\">{tex}BC = 7cm,{\\text{ }}\\angle B = 45^\\circ ,{\\text{ }}\\angle A = 105^\\circ .{\/tex}<\/span><strong> Then, construct a triangle whose sides are <\/strong><span class=\"math-tex\">{tex}\\frac{4}{3}{\/tex}<\/span><strong> times the corresponding sides of <\/strong><span class=\"math-tex\">{tex}\\Delta ABC.{\/tex}<\/span><\/p>\n<p><strong>Ans. <\/strong>Draw <span class=\"math-tex\">{tex}\\Delta ABC{\/tex}<\/span> in which <span class=\"math-tex\">{tex}BC = 7\\,cm,\\,\\angle B = 45^\\circ ,\\,\\angle A = 105^\\circ {\/tex}<\/span> and hence <span class=\"math-tex\">{tex}\\angle C = 30^\\circ .{\/tex}<\/span> Construction of similar triangle <span class=\"math-tex\">{tex}A&#8217;BC'{\/tex}<\/span> as shown below:<br \/>\n<img decoding=\"async\" id=\"Picture 14\" style=\"height: 265px; width: 268px;\" src=\"http:\/\/media-mycbseguide.s3.amazonaws.com\/images\/cbse\/10\/mathematics\/sp18\/image013.png\" \/><br \/>\n<strong>27. Prove that <\/strong><span class=\"math-tex\">{tex}\\frac{{\\cos \\theta &#8211; \\sin \\theta + 1}}{{\\cos \\theta + \\sin \\theta &#8211; 1}} = \\cos ec\\theta + \\cot \\theta {\/tex}<\/span><\/p>\n<p><strong>Ans. <\/strong><span class=\"math-tex\">{tex}LHS = \\frac{{\\cos \\theta &#8211; \\sin \\theta + 1}}{{\\cos \\theta + \\sin \\theta &#8211; 1}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} = \\frac{{\\cos \\theta &#8211; \\sin \\theta + 1}}{{\\cos \\theta + \\sin \\theta &#8211; 1}} \\times \\frac{{\\cos \\theta + \\sin \\theta + 1}}{{\\cos \\theta + \\sin \\theta + 1}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} = \\frac{{{{(\\cos \\theta + 1)}^2} &#8211; {{\\sin }^2}\\theta }}{{{{(\\cos \\theta + \\sin \\theta )}^2} &#8211; {1^2}}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} = \\frac{{{{\\cos }^2}\\theta + 1 + 2\\cos \\theta &#8211; {{\\sin }^2}\\theta }}{{{{\\cos }^2}\\theta + {{\\sin }^2}\\theta + 2\\sin \\theta \\cos \\theta &#8211; 1}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} = \\frac{{2{{\\cos }^2}\\theta + 2\\cos \\theta }}{{2\\sin \\theta \\cos \\theta }}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} = \\frac{{2\\cos \\theta (\\cos \\theta + 1)}}{{2\\sin \\theta \\cos \\theta }}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} = \\frac{{\\cos \\theta + 1}}{{\\sin \\theta }} = {\\text{cosec}}\\theta {\\text{ + cot}}\\theta {\\text{ = RHS}}{\/tex}<\/span><\/p>\n<p><strong>28. The angles of depression of the top and bottom of a building 50 metres high as observed from the top of a tower are <\/strong><span class=\"math-tex\">{tex}30^\\circ {\\text{ }}and{\\text{ }}60^\\circ ,{\/tex}<\/span><strong> respectively. Find the height of the tower and also the horizontal distance between the building and the tower.<\/strong><\/p>\n<p><strong>Ans. <\/strong><span class=\"math-tex\">{tex}In\\,\\Delta BTP \\Rightarrow \\tan \\,30^\\circ = \\frac{{TP}}{{BP}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow \\frac{1}{{\\sqrt 3 }} = \\frac{{TP}}{{BP}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}BP = TP\\sqrt 3 \\,&#8230;(i){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}In\\,\\Delta GTR,{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\tan \\,60^\\circ = \\frac{{TR}}{{GR}} \\Rightarrow \\sqrt 3 = \\frac{{TR}}{{GR}}{\/tex}<\/span><span class=\"math-tex\">{tex} \\Rightarrow GR = \\frac{{TR}}{{\\sqrt 3 }}\\,..(ii){\/tex}<\/span><br \/>\nNow, <span class=\"math-tex\">{tex}TP\\sqrt 3 = \\frac{{TR}}{{\\sqrt 3 }}(as\\,BP = GR){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow 3TP = TP + PR{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow 2TP = BG \\Rightarrow TP = \\frac{{50}}{2}m = 25m{\/tex}<\/span><br \/>\nNow, <span class=\"math-tex\">{tex}TR = TP + PR = (25 + 50)m{\/tex}<\/span><br \/>\nHeight of tower = TR = 75 m<br \/>\nDistance between building and tower <span class=\"math-tex\">{tex} = GR = \\frac{{TR}}{{\\sqrt 3 }}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow GR = \\frac{{75}}{{\\sqrt 3 }}m = 25\\sqrt 3 \\,m{\/tex}<\/span><br \/>\n<strong>29. Two dairy owners A and B sell flavoured milk filled to capacity in mugs of negligible thickness, which are cylindrical in shape with a raised hemispherical bottom. The mugs are 14 cm high and have diameter of 7 cm as shown in given figure. <\/strong><\/p>\n<p><strong>Both A and B sell flavoured milk at the rate of <\/strong><span class=\"math-tex\">{tex}Rs.\\,80{\/tex}<\/span><strong> per litre. The dairy owner A uses the formula <\/strong><span class=\"math-tex\">{tex}\\pi {r^2}h{\/tex}<\/span><strong> to find the volume of milk in the mug and charges <\/strong><span class=\"math-tex\">{tex}Rs.\\,\\;43.12{\/tex}<\/span><strong> for it. The dairy owner B is of the view that the price of actual quantity of milk should be charged. What according to him should be the price of one mug of milk? Which value is exhibited by the dairy owner B? <\/strong><span class=\"math-tex\">{tex}\\left( {use\\,\\pi = \\frac{{22}}{7}} \\right){\/tex}<\/span><br \/>\n<strong><img decoding=\"async\" id=\"Picture 6\" style=\"height: 106px; width: 90px;\" src=\"http:\/\/media-mycbseguide.s3.amazonaws.com\/images\/cbse\/10\/mathematics\/sp18\/image014.jpg\" \/><\/strong><\/p>\n<p><strong>Ans.<\/strong> Capacity of mug (actual quantity of milk)<span class=\"math-tex\">{tex} = \\pi {r^2}h &#8211; \\frac{2}{3}\\pi {r^3}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} = \\pi {r^2}\\left( {h &#8211; \\frac{2}{3}r} \\right){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} = \\frac{{22}}{7} \\times \\frac{7}{2} \\times \\frac{7}{2} \\times \\left( {14 &#8211; \\frac{2}{3} \\times \\frac{7}{2}} \\right){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} = \\frac{{2695}}{6}c{m^3}{\/tex}<\/span><br \/>\nAmount dairy owner B should charge for one mug of milk <span class=\"math-tex\">{tex} = \\frac{{2695}}{6} \\times \\frac{{80}}{{1000}} = Rs.\\,35.93{\/tex}<\/span><br \/>\nValue exhibited by dairy owner B: honesty (or any similar value)<\/p>\n<p><strong>30. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is <\/strong><span class=\"math-tex\">{tex}Rs.\\,18.{\/tex}<\/span><strong> Find the missing frequency k.<\/strong><\/p>\n<table class=\"mobile\" border=\"1\" cellspacing=\"0\" cellpadding=\"0\">\n<tbody>\n<tr>\n<td><strong>Daily pocket allowance (in Rs.)<\/strong><\/td>\n<td><strong>11-13<\/strong><\/td>\n<td><strong>13-15<\/strong><\/td>\n<td><strong>15-17<\/strong><\/td>\n<td><strong>17-19<\/strong><\/td>\n<td><strong>19-21<\/strong><\/td>\n<td><strong>21-23<\/strong><\/td>\n<td><strong>23-25<\/strong><\/td>\n<\/tr>\n<tr>\n<td><strong>Number of children<\/strong><\/td>\n<td><strong>3<\/strong><\/td>\n<td><strong>6<\/strong><\/td>\n<td><strong>9<\/strong><\/td>\n<td><strong>13<\/strong><\/td>\n<td><strong>k<\/strong><\/td>\n<td><strong>5<\/strong><\/td>\n<td><strong>4<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: center;\"><strong>OR<\/strong><br \/>\n<strong>The following frequency distribution shows the distance (in metres) thrown by 68 students in a Javelin throw competition.<\/strong><\/p>\n<table class=\"mobile\" border=\"1\" cellspacing=\"0\" cellpadding=\"0\">\n<tbody>\n<tr>\n<td><strong>Distance (in m)<\/strong><\/td>\n<td><strong>0-10<\/strong><\/td>\n<td><strong>10-20<\/strong><\/td>\n<td><strong>20-30<\/strong><\/td>\n<td><strong>30-40<\/strong><\/td>\n<td><strong>40-50<\/strong><\/td>\n<td><strong>50-60<\/strong><\/td>\n<td><strong>60-70<\/strong><\/td>\n<\/tr>\n<tr>\n<td><strong>Number of students<\/strong><\/td>\n<td><strong>4<\/strong><\/td>\n<td><strong>5<\/strong><\/td>\n<td><strong>13<\/strong><\/td>\n<td><strong>20<\/strong><\/td>\n<td><strong>14<\/strong><\/td>\n<td><strong>8<\/strong><\/td>\n<td><strong>4<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong>Draw a less than type Ogive for the given data and find the median distance thrown using this curve.<\/strong><\/p>\n<p><strong>Ans. <\/strong><\/p>\n<table class=\"mobile\" border=\"1\" cellspacing=\"0\" cellpadding=\"0\">\n<tbody>\n<tr>\n<td><strong>Daily pocket allowance (in Rs.)<\/strong><\/td>\n<td><strong>Number of children <\/strong><span class=\"math-tex\">{tex}\\left( {{f_i}} \\right){\/tex}<\/span><\/td>\n<td><strong>Mid-point <\/strong><span class=\"math-tex\">{tex}({x_i}){\/tex}<\/span><\/td>\n<td><span class=\"math-tex\">{tex}{u_i} = \\frac{{{x_i} &#8211; 18}}{2}{\/tex}<\/span><\/td>\n<td><span class=\"math-tex\">{tex}{f_i}{u_i}{\/tex}<\/span><\/td>\n<\/tr>\n<tr>\n<td>11 &#8211; 13<\/td>\n<td>3<\/td>\n<td>12<\/td>\n<td>-3<\/td>\n<td>-9<\/td>\n<\/tr>\n<tr>\n<td>13 \u2013 15<\/td>\n<td>6<\/td>\n<td>14<\/td>\n<td>-2<\/td>\n<td>-12<\/td>\n<\/tr>\n<tr>\n<td>15 &#8211; 17<\/td>\n<td>9<\/td>\n<td>16<\/td>\n<td>-1<\/td>\n<td>-9<\/td>\n<\/tr>\n<tr>\n<td>17 \u2013 19<\/td>\n<td>13<\/td>\n<td>18<\/td>\n<td>0<\/td>\n<td>0<\/td>\n<\/tr>\n<tr>\n<td>19 \u2013 21<\/td>\n<td>K<\/td>\n<td>20<\/td>\n<td>1<\/td>\n<td>k<\/td>\n<\/tr>\n<tr>\n<td>21 \u2013 23<\/td>\n<td>5<\/td>\n<td>22<\/td>\n<td>2<\/td>\n<td>10<\/td>\n<\/tr>\n<tr>\n<td>23 &#8211; 25<\/td>\n<td>4<\/td>\n<td>24<\/td>\n<td>3<\/td>\n<td>12<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span class=\"math-tex\">{tex}\\sum {f_i} = 40 + k{\/tex}<\/span><\/td>\n<td><\/td>\n<td><\/td>\n<td><span class=\"math-tex\">{tex}\\sum {f_i}{u_i} = k &#8211; 8{\/tex}<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><span class=\"math-tex\">{tex}Mean = \\bar x = a + h\\left( {\\frac{{\\sum {f_i}{u_i}}}{{\\sum {f_i}}}} \\right){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}18 = 18 + 2\\left( {\\frac{{k &#8211; 8}}{{40 + k}}} \\right){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow k = 8{\/tex}<\/span><br \/>\nOR<\/p>\n<table class=\"mobile\" border=\"1\" cellspacing=\"0\" cellpadding=\"0\">\n<tbody>\n<tr>\n<td>Less than<\/td>\n<td>Number of Students<\/td>\n<\/tr>\n<tr>\n<td>10<\/td>\n<td>4<\/td>\n<\/tr>\n<tr>\n<td>20<\/td>\n<td>9<\/td>\n<\/tr>\n<tr>\n<td>30<\/td>\n<td>22<\/td>\n<\/tr>\n<tr>\n<td>40<\/td>\n<td>42<\/td>\n<\/tr>\n<tr>\n<td>50<\/td>\n<td>56<\/td>\n<\/tr>\n<tr>\n<td>60<\/td>\n<td>64<\/td>\n<\/tr>\n<tr>\n<td>70<\/td>\n<td>68<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><img decoding=\"async\" id=\"Picture 15\" class=\"\" style=\"height: 239px; width: 240px;\" src=\"http:\/\/media-mycbseguide.s3.amazonaws.com\/images\/cbse\/10\/mathematics\/sp18\/image015.png\" \/><br \/>\nMedian distance is value of x that corresponds to Cumulative frequency <span class=\"math-tex\">{tex}\\frac{N}{2} = \\frac{{68}}{2} = 34{\/tex}<\/span><br \/>\nTherefore, Median distance = 36 m<\/p>\n<p>This is only initial part of the whole sample paper. Download Complete set of <strong>sample paper for class 10 Maths<\/strong><\/p>\n<h2><span class=\"ez-toc-section\" id=\"Sample_paper_for_class_10_Maths\"><\/span>Sample paper for class 10 Maths<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>It is sample paper for class 10 Maths. However, myCBSEguide provides the best sample papers for all the subjects. There are number of sample papers which you can download from myCBSEguide website.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"Sample_paper_for_class_10_all_subjects\"><\/span><strong>Sample paper for class 10 all subjects<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<ul>\n<li><a href=\"https:\/\/mycbseguide.com\/downloads\/cbse-class-10-science\/1176\/cbse-sample-papers\/2\/\"><strong>Science<\/strong><\/a><\/li>\n<li><strong><a href=\"https:\/\/mycbseguide.com\/downloads\/cbse-class-10-mathematics\/1202\/cbse-sample-papers\/2\/\">Mathematics<\/a><\/strong><\/li>\n<li><strong><a href=\"https:\/\/mycbseguide.com\/downloads\/cbse-class-10-social-science\/1896\/cbse-sample-papers\/2\/\">Social Science<\/a><\/strong><\/li>\n<li><strong><a href=\"https:\/\/mycbseguide.com\/downloads\/cbse-class-10-english-communicative\/1901\/cbse-sample-papers\/2\/\">English Communicative<\/a><\/strong><\/li>\n<li><strong><a href=\"https:\/\/mycbseguide.com\/downloads\/cbse-class-10-english-language-and-literature\/1903\/cbse-sample-papers\/2\/\">English Language and Literature<\/a><\/strong><\/li>\n<li><strong><a href=\"https:\/\/mycbseguide.com\/downloads\/cbse-class-10-hindi-course-a\/1905\/sample-papers\/23\/\">Hindi A<\/a><\/strong><\/li>\n<li><strong><a href=\"https:\/\/mycbseguide.com\/downloads\/cbse-class-10-hindi-course-b\/1907\/sample-papers\/23\/\">Hindi B<\/a><\/strong><\/li>\n<\/ul>\n<p>To download complete sample paper for class 10 Maths , Science, Social Science, Hindi and English; do check myCBSEguide app or website. myCBSEguide provides sample papers with solution, test papers for chapter-wise practice, NCERT solutions, NCERT Exemplar solutions, quick revision notes for ready reference, CBSE guess papers and CBSE important question papers. Sample Paper all are made available through <a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide&amp;referrer=utm_source%3Dmycbse_bottom%26utm_medium%3Dtext%26utm_campaign%3Dmycbseads\"><strong>the best app for CBSE students<\/strong><\/a> and myCBSEguide website.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Download CBSE sample paper for class 10 Maths for board exams available for download in myCBSEguide mobile app. The best app for CBSE students now provides class 10 maths sample paper includes all questions from Mathematics &#8211; Textbook for class X &#8211; NCERT Publication, Guidelines for Mathematics Laboratory in Schools, class X &#8211; CBSE Publication, &#8230; <a title=\"Sample paper for class 10 Maths 2018-19\" class=\"read-more\" href=\"https:\/\/mycbseguide.com\/blog\/sample-paper-for-class-10-maths\/\" aria-label=\"More on Sample paper for class 10 Maths 2018-19\">Read more<\/a><\/p>\n","protected":false},"author":2,"featured_media":8519,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1339,46,1998,2004,7,6],"tags":[448,1042,12,439,440,312,441],"class_list":["post-8723","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-cbse-sample-papers","category-cbse-class-10","category-class-10-sample-papers","category-maths-sample-papers-class-10-sample-papers","category-site-updates","category-tips-and-tricks","tag-cbse-class-10","tag-cbse-class-10-mathematics","tag-cbse-sample-papers","tag-cbse-sample-papers-2017-18","tag-cbse-sample-papers-2018","tag-cbse-solved-sample-papers","tag-class-10-maths-sample-papers"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.0 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Sample paper for class 10 Maths 2018-19 | myCBSEguide<\/title>\n<meta name=\"description\" content=\"Sample paper for class 10 Maths for board exam 2017-2018 as per the new CBSE board exam format. 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