{"id":2361,"date":"2026-11-13T16:34:55","date_gmt":"2026-11-13T11:04:55","guid":{"rendered":"https:\/\/xamcontent.com\/cisce\/?p=2361"},"modified":"2026-11-13T16:34:56","modified_gmt":"2026-11-13T11:04:56","slug":"icse-class-10-mathematics-2020-question-paper-with-solutions","status":"publish","type":"post","link":"https:\/\/xamcontent.com\/cisce\/icse-class-10-mathematics-2020-question-paper-with-solutions\/","title":{"rendered":"ICSE Class 10 Mathematics 2020 Question Paper with Solutions"},"content":{"rendered":"<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_68_1 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title \" >Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"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:\/\/xamcontent.com\/cisce\/icse-class-10-mathematics-2020-question-paper-with-solutions\/#ICSE_Mathematics_2020_%E2%80%94_Section_A\" title=\"ICSE Mathematics 2020 \u2014 Section A\">ICSE Mathematics 2020 \u2014 Section A<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/xamcontent.com\/cisce\/icse-class-10-mathematics-2020-question-paper-with-solutions\/#ICSE_Mathematics_2020_%E2%80%94_Section_B\" title=\"ICSE Mathematics 2020 \u2014 Section B\">ICSE Mathematics 2020 \u2014 Section B<\/a><\/li><\/ul><\/nav><\/div>\n<span class=\"rt-reading-time\" style=\"display: block;\"><span class=\"rt-label rt-prefix\">Reading Time: <\/span> <span class=\"rt-time\">35<\/span> <span class=\"rt-label rt-postfix\">minutes<\/span><\/span>\n<div class=\"xc-nd xc-icse-maths-pyq\">\n<div class=\"xc-intro-box\">\n<div class=\"xc-intro-label\">ICSE Class 10 Mathematics PYQ<\/div>\n<p class=\"xc-intro-text\">Practice the complete <strong>ICSE Class 10 Mathematics 2020 Question Paper with Solutions<\/strong>, including algebra, geometry, trigonometry, statistics, graphs, constructions and commercial mathematics. Use this solved paper to revise important concepts, practise board-style questions and strengthen your exam preparation.<\/p>\n<\/div>\n<div class=\"xc-info-grid\">\n<div class=\"xc-info-card\"><div class=\"xc-info-card-label\">Year<\/div><div class=\"xc-info-card-value\">2020<\/div><\/div>\n<div class=\"xc-info-card\"><div class=\"xc-info-card-label\">Maximum Marks<\/div><div class=\"xc-info-card-value\">80<\/div><\/div>\n<div class=\"xc-info-card\"><div class=\"xc-info-card-label\">Time Allowed<\/div><div class=\"xc-info-card-value\">Two hours and a half<\/div><\/div>\n<\/div>\n<div class=\"xc-set\">\n<div class=\"xc-set-header\"><div class=\"xc-set-icon\">\ud83d\udcc4<\/div><div><div class=\"xc-set-title\">Paper Instructions<\/div><div class=\"xc-set-sub\">ICSE Class 10 Mathematics 2020<\/div><\/div><\/div>\n<div class=\"xc-qtext\"><p>Attempt all questions from Section A and any four questions from Section B.<\/p><p>All working, including rough work, must be clearly shown and must be done on the same sheet as the rest of the answer.<\/p><p>Omission of essential working will result in loss of marks.<\/p><p>The intended marks for questions or parts of question are given in brackets [ ].<\/p><p>Mathematical tables are provided.<\/p><\/div>\n<\/div>\n<div class=\"xc-section-heading\">\n<div class=\"xc-section-heading-icon\">\u2713<\/div>\n<div class=\"xc-section-heading-body\"><div class=\"xc-section-heading-label\">Section A<\/div><h2 class=\"xc-section-heading-title\"><span class=\"ez-toc-section\" id=\"ICSE_Mathematics_2020_%E2%80%94_Section_A\"><\/span>ICSE Mathematics 2020 \u2014 Section A<span class=\"ez-toc-section-end\"><\/span><\/h2><div class=\"xc-section-heading-sub\">40 marks \u2022 Attempt all questions from this Section.<\/div><\/div>\n<div class=\"xc-section-heading-stat\"><div class=\"xc-section-heading-stat-num\">40<\/div><div class=\"xc-section-heading-stat-label\">Marks<\/div><\/div>\n<\/div>\n<div class=\"xc-set\">\n<div class=\"xc-set-header\"><div class=\"xc-set-icon\">1<\/div><div><div class=\"xc-set-title\">Question 1<\/div><div class=\"xc-set-sub\">ICSE Class 10 Mathematics 2020<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2020-q1-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>Solve the following Quadratic Equation :<\/p>\n<div class=\"mathjax-overflow\">\\[\nx^{2}-7 x+3=0\n\\]<\/div>\n<p>Give your answer correct to two decimal places. <strong>[3]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><div class=\"mathjax-overflow\">\\[\nx^{2}-7 x+3=0\n\\]<\/div>\n<p>Compare the equation by \\(a x^{2}+b x+c=0\\) then, \\(a=1, b=-7\\) and \\(c=3\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\nx &amp; =\\frac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a} \\\\\n&amp; =\\frac{7 \\pm \\sqrt{49-4 \\times 1 \\times 3}}{2 \\times 1} \\\\\n&amp; =\\frac{7 \\pm \\sqrt{37}}{2} \\\\\n&amp; =\\frac{7 \\pm \\sqrt{37}}{2}=\\frac{7 \\pm 6.08}{2}\n\\end{aligned}\n\\]<\/div>\n<p>\u2234<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\nx &amp; =\\frac{7+6.08}{2}=\\frac{13.08}{2} \\\\\n&amp; =6.54\n\\end{aligned}\n\\]<\/div>\n<p>and<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\nx &amp; =\\frac{7-6.08}{2}=\\frac{0.92}{2} \\\\\n&amp; =0.46\n\\end{aligned}\n\\]<\/div>\n<p>Hence, \\(x=6.54\\) (Approx.) and \\(x=0.46\\) (Approx).<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2020-q1-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>\\(\\quad\\) Given \\(\\mathbf{A}=\\left[\\begin{array}{cc}\\boldsymbol{x} &amp; \\mathbf{3} \\\\ \\boldsymbol{y} &amp; \\mathbf{3}\\end{array}\\right]\\) If \\(\\boldsymbol{A}^{\\mathbf{2}}=\\mathbf{3} \\boldsymbol{I}\\), where \\(\\boldsymbol{I}\\) is the identity matrix of order 2, find \\(x\\) and \\(y\\). <strong>[3]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\mathrm{A} &amp; =\\left[\\begin{array}{ll}\nx &amp; 3 \\\\\ny &amp; 3\n\\end{array}\\right] \\\\\nA^{2} &amp; =\\mathrm{A} \\times \\mathrm{A}\n\\end{aligned}\n\\]<\/div>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; =\\left[\\begin{array}{ll}\nx &amp; 3 \\\\\ny &amp; 3\n\\end{array}\\right]\\left[\\begin{array}{ll}\nx &amp; 3 \\\\\ny &amp; 3\n\\end{array}\\right] \\\\\n&amp; =\\left[\\begin{array}{ll}\nx^{2}+3 y &amp; 3 x+9 \\\\\nx y+3 y &amp; 3 y+9\n\\end{array}\\right] \\\\\nA^{2} &amp; =3 I \\text { (Given) } \\\\\n{\\left[\\begin{array}{ll}\nx^{2}+3 y &amp; 3 x+9 \\\\\nx y+3 y &amp; 3 y+9\n\\end{array}\\right]=} &amp; 3\\left[\\begin{array}{ll}\n1 &amp; 0 \\\\\n0 &amp; 1\n\\end{array}\\right]=\\left[\\begin{array}{ll}\n3 &amp; 0 \\\\\n0 &amp; 3\n\\end{array}\\right]\n\\end{aligned}\n\\]<\/div>\n<p>Comparing the elements of matrices<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; 3 x+9=0 \\Rightarrow x=-3 \\\\\n&amp; 3 y+9=3 \\Rightarrow y=-2\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2020-q1-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>Using ruler and compass construct a triangle \\(\\boldsymbol{A} \\boldsymbol{B} \\boldsymbol{C}\\) where \\(\\boldsymbol{A} \\boldsymbol{B}=\\mathbf{3 ~ c m}, \\boldsymbol{B} \\boldsymbol{C}=\\mathbf{4 ~ c m}\\) and \\(\\angle \\boldsymbol{A} \\boldsymbol{B} \\boldsymbol{C}\\) \\(=90^{\\circ}\\). Hence construct a circle circumscribing triangle \\(A B C\\). Measure and write down the radius of the circle. <strong>[4]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/c3f56170-7ce6-4654-8ea6-0da7169b9fde-01_357_341_1590_1165.jpg\" alt=\"ICSE Class 10 Mathematics 2020 Question 1(c) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>Steps of construction<\/p>\n<p>(i) Draw \\(B C=4 \\mathrm{~cm}\\)<\/p>\n<p>(ii) Make \\(\\angle P B C=90^{\\circ}\\)<\/p>\n<p>(iii) Taking centre \\(B\\) and radius 3 cm draw an arc which intersect BP at the point A.<\/p>\n<p>(iv) Join \\(A C, \\triangle A B C\\) is required triangle. \\(\\angle B=90^{\\circ}\\) \\(\\therefore A C\\) is the diameter of circle.<\/p>\n<p>(v) Draw perpendicular bisector of \\(A C\\).<\/p>\n<p>(vi) Taking centre \\(O\\) and radius equal to \\(B O\\) draw a circle which passes through vertex \\(A, B\\) and \\(C\\). Radius of circumscribe circle \\(=2.5 \\mathrm{~cm}\\).<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"xc-set\">\n<div class=\"xc-set-header\"><div class=\"xc-set-icon\">2<\/div><div><div class=\"xc-set-title\">Question 2<\/div><div class=\"xc-set-sub\">ICSE Class 10 Mathematics 2020<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2020-q2-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>Use factor theorem to factorise \\(\\mathbf{6} \\boldsymbol{x}^{\\mathbf{3}}+\\mathbf{1 7} \\boldsymbol{x}^{\\mathbf{2}}+\\mathbf{4} \\boldsymbol{x}-\\) 12 completely. <strong>[3]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\np(x) &amp; =6 x^{3}+17 x^{2}+4 x-12 \\\\\np(-2) &amp; =6(-2)^{3}+17(-2)^{2}+4(-2)-12 \\\\\n&amp; =-48+68-8-12 \\\\\n&amp; =68-68=0\n\\end{aligned}\n\\]<\/div>\n<p>\\(\\therefore(x+2)\\) is a factor of given polynomial \\(p(x)\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; x + 2 \\longdiv { 6 x ^ { 3 } + 1 7 x ^ { 2 } + 4 x &#8211; 1 2 } ( 6 x ^ { 2 } + 5 x &#8211; 6 \\\\\n&amp; \\begin{array}{c}\n6 x^{3}+12 x^{2} \\\\\n\\frac{(-)}{(-)} \\\\\n5 x^{2}+4 x-12 \\\\\n5 x^{2}+10 x \\\\\n\\frac{(-) \\quad(-)}{-6 x-12} \\\\\n-6 x-12 \\\\\n\\frac{(+)(+)}{0}\n\\end{array} \\\\\n&amp; \\therefore 6 x^{3}+17 x^{2}+4 x-12 \\\\\n&amp; =(x+2)\\left(6 x^{2}+5 x-6\\right) \\\\\n&amp; =(x+2)\\left\\{6 x^{2}+9 x-4 x-6\\right\\} \\\\\n&amp; =(x+2)\\{3 x(2 x+3)-2(2 x+3)\\} \\\\\n&amp; =(x+2)(2 x+3)(3 x-2)\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2020-q2-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>Solve the following inequation and represent the solution set on the number line. <strong>[3]<\/strong><\/p>\n<div class=\"mathjax-overflow\">\\[\n\\frac{3 x}{5}+2&lt;x+4 \\leq \\frac{x}{2}+5, x \\in R\n\\]<\/div><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><div class=\"mathjax-overflow\">\\[\n\\frac{3 x}{5}+2&lt;x+4 \\leq \\frac{x}{2}+5, \\quad x \\in R\n\\]<\/div>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; \\frac{3 x}{5}+2&lt;x+14 \\\\\n&amp; \\frac{3 x}{5}-x&lt;14-2 \\\\\n&amp; \\frac{3 x-5 x}{5}&lt;12 \\\\\n&amp; \\text { Multiply by } 5 \\\\\n&amp; -2 x&lt;12 \\times 5 \\\\\n&amp; x&gt;\\frac{60}{-2} \\\\\n&amp; x&gt;-30 \\\\\n&amp; \\therefore x&gt;-30&lt;x \\leq 2\n\\end{aligned}\n\\]<\/div>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/c3f56170-7ce6-4654-8ea6-0da7169b9fde-02_67_664_2513_320.jpg\" alt=\"ICSE Class 10 Mathematics 2020 Question 2(b) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2020-q2-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>Draw a Histogram for the given data, using a graph paper :<\/p>\n<div class=\"math-responsive\"><table><thead><tr><th>Weekly Wages (in \u20b9)<\/th><th>No. of People<\/th><\/tr><\/thead><tbody><tr><td>3000-4000<\/td><td>4<\/td><\/tr><tr><td>4000-5000<\/td><td>9<\/td><\/tr><tr><td>5000-6000<\/td><td>18<\/td><\/tr><tr><td>6000-7000<\/td><td>6<\/td><\/tr><tr><td>7000-8000<\/td><td>7<\/td><\/tr><tr><td>8000-9000<\/td><td>2<\/td><\/tr><tr><td>9000-10000<\/td><td>4<\/td><\/tr><\/tbody><\/table><\/div>\n<p>Estimate the mode from the graph. <strong>[4]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/c3f56170-7ce6-4654-8ea6-0da7169b9fde-02_832_785_301_1019.jpg\" alt=\"ICSE Class 10 Mathematics 2020 Question 2(c) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>= Mode -5400 (approx)<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"xc-set\">\n<div class=\"xc-set-header\"><div class=\"xc-set-icon\">3<\/div><div><div class=\"xc-set-title\">Question 3<\/div><div class=\"xc-set-sub\">ICSE Class 10 Mathematics 2020<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2020-q3-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>In the figure given below, O is the centre of the circle and \\(A B\\) is a diameter. If \\(\\boldsymbol{A} \\boldsymbol{C}=\\boldsymbol{B} \\boldsymbol{D}\\) and \\(\\angle \\boldsymbol{A} \\boldsymbol{O} \\boldsymbol{C}=\\mathbf{7 2} \\boldsymbol{\u00b0}^{\\boldsymbol{\\circ}}\\). Find:<\/p>\n<p>(i) \\(\\angle \\mathrm{ABC}\\)<\/p>\n<p>(ii) \u2220BAD<\/p>\n<p>(iii) \u2220ABD <strong>[3]<\/strong><\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/c3f56170-7ce6-4654-8ea6-0da7169b9fde-02_307_313_1405_1259.jpg\" alt=\"ICSE Class 10 Mathematics 2020 Question 3(a) figure\" loading=\"lazy\" decoding=\"async\"><\/div><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>\\(A B\\) is the diameter of circle (i)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{array}{ll}\n\\therefore \\quad \\angle A D B=90^{\\circ} &amp; \\text { (semicircle angle) } \\\\\n\\angle A B C=\\frac{1}{2} \\angle A O C &amp; \\text { (same arc angles at }\n\\end{array}\n\\]<\/div>\n<p>(same arc angles at circumference and centres)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\therefore \\quad \\angle A B C=\\frac{1}{2} \\times 72^{\\circ}=36^{\\circ}\n\\]<\/div>\n<p>(ii)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{array}{rlrl}\nA C &amp; =B D &amp; \\\\\n\\therefore &amp; &amp; \\angle A B C &amp; =\\angle B A D \\text { (equal arcs make equal } \\\\\n&amp; &amp; &amp; \\text { angle) } \\\\\n\\therefore &amp; \\angle B A D &amp; =36^{\\circ} &amp;\n\\end{array}\n\\]<\/div>\n<p>(iii) In \\(\\triangle A B D\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\angle A B D+\\angle B A D+\\angle A D B=180^{\\circ}\n\\]<\/div>\n<p>(sum of all angles of \\(\\Delta\\) )<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\angle A B D+36^{\\circ} &amp; +90^{\\circ}=180^{\\circ} \\\\\n\\angle A B D &amp; =180^{\\circ}-126^{\\circ} \\\\\n\\angle A B D &amp; =54^{\\circ}\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2020-q3-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>Prove that : <strong>[3]<\/strong><\/p>\n<div class=\"mathjax-overflow\">\\[\n\\frac{\\sin A}{1+\\cot A}-\\frac{\\cos A}{1+\\tan A}=\\sin A-\\cos A\n\\]<\/div><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\text { L.H.S. }= &amp; \\frac{\\sin A}{1+\\cot A}-\\frac{\\cos A}{1+\\tan A} \\\\\n= &amp; \\frac{\\sin A}{1+\\frac{\\cos A}{\\sin A}}-\\frac{\\cos A}{1+\\frac{\\sin A}{\\cos A}} \\\\\n= &amp; \\frac{\\sin ^{2} A}{\\sin A+\\cos A}-\\frac{\\cos ^{2} A}{\\cos A+\\sin A} \\\\\n= &amp; \\frac{\\sin ^{2} A-\\cos ^{2} A}{\\sin A+\\cos A} \\\\\n= &amp; \\frac{\\left(a^{2}-b^{2}=(a+b)(a-b)\\right]}{(\\sin A+\\cos A)(\\sin A-\\cos A)} \\\\\n= &amp; \\sin \\mathrm{A}-\\cos \\mathrm{A} \\\\\n= &amp; \\text { Hence proved. }\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2020-q3-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>In what ratio is the line joining \\(\\boldsymbol{P}(\\mathbf{5}, \\mathbf{3})\\) and \\(Q(-5,3)\\) divided by the \\(y\\)-axis ? Also find the coordinates of the point of intersection. <strong>[4]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Let the line joining the \\((5,3)\\) and (-5, 3) divided by the \\(y\\)-axis in the ratio \\(k: 1\\). (-5, 3)<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/c3f56170-7ce6-4654-8ea6-0da7169b9fde-03_395_453_1114_399.jpg\" alt=\"ICSE Class 10 Mathematics 2020 Question 3(c) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>Let the coordinate of the point of intersection \\(R\\) be \\((0, y)\\).<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\nR_{x} &amp; =\\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}} \\\\\n0 &amp; =\\frac{1 \\times 5+k(-5)}{1+k} \\\\\n5-5 k=0 &amp; \\Rightarrow k=1 \\\\\nR_{y} &amp; =\\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}} \\\\\ny &amp; =\\frac{1 \\times 3+k \\times 3}{1+k} \\\\\ny &amp; =\\frac{3+3}{1+1}=3 \\quad[\\mathrm{Put}, k=1] \\\\\n\\text { Ratio } &amp; =k: 1=1: 1\n\\end{aligned}\n\\]<\/div>\n<p>Point of intersection of the line (0, 3).<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"xc-set\">\n<div class=\"xc-set-header\"><div class=\"xc-set-icon\">4<\/div><div><div class=\"xc-set-title\">Question 4<\/div><div class=\"xc-set-sub\">ICSE Class 10 Mathematics 2020<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2020-q4-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>A solid spherical ball of radius 6 cm is melted and recast into 64 identical spherical marbles. Find the radius of each marble. <strong>[3]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Volume of solid spherical ball \\(=\\frac{4}{3} \\pi r^{3}\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; =\\frac{4}{3} \\pi(6)^{3} \\\\\n&amp; =\\frac{4}{3} \\pi \\times 216 \\\\\n&amp; =\\frac{216 \\times 4}{3} \\pi \\mathrm{~cm}^{3}\n\\end{aligned}\n\\]<\/div>\n<p>Let the radius of spherical marble be \\(R\\) cm. Then 64 \u00d7 volume of spherical marbles = volume of spherical ball<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n64 \\times \\frac{4}{3} \\pi R^{3} &amp; =\\frac{4 \\times 216}{3} \\pi \\mathrm{~cm}^{3} \\\\\nR^{3} &amp; =\\frac{216}{64} \\\\\nR &amp; =\\sqrt[3]{\\frac{216}{64}}=\\frac{6}{4}=\\frac{3}{2} \\mathrm{~cm} \\\\\n&amp; =1.5 \\mathrm{~cm}\n\\end{aligned}\n\\]<\/div>\n<p>\u2234 Radius of spherical marble is 1.5 cm.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2020-q4-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>Each of the letters of the word &#8216;AUTHORIZES&#8217; is written on identical circular discs and put in a bag. They are well shuffled. If a disc is drawn at random from the bag, what is the probability that the letter is :<\/p>\n<p>(i) a vowel ?<\/p>\n<p>(ii) one of the first 9 letters of the English alphabet which appears in the given word ?<\/p>\n<p>(iii) one of the last 9 letters of the English alphabet which appears in the given word ? <strong>[3]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>&#8216;AUTHORIZES&#8217; Total number of disc \\(n(\\mathrm{~S})=10\\)<\/p>\n<p>(i) Number of discs a letter of vowel = 5<\/p>\n<p>(A, U, O, I, E) \\(\\therefore n(\\mathrm{E})=5\\) Probability of a disc is vowel<\/p>\n<div class=\"mathjax-overflow\">\\[\n=\\frac{n(\\mathrm{E})}{n(\\mathrm{~S})}=\\frac{5}{10}=\\frac{1}{2}\n\\]<\/div>\n<p>(ii) One of the first 9 letter in english alphabet \\(=4(A, E, H, \\mathrm{I})\\)<\/p>\n<p>\u2234<\/p>\n<div class=\"mathjax-overflow\">\\[\nn(E)=4\n\\]<\/div>\n<p>Probability of a disc is drawn is written one of first 9 letters of english alphabet<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; =\\frac{n(\\mathrm{E})}{n(\\mathrm{~S})}=\\frac{4}{10} \\\\\n&amp; =\\frac{2}{5}\n\\end{aligned}\n\\]<\/div>\n<p>(iii) One of the last \\(9(R, S, T, U, Z)\\) letters of english alphabet<\/p>\n<div class=\"mathjax-overflow\">\\[\nn(E)=5\n\\]<\/div>\n<p>Probability of a disc is drawn is written one of the last 9 letters of english alphabet<\/p>\n<div class=\"mathjax-overflow\">\\[\n=\\frac{n(\\mathrm{E})}{n(\\mathrm{~S})}=\\frac{5}{10}\n\\]<\/div>\n<div class=\"mathjax-overflow\">\\[\n=\\frac{1}{2}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2020-q4-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>Mr. Bedi visits the market and buys the following articles : Medicines costing \u20b9 950, GST @ 5% A Pair of shoes costing \u20b9 3000, GST @ 18% A Laptop bag costing \u20b9 1000 with a discount of 30% GST @ 18%<\/p>\n<p>(i) Calculate the total amount of GST paid.<\/p>\n<p>(ii) The total bill amount including GST paid by Mr. Bedi. <strong>[4]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Medicine costing \\(=\\text{\u20b9} 950\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\text { GST @ } 5 \\%=\\frac{5}{100} \\times 950=\\text{\u20b9} 47.50\n\\]<\/div>\n<p>A pair of shoes costing \\(=\\text{\u20b9} 3000\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\text { GST @ } 18 \\%=\\frac{18}{100} \\times 3000=\\text{\u20b9} 540\n\\]<\/div>\n<p>Laptop bag costing \u20b9 1000 with a discount of 30%<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\therefore \\quad \\text { Net } \\text { cost } &amp; =1000-\\frac{30}{100} \\times 1000 \\\\\n&amp; =\\text{\u20b9} 700\n\\end{aligned}\n\\]<\/div>\n<p>GST @ \\(18 \\%=\\frac{18}{100} \\times 700=\\text{\u20b9} 126\\)<\/p>\n<p>(i) Total amount of GST = \u20b9 47.50 + 540 + 126<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\text { = \\text{\u20b9} } 713.50\n\\]<\/div>\n<p>(ii) Total bill amount including GST<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; =\\text{\u20b9} 950+3000+700+713.50 \\\\\n&amp; =\\text{\u20b9} 4650+713.50 \\\\\n&amp; =\\text{\u20b9} 5363.50\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<\/div>\n<div class=\"xc-section-heading\">\n<div class=\"xc-section-heading-icon\">\u270d\ufe0f<\/div>\n<div class=\"xc-section-heading-body\"><div class=\"xc-section-heading-label\">Section B<\/div><h2 class=\"xc-section-heading-title\"><span class=\"ez-toc-section\" id=\"ICSE_Mathematics_2020_%E2%80%94_Section_B\"><\/span>ICSE Mathematics 2020 \u2014 Section B<span class=\"ez-toc-section-end\"><\/span><\/h2><div class=\"xc-section-heading-sub\">40 marks \u2022 Attempt any four questions from this Section.<\/div><\/div>\n<div class=\"xc-section-heading-stat\"><div class=\"xc-section-heading-stat-num\">40<\/div><div class=\"xc-section-heading-stat-label\">Marks to Attempt<\/div><\/div>\n<\/div>\n<div class=\"xc-set\">\n<div class=\"xc-set-header\"><div class=\"xc-set-icon\">5<\/div><div><div class=\"xc-set-title\">Question 5<\/div><div class=\"xc-set-sub\">ICSE Class 10 Mathematics 2020<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2020-q5-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>A company with 500 shares of nominal value \u20b9 120 declares an annual dividend of \\(15 \\%\\). Calculate :<\/p>\n<p>(i) the total amount of dividend paid by the company.<\/p>\n<p>(ii) annual income of Mr. Sharma who holds 80 shares of the company.<\/p>\n<p>If the return percent of Mr. Sharma from his shares is 10%. Find the market value of each share. <strong>[3]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Number of shares \\(=500\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\text { Nominal value } &amp; =\\text{\u20b9} 120 \\\\\n\\text { Annual dividend } &amp; =15 \\%\n\\end{aligned}\n\\]<\/div>\n<p>(i) Total amount of dividend paid by the company = Rate of dividend \u00d7 Nominal value of share \u00d7 Number of share<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; =\\frac{15}{100} \\times 120 \\times 500 \\\\\n&amp; =15 \\times 600=\\text{\u20b9} 9000\n\\end{aligned}\n\\]<\/div>\n<p>\u2234 Total amount paid by the company = \u20b9 9000<\/p>\n<p>(ii) Mr sharma&#8217;s Annual Income = Rate of dividend \u00d7 Nominal value of share \u00d7 Number of share<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; =\\frac{15}{100} \\times 120 \\times 80 \\\\\n&amp; =15 \\times 96=\\text{\u20b9} 1440\n\\end{aligned}\n\\]<\/div>\n<p>\u2234 Mr sharma&#8217;s Annual Income \\(=\\text{\u20b9} 1440\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\text { Return of Investment } &amp; =\\frac{\\text { Income }}{\\text { Investment }} \\times 100 \\\\\n10 &amp; =\\frac{1440}{\\text { Investment }} \\times 100\n\\end{aligned}\n\\]<\/div>\n<p>\\(\\therefore \\quad\\) Investment \\(=\\frac{144000}{10}=\\text{\u20b9} 14400\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\text { Market value of share } &amp; =\\frac{\\text { Investment }}{\\text { No. of share }} \\\\\n&amp; =\\frac{14400}{80}\n\\end{aligned}\n\\]<\/div>\n<p>\u2234 Market value of share \\(=\\text{\u20b9} 180\\)<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2020-q5-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>The mean of the following data is 16. Calculate the value of \\(\\boldsymbol{f}\\). <strong>[3]<\/strong><\/p>\n<div class=\"math-responsive\"><table><thead><tr><th>Marks<\/th><th>5<\/th><th>10<\/th><th>15<\/th><th>20<\/th><th>25<\/th><\/tr><\/thead><tbody><tr><td>No. of Students<\/td><td>3<\/td><td>7<\/td><td>\\(f\\)<\/td><td>9<\/td><td>6<\/td><\/tr><\/tbody><\/table><\/div><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><div class=\"math-responsive\"><table><thead><tr><th>Marks<\/th><th>No. of Students<\/th><th><\/th><\/tr><\/thead><tbody><tr><td>\\((x)\\)<\/td><td>(f)<\/td><td>\\(f \\times x\\)<\/td><\/tr><tr><td>5<\/td><td>3<\/td><td>15<\/td><\/tr><tr><td>10<\/td><td>7<\/td><td>70<\/td><\/tr><tr><td>15<\/td><td>\\(f\\)<\/td><td>\\(15 f\\)<\/td><\/tr><tr><td>20<\/td><td>9<\/td><td>180<\/td><\/tr><tr><td>25<\/td><td>6<\/td><td>150<\/td><\/tr><tr><td><\/td><td>\\(25+f\\)<\/td><td>\\(415+15 f\\)<\/td><\/tr><\/tbody><\/table><\/div>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\text { Mean } &amp; =16 \\\\\n\\frac{\\Sigma f x}{\\Sigma f} &amp; =16 \\\\\n\\frac{415+15 f}{25+f} &amp; =16 \\\\\n415+15 f &amp; =400+16 f \\\\\n16 f-15 f &amp; =415-400\n\\end{aligned}\n\\]<\/div>\n<p>\u2234<\/p>\n<div class=\"mathjax-overflow\">\\[\nf=15\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2020-q5-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>The \\(\\mathbf{4}^{\\text {th }}, \\mathbf{6}^{\\text {th }}\\) and the last term of a geometric progression are 10, 40 and 640 respectively. If the common ratio is positive, find the first term, common ratio and the number of terms of the series. <strong>[4]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Let the first term of G.P. be \\(a\\), common ratio be \\(r\\) and number of terms be \\(n\\).<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\na_{4} &amp; =10, a_{6}=40 \\text { and } a_{n}=640 \\text { (given) } \\\\\na_{4} &amp; =10 \\\\\na r^{3} &amp; =10 \\qquad \\text{&#8230;(i)}\n\\end{aligned}\n\\]<\/div>\n<p>Again,<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\na_{6} &amp; =40 \\\\\na r^{5} &amp; =40\n\\end{aligned}\n\\]<\/div>\n<p>From eqs. (i) \\&amp; (ii)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\frac{a r^{5}}{a r^{3}} &amp; =\\frac{40}{10} \\\\\n\\Rightarrow r^{2} &amp; =4 \\\\\n\\therefore \\quad r &amp; = \\pm 2 \\\\\n\\therefore \\quad r &amp; =2\n\\end{aligned}\n\\]<\/div>\n<p>( \\(r&gt;0\\) ) From eq. (i)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\na r^{3} &amp; =10 \\\\\na(2)^{3} &amp; =10 \\\\\na &amp; =\\frac{10}{8}=\\frac{5}{4}\n\\end{aligned}\n\\]<\/div>\n<p>Again,<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\na_{n} &amp; =640 \\\\\na r^{n-1} &amp; =640 \\\\\n\\frac{5}{4}(2)^{n-1} &amp; =640\n\\end{aligned}\n\\]<\/div>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; 2^{n-1}=\\frac{640 \\times 4}{5} \\\\\n&amp; 2^{n-1}=128 \\times 4=2^{9}\n\\end{aligned}\n\\]<\/div>\n<p>Comparing the power<\/p>\n<div class=\"mathjax-overflow\">\\[\nn-1=9 \\Rightarrow n=10\n\\]<\/div>\n<p>\u2234 First term of G.P. \\(=\\frac{5}{4}\\), common ratio \\(=2\\) and number of terms \\(=10\\)<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"xc-set\">\n<div class=\"xc-set-header\"><div class=\"xc-set-icon\">6<\/div><div><div class=\"xc-set-title\">Question 6<\/div><div class=\"xc-set-sub\">ICSE Class 10 Mathematics 2020<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2020-q6-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>If \\(\\boldsymbol{A}=\\left[\\begin{array}{ll}\\mathbf{3} &amp; \\mathbf{0} \\\\ \\mathbf{5} &amp; \\mathbf{1}\\end{array}\\right]\\) and \\(\\boldsymbol{B}=\\left[\\begin{array}{cc}-\\mathbf{4} &amp; \\mathbf{2} \\\\ \\mathbf{1} &amp; \\mathbf{0}\\end{array}\\right]\\) Find \\(A^{2}-2 A B+B^{2}\\). <strong>[3]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>\\(A=\\left[\\begin{array}{ll}3 &amp; 0 \\\\ 5 &amp; 1\\end{array}\\right]\\) and \\(B=\\left[\\begin{array}{cc}-4 &amp; 2 \\\\ 1 &amp; 0\\end{array}\\right]\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\nA^{2} &amp; =A \\times A=\\left[\\begin{array}{ll}\n3 &amp; 0 \\\\\n5 &amp; 1\n\\end{array}\\right] \\times\\left[\\begin{array}{ll}\n3 &amp; 0 \\\\\n5 &amp; 1\n\\end{array}\\right] \\\\\n&amp; =\\left[\\begin{array}{cc}\n9+0 &amp; 0+0 \\\\\n15+5 &amp; 0+1\n\\end{array}\\right] \\\\\n\\therefore \\quad A^{2} &amp; =\\left[\\begin{array}{cc}\n9 &amp; 0 \\\\\n20 &amp; 1\n\\end{array}\\right] \\\\\nA B &amp; =\\left[\\begin{array}{ll}\n3 &amp; 0 \\\\\n5 &amp; 1\n\\end{array}\\right]\\left[\\begin{array}{cc}\n-4 &amp; 2 \\\\\n1 &amp; 0\n\\end{array}\\right]\n\\end{aligned}\n\\]<\/div>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp;=\\left[\\begin{array}{cc}\n-12+0 &amp; 6+0 \\\\\n-20+1 &amp; 10+0\n\\end{array}\\right] \\\\\n&amp; A B=\\left[\\begin{array}{cc}\n-12 &amp; 6 \\\\\n-19 &amp; 10\n\\end{array}\\right] \\\\\n&amp; \\quad B^{2}=B \\times B=\\left[\\begin{array}{cc}\n-4 &amp; 2 \\\\\n1 &amp; 0\n\\end{array}\\right]\\left[\\begin{array}{cc}\n-4 &amp; 2 \\\\\n1 &amp; 0\n\\end{array}\\right] \\\\\n&amp; \\quad=\\left[\\begin{array}{cc}\n16+2 &amp; -8+0 \\\\\n-4+0 &amp; 2+0\n\\end{array}\\right] \\\\\n&amp; \\therefore B^{2}=\\left[\\begin{array}{cc}\n18 &amp; -8 \\\\\n-4 &amp; 2\n\\end{array}\\right] \\\\\n&amp; \\therefore A^{2}-2 A B+B^{2} \\\\\n&amp; {\\left[\\begin{array}{cc}\n9 &amp; 0 \\\\\n20 &amp; 1\n\\end{array}\\right]-2\\left[\\begin{array}{cc}\n-12 &amp; 6 \\\\\n-19 &amp; 10\n\\end{array}\\right]+\\left[\\begin{array}{cc}\n18 &amp; -8 \\\\\n-4 &amp; 2\n\\end{array}\\right] } \\\\\n&amp; {\\left[\\begin{array}{cc}\n9 &amp; 0 \\\\\n20 &amp; 1\n\\end{array}\\right]+\\left[\\begin{array}{cc}\n24 &amp; -12 \\\\\n38 &amp; -20\n\\end{array}\\right]+\\left[\\begin{array}{cc}\n18 &amp; -8 \\\\\n-4 &amp; 2\n\\end{array}\\right] } \\\\\n&amp; {\\left[\\begin{array}{cc}\n9+24+18 &amp; 0-12-8 \\\\\n20+38-4 &amp; 1-20+2\n\\end{array}\\right]=\\left[\\begin{array}{cc}\n51 &amp; -20 \\\\\n54 &amp; -17\n\\end{array}\\right] }\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2020-q6-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>In the given figure \\(\\boldsymbol{A} \\boldsymbol{B}=\\mathbf{9 ~ c m}, \\boldsymbol{P} \\boldsymbol{A}=\\mathbf{7 . 5 ~ c m}\\) and \\(P C=5 \\mathrm{~cm}\\). Chords \\(\\boldsymbol{A} \\boldsymbol{D}\\) and \\(\\boldsymbol{B} \\boldsymbol{C}\\) intersect at \\(\\boldsymbol{P}\\).<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/c3f56170-7ce6-4654-8ea6-0da7169b9fde-05_327_321_1324_422.jpg\" alt=\"ICSE Class 10 Mathematics 2020 Question 6(b) figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>(i) Prove that \\(\\triangle P A B \\sim \\triangle P C D\\).<\/p>\n<p>(ii) Find the length of \\(C D\\).<\/p>\n<p>(iii) Find area of \\(\\triangle P A B\\) : area of \\(\\triangle P C D\\). <strong>[3]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>(i) \\(\\triangle P A B\\) and \\(\\triangle P C D\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\angle A P B=\\angle C P D\n\\]<\/div>\n<p>(Vertically opposite angle)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\angle A B P=\\angle P D C\n\\]<\/div>\n<p>(same arc \\(\\stackrel{\\triangle}{\\text { AC }}\\) angles)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\therefore \\quad \\triangle P A B \\sim \\triangle P C D\n\\]<\/div>\n<p>(AA similarly test)<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/c3f56170-7ce6-4654-8ea6-0da7169b9fde-05_336_327_1455_1253.jpg\" alt=\"ICSE Class 10 Mathematics 2020 Question 6(b) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>(ii)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\frac{P A}{P C}=\\frac{A B}{C D}=\\frac{P B}{P D}\n\\]<\/div>\n<p>(Property of similar triangle)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\frac{P A}{P C} &amp; =\\frac{A B}{C D} \\\\\n\\frac{7.5}{5} &amp; =\\frac{9}{C D}\n\\end{aligned}\n\\]<\/div>\n<p>\u2234<\/p>\n<div class=\"mathjax-overflow\">\\[\nC D=\\frac{9 \\times 5}{7.5}=\\frac{9 \\times 5 \\times 10}{75}\n\\]<\/div>\n<p>\u2234<\/p>\n<div class=\"mathjax-overflow\">\\[\nC D=6 \\mathrm{~cm}\n\\]<\/div>\n<p>(iii)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\frac{a r \\triangle P A B}{a r \\triangle P C D} &amp; =\\frac{A P^{2}}{C P^{2}}=\\left(\\frac{A P}{C P}\\right)^{2} \\\\\n&amp; =\\left(\\frac{7.5}{5}\\right)^{2} \\\\\n&amp; =\\frac{9}{4}\n\\end{aligned}\n\\]<\/div>\n\\(\\therefore \\operatorname{ar} \\triangle P A B: \\operatorname{ar} \\triangle P C D=9: 4\\)<\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2020-q6-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>From the top of a cliff, the angle of depression of the top and bottom of a tower are observed to be 45\u00b0 and 60\u00b0 respectively. If the height of the tower is 20 m. Find :<\/p>\n<p>(i) the height of the cliff.<\/p>\n<p>(ii) the distance between the cliff and the tower. <strong>[4]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/c3f56170-7ce6-4654-8ea6-0da7169b9fde-06_648_542_265_332.jpg\" alt=\"ICSE Class 10 Mathematics 2020 Question 6(c) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>\\(O P\\) is the cliff and \\(A B\\) is the tower. In \\(\\triangle P B C\\),<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\tan B &amp; =\\frac{C P}{B C} \\\\\n\\tan 45^{\\circ} &amp; =\\frac{C P}{B C} \\\\\n1 &amp; =\\frac{C P}{B C}\n\\end{aligned}\n\\]<\/div>\n<p>\u2234<\/p>\n<div class=\"mathjax-overflow\">\\[\nC P=B C \\qquad \\text{&#8230;(i)}\n\\]<\/div>\n<p>In \\(\\triangle O A P\\),<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\tan A= &amp; \\frac{O P}{O A} \\\\\n&amp; \\left(\\because \\angle Q P A=\\angle P A O=60^{\\circ}\\right) \\\\\n\\tan 60^{\\circ}= &amp; \\frac{O C+C P}{O A} \\\\\n\\sqrt{3}= &amp; \\frac{A B+C P}{O A} \\\\\n\\sqrt{3}= &amp; \\frac{20+C P}{B C} \\quad(O A=B C) \\\\\n\\sqrt{3}= &amp; \\frac{20+C P}{C P} \\quad(B C=C P) \\\\\n\\sqrt{3} C P= &amp; 20+C P \\\\\n(\\sqrt{3}-1) C P= &amp; 20\n\\end{aligned}\n\\]<\/div>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\nC P &amp; =\\frac{20}{\\sqrt{3}-1} \\mathrm{~m} \\\\\nC P &amp; =\\frac{20}{\\sqrt{3}-1} \\times \\frac{\\sqrt{3}+1}{\\sqrt{3}+1} \\\\\nC P &amp; =\\frac{20(\\sqrt{3}+1)}{3-1} \\\\\n&amp; =10(\\sqrt{3}+1) \\mathrm{m} \\\\\nC P &amp; =10(1.732+1) \\mathrm{m} \\\\\nC P &amp; =27.32 \\mathrm{~m}\n\\end{aligned}\n\\]<\/div>\n<p>(i) height of the cliff<\/p>\n<div class=\"mathjax-overflow\">\\[\nO P=O C+C P\n\\]<\/div>\n<div class=\"mathjax-overflow\">\\[\n=20+27.32\n\\]<\/div>\n<p>\u2234 Height of the cliff \\(=47.32 \\mathrm{~m}\\)<\/p>\n<p>(ii) Distance between the cliff and the tower<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; O A=B C=C P \\\\\n&amp; O A=27.32 \\mathrm{~m}\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<\/div>\n<div class=\"xc-set\">\n<div class=\"xc-set-header\"><div class=\"xc-set-icon\">7<\/div><div><div class=\"xc-set-title\">Question 7<\/div><div class=\"xc-set-sub\">ICSE Class 10 Mathematics 2020<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2020-q7-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>Find the value of &#8216; \\(\\boldsymbol{p}\\) &#8216; if the lines, \\(\\mathbf{5} \\boldsymbol{x}-\\mathbf{3} \\boldsymbol{y}+\\mathbf{2}=\\mathbf{0}\\) and \\(6 x-p y+7=0\\) are perpendicular to each other. Hence, find the equation of a line passing through (-2, -1) and parallel to \\(\\mathbf{6} \\boldsymbol{x}-\\boldsymbol{p} \\boldsymbol{y}+\\mathbf{7}=\\) 0. <strong>[3]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Slope of line \\(5 x-3 y+2=0\\) is<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; =-\\frac{\\text { coefficent of } x}{\\text { coefficent of } y} \\\\\n&amp; =\\frac{-5}{-3}=\\frac{5}{3}\n\\end{aligned}\n\\]<\/div>\n<p>Slope of line \\(6 x-p y+7=0\\) is \\(=\\frac{-6}{-p}=\\frac{6}{p}\\) Lines are perpendicular each other \u2234 Product of slopes = -1<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\frac{5}{3} \\times \\frac{6}{p} &amp; =-1 \\\\\n3 p &amp; =-30 \\Rightarrow p=-10\n\\end{aligned}\n\\]<\/div>\n<p>\u2234<\/p>\n<div class=\"mathjax-overflow\">\\[\np=-10\n\\]<\/div>\n<p>Given line \\(6 x-p y+7=0\\) is \\(6 x+10 y+7=0\\) Equation of the line, parallel to \\(6 x+10 y+7=0\\) is \\(6 x+10 y+k=0\\) Since, line passes through (-2,-1),<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; \\therefore 6(-2)+10(-1)+k=0 \\\\\n&amp; \\quad-12-10+k=0 \\Rightarrow k=22\n\\end{aligned}\n\\]<\/div>\n<p>Equation of the line parallel to \\(6 x+10 y+7\\) is \\(6 x+10 y+22=0\\) or \\(3 x+5 y+11=0\\)<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2020-q7-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>Using properties of proportion find \\(x: y\\), given : <strong>[3]<\/strong><\/p>\n<div class=\"mathjax-overflow\">\\[\n\\frac{x^{2}+2 x}{2 x+4}=\\frac{y^{2}+3 y}{3 y+9}\n\\]<\/div><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><div class=\"mathjax-overflow\">\\[\n\\frac{x^{2}+2 x}{2 x+4}=\\frac{y^{2}+3 y}{3 y+9}\n\\]<\/div>\n<p>(using compodendo dividendo \\(\\frac{a}{b}=\\frac{c}{d}\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\left.\\Rightarrow \\frac{a+b}{a-b}=\\frac{c+d}{c-d}\\right)\n\\]<\/div>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\frac{x^{2}+2 x+2 x+4}{x^{2}+2 x-2 x-4} &amp; =\\frac{y^{2}+3 y+3 y+9}{y^{2}+3 y-3 y-9} \\\\\n\\frac{x^{2}+4 x+4}{x^{2}-4} &amp; =\\frac{y^{2}+6 y+9}{y^{2}-9} \\\\\n\\frac{(x+2)^{2}}{(x+2)(x-2)} &amp; =\\frac{(y+3)^{2}}{(y+3)(y-3)} \\\\\n\\frac{x+2}{x-2} &amp; =\\frac{y+3}{y-3}\n\\end{aligned}\n\\]<\/div>\n<p>(using compodendo &#8211; dividendo)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\frac{x+2+x-2}{x+2-x+2} &amp; =\\frac{y+3+y-3}{y+3-y+3} \\\\\n\\frac{2 x}{4} &amp; =\\frac{2 y}{6} \\\\\n\\frac{x}{2} &amp; =\\frac{y}{3} \\\\\n\\frac{x}{y} &amp; =\\frac{2}{3}\n\\end{aligned}\n\\]<\/div>\n<p>\u2234<\/p>\n<div class=\"mathjax-overflow\">\\[\nx: y=2: 3\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2020-q7-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>In the given figure \\(\\boldsymbol{T} \\boldsymbol{P}\\) and \\(\\boldsymbol{T} \\boldsymbol{Q}\\) are two tangents to the circle with centre \\(\\boldsymbol{O}\\), touching at \\(\\boldsymbol{A}\\) and \\(\\boldsymbol{C}\\) respectively. If \\(\\angle B C Q=55^{\\circ}\\) and \\(\\angle B A P=60^{\\circ}\\), find :<\/p>\n<p>(i) \\(\\angle O B A\\) and \\(\\angle O B C\\)<\/p>\n<p>(ii) \\(\\angle A O C\\)<\/p>\n<p>(iii) \\(\\angle A T C\\) <strong>[4]<\/strong><\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/c3f56170-7ce6-4654-8ea6-0da7169b9fde-06_403_594_1136_1119.jpg\" alt=\"ICSE Class 10 Mathematics 2020 Question 7(c) figure\" loading=\"lazy\" decoding=\"async\"><\/div><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Given, \\(\\angle B C Q=55^{\\circ}\\) and \\(\\angle B A P=60^{\\circ}\\)<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/c3f56170-7ce6-4654-8ea6-0da7169b9fde-07_404_587_1515_291.jpg\" alt=\"ICSE Class 10 Mathematics 2020 Question 7(c) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>(i) \u2235<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\angle P A O=90^{\\circ}\n\\]<\/div>\n<p>(As PAT is the tangent) \u2234<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\angle O A B=90^{\\circ}-60^{\\circ}=30^{\\circ}\n\\]<\/div>\n<p>In \\(\\triangle O A B, O A=O B\\) (radii of circle)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\angle O B A=\\angle O A B=30^{\\circ}\n\\]<\/div>\n<p>Similarly<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\therefore \\angle Q C O=90^{\\circ}\n\\]<\/div>\n<p>(As QCT is the tangent) \u2234<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\angle O C B=90^{\\circ}-55^{\\circ}=35^{\\circ}\n\\]<\/div>\n<p>In \\(\\triangle O C B, O C=O B\\) (radii of circle)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\angle O B C=\\angle O C B=35^{\\circ}\n\\]<\/div>\n<p>(ii) Now,<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\angle A B C &amp; =\\angle O B A+\\angle O B C \\\\\n&amp; =30^{\\circ}+35^{\\circ}=65^{\\circ}\n\\end{aligned}\n\\]<\/div>\n<p>Again, \\(\\angle A O C=2 \\angle A B C=2 \\times 65^{\\circ}=130^{\\circ}\\) ( \u2235 Angle subtended at the center of the circle by an arc is twice the angle subtended at the circle) (iii)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\angle A O C+\\angle A T C &amp; =180^{\\circ} \\\\\n130^{\\circ}+\\angle A T C &amp; =180^{\\circ} \\\\\n\\therefore \\quad \\angle A T C &amp; =180^{\\circ}-130^{\\circ}=50^{\\circ} \\\\\n\\therefore \\quad \\angle A T C &amp; =50^{\\circ}\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<\/div>\n<div class=\"xc-set\">\n<div class=\"xc-set-header\"><div class=\"xc-set-icon\">8<\/div><div><div class=\"xc-set-title\">Question 8<\/div><div class=\"xc-set-sub\">ICSE Class 10 Mathematics 2020<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2020-q8-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>What must be added to the polynomial \\(\\mathbf{2} \\boldsymbol{x}^{\\mathbf{3}}-\\) \\(\\mathbf{3} \\boldsymbol{x}^{\\mathbf{2}}-\\mathbf{8} \\boldsymbol{x}\\), so that it leaves a remainder 10 when divided by \\(\\mathbf{2} \\boldsymbol{x}+\\mathbf{1}\\) ? <strong>[3]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\np(x) &amp; =2 x^{3}-3 x^{2}-8 x \\\\\nq(x) &amp; =2 x+1 \\\\\n\\text { Remainder } &amp; =10\n\\end{aligned}\n\\]<\/div>\n<p>Let \\(k\\) be added to get remainder 10 when divided by \\((2 x+1)\\).<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; p(x)=2 x^{3}-3 x^{2}-8 x+k \\\\\n&amp; \\text { Remainder }=10 \\\\\n&amp; p\\left(-\\frac{1}{2}\\right)=10 \\\\\n&amp; 2\\left(-\\frac{1}{2}\\right)^{3}-3\\left(-\\frac{1}{2}\\right)^{2}-8\\left(-\\frac{1}{2}\\right)+k=10 \\\\\n&amp;-2 \\times \\frac{1}{8}-3 \\times \\frac{1}{4}+4+k=10 \\\\\n&amp;-1+4+k=10 \\\\\n&amp; k=7\n\\end{aligned}\n\\]<\/div>\n<p>Hence, 7 be added to get remainder 10 when divided by \\(2 x+1\\) to given polynomial.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2020-q8-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>Mr. Sona has a recurring deposit account and deposits \u20b9 750 per month for 2 years. If he gets \u20b9 19125 at the time of maturity, find the rate of interest. <strong>[3]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Maturity amount (M.A.) = \u20b9 19125 Monthly deposit (P)= \u20b9 750 Time \\((n)=2\\) years \\(=24\\) months<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; \\text { M.A. }=P \\times n+P \\times \\frac{r}{100} \\times \\frac{n(n+1)}{2} \\times \\frac{1}{12} \\\\\n&amp; 750 \\times 24+\\frac{750 \\times r}{2400}(24)(24+1)=19125 \\\\\n&amp; 18000+\\frac{750}{4} r=19125 \\\\\n&amp; \\frac{750 r}{4}=19125-18000 \\\\\n&amp; r=\\frac{1125 \\times 4}{750}=\\frac{4500}{750} \\\\\n&amp; \\begin{array}{l}\nr=6 \\\\\n\\therefore \\quad \\text { Rate of interest }=6 \\%\n\\end{array}\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2020-q8-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>Use graph paper for this question. Take \\(\\mathbf{1 ~ c m}=\\mathbf{1}\\) unit on both \\(\\boldsymbol{x}\\) and \\(\\boldsymbol{y}\\) axes.<\/p>\n<p>(i) Plot the following points on your graph sheets. \\(A(-4,0), B(-3,2), C(0,4), D(4,1)\\) and E \\((7,3)\\)<\/p>\n<p>(ii) Reflect the points \\(B, C, D\\) and \\(E\\) on the \\(x\\)-axis and name them as \\(\\boldsymbol{B}^{\\prime}, \\boldsymbol{C}^{\\prime}, \\boldsymbol{D}^{\\prime}\\) and \\(\\boldsymbol{E}^{\\prime}\\) respectively.<\/p>\n<p>(iii) Join the points \\(\\boldsymbol{A}, \\boldsymbol{B}, \\boldsymbol{C}, \\boldsymbol{D}, \\boldsymbol{E}, \\boldsymbol{E}^{\\prime}, \\boldsymbol{D}^{\\prime}, \\boldsymbol{C}^{\\prime}, \\boldsymbol{B}^{\\prime}\\) and \\(\\boldsymbol{A}\\) in order.<\/p>\n<p>(iv) Name the closed figure formed. <strong>[4]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>(i)<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/c3f56170-7ce6-4654-8ea6-0da7169b9fde-08_991_1081_275_333.jpg\" alt=\"ICSE Class 10 Mathematics 2020 Question 8(c) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>(ii) B'(-3, -2), C'(0, -4), D'(4, -1) and E'(7, -3) respectively are the reflected points through the x-axis.<\/p>\n<p>(iii) On joining the points A, B, C, D, E, E&#8217;, D&#8217;, C&#8217;, B&#8217; and A in order, we get a closed figure.<\/p>\n<p>(iv) Closed figure formed is nine sided polygon or nonagon, polygon fish or kite.<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"xc-set\">\n<div class=\"xc-set-header\"><div class=\"xc-set-icon\">9<\/div><div><div class=\"xc-set-title\">Question 9<\/div><div class=\"xc-set-sub\">ICSE Class 10 Mathematics 2020<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2020-q9-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>40 Students enter for a game of shot-put competition. The distance thrown (in metres) is recorded below :<\/p>\n<div class=\"math-responsive\"><table><thead><tr><th>Distance in \\(\\boldsymbol{m}\\)<\/th><th>12-13<\/th><th>13-14<\/th><th>14-15<\/th><th>15-16<\/th><th>16-17<\/th><th>17-18<\/th><th>18-19<\/th><\/tr><\/thead><tbody><tr><td>Number of Students<\/td><td>3<\/td><td>9<\/td><td>12<\/td><td>9<\/td><td>4<\/td><td>2<\/td><td>1<\/td><\/tr><\/tbody><\/table><\/div>\n<p>Use a graph paper to draw an ogive for the above distribution. Use a scale of \\(2 \\mathrm{~cm}=1 \\mathrm{~m}\\) on one axis and 2 cm \\(=5\\) students on the other axis. Hence, using your graph find :<\/p>\n<p>(iii) number of students who cover a distance which is above \\(\\mathbf{1 6} \\frac{\\mathbf{1}}{\\mathbf{2}} \\mathrm{m}\\).<\/p>\n<p>(i) the median.<\/p>\n<p>(ii) upper qartile. <strong>[6]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/c3f56170-7ce6-4654-8ea6-0da7169b9fde-08_740_887_1867_341.jpg\" alt=\"ICSE Class 10 Mathematics 2020 Question 9(a) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<div class=\"math-responsive\"><table><thead><tr><th>Distance in ( \\(\\boldsymbol{m}\\) )<\/th><th>Number of students<\/th><th>Less than<\/th><th>Cumulative frequency<\/th><\/tr><\/thead><tbody><tr><td>12-13<\/td><td>3<\/td><td>13<\/td><td>3<\/td><\/tr><tr><td>13-14<\/td><td>9<\/td><td>14<\/td><td>12<\/td><\/tr><tr><td>14-15<\/td><td>12<\/td><td>15<\/td><td>24<\/td><\/tr><tr><td>15-16<\/td><td>9<\/td><td>16<\/td><td>33<\/td><\/tr><tr><td>16-17<\/td><td>4<\/td><td>17<\/td><td>37<\/td><\/tr><tr><td>17-18<\/td><td>2<\/td><td>18<\/td><td>39<\/td><\/tr><tr><td>18-19<\/td><td>1<\/td><td>19<\/td><td>40<\/td><\/tr><\/tbody><\/table><\/div>\n<p>(i) Median = 14.6\u00b1 0.2<\/p>\n<p>(ii) Upper quartile \\(=15.75 \\pm 0.2\\)<\/p>\n<p>(iii) Number of students above \\(16 \\frac{1}{2} \\mathrm{~m}=40-35=5\\)<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2020-q9-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; \\text { If } x=\\frac{\\sqrt{2 a+1}+\\sqrt{2 a-1}}{\\sqrt{2 a+1}-\\sqrt{2 a-1}} \\text {, prove that } x^{2}-4 a x+1 \\\\\n&amp; =0 .\n\\end{aligned}\n\\]<\/div><p><strong>[4]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; x=\\frac{\\sqrt{2 a+1}+\\sqrt{2 a-1}}{\\sqrt{2 a+1}-\\sqrt{2 a-1}} \\\\\n&amp; \\frac{x+1}{x-1}=\\frac{\\sqrt{2 a+1}+\\sqrt{2 a-1}+\\sqrt{2 a+1}-\\sqrt{2 a-1}}{\\sqrt{2 a+1}+\\sqrt{2 a-1}-\\sqrt{2 a+1}+\\sqrt{2 a-1}}\n\\end{aligned}\n\\]<\/div>\n<p>(Apply compodendo-dividendo \\(\\frac{A}{B}=\\frac{C}{D}\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; \\left.\\Rightarrow \\frac{A+B}{A-B}=\\frac{C+D}{C-D}\\right) \\\\\n\\frac{x+1}{x-1}=\\frac{2 \\sqrt{2 a+1}}{2 \\sqrt{2 a-1}}=\\frac{\\sqrt{2 a+1}}{\\sqrt{2 a-1}} &amp;\n\\end{aligned}\n\\]<\/div>\n<p>Squaring both sides<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; \\frac{(x+1)^{2}}{(x-1)^{2}}=\\left(\\frac{\\sqrt{2 a+1}}{\\sqrt{2 a-1}}\\right)^{2} \\\\\n&amp; \\frac{x^{2}+2 x+1}{x^{2}-2 x+1}=\\frac{2 a+1}{2 a-1}\n\\end{aligned}\n\\]<\/div>\n<p>Apply compodendo &#8211; dividendo<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\frac{x^{2}+2 x+1+x^{2}-2 x+1}{x^{2}+2 x+1-x^{2}+2 x-1} &amp; =\\frac{2 a+1+2 a-1}{2 a+1-2 a+1} \\\\\n\\frac{2\\left(x^{2}+1\\right)}{4 x} &amp; =\\frac{4 a}{2} \\\\\n\\frac{x^{2}+1}{2 x} &amp; =\\frac{2 a}{1} \\\\\nx^{2}+1 &amp; =4 a x \\\\\nx^{2}-4 a x+1 &amp; =0 \\quad \\text { Hence proved. }\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<\/div>\n<div class=\"xc-set\">\n<div class=\"xc-set-header\"><div class=\"xc-set-icon\">10<\/div><div><div class=\"xc-set-title\">Question 10<\/div><div class=\"xc-set-sub\">ICSE Class 10 Mathematics 2020<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2020-q10-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>If the \\(\\mathbf{6}^{\\text {th }}\\) term of an A.P. is equal to four times its first term and the sum of fist six terms is 75, find the first term and the common difference. <strong>[3]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Let the first term of an A.P. be \\(a\\) and common difference be \\(d\\) respectively<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\na_{6} &amp; =4 \\times a \\text { (given) } \\\\\na_{6} &amp; =4 a  \\qquad \\text{&#8230;(i)}\\\\\na+5 d &amp; =4 a \\quad\\left\\{a_{n}=a+(n-1) d\\right\\} \\\\\n3 a &amp; =5 d \\\\\na &amp; =\\frac{5}{3} d  \\qquad \\text{&#8230;(ii)}\\\\\nS_{6} &amp; =75 \\text { (given) } \\\\\n\\frac{n}{2}\\left[a+a_{6}\\right] &amp; =75 \\\\\n\\frac{6}{2}\\left(a+a_{6}\\right) &amp; =75 \\\\\na+a_{6} &amp; =\\frac{75}{3} \\\\\na+4 a &amp; =25 \\\\\na &amp; =\\frac{25}{5}=5 \\\\\na &amp; =5\n\\end{aligned}\n\\]<\/div>\n<p>From equation (i)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; &amp; a &amp; =\\frac{5}{3} d \\\\\n&amp; &amp; 5 &amp; =\\frac{5}{3} d \\\\\n\\therefore &amp; &amp; d &amp; =\\frac{5 \\times 3}{5} \\times 3 \\\\\n\\therefore &amp; &amp; d &amp; =3\n\\end{aligned}\n\\]<\/div>\n<p>First term of an A.P. \\(=5\\) and common difference = 3<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2020-q10-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>The difference of two natural numbers is 7 and their product is 450. Find the numbers. <strong>[3]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Let the numbers be \\(x\\) and \\(y\\). According the given condition<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\nx-y=7 &amp; \\Rightarrow y=x-7[x&gt;y]  \\qquad \\text{&#8230;(i)}\\\\\nx y &amp; =450 \\qquad \\text{&#8230;(i)}\n\\end{aligned}\n\\]<\/div>\n<p>From (i) \\&amp; (ii)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\nx(x-7) &amp; =450 \\\\\nx^{2}-7 x-450 &amp; =0 \\\\\nx^{2}-25 x+18 x-450 &amp; =0\n\\end{aligned}\n\\]<\/div>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\nx(x-25)+18(x-25) &amp; =0 \\\\\n(x-25)(x+18) &amp; =0\n\\end{aligned}\n\\]<\/div>\n<p>If \\(x+18=0 \\Rightarrow x=-18\\) it is not a natural number.<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\text { If } x-25=0 \\Rightarrow x=25, y=25-7=18\n\\]<\/div>\n<p>Thus, numbers 25 and 18.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2020-q10-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>Use ruler and compass for this question. Construct a circle of radius 4.5 cm . Draw a chord \\(\\boldsymbol{A} \\boldsymbol{B}=\\mathbf{6 ~ c m}\\).<\/p>\n<p>(i) Find the locus of points equidistant from \\(\\boldsymbol{A}\\) and \\(\\boldsymbol{B}\\).<\/p>\n<p>Mark the point where it meets the circle as \\(\\boldsymbol{D}\\).<\/p>\n<p>(ii) Join \\(\\boldsymbol{A} \\boldsymbol{D}\\) and find the locus of points which are equidistant from \\(A D\\) and \\(A B\\). Mark the point where it meets the circle as \\(\\boldsymbol{C}\\).<\/p>\n<p>(iii) Join \\(\\boldsymbol{B} \\boldsymbol{C}\\) and \\(\\boldsymbol{C} \\boldsymbol{D}\\), Mesuare and write down the length of side \\(\\boldsymbol{C} \\boldsymbol{D}\\) of the quadrilateral \\(\\boldsymbol{A} \\boldsymbol{B} \\boldsymbol{C} \\boldsymbol{D}\\). <strong>[4]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/c3f56170-7ce6-4654-8ea6-0da7169b9fde-10_412_320_533_498.jpg\" alt=\"ICSE Class 10 Mathematics 2020 Question 10(c) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>\\(B C=C D=5.1 \\mathrm{~cm}\\) (Approx.)<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"xc-set\">\n<div class=\"xc-set-header\"><div class=\"xc-set-icon\">11<\/div><div><div class=\"xc-set-title\">Question 11<\/div><div class=\"xc-set-sub\">ICSE Class 10 Mathematics 2020<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2020-q11-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>A model of a high rise building is made to a scale of 1 : 50.<\/p>\n<p>(i) If the height of the model is 0.8 m, find the height of the actual building.<\/p>\n<p>(ii) If the floor area of a flat in the building is \\(\\mathbf{2 0 ~ \\mathbf { m } ^ { 2 }}\\), find the floor area of that in the model. <strong>[3]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Scale \\(=1: 50\\) Height of the model 0.8 m \u2234 Height of the building \\(=0.8 \\times 50 \\mathrm{~m}=40 \\mathrm{~m}\\) Floor area of building \\(=20 \\mathrm{~m}^{2}\\) Floor area of the model \\(=\\frac{20}{50 \\times 50} \\mathrm{~m}^{2}\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; =\\frac{1}{125} \\mathrm{~m}^{2} \\\\\n&amp; =0.008 \\mathrm{~m}^{2}\n\\end{aligned}\n\\]<\/div>\n<p>\u2234 Floor area of the model<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; =0.008 \\times 10000 \\mathrm{~cm}^{2} \\\\\n&amp; =80 \\mathrm{~cm}^{2}\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2020-q11-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>From a solid wooden cylinder of height 28 cm and diameter 6 cm, two conical cavities are hollowed out. The diameters of the cones are also of 6 cm and height 10.5 cm. Taking \\(\\pi=\\frac{22}{7}\\) find the volume of the remaining solid. <strong>[3]<\/strong><\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/c3f56170-7ce6-4654-8ea6-0da7169b9fde-10_557_370_1551_402.jpg\" alt=\"ICSE Class 10 Mathematics 2020 Question 11(b) figure\" loading=\"lazy\" decoding=\"async\"><\/div><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Dimension of cylinder Height \\((h)=28 \\mathrm{~cm}\\) Diameter \\((2 r)=6 \\mathrm{~cm}\\)<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/c3f56170-7ce6-4654-8ea6-0da7169b9fde-10_555_369_674_1232.jpg\" alt=\"ICSE Class 10 Mathematics 2020 Question 11(b) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>Dimension of conical cavities Height \\((\\mathrm{H})=10.5 \\mathrm{~cm}\\) diameter \\((2 r)=6 \\mathrm{~cm}\\) Volume of remaining solid = Volume of cylinder &#8211; 2 \u00d7 Volume of cavity<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; =\\pi r^{2} h-2 \\times \\frac{1}{3} \\pi r^{2} H \\\\\n&amp; =\\pi r^{2}\\left(h-\\frac{2}{3} \\mathrm{H}\\right) \\\\\n&amp; =\\frac{22}{7} \\times 3 \\times 3\\left(28-\\frac{2}{3} \\times 10.5\\right) \\\\\n&amp; =\\frac{22}{7} \\times 3 \\times 3(28-7) \\\\\n&amp; =\\frac{22}{7} \\times 3 \\times 3 \\times 21 \\\\\n&amp; =22 \\times 3 \\times 3 \\times 3=594 \\mathrm{~cm}^{3}\n\\end{aligned}\n\\]<\/div>\n<p>Hence, volume of remaining solid \\(=594 \\mathrm{~cm}^{3}\\)<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2020-q11-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>Prove the identity <strong>[4]<\/strong><\/p>\n<div class=\"mathjax-overflow\">\\[\n\\left(\\frac{1-\\tan \\theta}{1-\\cot \\theta}\\right)^{2}=\\tan ^{2} \\theta\n\\]<\/div><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; \\left(\\frac{1-\\tan \\theta}{1-\\cot \\theta}\\right)^{2}=\\tan ^{2} \\theta \\\\\n\\text { L.H.S. } &amp; =\\left(\\frac{1-\\tan \\theta}{1-\\cot \\theta}\\right)^{2} \\\\\n&amp; =\\left(\\frac{1-\\tan \\theta}{1-\\frac{1}{\\tan \\theta}}\\right)^{2}\n\\end{aligned}\n\\]<\/div>\n<div class=\"mathjax-overflow\">\\[\n=\\left(\\frac{1-\\tan \\theta}{\\frac{\\tan \\theta-1}{\\tan \\theta}}\\right)^{2}\n\\]<\/div>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; =\\left\\{\\frac{\\tan \\theta(1-\\tan \\theta)}{-(1-\\tan \\theta)}\\right\\}^{2} \\\\\n&amp; =\\left\\{\\frac{\\tan \\theta}{-1}\\right\\}^{2}=\\tan ^{2} \\theta\n\\end{aligned}\n\\]<\/div>\n<p>= R.H.S. Hence proved. \\(\\square\\)<\/p><\/div>\n<\/div>\n<\/div>\n<\/div>\n\n","protected":false},"excerpt":{"rendered":"<p><span class=\"rt-reading-time\" style=\"display: block;\"><span class=\"rt-label rt-prefix\">Reading Time: <\/span> <span class=\"rt-time\">35<\/span> <span class=\"rt-label rt-postfix\">minutes<\/span><\/span> ICSE Class 10 Mathematics PYQ Practice the complete ICSE Class 10 Mathematics 2020 Question Paper with Solutions, including algebra, geometry, [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_lmt_disableupdate":"","_lmt_disable":"","site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"var(--ast-global-color-4)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"iawp_total_views":3,"footnotes":""},"categories":[6],"tags":[],"class_list":["post-2361","post","type-post","status-publish","format-standard","hentry","category-class-10-icse"],"modified_by":"sanjjeett","_links":{"self":[{"href":"https:\/\/xamcontent.com\/cisce\/wp-json\/wp\/v2\/posts\/2361"}],"collection":[{"href":"https:\/\/xamcontent.com\/cisce\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/xamcontent.com\/cisce\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/xamcontent.com\/cisce\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/xamcontent.com\/cisce\/wp-json\/wp\/v2\/comments?post=2361"}],"version-history":[{"count":3,"href":"https:\/\/xamcontent.com\/cisce\/wp-json\/wp\/v2\/posts\/2361\/revisions"}],"predecessor-version":[{"id":2365,"href":"https:\/\/xamcontent.com\/cisce\/wp-json\/wp\/v2\/posts\/2361\/revisions\/2365"}],"wp:attachment":[{"href":"https:\/\/xamcontent.com\/cisce\/wp-json\/wp\/v2\/media?parent=2361"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/xamcontent.com\/cisce\/wp-json\/wp\/v2\/categories?post=2361"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/xamcontent.com\/cisce\/wp-json\/wp\/v2\/tags?post=2361"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}