{"id":2381,"date":"2026-11-13T16:45:00","date_gmt":"2026-11-13T11:15:00","guid":{"rendered":"https:\/\/xamcontent.com\/cisce\/?p=2381"},"modified":"2026-11-13T16:45:02","modified_gmt":"2026-11-13T11:15:02","slug":"icse-class-10-mathematics-2019-question-paper-with-solutions","status":"publish","type":"post","link":"https:\/\/xamcontent.com\/cisce\/icse-class-10-mathematics-2019-question-paper-with-solutions\/","title":{"rendered":"ICSE Class 10 Mathematics 2019 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-2019-question-paper-with-solutions\/#ICSE_Mathematics_2019_%E2%80%94_Section_A\" title=\"ICSE Mathematics 2019 \u2014 Section A\">ICSE Mathematics 2019 \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-2019-question-paper-with-solutions\/#ICSE_Mathematics_2019_%E2%80%94_Section_B\" title=\"ICSE Mathematics 2019 \u2014 Section B\">ICSE Mathematics 2019 \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\">33<\/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 2019 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\">2019<\/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 2019<\/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_2019_%E2%80%94_Section_A\"><\/span>ICSE Mathematics 2019 \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 2019<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2019-q1-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>Solve the following inequation and write down the solution set:<\/p>\n<div class=\"mathjax-overflow\">\\[\n11 x-4&lt;15 x+4 \\leq 13 x+14, x \\in \\mathbf{W}\n\\]<\/div>\n<p>Represent the solution on a real number line. <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>\\(11 x-4&lt;15 x+4 \\leq 13 x+14, x \\in W\\)\n<div class=\"math-responsive\"><table><tbody><tr><td>\\(11 x-4&lt;15 x+4\\)<\/td><td>\\(15 x+4 \\leq 13 x+14\\)<\/td><\/tr><tr><td>\\(11 x-15 x&lt;4+4\\)<\/td><td>\\(15 x-13 x \\leq 14-4\\)<\/td><\/tr><tr><td>\\(-4 x&lt;8\\)<\/td><td>\\(2 x \\leq 10\\)<\/td><\/tr><tr><td>\\(x&gt;\\frac{8}{-4}\\)<\/td><td>\\(x \\leq \\frac{10}{2}\\)<\/td><\/tr><tr><td>\\(x&gt;-2\\)<\/td><td>\\(x \\leq 5\\)<\/td><\/tr><tr><td>\\(-2&lt;x \\leq 5\\)<\/td><td><\/td><\/tr><\/tbody><\/table><\/div>\n<p>But \\(x \\in W\\), therefore required solution set is \\{0, 1, 2, 3, 4, 5\\}. The solution set on real number line is shown below.<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/1b794c1d-17ab-4ff8-87dd-647eb47e1e61-01_65_791_1167_1020.jpg\" alt=\"ICSE Class 10 Mathematics 2019 Question 1(a) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2019-q1-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>A man invests \u20b9 4500 in shares of a company which is paying 7.5% dividend. If \u20b9 100 shares are available at a discount of 10%. Find:<\/p>\n<p>(i) Number of shares he purchases.<\/p>\n<p>(ii) His annual income. <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>Investment \\(=\\text{\u20b9} 4500\\), Face value of shares \\(=\\text{\u20b9} 100\\) Rate of dividend \\(=7.5 \\%, \\quad\\) Discount \\(=10 \\%\\)<\/p>\n<p>(i) Market value of 1 share \\(=100-10 \\%=100-10\\) = \u20b9 90<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\text { Number of shares } &amp; =\\frac{\\text { Investment }}{\\text { M.V. of 1 share }} \\\\\n&amp; =\\frac{4500}{90}=50\n\\end{aligned}\n\\]<\/div>\n<p>Number of shares he purchases \\(=50\\)<\/p>\n<p>(ii) Annual Income = Rate of Dividend \u00d7 Face value of 1 share \u00d7 no. of shares<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; =\\frac{7.5}{100} \\times 100 \\times 50 \\\\\n\\text { Annual Income } &amp; =\\text{\u20b9} 375\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2019-q1-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>In a class of 40 students, marks obtained by the students in a class test (out of 10) are given below:<\/p>\n<div class=\"math-responsive\"><table><thead><tr><th>Marks<\/th><th>1<\/th><th>2<\/th><th>3<\/th><th>4<\/th><th>5<\/th><th>6<\/th><th>7<\/th><th>8<\/th><th>9<\/th><th>10<\/th><\/tr><\/thead><tbody><tr><td>Number of students<\/td><td>1<\/td><td>2<\/td><td>3<\/td><td>3<\/td><td>6<\/td><td>10<\/td><td>5<\/td><td>4<\/td><td>3<\/td><td>3<\/td><\/tr><\/tbody><\/table><\/div>\n<p>Calculate the following for the given distribution:<\/p>\n<p>(i) Median<\/p>\n<p>(ii) Mode <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=\"math-responsive\"><table><tbody><tr><td>Marks ( \\(\\boldsymbol{x}\\) )<\/td><td>Number of students (f)<\/td><td>c.f.<\/td><\/tr><tr><td>1<\/td><td>1<\/td><td>1<\/td><\/tr><tr><td>2<\/td><td>2<\/td><td>3<\/td><\/tr><tr><td>3<\/td><td>3<\/td><td>6<\/td><\/tr><tr><td>4<\/td><td>3<\/td><td>9<\/td><\/tr><tr><td>5<\/td><td>6<\/td><td>15<\/td><\/tr><tr><td>6<\/td><td>10<\/td><td>25<\/td><\/tr><tr><td>7<\/td><td>5<\/td><td>30<\/td><\/tr><tr><td>8<\/td><td>4<\/td><td>34<\/td><\/tr><tr><td>9<\/td><td>3<\/td><td>37<\/td><\/tr><tr><td>10<\/td><td>3<\/td><td>40<\/td><\/tr><\/tbody><\/table><\/div>\n<p>As, \\(n\\) is even. \u2234 Median \\(=\\frac{\\left(\\frac{n}{2}\\right)^{\\text {th }}+\\left(\\frac{n}{2}+1\\right)^{\\text {th }}}{2}\\) term<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; =\\frac{\\left(\\frac{40}{2}\\right)^{\\text {th }}+\\left(\\frac{40}{2}+1\\right)^{\\text {th }}}{2} \\text { term } \\\\\n&amp; =\\frac{20^{\\text {th }}+21^{\\text {th }}}{2} \\text { term } \\\\\n&amp; \\text { Median }=\\frac{6+6}{2}=6 \\text { marks }\n\\end{aligned}\n\\]<\/div>\n<p>Here, maximum frequency is 10 whose value is 6<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{array}{ll}\n\\therefore &amp; \\text { Mode }=6 \\text { marks } \\\\\n\\therefore &amp; \\text { Median }=\\text { Mode }=6 \\text { marks }\n\\end{array}\n\\]<\/div><\/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 2019<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2019-q2-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>Using the factor theorem, show that \\((\\boldsymbol{x}-\\mathbf{2})\\) is a factor of \\(x^{3}+x^{2}-4 x-4\\). Hence, factorise the polynomial 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><p>Let \\(f(x)=x^{3}+x^{2}-4 x-4\\) Put \\(x-2=0\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\nx=2\n\\]<\/div>\n<p>\u2234<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\nf(2) &amp; =(2)^{3}+(2)^{2}-4(2)-4 \\\\\n&amp; =8+4-8-4\n\\end{aligned}\n\\]<\/div>\n<p>Remainder \\(=f(2)=0\\) \\(\\therefore \\quad(x-2)\\) is a factor of \\(f(x)=x^{3}+x^{2}-4 x-4\\) Now, \\(f(x)=x^{3}+x^{2}-4 x-4\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; =x^{2}(x+1)-4(x+1) \\\\\n&amp; =(x+1)\\left(x^{2}-4\\right) \\\\\n&amp; =(x+1)\\left\\{(x)^{2}-(2)^{2}\\right\\} \\\\\n&amp; =(x+1)(x+2)(x-2)\n\\end{aligned}\n\\]<\/div>\n<p>Alternative Method: As, \\((x-2)\\) is a factor of \\(x^{3}+x^{2}-4 x-4\\).<\/p>\n<div class=\"math-responsive\"><table><tbody><tr><td>\\(x^{2}+3 x+2\\)<\/td><td><\/td><\/tr><tr><td>\\(x-2\\)<\/td><td><\/td><\/tr><tr><td><\/td><td>\\(x^{3}+x^{2}-4 x-4\\)<\/td><\/tr><tr><td><\/td><td>\\(x^{3}-2 x^{2}\\)<\/td><\/tr><tr><td><\/td><td>\\((-) \\quad(+)\\)<\/td><\/tr><tr><td>\\(3 x^{2}-4 x-4\\)<\/td><td><\/td><\/tr><tr><td><\/td><td>\\(3 x^{2}-6 x\\)<\/td><\/tr><tr><td><\/td><td>\\((-) \\quad(+)\\)<\/td><\/tr><\/tbody><\/table><\/div>\n<div class=\"math-responsive\"><table><tbody><tr><td>\\(2 x-4\\)<\/td><\/tr><tr><td>\\(2 x-4\\)<\/td><\/tr><tr><td>(-) (+)<\/td><\/tr><tr><td>\u00d7<\/td><\/tr><\/tbody><\/table><\/div>\n<div class=\"mathjax-overflow\">\\[\n\\therefore \\quad \\begin{aligned}\n&amp; x^{3}+x^{2}-4 x-4=(x-2)\\left(x^{2}+3 x+2\\right) \\\\\n&amp; =(x-2)\\left\\{x^{2}+2 x+x+2\\right\\} \\\\\n&amp; =(x-2)\\{x(x+2)+1(x+2)\\} \\\\\n&amp; =(x-2)(x+2)(x+1)\n\\end{aligned}\n\\]<\/div>\n<div class=\"mathjax-overflow\">\\[\n\\therefore x^{3}+x^{2}-4 x-4=(x-2)(x+1)(x+2)\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2019-q2-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(\\operatorname{cosec} \\theta-\\sin \\theta)(\\sec \\theta-\\cos \\theta)(\\tan \\theta-\\cot \\theta)=1 \\text { }\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; =(\\operatorname{cosec} \\theta-\\sin \\theta)(\\sec \\theta-\\cos \\theta)(\\tan \\theta+\\cot \\theta) \\\\\n&amp; =\\left(\\frac{1}{\\sin \\theta}-\\sin \\theta\\right)\\left(\\frac{1}{\\cos \\theta}-\\cos \\theta\\right)\\left(\\frac{\\sin \\theta}{\\cos \\theta}+\\frac{\\cos \\theta}{\\sin \\theta}\\right) \\\\\n&amp; =\\left(\\frac{1-\\sin ^{2} \\theta}{\\sin \\theta}\\right)\\left(\\frac{1-\\cos ^{2} \\theta}{\\cos \\theta}\\right)\\left(\\frac{\\sin ^{2} \\theta+\\cos ^{2} \\theta}{\\sin \\theta \\cos \\theta}\\right) \\\\\n&amp; =\\left(\\frac{\\cos ^{2} \\theta}{\\sin \\theta}\\right)\\left(\\frac{\\sin ^{2} \\theta}{\\cos \\theta}\\right)\\left(\\frac{1}{\\sin \\theta \\cos \\theta}\\right) \\\\\n&amp; =\\frac{\\cos ^{2} \\theta \\cdot \\sin ^{2} \\theta}{\\cos ^{2} \\theta \\sin ^{2} \\theta} \\\\\n&amp; =1=\\text { R.H.S. }\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2019-q2-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>In an Arithmetic Progression (A.P.) the fourth and sixth terms are 8 and 14 respectively. Find the:<\/p>\n<p>(i) first term<\/p>\n<p>(ii) common difference<\/p>\n<p>(iii) sum of the first 20 terms. <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 an A.P. be &#8216; \\(a\\) &#8216; and common difference be &#8216; \\(d\\) &#8216;. Now \\(a_{4}=8\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; a+(4-1) d=8 \\quad\\left[a_{n}=a+(n-1) d\\right] \\\\\n&amp; a+3 d=8  \\qquad \\text{&#8230;(i)}\\\\\n&amp; \\text { and } a^{6}=14 \\\\\n&amp; a+(6-1)=14 \\\\\n&amp; a+5 d=14 \\qquad \\text{&#8230;(ii)}\n\\end{aligned}\n\\]<\/div>\n<p>By subtracting eq. (i) from eq. (ii)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{gathered}\na+5 d=14 \\\\\n\\begin{array}{r}\na+3 d=8 \\\\\n(-) \\quad(-) \\quad(-) \\\\\n\\hline 2 d-6\n\\end{array} \\\\\nd=\\frac{6}{2}=3\n\\end{gathered}\n\\]<\/div>\n<p>From eq. (i)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; a+3(3)=8 \\\\\n&amp; a=8-9=-1\n\\end{aligned}\n\\]<\/div>\n\\(\\therefore \\mathrm{a}=-1\\)\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\nS_{n} &amp; =\\frac{n}{2}\\{2 a+(n-1) d\\} \\\\\nS_{20} &amp; =\\frac{20}{2}\\{2 \\times-1+(20-1) 3\\} \\\\\n&amp; =10(-2+57) \\\\\n&amp; =10 \\times 55 \\\\\nS_{20} &amp; =550\n\\end{aligned}\n\\]<\/div>\n<p>(i) Hence, First term of an A.P. \\(=-1\\)<\/p>\n<p>(ii) Common difference of an A.P. \\(=3\\)<\/p>\n<p>(iii) Sum of first 20 terms = 550<\/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 2019<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2019-q3-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>Simplify <strong>[3]<\/strong><\/p>\n<div class=\"mathjax-overflow\">\\[\n\\sin A\\left[\\begin{array}{cc}\n\\sin A &amp; -\\cos A \\\\\n\\cos A &amp; \\sin A\n\\end{array}\\right]+\\cos A\\left[\\begin{array}{cc}\n\\cos A &amp; \\sin A \\\\\n-\\sin A &amp; \\cos A\n\\end{array}\\right]\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; \\sin \\mathrm{A}\\left[\\begin{array}{cc}\n\\sin \\mathrm{A} &amp; -\\cos \\mathrm{A} \\\\\n\\cos \\mathrm{~A} &amp; \\sin \\mathrm{~A}\n\\end{array}\\right]+\\cos \\mathrm{A}\\left[\\begin{array}{cc}\n\\cos \\mathrm{A} &amp; \\sin \\mathrm{~A} \\\\\n-\\sin \\mathrm{A} &amp; \\cos \\mathrm{~A}\n\\end{array}\\right] \\\\\n&amp; =\\left[\\begin{array}{cc}\n\\sin ^{2} \\mathrm{~A} &amp; -\\sin \\mathrm{A} \\cos \\mathrm{~A} \\\\\n\\sin \\mathrm{~A} \\cos \\mathrm{~A} &amp; \\sin ^{2} \\mathrm{~A}\n\\end{array}\\right]+ \\\\\n&amp; =\\left[\\begin{array}{cc}\n\\cos ^{2} \\mathrm{~A} &amp; \\cos \\mathrm{~A} \\sin \\mathrm{~A} \\\\\n-\\sin \\mathrm{A} \\cos \\mathrm{~A} &amp; \\cos ^{2} \\mathrm{~A}\n\\end{array}\\right] \\\\\n&amp; =\\left[\\begin{array}{ll}\n1 &amp; 0 \\\\\n0 &amp; 1\n\\end{array}\\right]=\\mathrm{I}_{2}\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2019-q3-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>M and N are two points on the X axis and Y axis respectively. P(3, 2) divides the line segment MN in the ratio 2 : 3. Find:<\/p>\n<p>(i) the coordinates of M and N.<\/p>\n<p>(ii) slope of the line MN. <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 coordinate of the points M and N be \\((x, 0)\\) and \\((0, y)\\) respectively. \\(\\mathrm{MP}=2\\) units and \\(\\mathrm{PN}=3\\) units (given)<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/1b794c1d-17ab-4ff8-87dd-647eb47e1e61-03_365_487_1874_339.jpg\" alt=\"ICSE Class 10 Mathematics 2019 Question 3(b) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>(i)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; P_{x}=\\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}} \\\\\n&amp; 3=\\frac{2 \\times 0+3 \\times x}{2+3} \\\\\n&amp; 3 \\times 5=3 x \\\\\n&amp; \\therefore \\quad x=5\n\\end{aligned}\n\\]<\/div>\n<p>Hence, coordinate of point M is (5, 0).<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; P_{y}=\\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}} \\\\\n&amp; 2=\\frac{2 \\times y+3 \\times 0}{2+3} \\\\\n&amp; 2 \\times 5=2 y\n\\end{aligned}\n\\]<\/div>\n<div class=\"mathjax-overflow\">\\[\n\\therefore \\quad y=5\n\\]<\/div>\n<p>Hence, coordinate of the point N is (0, 5).<\/p>\n<p>(ii) Slope of line MN<\/p>\n<div class=\"mathjax-overflow\">\\[\n(m)=\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\\frac{5-0}{0-5}=\\frac{5}{-5}\n\\]<\/div>\n<p>\u2234 Slope of line \\(\\mathrm{MN}=-1\\)<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2019-q3-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>A solid metallic sphere of radius 6 cm is melted and made into a solid cylinder of height 32 cm. Find the:<\/p>\n<p>(i) radius of the cylinder.<\/p>\n<p>(ii) curved surface area of the cylinder. [Take \\(\\pi=3.1\\) ] <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>Radius of sphere \\((R)=6 \\mathrm{~cm}\\) Height of cylinder \\((h)=32 \\mathrm{~cm}\\) Given, \\(\\pi=3.1\\)<\/p>\n<p>(i) Volume of cylinder = Volume of sphere<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\pi r^{2} h=\\frac{4}{3} \\pi R^{3} \\quad \\text { [where, } r \\text { is the radius of cylinder] }\n\\]<\/div>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; r^{2} h=\\frac{4}{3} R^{3} \\\\\n&amp; r^{2} \\times 32=\\frac{4}{3} \\times(6)^{3} \\\\\n&amp; r^{2}=\\frac{4 \\times 6 \\times 6 \\times 6}{3 \\times 32} \\\\\n&amp; r^{2}=9 \\quad \\therefore \\quad r=3\n\\end{aligned}\n\\]<\/div>\n<p>Radius of cylinder \\(=3 \\mathrm{~cm}\\)<\/p>\n<p>(ii) Curved surface area of cylinder \\(=2 \\pi r h\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n=2 \\times 3.1 \\times 3 \\times 32=18.6 \\times 32=595.2 \\mathrm{~cm}^{2}\n\\]<\/div>\n<p>\u2234 C.S.A. of cylinder \\(=595.2 \\mathrm{~cm}^{2}\\)<\/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 2019<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2019-q4-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>The following numbers, \\(\\boldsymbol{k}+\\mathbf{3}, \\boldsymbol{k}+\\mathbf{2 , 3} \\boldsymbol{k}-\\mathbf{7}\\) and \\(\\mathbf{2} \\boldsymbol{k}-\\mathbf{3}\\) are in proportion. Find \\(\\boldsymbol{k}\\). <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>\\((k+3),(k+2),(3 k-7)\\) and \\((2 k-3)\\) are in proportion.<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\therefore \\quad \\begin{aligned}\n&amp; \\frac{k+3}{k+2}=\\frac{3 k-7}{2 k-3} \\\\\n&amp; \\quad \\Rightarrow(k+3)(2 k-3)=(k+2)(3 k-7) \\\\\n&amp; \\quad \\Rightarrow 2 k^{2}+3 k-9=3 k^{2}-k-14 \\\\\n&amp; \\quad \\Rightarrow k^{2}-4 k-5=0\n\\end{aligned}\n\\]<\/div>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; \\Rightarrow k^{2}-5 k+k-5=0 \\\\\n&amp; \\Rightarrow k(k-5)+1(k-5)=0 \\\\\n&amp; \\Rightarrow(k-5)(k+1)=0 \\\\\n&amp; k-5=0 \\Rightarrow k=5 \\\\\n&amp; \\text { or } k+1=0 \\Rightarrow k=-1 \\\\\n&amp; \\text { So, } k=5 \\text { and }-1\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2019-q4-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>Solve for \\(\\boldsymbol{x}\\) the quadratic equation \\(\\boldsymbol{x}^{\\mathbf{2}}-\\mathbf{4} \\boldsymbol{x}-\\mathbf{8}=\\) 0. Give your answer correct to three significant figures. <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&amp; x^{2}-4 x-8=0 \\\\\n&amp; x^{2}-4 x+4-4-8=0 \\\\\n&amp; \\left(x^{2}-4 x+4\\right)-12=0 \\\\\n&amp; (x-2)^{2}=12\n\\end{aligned}\n\\]<\/div>\n<p>\\(x-2= \\pm \\sqrt{12}\\) [Taking square root on both sides]<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; x-2= \\pm 2 \\sqrt{3} \\\\\n&amp; x-2= \\pm 2(1.732) \\\\\n&amp; x=2+3.464 \\text { and } x=2-3.464 \\\\\n&amp; x=2 \\pm 3.464 \\\\\n&amp; x=5.464 \\text { and } x=-1.464 \\\\\n\\therefore \\quad &amp; x=5.46 \\text { and } x=-1.46\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2019-q4-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>Use ruler and compass only for answering this question. Draw a circle of radius 4 cm. Mark the centre as O. Mark a point P outside the circle at a distance of 7 cm from the centre. Construct two tangents to the circle from the external point P. Measure and write down the length of any one tangent. <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>Steps of construction<\/p>\n<p>(i) Firstly, draw a circle of 4 cm having centre O.<\/p>\n<p>(ii) Draw a line segment \\(\\mathrm{OP}=7 \\mathrm{~cm}\\).<\/p>\n<p>(iii) Draw a perpendicular bisector on OP which intersect at C.<\/p>\n<p>(vi) With C as centre and OC as radius, draw a circle. This circle intersect the given circle at points D and E.<\/p>\n<p>(v) Join PD and PE, which are the required tangents.<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/1b794c1d-17ab-4ff8-87dd-647eb47e1e61-04_381_373_759_1230.jpg\" alt=\"ICSE Class 10 Mathematics 2019 Question 4(c) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>(vi) \\(\\mathrm{PD}=5.7 \\mathrm{~cm}\\)<\/p><\/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_2019_%E2%80%94_Section_B\"><\/span>ICSE Mathematics 2019 \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 2019<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2019-q5-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>There are 25 discs numbered 1 to 25. They are put in a closed box and shaken thoroughly.A disc is drawn at random from the box. Find the probability that the number on the disc is:<\/p>\n<p>(i) an odd number.<\/p>\n<p>(ii) divisible by 2 and 3 both.<\/p>\n<p>(iii) a number less than 16. <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>Total number of discs, \\(n(S)=25\\)<\/p>\n<p>(i) An odd number on disc<\/p>\n<p>\\(=\\{1,3,5,7,9,11,13,15,17,19,21,23,25\\}\\) \\(n(E)=13\\) Probability of getting odd number on disc<\/p>\n<div class=\"mathjax-overflow\">\\[\n=\\frac{\\text { Odd number on disc }}{\\text { Total number of disc }}\n\\]<\/div>\n<div class=\"mathjax-overflow\">\\[\n=\\frac{n(E)}{n(S)}=\\frac{13}{25}\n\\]<\/div>\n<p>(ii) Number on disc, which is divisible by 2 and 3 both \\(=\\{6,12,18,24\\}\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\nn(E)=4\n\\]<\/div>\n<p>Probability of a number divisible by 2 and 3<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; \\text { both } P(E)=\\frac{n(E)}{n(S)} \\\\\n&amp; P(E)=\\frac{4}{25}\n\\end{aligned}\n\\]<\/div>\n<p>(iii) Number of disc having a number less than 16,<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; n(E)=15 \\\\\n&amp; P(E)=\\frac{n(E)}{n(S)}=\\frac{15}{25}=\\frac{3}{5}\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2019-q5-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>Rekha opened a recurring deposit account for 20 months. The rate of interest is 9% per annum and Rekha receives \u20b9 441 as interest at the time of maturity. Find the amount Rekha deposited each month. <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>Interest on recurring deposit account<\/p>\n<div class=\"mathjax-overflow\">\\[\n=\\frac{P \\cdot n(n+1) \\times r}{2400}\n\\]<\/div>\n<p>where \\(P=\\) Amount deposit in each month \\(n=\\) number of months \\(=20\\) months \\(r=\\) Rate of Interest \\(=9 \\%\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; I=\\frac{P . n(n+1) r}{2400} \\\\\n&amp; 441=\\frac{P(20)(20+1) \\times 9}{2400} \\\\\n&amp; P=\\frac{441 \\times 2400}{20 \\times 21 \\times 9} \\\\\n&amp; P=\\text{\u20b9} 280\n\\end{aligned}\n\\]<\/div>\n<p>Hence, Rekha deposited each month of amount \u20b9 280.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2019-q5-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>Use a graph sheet for this question. Take \\(\\mathbf{1 ~ c m}=\\mathbf{1}\\) unit along both \\(\\boldsymbol{x}\\) and \\(\\boldsymbol{y}\\) axis.<\/p>\n<p>(i) Plot the following points:<\/p>\n<p>A(0, 5), B(3, 0), C(1, 0) and D(1, -5)<\/p>\n<p>(ii) Reflect the points B, C and D on the y axis and name them as B&#8217;, C&#8217; and D&#8217; respectively.<\/p>\n<p>(iii) Write down the coordinates of B&#8217;, C&#8217; and D&#8217;.<\/p>\n<p>(iv) Join the points A, B, C, D, D&#8217;, C&#8217;, B&#8217;, A in order and give a name to the closed figure ABCDD&#8217;CB&#8217;A. <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) Plot the given points on graph paper.<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/1b794c1d-17ab-4ff8-87dd-647eb47e1e61-05_579_645_397_261.jpg\" alt=\"ICSE Class 10 Mathematics 2019 Question 5(c) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>(ii) The reflection of the points B, C and D on \\(y\\)-axis are B&#8217;, C&#8217; and D&#8217; are shown in the figure.<\/p>\n<p>(iii) B'(-3,0), C'(-1,0) and D'(-1,-5)<\/p>\n<p>(iv) Arrow<\/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 2019<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2019-q6-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>In the given figure, \\(\\angle \\mathbf{P Q R}=\\angle \\mathbf{P S T}=\\mathbf{9 0}^{\\circ}, \\mathbf{P Q}\\) \\(=5 \\mathrm{~cm}\\) and \\(\\mathrm{PS}=2 \\mathrm{~cm}\\).<\/p>\n<p>(i) Prove that \\(\\triangle \\mathrm{PQR} \\sim \\Delta \\mathrm{PST}\\).<\/p>\n<p>(ii) Find Area of \\(\\triangle \\mathrm{PQR}\\) : Area of quadrilateral SRQT. <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\/1b794c1d-17ab-4ff8-87dd-647eb47e1e61-05_330_600_1399_283.jpg\" alt=\"ICSE Class 10 Mathematics 2019 Question 6(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>(i) In \\(\\triangle \\mathrm{PQR}\\) and \u25b3PST,<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{array}{rlr}\n\\angle \\mathrm{RPQ} &amp; =\\angle \\mathrm{TPS} &amp; \\text { (common) } \\\\\n\\angle \\mathrm{PQR} &amp; =\\angle \\mathrm{PST}=90^{\\circ} &amp; \\text { (given) } \\\\\n\\therefore \\quad &amp; \\triangle \\mathrm{PQR} &amp; \\sim \\triangle \\mathrm{PST} \\text { (AA similarity test) }\n\\end{array}\n\\]<\/div>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/1b794c1d-17ab-4ff8-87dd-647eb47e1e61-05_270_499_958_1166.jpg\" alt=\"ICSE Class 10 Mathematics 2019 Question 6(a) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>(ii)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; \\frac{\\operatorname{ar} \\triangle \\mathrm{PQR}}{\\operatorname{ar} \\triangle \\mathrm{PST}}=\\frac{\\mathrm{PQ}^{2}}{\\mathrm{PS}^{2}} \\\\\n&amp; \\frac{\\operatorname{ar} \\triangle \\mathrm{PQR}}{\\operatorname{ar} \\triangle \\mathrm{PST}}=\\left(\\frac{5}{2}\\right)^{2}=\\frac{25}{4}\n\\end{aligned}\n\\]<\/div>\n<p>\\(\\operatorname{ar} \\Delta \\mathrm{PQR}=25 k\\) and ar \\(\\Delta \\mathrm{PST}=4 k\\) (where \\(k \\neq 0\\) ) Area of quad. SRQT \\(=\\operatorname{ar} \\triangle \\mathrm{PQR}-\\operatorname{ar} \\triangle \\mathrm{PST}\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; =25 k-4 k=21 k \\\\\n\\operatorname{ar} \\triangle \\mathrm{PQR}: \\text { ar quad. } \\mathrm{SRQT} &amp; =25 k: 21 k \\\\\n&amp; =25: 21\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2019-q6-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>The first and last term of a Geometrical Progression (G.P.) are 3 and 96 respectively. If the common ratio is 2, find:<\/p>\n<p>(i) &#8216; \\(\\boldsymbol{n}\\) &#8216; the number of terms of the G.P.<\/p>\n<p>(ii) Sum of the \\(\\boldsymbol{n}\\) terms. <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>First term of G.P. \\((a)=3\\) last term of G.P. \\(\\left(a_{n}\\right)=96\\) and common ratio \\(r=2\\)<\/p>\n<p>(i) Let the number in G.P. be \\(n\\), then<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; a_{n}=96 \\\\\n&amp; a r^{n-1}=96 \\\\\n&amp; 3(2)^{n-1}=96 \\\\\n&amp; 2^{n-1}=\\frac{96}{3}=32 \\\\\n&amp; 2^{\\mathrm{n}-1}=2^{5}\n\\end{aligned}\n\\]<\/div>\n<p>Comparing the powers, we get<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; n-1=5 \\\\\n&amp; n=6\n\\end{aligned}\n\\]<\/div>\n<p>Number of terms in G.P. \\(=6\\) (ii)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; S_{n}=\\frac{a\\left(r^{n}-1\\right)}{r-1} \\\\\n&amp; =\\frac{3\\left(2^{6}-1\\right)}{2-1} \\\\\n&amp; =3(64-1) \\\\\n&amp; =3 \\times 63=189\n\\end{aligned}\n\\]<\/div>\n<p>Sum of \\(n\\) terms of G.P. \\(=189\\)<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2019-q6-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>A hemispherical and a conical hole is scooped out of a solid wooden cylinder. Find the volume of the remaining solid where the measurements are as follows: The height of the solid cylinder is 7 cm, radius of each of hemisphere, cone and cylinder is 3 cm. Height of cone is 3 cm. Give your answer correct to the nearest whole number. \\(\\mid\\) Take \\(\\left.\\pi=\\frac{22}{7} \\right\\rvert\\,\\). <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\/1b794c1d-17ab-4ff8-87dd-647eb47e1e61-05_436_375_271_1231.jpg\" alt=\"ICSE Class 10 Mathematics 2019 Question 6(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, height of cylinder \\((H)=7 \\mathrm{~cm}\\) Radius of hemisphere \\((r)=3 \\mathrm{~cm}\\) Height of cone \\((h)=3 \\mathrm{~cm}\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\pi=\\frac{22}{7}\n\\]<\/div>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/1b794c1d-17ab-4ff8-87dd-647eb47e1e61-06_437_380_870_395.jpg\" alt=\"ICSE Class 10 Mathematics 2019 Question 6(c) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>Volume of remaining figure = (Volume of cylinder &#8211; Volume of cone &#8211; Volume of hemisphere)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; =\\pi r^{2} H-\\left(\\frac{1}{3} \\pi r^{2} h+\\frac{2}{3} \\pi r^{3}\\right) \\\\\n&amp; =\\pi r^{2} H-\\frac{1}{3} \\pi r^{2} h-\\frac{2}{3} \\pi r^{3} \\\\\n&amp; =\\pi r^{2}\\left(H-\\frac{1}{3} h-\\frac{2}{3} r\\right) \\\\\n&amp; =\\frac{22}{7} \\times 3 \\times 3\\left(7-\\frac{3}{3}-\\frac{2}{3} \\times 3\\right) \\\\\n&amp; =\\frac{22 \\times 3 \\times 3}{7}(7-1-2) \\\\\n&amp; =\\frac{198}{7} \\times 4 \\\\\n&amp; =\\frac{792}{7} \\mathrm{~cm}^{3}\n\\end{aligned}\n\\]<\/div>\n<p>\u2234 Volume of remaining solid \\(=113 \\frac{1}{7} \\mathrm{~cm}^{3}\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n=113 \\mathrm{~cm}^{3}\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 2019<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2019-q7-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>In the given figure AC is a tangent to the circle with centre O. If \\(\\angle \\mathbf{A D B}=\\mathbf{5 5}^{\\circ}\\), find \\(\\boldsymbol{x}\\) and \\(\\boldsymbol{y}\\). Give reasons for your answers. <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\/1b794c1d-17ab-4ff8-87dd-647eb47e1e61-06_529_388_271_1222.jpg\" alt=\"ICSE Class 10 Mathematics 2019 Question 7(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>In \\(\\triangle \\mathrm{ABD}\\),<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/1b794c1d-17ab-4ff8-87dd-647eb47e1e61-06_432_374_1410_1230.jpg\" alt=\"ICSE Class 10 Mathematics 2019 Question 7(a) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; \\angle \\mathrm{A}+\\angle \\mathrm{D}+\\angle \\mathrm{B}=180^{\\circ} \\quad(\\text { sum of all angles of } \\Delta) \\\\\n&amp; 90^{\\circ}+55^{\\circ}+\\angle \\mathrm{B}=180^{\\circ}\n\\end{aligned}\n\\]<\/div>\n<p>(AC is the tangent of circle)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{array}{ll}\n\\angle B=180^{\\circ}-145^{\\circ} \\\\\n\\therefore &amp; \\angle B=35^{\\circ} \\\\\n&amp; \\angle A O E=2 \\angle A B E\n\\end{array}\n\\]<\/div>\n<p>(same arc make angles at center and circumference)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\therefore \\quad \\begin{aligned}\n&amp; y=2 \\times 35^{\\circ} \\\\\n&amp; y=70^{\\circ}=\\angle \\mathrm{AOE} \\\\\n&amp; \\text { In } \\triangle \\mathrm{AOC}, \\\\\n&amp; \\angle \\mathrm{~A}+\\angle \\mathrm{O}+\\angle \\mathrm{C}=180^{\\circ} \\\\\n&amp; 90^{\\circ}+70^{\\circ}+x^{\\circ}=180^{\\circ} \\\\\n&amp; x=180-160^{\\circ} \\\\\n&amp; x=20^{\\circ}\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2019-q7-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>The model of a building is constructed with the scale factor 1 : 30.<\/p>\n<p>(i) If the height of the model is 80 cm, find the actual height of the building in meters.<\/p>\n<p>(ii) If the actual volume of a tank at the top of the building is \\(\\mathbf{2 7} \\mathbf{~ m}^{\\mathbf{3}}\\), find the volume of the tank on the top of 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>(i) \\(\\frac{\\text { Height of model }}{\\text { Actual height of the building }}=\\frac{1}{30}\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\frac{80 \\mathrm{~cm}}{\\text { Actual height of the building }}=\\frac{1}{30}\n\\]<\/div>\n<p>Actual height of the building \\(=80 \\times 30 \\mathrm{~cm}\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n=2400 \\mathrm{~cm}=24 \\mathrm{~m}\n\\]<\/div>\n<p>(ii)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; \\frac{\\text { Volume of the tank on the top of model }}{\\text { Actual volume of the tank at the top of building }} \\\\\n&amp; \\qquad=\\left(\\frac{1}{30}\\right)^{3}\n\\end{aligned}\n\\]<\/div>\n\\(\\frac{\\text { Volume of the tank on the top of model }}{27 \\mathrm{~m}^{3}}\\)\n<div class=\"mathjax-overflow\">\\[\n=\\frac{1}{27000}\n\\]<\/div>\n<p>Volume of the tank on the top of model \\(=\\frac{27}{27000}\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n=0.001 \\mathrm{~m}^{3}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2019-q7-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>Given \\(\\left|\\begin{array}{cc}\\mathbf{4} &amp; \\mathbf{2} \\\\ -\\mathbf{1} &amp; \\mathbf{1}\\end{array}\\right|\\), where M is a matrix and I is unit matrix of order 2\u00d72.<\/p>\n<p>(i) State the order of matrix M.<\/p>\n<p>(ii) Find the matrix M. <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>Given, \\(\\left[\\begin{array}{cc}4 &amp; 2 \\\\ -1 &amp; 1\\end{array}\\right] \\mathrm{M}=6 \\mathrm{I}\\)<\/p>\n<p>(i) \\(\\left[\\begin{array}{cc}4 &amp; 2 \\\\ -1 &amp; 1\\end{array}\\right]\\) is the order of \\(2 \\times 2\\)<\/p>\n<p>and I is also the order of 2 \u00d7 2 \u2234 M will be order of 2 \u00d7 2.<\/p>\n<p>(ii) Let \\(\\mathrm{M}=\\left[\\begin{array}{cc}a &amp; b \\\\ c &amp; d\\end{array}\\right]\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\left[\\begin{array}{cc}\n4 &amp; 2 \\\\\n-1 &amp; 1\n\\end{array}\\right]\\left[\\begin{array}{ll}\na &amp; b \\\\\nc &amp; d\n\\end{array}\\right]=6\\left[\\begin{array}{ll}\n1 &amp; 0 \\\\\n0 &amp; 1\n\\end{array}\\right]\n\\]<\/div>\n<div class=\"mathjax-overflow\">\\[\n\\begin{gather*}\n{\\left[\\begin{array}{cc}\n4 a+2 c &amp; 4 b+2 d \\\\\n-a+c &amp; -b+d\n\\end{array}\\right]=\\left[\\begin{array}{ll}\n6 &amp; 0 \\\\\n0 &amp; 6\n\\end{array}\\right]} \\\\\n4 a+2 c=6  \\qquad \\text{&#8230;(i)}\\\\\n4 b+2 d=0  \\qquad \\text{&#8230;(i)}\\\\\n-a+c=0  \\qquad \\text{&#8230;(iii)}\\\\\n-b+d=6 \\qquad \\text{&#8230;(iv)}\n\\end{gather*}\n\\]<\/div>\n<p>From (i) and (iii)<\/p>\n<div class=\"mathjax-overflow\">\\[\na=c=1\n\\]<\/div>\n<p>From (ii) and (iv)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; b=-2 \\text { and } d=4 \\\\\n\\therefore &amp; \\mathrm{M}=\\left[\\begin{array}{cc}\n1 &amp; -2 \\\\\n1 &amp; 4\n\\end{array}\\right]\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 2019<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2019-q8-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>The sum of the first three terms of an Arithmetic Progression (A.P.) is 42 and the product of the first and third term is 52. 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 three terms be \\((a-d), a\\) and \\((a+d)\\). Sum of the first three terms \\(=42\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\na-d+a+a+d &amp; =42 \\\\\n3 a &amp; =42\n\\end{aligned}\n\\]<\/div>\n<p>\u2234<\/p>\n<div class=\"mathjax-overflow\">\\[\na=\\frac{42}{3}=14\n\\]<\/div>\n<p>Product of the first and third term \\(=52\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n(a-d)(a+d) &amp; =52 \\\\\na^{2}-d^{2} &amp; =52 \\\\\n(14)^{2}-d^{2} &amp; =52 \\\\\nd^{2}=196-52 &amp; =144\n\\end{aligned}\n\\]<\/div>\n\\(\\therefore \\quad d= \\pm \\sqrt{144}= \\pm 12\\)\n<p>\\(\\therefore \\quad\\) First term \\(=2\\) and common difference \\(= \\pm 12\\)<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2019-q8-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>The vertices of a \\(\\triangle \\mathrm{ABC}\\) are A(3, 8), B(-1, 2) and C(6, -6), Find:<\/p>\n<p>(i) Slope of BC.<\/p>\n<p>(ii) Equation of a line perpendicular to BC and passing through A. <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) Slope of \\(\\mathrm{BC}=\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\\frac{-6-2}{6+1}\\)<\/p>\n<p>Slope of \\(\\mathrm{BC}=\\frac{-8}{7}\\)<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/1b794c1d-17ab-4ff8-87dd-647eb47e1e61-07_412_412_1532_1215.jpg\" alt=\"ICSE Class 10 Mathematics 2019 Question 8(b) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>(ii) Slope of \\(\\mathrm{BC} \\times\\) Slope of \\(\\mathrm{AD}=-1 \\quad(\\mathrm{BC} \\perp \\mathrm{AD})\\)<\/p>\n<p>\\(\\frac{-8}{7} \\times\\) Slope of \\(\\mathrm{AD}(m)=-1\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\nm=\\frac{7}{8}\n\\]<\/div>\n<p>Equation of line perpendicular to BD i.e,. AD is<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\therefore \\quad \\begin{aligned}\n&amp; y-y_{1}=m\\left(x-x_{1}\\right) \\\\\n&amp; y-8=\\frac{7}{8}(x-3) \\\\\n&amp; 8 y-64=7 x-21 \\\\\n&amp; 7 x-8 y+43=0\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2019-q8-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>Using ruler and a compass only construct a semi-circle with diameter \\(\\mathbf{B C}=\\mathbf{7 ~ c m}\\). Locate a point A on the circumference of the semicircle such that A is equidistant from B and C. Complete the cyclic quadrilateral ABCD, such that D is equidistant from AB and BC. Measure \u2220ADC and write it down. <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>Firstly, draw a semicircle of radius 3.5 cm having centre O.<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/1b794c1d-17ab-4ff8-87dd-647eb47e1e61-08_406_445_355_361.jpg\" alt=\"ICSE Class 10 Mathematics 2019 Question 8(c) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>Justification : A is the equidistant from the point B and C. \u2234 A is lying on perpendicular bisector of BC. D is the equidistant from the side AB and BC. In \\(\\triangle \\mathrm{AOB}\\),<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; \\mathrm{AO}=\\mathrm{OB} \\text { and } \\angle \\mathrm{AOB}=90^{\\circ} \\\\\n\\therefore \\quad &amp; \\angle \\mathrm{ABO}=45^{\\circ}\n\\end{aligned}\n\\]<\/div>\n<p>In cyclic quadrilateral ABCD, \\(\\angle \\mathrm{ABC}+\\angle \\mathrm{ADC}=180^{\\circ}\\) (Opposite angles of cyclic quadrilateral)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; 45^{\\circ}+\\angle \\mathrm{ADC}=180^{\\circ} \\\\\n&amp; \\angle \\mathrm{ADC}=180^{\\circ}-45^{\\circ} \\\\\n\\therefore \\quad &amp; \\angle \\mathrm{ADC}=135^{\\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\">9<\/div><div><div class=\"xc-set-title\">Question 9<\/div><div class=\"xc-set-sub\">ICSE Class 10 Mathematics 2019<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2019-q9-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>The data on the number of patients attending a hospital in a month are given below. Find the average (mean) number of patients attending the hospital in a month by using the shortcut method. Take the assumed mean as 45. Give your answer correct to 2 decimal places. <strong>[3]<\/strong><\/p>\n<div class=\"math-responsive\"><table><tbody><tr><td>Number of patients<\/td><td>\\(10-20\\)<\/td><td>\\(20-30\\)<\/td><td>\\(30-40\\)<\/td><td>\\(40-50\\)<\/td><td>\\(50-60\\)<\/td><td>\\(60-70\\)<\/td><\/tr><tr><td>Number of days<\/td><td>5<\/td><td>2<\/td><td>7<\/td><td>9<\/td><td>2<\/td><td>5<\/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><p>Here class width \\(h=10\\),<\/p>\n<div class=\"math-responsive\"><table><tbody><tr><td>Number of patients<\/td><td>Number of days \\(\\boldsymbol{(} \\boldsymbol{f} \\boldsymbol{)}\\)<\/td><td>\\(x\\)<\/td><td>\\(d=\\frac{\\boldsymbol{x}-\\boldsymbol{A}}{\\boldsymbol{h}}\\)<\/td><td>\\(f d\\)<\/td><\/tr><tr><td>\\(10-20\\)<\/td><td>5<\/td><td>15<\/td><td>-3<\/td><td>-15<\/td><\/tr><tr><td>\\(20-30\\)<\/td><td>2<\/td><td>25<\/td><td>-2<\/td><td>-4<\/td><\/tr><tr><td>\\(30-40\\)<\/td><td>7<\/td><td>35<\/td><td>-1<\/td><td>-7<\/td><\/tr><tr><td>\\(40-50\\)<\/td><td>9<\/td><td>\\(45=A\\)<\/td><td>0<\/td><td>0<\/td><\/tr><tr><td>\\(50-60\\)<\/td><td>2<\/td><td>55<\/td><td>1<\/td><td>2<\/td><\/tr><tr><td>\\(60-70\\)<\/td><td>5<\/td><td>65<\/td><td>2<\/td><td>10<\/td><\/tr><tr><td><\/td><td>\\(\\Sigma f=30\\)<\/td><td><\/td><td><\/td><td>\\(\\Sigma f d=-14\\)<\/td><\/tr><\/tbody><\/table><\/div>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; \\bar{x}=A+\\frac{\\sum f d}{\\sum f} \\times h \\\\\n&amp; =45+\\frac{(-14)}{30} \\times 10 \\\\\n&amp; =45-4.666 \\\\\n&amp; =40.334 \\\\\n&amp; \\bar{x}=40.33\n\\end{aligned}\n\\]<\/div>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; \\frac{2 \\sqrt{5 x}}{2 \\sqrt{2 x-6}}=\\frac{5}{3} \\\\\n&amp; \\frac{\\sqrt{5 x}}{\\sqrt{2 x-6}}=\\frac{5}{3} \\quad \\text { [squaring both sides] } \\\\\n&amp; \\frac{5 x}{2 x-6}=\\frac{25}{9} \\\\\n&amp; 50 x-150=45 x\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2019-q9-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>Using properties of proportion solve for \\(\\boldsymbol{x}\\), given <strong>[3]<\/strong><\/p>\n<div class=\"mathjax-overflow\">\\[\n\\frac{\\sqrt{5 x}+\\sqrt{2 x-6}}{\\sqrt{5 x}-\\sqrt{2 x-6}}=4\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{\\sqrt{5 x}+\\sqrt{2 x-6}}{\\sqrt{5 x}+\\sqrt{2 x-6}}=4\n\\]<\/div>\n<p>By componendo-dividendo<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\frac{\\sqrt{5 x}+\\sqrt{2 x-6}+\\sqrt{5 x}-\\sqrt{2 x-6}}{\\sqrt{5 x}+\\sqrt{2 x-6}-\\sqrt{5 x}+\\sqrt{2 x-6}}=\\frac{4+1}{4-1}\n\\]<\/div>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; 5 x=150 \\\\\n&amp; x=\\frac{150}{5} \\\\\n\\therefore \\quad &amp; x=30\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2019-q9-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>Sachin invests \u20b9 8500 in 10%, \u20b9 100 shares at \u20b9 170. He sells the shares when the price of each share rises by \u20b9 30. He invests the proceeds in 12%, \u20b9 100 shares at \u20b9 125. Find:<\/p>\n<p>(i) the sale proceeds.<\/p>\n<p>(ii) the number of \u20b9 125 shares he buys.<\/p>\n<p>(iii) the change in his annual income. <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>Total investment \\(=\\text{\u20b9} 8500\\) Market value of each share \\(=\\text{\u20b9} 170\\) Number of shares purchased \\(=\\frac{8500}{170}=50\\) Dividend received \\(=\\text{\u20b9}\\left(\\frac{10}{100}\\right) \\times 50 \\times 100=\\) \u20b9500 Now, market value of each share \\(=\\text{\u20b9}(170+30)\\) = \u20b9 200 Amount received on selling \\(=\\text{\u20b9}(50 \\times 200)=\\) \u20b910000 Market value of new shares \\(=\\text{\u20b9} 125\\) each Number of shares purchased \\(=\\frac{10000}{125}=80\\) Dividend received \\(=\\left(\\frac{12}{100}\\right) \\times 80 \\times 100=\\text{\u20b9} 960\\) Change in income \\(=\\text{\u20b9}(960-500)=\\text{\u20b9} 460\\) Selling price of these shares = Market value of one share \u00d7 Number of shares<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; =200 \\times 50 \\\\\n&amp; =\\text{\u20b9} 10,000\n\\end{aligned}\n\\]<\/div>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\text { Number of shares purchased } &amp; =\\frac{10000}{125} \\\\\n&amp; =80\n\\end{aligned}\n\\]<\/div>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\text { Annual Income of these shares } &amp; =\\frac{12}{100} \\times 100 \\times 80 \\\\\n&amp; =\\text{\u20b9} 960\n\\end{aligned}\n\\]<\/div>\n<p>The change in his Annual Income \\(=\\text{\u20b9}(960-\\) 500) = \u20b9 460<\/p>\n<p>(i) Sale proceeds \\(=\\text{\u20b9} 10000\\)<\/p>\n<p>(ii) Number of \u20b9 125 shares = 80<\/p>\n<p>(iii) Change in his annual Income \\(=\\text{\u20b9} 460\\)<\/p><\/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 2019<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2019-q10-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>Use graph paper for this question. The marks obtained by 120 students in an English test are given below:<\/p>\n<div class=\"math-responsive\"><table><tbody><tr><td>Marks<\/td><td>\\(0-10\\)<\/td><td>\\(10-20\\)<\/td><td>\\(20-30\\)<\/td><td>\\(30-40\\)<\/td><td>\\(40-50\\)<\/td><td>\\(50-60\\)<\/td><td>\\(60-70\\)<\/td><td>\\(70-80\\)<\/td><td>\\(80-90\\)<\/td><td>\\(90-100\\)<\/td><\/tr><tr><td>No. of students<\/td><td>5<\/td><td>9<\/td><td>16<\/td><td>22<\/td><td>26<\/td><td>18<\/td><td>11<\/td><td>6<\/td><td>4<\/td><td>3<\/td><\/tr><\/tbody><\/table><\/div>\n<p>Draw the ogive and hence, estimate:<\/p>\n<p>(i) the median marks.<\/p>\n<p>(ii) the number of students who did not pass the test if the pass percentage was 50.<\/p>\n<p>(iii) the upper quartile marks. <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=\"math-responsive\"><table><thead><tr><th>Marks<\/th><th>No. of students<\/th><th>Less than<\/th><th>c.f. Less Than Type<\/th><\/tr><\/thead><tbody><tr><td>\\(0-10\\)<\/td><td>5<\/td><td>10<\/td><td>5<\/td><\/tr><tr><td>\\(10-20\\)<\/td><td>9<\/td><td>20<\/td><td>14<\/td><\/tr><tr><td>\\(20-30\\)<\/td><td>16<\/td><td>30<\/td><td>30<\/td><\/tr><tr><td>\\(30-40\\)<\/td><td>22<\/td><td>40<\/td><td>52<\/td><\/tr><tr><td>\\(40-50\\)<\/td><td>26<\/td><td>50<\/td><td>78<\/td><\/tr><tr><td>\\(50-60\\)<\/td><td>18<\/td><td>60<\/td><td>96<\/td><\/tr><tr><td>\\(60-70\\)<\/td><td>11<\/td><td>70<\/td><td>107<\/td><\/tr><tr><td>\\(70-80\\)<\/td><td>6<\/td><td>80<\/td><td>113<\/td><\/tr><tr><td>\\(80-90\\)<\/td><td>4<\/td><td>90<\/td><td>117<\/td><\/tr><tr><td>\\(90-100\\)<\/td><td>3<\/td><td>100<\/td><td>120<\/td><\/tr><tr><td><\/td><td>120<\/td><td><\/td><td><\/td><\/tr><\/tbody><\/table><\/div>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/1b794c1d-17ab-4ff8-87dd-647eb47e1e61-09_897_781_1171_1020.jpg\" alt=\"ICSE Class 10 Mathematics 2019 Question 10(a) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>Here, \\(\\mathrm{N}=120\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\frac{\\mathrm{N}}{2}=\\frac{120}{2}=60\n\\]<\/div>\n<p>(i) Median = 42 marks<\/p>\n<p>(ii) The number of student who did not pass the test if the pass percentage was 50 is 78.<\/p>\n<p>(iii) The upper quartile marks is 57.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2019-q10-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>A man observes the angle of elevation of the top of the tower to be 45\u00b0. He walks towards it in a horizontal line through its base. On covering 20 m the angle of elevation changes to 60\u00b0. Find the height of the tower correct to 2 significant figures. <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 height of tower be \\(h \\mathrm{~m}\\). Angle of elevation of the top of tower is 45\u00b0 and 60\u00b0 from the points A and B respectively.<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/1b794c1d-17ab-4ff8-87dd-647eb47e1e61-10_401_530_269_318.jpg\" alt=\"ICSE Class 10 Mathematics 2019 Question 10(b) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<div class=\"mathjax-overflow\">\\[\n\\therefore \\quad \\mathrm{AB}=20 \\mathrm{~m}\n\\]<\/div>\n<p>Let OB be \\(x \\mathrm{~m}\\), and OP is the tower having height \\(=h \\mathrm{~m}\\).<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; \\text { In } \\triangle \\mathrm{AOP}, \\angle \\mathrm{O}=90^{\\circ} \\\\\n&amp; \\tan \\mathrm{A}=\\frac{\\mathrm{OP}}{\\mathrm{AO}} \\\\\n&amp; \\tan 45^{\\circ}=\\frac{h}{x+20} \\\\\n&amp; 1=\\frac{h}{x+20} \\\\\n&amp; x+20=h \\qquad \\text{&#8230;(i)}\n\\end{aligned}\n\\]<\/div>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; \\text { In } \\triangle \\mathrm{BOP}, \\angle \\mathrm{O}=90^{\\circ} \\\\\n&amp; \\tan \\mathrm{B}=\\frac{\\mathrm{OP}}{\\mathrm{OB}} \\\\\n&amp; \\tan 60^{\\circ}=\\frac{h}{x} \\\\\n&amp; \\sqrt{3}=\\frac{h}{x} \\\\\n&amp; h=\\sqrt{3} x \\qquad \\text{&#8230;(ii)}\n\\end{aligned}\n\\]<\/div>\n<p>from eq. (i) and (ii)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; h=\\sqrt{3}(h-20) \\Rightarrow h(\\sqrt{3}-1)=20 \\sqrt{3} \\\\\n&amp; h=\\frac{20 \\sqrt{3}}{\\sqrt{3}-1} \\\\\n&amp; h=\\frac{20 \\sqrt{3}}{\\sqrt{3}-1} \\times \\frac{\\sqrt{3}+1}{\\sqrt{3}+1} \\\\\n&amp; h=\\frac{20 \\sqrt{3}(\\sqrt{3}+1)}{3-1} \\\\\n&amp; h=\\frac{20(3+\\sqrt{3})}{2} \\\\\n&amp; h=10(3+\\sqrt{3}) \\\\\n&amp; h=10(3+1.73) \\\\\n&amp; h=47.3 \\mathrm{~m}\n\\end{aligned}\n\\]<\/div>\n<p>\u2234 Height of the tower \\(=47 \\mathrm{~m}\\)<\/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 2019<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2019-q11-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>Using the Remainder Theorem, find the remainders obtained when \\(\\boldsymbol{x}^{\\mathbf{3}}+(\\mathbf{R} \\boldsymbol{x}+\\mathbf{8}) \\boldsymbol{x}+\\mathbf{R}\\) is divided by \\(\\boldsymbol{x}+\\mathbf{1}\\) and \\(\\boldsymbol{x &#8211; 2}\\). Hence, find \\(\\boldsymbol{k}\\) if the sum of the two remainders is 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><p>Let \\(\\mathrm{R}_{1}\\) and \\(\\mathrm{R}_{2}\\) are remainders when given polynomial is divided by \\((x+1)\\) and \\((x-2)\\) respectively. Let \\(P(x)=x 3+(k x+8) \\mathrm{x}+8\\) Remainder<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; \\mathrm{R}_{1}=P(-1)=(-1)^{3}+(\\mathrm{K}(-1)+8)(-1)+\\mathrm{k} \\\\\n&amp; R_{1}=P(-1)=-1+k-8+k=2 k-9 \\\\\n&amp; \\mathrm{R}_{1}=2 k-9\n\\end{aligned}\n\\]<\/div>\n<p>Remainder<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; \\mathrm{R}_{2}=\\mathrm{P}(2)=(2)^{3}+\\{k(2)+8\\}(2)+k \\\\\n&amp; \\mathrm{R}_{2}=\\mathrm{P}(2)=8+4 k+16+k \\\\\n&amp; \\mathrm{R}_{2}=5 k+24 \\\\\n&amp; \\mathrm{R}_{1}+\\mathrm{R}_{2}=1 \\\\\n&amp; 2 k-9+5 k+24=1 \\\\\n&amp; 7 k=-14 \\\\\n&amp; k=-2 \\quad \\therefore k=-2\n\\end{aligned}\n\\]<\/div>\n<p>(Given)<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2019-q11-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>The product of two consecutive natural numbers which are multiples of 3 is equal to 810. Find the two 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 two consecutive natural numbers that are multiple of 3 be \\(3 x\\) and \\(3 x+3\\). Given : Product of two consecutive numbers = 810<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; 3 x(3 x+3)=810 \\\\\n&amp; 9 x(x+1)=810 \\\\\n&amp; x(x+1)=\\frac{810}{9}\n\\end{aligned}\n\\]<\/div>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; x^{2}+x-90=0 \\\\\n&amp; x^{2}+10 x-9 x-90=0 \\\\\n&amp; x(x+10)-9(x+10)=0 \\\\\n&amp; (x+10)(x-9)=0\n\\end{aligned}\n\\]<\/div>\n<p>If, \\(x+10=0 \\Rightarrow x=-10\\) It is not natural number<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\text { If, } x-9=0 \\Rightarrow x=9\n\\]<\/div>\n<p>\u2234 Numbers are \\(3 x=3 \\times 9=27\\) and \\(3 x+3=27+3=30\\)<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2019-q11-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>In the given figure, ABCDE is a pentagon inscribed in a circle such that AC is a diameter and side \\(\\mathbf{B C}\\left|\\mid \\mathbf{A E}\\right.\\). If \\(\\angle \\mathbf{B A C}=\\mathbf{5 0}^{\\circ}\\), find giving reasons:<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/1b794c1d-17ab-4ff8-87dd-647eb47e1e61-10_387_365_549_1235.jpg\" alt=\"ICSE Class 10 Mathematics 2019 Question 11(c) figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>(i) \\(\\angle \\mathrm{ACB}\\)<\/p>\n<p>(ii) \u2220EDC<\/p>\n<p>(iii) \u2220BEC<\/p>\n<p>Hence, prove that BE is also a diameter. <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) AC is a diameter of circle.<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/1b794c1d-17ab-4ff8-87dd-647eb47e1e61-11_387_366_546_400.jpg\" alt=\"ICSE Class 10 Mathematics 2019 Question 11(c) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n\\(\\therefore \\quad \\angle \\mathrm{ABC}=90^{\\circ}\\)\n<p>In \\(\\triangle \\mathrm{ABC}\\),<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\angle \\mathrm{BAC}+\\angle \\mathrm{ABC}+\\angle \\mathrm{ACB}=180^{\\circ}\n\\]<\/div>\n<p>(sum of angles of \\(\\Delta\\) )<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{array}{ll}\n&amp; 50^{\\circ}+90^{\\circ}+\\angle \\mathrm{ACB}=180^{\\circ} \\\\\n\\therefore &amp; \\angle \\mathrm{ACB}=180^{\\circ}-140^{\\circ} \\\\\n\\therefore &amp; \\angle \\mathrm{ACB}=40^{\\circ}\n\\end{array}\n\\]<\/div>\n<p>(ii) BC||AE<\/p>\n<p>\\(\\therefore \\angle \\mathrm{EAC}=\\angle \\mathrm{ACB}\\) (Alternative angles)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\angle \\mathrm{EAC}=40^{\\circ}\n\\]<\/div>\n<p>Since, ACDE is a cyclic quadrilateral.<\/p>\n\\(\\therefore \\quad \\angle \\mathrm{EAC}+\\angle \\mathrm{EDC}=180^{\\circ}\\)\n<p>(Sum of opposite angles of cyclic quadrilateral is 180\u00b0)<\/p>\n\\(\\therefore \\angle \\mathrm{EDC}=180^{\\circ}-40^{\\circ}\\)\n<div class=\"mathjax-overflow\">\\[\n\\angle \\mathrm{EDC}=140^{\\circ}\n\\]<\/div>\n<p>(iii) \\(\\angle \\mathrm{BEC}=\\angle \\mathrm{BAC}\\) (same arc (BC))<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\therefore \\quad \\begin{aligned}\n&amp; \\angle \\mathrm{BEC}=50^{\\circ} \\\\\n&amp; \\angle \\mathrm{BAE}=\\angle \\mathrm{BAC}+\\angle \\mathrm{CAE}=50^{\\circ}+40^{\\circ}=90^{\\circ}\n\\end{aligned}\n\\]<\/div>\n<p>\u2234 BE is a diameter, as the angle subtended by a diameter on any point of circle is 90\u00b0.<\/p><\/div>\n<\/div>\n<\/div>\n<\/div>\n\n\n\n\n<p><\/p>\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\">33<\/span> <span class=\"rt-label rt-postfix\">minutes<\/span><\/span> ICSE Class 10 Mathematics PYQ Practice the complete ICSE Class 10 Mathematics 2019 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 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