{"id":2401,"date":"2026-11-13T17:44:20","date_gmt":"2026-11-13T12:14:20","guid":{"rendered":"https:\/\/xamcontent.com\/cisce\/?p=2401"},"modified":"2026-11-13T17:44:21","modified_gmt":"2026-11-13T12:14:21","slug":"icse-class-10-mathematics-2018-question-paper-with-solutions","status":"publish","type":"post","link":"https:\/\/xamcontent.com\/cisce\/icse-class-10-mathematics-2018-question-paper-with-solutions\/","title":{"rendered":"ICSE Class 10 Mathematics 2018 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-2018-question-paper-with-solutions\/#ICSE_Mathematics_2018_%E2%80%94_Section_A\" title=\"ICSE Mathematics 2018 \u2014 Section A\">ICSE Mathematics 2018 \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-2018-question-paper-with-solutions\/#ICSE_Mathematics_2018_%E2%80%94_Section_B\" title=\"ICSE Mathematics 2018 \u2014 Section B\">ICSE Mathematics 2018 \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\">34<\/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 2018 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\">2018<\/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 2018<\/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_2018_%E2%80%94_Section_A\"><\/span>ICSE Mathematics 2018 \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 2018<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2018-q1-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>Find the value of &#8216; \\(\\boldsymbol{x}\\) &#8216; and &#8216; \\(\\boldsymbol{y}\\) &#8216; if : <strong>[3]<\/strong><\/p>\n<div class=\"mathjax-overflow\">\\[\n2\\left[\\begin{array}{cc}\nx &amp; 7 \\\\\n9 &amp; y-5\n\\end{array}\\right]+\\left[\\begin{array}{cc}\n6 &amp; -7 \\\\\n4 &amp; 5\n\\end{array}\\right]=\\left[\\begin{array}{cc}\n10 &amp; 7 \\\\\n22 &amp; 15\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{gathered}\n2\\left[\\begin{array}{cc}\nx &amp; 7 \\\\\n9 &amp; y-5\n\\end{array}\\right]+\\left[\\begin{array}{cc}\n6 &amp; -7 \\\\\n4 &amp; 5\n\\end{array}\\right]=\\left[\\begin{array}{cc}\n10 &amp; 7 \\\\\n22 &amp; 5\n\\end{array}\\right] \\\\\n{\\left[\\begin{array}{cc}\n2 x &amp; 14 \\\\\n18 &amp; 2 y-10\n\\end{array}\\right]+\\left[\\begin{array}{cc}\n6 &amp; -7 \\\\\n4 &amp; 5\n\\end{array}\\right]=\\left[\\begin{array}{cc}\n10 &amp; 7 \\\\\n22 &amp; 15\n\\end{array}\\right]}\n\\end{gathered}\n\\]<\/div>\n<p>or<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\left[\\begin{array}{cc}\n2 x+6 &amp; 7 \\\\\n22 &amp; 2 y-5\n\\end{array}\\right]=\\left[\\begin{array}{cc}\n10 &amp; 7 \\\\\n22 &amp; 15\n\\end{array}\\right]\n\\]<\/div>\n<p>By equating \\(2 x+6=10\\) or \\(x=2\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n2 y-5=15 \\text { or } y=10\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2018-q1-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>Sonia had a recurring deposit account in a bank and deposited \u20b9 600 per month for \\(\\mathbf{2} \\frac{\\mathbf{1}}{\\mathbf{2}}\\) years. If the rate of interest was 10% p.a., find the maturity value of this account. <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 &amp; =\\text{\u20b9} 600 \\\\\nn &amp; =2 \\frac{1}{2} \\text { year }=30 \\text { months } \\\\\nr &amp; =10 \\% \\text { p.a. } \\\\\n\\text { M.V. } &amp; =? \\\\\n\\text { M.V. } &amp; =p \\times n+\\frac{p \\times n(n+1) \\times r}{2,400}\n\\end{aligned}\n\\]<\/div>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; =600 \\times 30+\\frac{600 \\times 30 \\times 31 \\times 10}{2,400} \\\\\n&amp; =18,000+2,325 \\\\\n&amp; =\\text{\u20b9} 20,325\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2018-q1-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>Cards bearing numbers 2, 4, 6, 8, 10, 12, 14, 16, 18 and 20 are kept in a bag. A card is drawn at random from the bag. Find the probability of getting a card which is :<\/p>\n<p>(i) a prime number.<\/p>\n<p>(ii) a number divisible by 4 .<\/p>\n<p>(iii) a number that is a multiple of 6.<\/p>\n<p>(iv) an odd number. <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=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\text { Prime number } &amp; =\\{2\\} \\\\\n\\text { No. of favourable cards } &amp; =1 \\\\\n\\text { Total number of cards } &amp; =10\n\\end{aligned}\n\\]<\/div>\n<p>Hence, probability of a getting a prime number card<\/p>\n<div class=\"mathjax-overflow\">\\[\n=\\frac{\\text { Number of favourable cards }}{\\text { Total no. of cards }}=\\frac{1}{10}\n\\]<\/div>\n<p>(ii) Number divisible by \\(4=\\{4,8,12,16,20\\}\\)<\/p>\n<p>No. of favourable cards \\(=5\\) Hence, probability of a getting card, where number divisible by 4 .<\/p>\n<div class=\"mathjax-overflow\">\\[\n=\\frac{\\text { Number of favourable cards }}{\\text { Total no. of cards }}=\\frac{5}{10}=\\frac{1}{2}\n\\]<\/div>\n<p>(iii) Number which are multiple of \\(6=\\) \\{6, 12, 18\\}<\/p>\n<p>No. of favourable cards \\(=3\\) Hence, probability of getting card, which is multiple of 6.<\/p>\n<div class=\"mathjax-overflow\">\\[\n=\\frac{\\text { Number of favourable cards }}{\\text { Total No. of cards }}=\\frac{3}{10}\n\\]<\/div>\n<p>(iv) Odd number = 0 (No. odd card)<\/p>\n<p>No. of favourable cards \\(=0\\) Hence, probability of getting an odd number card<\/p>\n<div class=\"mathjax-overflow\">\\[\n=\\frac{\\text { Number of favourable cards }}{\\text { Total No. of cards }}=\\frac{0}{10}=0\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 2018<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2018-q2-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. Find the<\/p>\n<p>(i) radius of the cylinder<\/p>\n<p>(ii) volume of cylinder. (use \\(\\pi=\\frac{\\mathbf{2 2}}{\\mathbf{7}}\\) ) <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>Given<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\text { circumference } &amp; =132 \\mathrm{~cm} \\\\\n\\text { height } &amp; =25 \\mathrm{~cm}\n\\end{aligned}\n\\]<\/div>\n<p>(i)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\text { circumference } &amp; =2 \\pi r \\\\\n132 &amp; =2 \\pi r \\\\\n132 &amp; =\\frac{2 \\times 22}{7} \\times r \\\\\nr &amp; =\\frac{132 \\times 7}{2 \\times 22}=21 \\mathrm{~cm}\n\\end{aligned}\n\\]<\/div>\n<p>Hence, radius of cylinder \\(=21 \\mathrm{~cm}\\).<\/p>\n<p>(ii) volume of cylinder \\(=\\pi r^{2} h\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; =\\frac{22}{7} \\times 21 \\times 21 \\times 25 \\\\\n&amp; =34,650 \\mathrm{~cm}^{3}\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2018-q2-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>If \\((\\boldsymbol{k}-\\mathbf{3}),(\\mathbf{2} \\boldsymbol{k}+\\mathbf{1})\\) and \\((\\mathbf{4} \\boldsymbol{k}+\\mathbf{3})\\) are three consecutive terms of an A.P., find the value of \\(\\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>For three consecutive terms of an A.P., the common difference should be same, i.e,.<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n(2 k+1)-(k-3) &amp; =(4 k+3)-(2 k+1) \\\\\n2 k+1-k+3 &amp; =4 k+3-2 k-1 \\\\\nk+4 &amp; =2 k+2 \\\\\nk &amp; =2\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2018-q2-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>\\(P Q R S\\) is a cyclic quadrilateral. Given \\(\\angle Q P S=\\) \\(73^{\\circ}, \\angle P Q S=55^{\\circ}\\) and \\(\\angle P S R=82^{\\circ}\\), calculate :<\/p>\n<p>(i) \\(\\angle Q R S\\)<\/p>\n<p>(ii) \\(\\angle R Q S\\)<\/p>\n<p>(iii) \\(\\angle P R Q\\) <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\/12126fb9-3243-49d7-ba04-c01fd01e9dfc-2_374_361_861_403.jpg\" alt=\"ICSE Class 10 Mathematics 2018 Question 2(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>(i) Since, sum of opposite pairs of angles in cyclic quadrilaterals is 180\u00b0.<\/p>\n<p>Hence,<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\angle Q R S+\\angle Q P S &amp; =180^{\\circ} \\\\\n\\angle Q R S+73^{\\circ} &amp; =180^{\\circ} \\\\\n\\angle Q R S &amp; =180^{\\circ}-73^{\\circ} \\\\\n&amp; =107^{\\circ}\n\\end{aligned}\n\\]<\/div>\n<p>(ii)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\angle P Q R+\\angle P S R=180^{\\circ}\n\\]<\/div>\n<p>(PQRS is a cyclic quadrilateral)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\angle P Q R+82^{\\circ} &amp; =180^{\\circ} \\\\\n\\angle P Q R &amp; =180^{\\circ}-82 \\\\\n&amp; =98^{\\circ}\n\\end{aligned}\n\\]<\/div>\n\\(\\because \\quad \\angle P Q S+\\angle R Q S=\\angle P Q R\\)\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n55^{\\circ}+\\angle R Q S &amp; =98^{\\circ} \\\\\n\\angle R Q S &amp; =98^{\\circ}-55^{\\circ} \\\\\n&amp; =43^{\\circ}\n\\end{aligned}\n\\]<\/div>\n<p>(iii) Join \\(P\\) to \\(R\\)<\/p>\n<p>In \\(\\triangle P S Q\\),<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\angle S P Q+\\angle P Q S+\\angle P S Q &amp; =180^{\\circ} \\\\\n73^{\\circ}+55^{\\circ}+\\angle P S Q &amp; =180^{\\circ} \\\\\n\\angle P S Q &amp; =180^{\\circ}-128^{\\circ} \\\\\n&amp; =52^{\\circ}\n\\end{aligned}\n\\]<\/div>\n<p>Now, \\(\\angle P R Q=\\angle P S Q(\\because\\) Angle in same segment \\()\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\therefore \\quad \\angle P R Q=52^{\\circ}\n\\]<\/div>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/12126fb9-3243-49d7-ba04-c01fd01e9dfc-2_371_359_1500_1236.jpg\" alt=\"ICSE Class 10 Mathematics 2018 Question 2(c) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div><\/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 2018<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2018-q3-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>If \\((\\boldsymbol{x}+\\mathbf{2})\\) and \\((\\boldsymbol{x}+\\mathbf{3})\\) are factors of \\(\\boldsymbol{x}^{\\mathbf{3}}+\\boldsymbol{a} \\boldsymbol{x}+\\boldsymbol{b}\\), find the values of &#8216; \\(\\boldsymbol{a}\\) &#8216; and &#8216; \\(\\boldsymbol{b}\\) &#8216;. <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>\\(x^{3}+a x+b\\)\n<p>\\((x+2)\\) is a factor of given polynomial then \\(x=-2\\) will satisfy the polynomial<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n(-2)^{3}+a(-2)+b &amp; =0 \\\\\n-8-2 a+b &amp; =0 \\\\\nb-2 a &amp; =8 \\qquad \\text{&#8230;(i)}\n\\end{aligned}\n\\]<\/div>\n<p>\\((x+3)\\) is also a factor of given polynomial then \\(x=-3\\) will satisfy the polynomial<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n(-3)^{3}+a(-3)+b &amp; =0 \\\\\n-27-3 a+b &amp; =0 \\\\\nb-3 a &amp; =27 \\qquad \\text{&#8230;(ii)}\n\\end{aligned}\n\\]<\/div>\n<p>On solving (i) and (ii), we get<\/p>\n<div class=\"mathjax-overflow\">\\[\na=-19, b=-30\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2018-q3-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>Prove that \\(\\sqrt{\\boldsymbol{\\operatorname { s e c }}^{\\mathbf{2}} \\theta+\\boldsymbol{\\operatorname { c o s e c }}^{\\mathbf{2}} \\theta}=\\boldsymbol{\\operatorname { t a n }} \\theta+\\boldsymbol{\\operatorname { c o t }} \\theta\\) <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\\text { L.H.S. } &amp; =\\sqrt{\\sec ^{2} \\theta+\\operatorname{cosec}^{2} \\theta} \\\\\n&amp; =\\sqrt{1+\\tan ^{2} \\theta+1+\\cot ^{2} \\theta}\n\\end{aligned}\n\\]<\/div>\n<p>[Since, \\(1+\\tan ^{2} \\theta=\\sec ^{2} \\theta\\) and \\(1+\\cot ^{2} \\theta=\\operatorname{cosec}^{2} \\theta\\) ]<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; =\\sqrt{\\tan ^{2} \\theta+\\cot ^{2} \\theta+2} \\\\\n&amp; =\\sqrt{\\tan ^{2} \\theta+\\cot ^{2} \\theta+2 \\tan \\theta \\cdot \\cot \\theta} \\\\\n&amp; \\quad \\quad[\\because \\tan \\theta \\cdot \\cot \\theta=1] \\\\\n&amp; =\\sqrt{(\\tan \\theta+\\cot \\theta)^{2}} \\\\\n&amp; =\\tan \\theta+\\cot \\theta=\\text { R.H.S. }\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2018-q3-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>Using a graph paper draw a histogram for the given distribution showing the number of runs scored by 50 batsmen. Estimate the mode of the data : <strong>[4]<\/strong><\/p>\n<div class=\"math-responsive\"><table><thead><tr><th>Runs scored<\/th><th>3000-4000<\/th><th>4000-5000<\/th><th>5000-6000<\/th><th>6000-7000<\/th><th>7000-8000<\/th><th>8000-9000<\/th><th>9000-10000<\/th><\/tr><\/thead><tbody><tr><td>No. of batsmen<\/td><td>4<\/td><td>18<\/td><td>9<\/td><td>6<\/td><td>7<\/td><td>2<\/td><td>4<\/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=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/12126fb9-3243-49d7-ba04-c01fd01e9dfc-3_894_848_1150_188.jpg\" alt=\"ICSE Class 10 Mathematics 2018 Question 3(c) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<div class=\"mathjax-overflow\">\\[\n\\text { Mode = } 4600\n\\]<\/div><\/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 2018<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2018-q4-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>Solve the following inequation, write down the solution set and represent it on the real number line : <strong>[3]<\/strong><\/p>\n<div class=\"mathjax-overflow\">\\[\n-2+10 x \\leq 13 x+10&lt;24+10 x, x \\in Z\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><p>or<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n-2+10 x &amp; \\leq 13 x+10&lt;24+10 x \\\\\n-2 &amp; \\leq 3 x+10&lt;24\n\\end{aligned}\n\\]<\/div>\n<p>or<\/p>\n<div class=\"mathjax-overflow\">\\[\n-12 \\leq 3 x&lt;14\n\\]<\/div>\n<p>or<\/p>\n<div class=\"mathjax-overflow\">\\[\n-4 \\leq x&lt;\\frac{14}{3}\n\\]<\/div>\n<p>or<\/p>\n<div class=\"mathjax-overflow\">\\[\n-4 \\leq x&lt;4 \\frac{2}{3}\n\\]<\/div>\n<p>\u2234 The solution set is \\(x \\in[-4,4]\\) and \\(x \\in Z\\)<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/12126fb9-3243-49d7-ba04-c01fd01e9dfc-3_64_800_568_1024.jpg\" alt=\"ICSE Class 10 Mathematics 2018 Question 4(a) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2018-q4-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>If the straight lines \\(\\mathbf{3} \\boldsymbol{x}-\\mathbf{5} \\boldsymbol{y}=\\mathbf{7}\\) and \\(\\mathbf{4} \\boldsymbol{x}+\\boldsymbol{a} \\boldsymbol{y}+\\) \\(\\mathbf{9}=\\mathbf{0}\\) are perpendicular to one another, find value of \\(\\boldsymbol{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>Eqn. (i)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n3 x-5 y &amp; =7 \\\\\ny &amp; =\\frac{3}{5} x-\\frac{7}{5}\n\\end{aligned}\n\\]<\/div>\n<p>\u2234<\/p>\n<div class=\"mathjax-overflow\">\\[\nm_{1}=\\frac{3}{5}\n\\]<\/div>\n<p>Eqn. (ii)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n4 x+a y+9 &amp; =0 \\\\\ny &amp; =\\frac{-4}{a} x-\\frac{9}{a}\n\\end{aligned}\n\\]<\/div>\n<p>\u2234<\/p>\n<div class=\"mathjax-overflow\">\\[\nm_{2}=\\frac{-4}{a}\n\\]<\/div>\n<p>Both lines are perpendicular to each other. \u2234<\/p>\n<div class=\"mathjax-overflow\">\\[\nm_{1} \\times m_{2}=-1\n\\]<\/div>\n<p>or<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\frac{3}{5} \\times\\left(\\frac{-4}{a}\\right)=-1\n\\]<\/div>\n<p>or<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\frac{-12}{5 a} &amp; =-1 \\\\\na &amp; =\\frac{12}{5}\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2018-q4-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>Solve \\(\\boldsymbol{x}^{\\mathbf{2}}+\\mathbf{7} \\boldsymbol{x}=\\mathbf{7}\\) and give your answer correct to two decimal places. <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{array}{r}\nx^{2}+7 x=7 \\\\\nx^{2}+7 x-7=0\n\\end{array}\n\\]<\/div>\n<p>Compare the equation to<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\na x^{2}+b x+c &amp; =0 \\\\\na &amp; =1, b=7, c=-7 \\\\\nD &amp; =b^{2}-4 a c \\\\\n&amp; =(7)^{2}-4 \\times 1 \\times(-7) \\\\\n&amp; =49+28 \\\\\n&amp; =77\n\\end{aligned}\n\\]<\/div>\n<p>Roots are real, different and irrational number, or<\/p>\n<div class=\"mathjax-overflow\">\\[\nx=\\frac{-b \\pm \\sqrt{D}}{2 a}\n\\]<\/div>\n<p>or<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; x=\\frac{-7 \\pm \\sqrt{77}}{2} \\\\\n&amp; x=\\frac{-7 \\pm 8.7749}{2}\n\\end{aligned}\n\\]<\/div>\n<p>Taking (+) sign<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\nx &amp; =\\frac{-7+8.7749}{2} \\\\\n&amp; =\\frac{1.7749}{2}=0.89\n\\end{aligned}\n\\]<\/div>\n<p>Taking (-) sign<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\nx &amp; =\\frac{-7-8.7749}{2} \\\\\n&amp; =-\\frac{15.7749}{2} \\\\\n&amp; =-7.89\n\\end{aligned}\n\\]<\/div>\n<p>40 Marks<\/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_2018_%E2%80%94_Section_B\"><\/span>ICSE Mathematics 2018 \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 2018<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2018-q5-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>The \\(\\mathbf{4}^{\\text {th }}\\) term of a G.P. is 16 and the \\(\\mathbf{7}^{\\text {th }}\\) term is 128. Find the first term and common ratio of the series. <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>Given,<\/p>\n<div class=\"mathjax-overflow\">\\[\nT_{4}=16 \\text { and } T_{7}=128\n\\]<\/div>\n<p>Let first term of G.P. be a and common ratio be \\(r\\). so<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; T_{4}=a r^{3}=16  \\qquad \\text{&#8230;(i)}\\\\\n&amp; T_{7}=a r^{6}=128 \\qquad \\text{&#8230;(ii)}\n\\end{aligned}\n\\]<\/div>\n<p>Dividing (ii) by (i), we get<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\frac{a r^{6}}{a r^{3}} &amp; =\\frac{128}{16} \\\\\nr^{3} &amp; =8=2^{3} \\\\\nr &amp; =2\n\\end{aligned}\n\\]<\/div>\n<p>From (i)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\na .(2)^{3} &amp; =16 \\\\\na \\times 8 &amp; =16 \\\\\na &amp; =2\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2018-q5-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>A man invests \u20b9 22,500 in \u20b9 50 shares available at 10% discount. If the dividend paid by the company is 12%, calculate :<\/p>\n<p>(i) The number of shares purchased.<\/p>\n<p>(ii) The annual dividend received.<\/p>\n<p>(iii) The rate of return he gets on his investment. Give your answer correct to the nearest whole number. <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>Given, Total investment \\(=\\text{\u20b9} 22,500\\), Face value \\(=\\text{\u20b9} 50\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\text { Market value }=\\text{\u20b9}\\left(50-\\frac{10}{100} \\times 50\\right)=\\text{\u20b9} 45\n\\]<\/div>\n<p>( \\(\\because 10 \\%\\) discount) Dividend \\(=12 \\%\\) (i)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\text { No. of shares }=\\frac{\\text { Investment }}{\\text { Market value of share }}\n\\]<\/div>\n<div class=\"mathjax-overflow\">\\[\n=\\frac{22500}{45}=500\n\\]<\/div>\n<p>(ii) Annual dividend =No. of shares \u00d7 Nominal value of share \u00d7 dividend rate<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; =500 \\times 50 \\times \\frac{12}{100} \\\\\n&amp; =\\text{\u20b9} 3,000\n\\end{aligned}\n\\]<\/div>\n<p>(iii) Return percentage<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; =\\frac{\\text { Total dividend }}{\\text { Total investment }} \\times 100 \\\\\n&amp; =\\frac{3,000}{22,500} \\times 100 \\\\\n&amp; =13.33 \\%=13 \\%\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2018-q5-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>Use graph paper for this question (Take 2 cm \\(=\\mathbf{1}\\) unit along both \\(\\boldsymbol{x}\\) and \\(\\boldsymbol{y}\\) axis). \\(A B C D\\) is a quadrilateral whose vertices are \\(A(2,2), B(2,-2), C(0-1)\\) and \\(D(0,1)\\)<\/p>\n<p>(i) Reflect quadrilateral \\(A B C D\\) on the \\(y\\)-axis and name it as \\(A^{\\prime} B^{\\prime} C D\\).<\/p>\n<p>(ii) Write down the coordinates of \\(A^{\\prime}\\) and \\(B^{\\prime}\\).<\/p>\n<p>(iii) Name two points which are invariant under the above reflection.<\/p>\n<p>(iv) Name the polygon \\(A^{\\prime} B^{\\prime} C 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><p>(i)<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/12126fb9-3243-49d7-ba04-c01fd01e9dfc-4_511_601_924_1170.jpg\" alt=\"ICSE Class 10 Mathematics 2018 Question 5(c) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>(ii) Co-ordinates of \\(A^{\\prime} \\rightarrow(-2,2)\\)<\/p>\n<p>Co-ordinates of \\(B^{\\prime} \\rightarrow(-2,-2)\\)<\/p>\n<p>(iii) Two invariant points are \\(C(0,-1)\\) and \\(D(0,1)\\)<\/p>\n<p>(iv) \\(A^{\\prime} B^{\\prime} C D\\) is a trapezium quadrilateral<\/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 2018<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2018-q6-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>Using properties of proportion, solve for \\(\\boldsymbol{x}\\). Given that \\(\\boldsymbol{x}\\) is positive : <strong>[3]<\/strong><\/p>\n<div class=\"mathjax-overflow\">\\[\n\\frac{2 x+\\sqrt{4 x^{2}-1}}{2 x-\\sqrt{4 x^{2}-1}}=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{2 x+\\sqrt{4 x^{2}-1}}{2 x-\\sqrt{4 x^{2}-1}}=4\n\\]<\/div>\n<p>Applying componendo \\&amp; dividendo<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\frac{2 x+\\sqrt{4 x^{2}-1}+2 x-\\sqrt{4 x^{2}-1}}{2 x+\\sqrt{4 x^{2}-1}-2 x+\\sqrt{4 x^{2}-1}} &amp; =\\frac{4+1}{4-1} \\\\\n\\frac{4 x}{2 \\sqrt{4 x^{2}-1}} &amp; =\\frac{5}{3} \\\\\n12 x &amp; =10 \\sqrt{4 x^{2}-1}\n\\end{aligned}\n\\]<\/div>\n<p>Squaring both sides,<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n144 x^{2} &amp; =100\\left(4 x^{2}-1\\right) \\\\\n36 x^{2} &amp; =25\\left(4 x^{2}-1\\right) \\\\\n36 x^{2} &amp; =100 x^{2}-25 \\\\\n100 x^{2}-36 x^{2} &amp; =25 \\\\\n64 x^{2} &amp; =25 \\\\\nx^{2} &amp; =\\frac{25}{64} \\\\\nx &amp; = \\pm \\frac{5}{8}\n\\end{aligned}\n\\]<\/div>\n<p>\u2235 x is positive, hence \\(x=\\frac{5}{8}\\)<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2018-q6-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>If \\(A=\\left[\\begin{array}{cc}2 &amp; 3 \\\\ 5 &amp; 7\\end{array}\\right], B=\\left[\\begin{array}{cc}0 &amp; 4 \\\\ -1 &amp; 7\\end{array}\\right]\\), and \\(C=\\left[\\begin{array}{cc}1 &amp; 0 \\\\ -1 &amp; 4\\end{array}\\right]\\), find \\(\\boldsymbol{A} \\boldsymbol{C}+\\boldsymbol{B}^{\\mathbf{2}}-\\mathbf{1 0} \\boldsymbol{C}\\). <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\\text { (b) } \\begin{aligned}\n&amp; A C=\\left[\\begin{array}{ll}\n2 &amp; 3 \\\\\n5 &amp; 7\n\\end{array}\\right]\\left[\\begin{array}{cc}\n1 &amp; 0 \\\\\n-1 &amp; 4\n\\end{array}\\right] \\\\\n&amp;=\\left[\\begin{array}{ll}\n2 \\times 1+3 \\times-1 &amp; 2 \\times 0+3 \\times 4 \\\\\n5 \\times 1+7 \\times-1 &amp; 5 \\times 0+7 \\times 4\n\\end{array}\\right] \\\\\n&amp;=\\left[\\begin{array}{ll}\n2-3 &amp; 0+12 \\\\\n5-7 &amp; 0+28\n\\end{array}\\right]=\\left[\\begin{array}{ll}\n-1 &amp; 12 \\\\\n-2 &amp; 28\n\\end{array}\\right] \\\\\n&amp; B^{2}=\\left[\\begin{array}{cc}\n0 &amp; 4 \\\\\n-1 &amp; 7\n\\end{array}\\right]\\left[\\begin{array}{cc}\n0 &amp; 4 \\\\\n-1 &amp; 7\n\\end{array}\\right] \\\\\n&amp;=\\left[\\begin{array}{cc}\n0 \\times 0+4 \\times-1 &amp; 0 \\times 4+4 \\times 7 \\\\\n-1 \\times 0+7 \\times-1 &amp; -1 \\times 4+7 \\times 7\n\\end{array}\\right] \\\\\n&amp;=\\left[\\begin{array}{cc}\n0-4 &amp; 0+28 \\\\\n0-7 &amp; -4+49\n\\end{array}\\right]=\\left[\\begin{array}{cc}\n-4 &amp; 28 \\\\\n-7 &amp; 45\n\\end{array}\\right] \\\\\n&amp; 10 C=10\\left[\\begin{array}{cc}\n1 &amp; 0 \\\\\n-1 &amp; 4\n\\end{array}\\right]=\\left[\\begin{array}{cc}\n10 &amp; 0 \\\\\n-10 &amp; 40\n\\end{array}\\right] \\\\\n&amp; \\therefore A C+B^{2}-10 C=\\left[\\begin{array}{cc}\n-1 &amp; 12 \\\\\n-2 &amp; 28\n\\end{array}\\right]+\\left[\\begin{array}{cc}\n-4 &amp; 28 \\\\\n-7 &amp; 45\n\\end{array}\\right]-\\left[\\begin{array}{cc}\n10 &amp; 0 \\\\\n-10 &amp; 40\n\\end{array}\\right] \\\\\n&amp;=\\left[\\begin{array}{cc}\n-1-4-10 &amp; 12+28-0 \\\\\n-2-7+10 &amp; 28+45-40\n\\end{array}\\right] \\\\\n&amp;=\\left[\\begin{array}{cc}\n-15 &amp; 40 \\\\\n1 &amp; 33\n\\end{array}\\right]\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2018-q6-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>Prove that \\((\\mathbf{1}+\\boldsymbol{\\operatorname { c o t }} \\theta-\\boldsymbol{\\operatorname { c o s e c }} \\theta)(\\mathbf{1}+\\boldsymbol{\\operatorname { t a n }} \\theta+\\) \\(\\boldsymbol{\\operatorname { s e c }} \\theta)=\\mathbf{2}\\) <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>L.H.S. \\(=(1+\\cot \\theta-\\operatorname{cosec} \\theta)(1+\\tan \\theta+\\sec \\theta)\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; =\\left(1+\\frac{\\cos \\theta}{\\sin \\theta}-\\frac{1}{\\sin \\theta}\\right)\\left(1+\\frac{\\sin \\theta}{\\cos \\theta}+\\frac{1}{\\cos \\theta}\\right) \\\\\n&amp; =\\left(\\frac{\\sin \\theta+\\cos \\theta-1}{\\sin \\theta}\\right)\\left(\\frac{\\cos \\theta+\\sin \\theta+1}{\\cos \\theta}\\right) \\\\\n&amp; =\\frac{(\\sin \\theta+\\cos \\theta-1)(\\sin \\theta+\\cos \\theta+1)}{\\sin \\theta \\cos \\theta} \\\\\n&amp; =\\frac{(\\sin \\theta+\\cos \\theta)^{2}-(1)^{2}}{\\sin \\theta \\cos \\theta} \\\\\n&amp; {\\left[\\because a^{2}-b^{2}=(a-b)(a+b)\\right]} \\\\\n&amp; =\\frac{\\left(\\sin ^{2} \\theta+\\cos ^{2} \\theta+2 \\sin \\theta \\cos \\theta\\right)-1}{\\sin \\theta \\cdot \\cos \\theta}\n\\end{aligned}\n\\]<\/div>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; =\\frac{(1+2 \\sin \\theta \\cos \\theta)-1}{\\sin \\theta \\cdot \\cos \\theta} \\\\\n&amp; =\\frac{2 \\sin \\theta \\cos \\theta}{\\sin \\theta \\cos \\theta}=2=\\text { R.H.S.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\">7<\/div><div><div class=\"xc-set-title\">Question 7<\/div><div class=\"xc-set-sub\">ICSE Class 10 Mathematics 2018<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2018-q7-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>Find the value of \\(\\boldsymbol{k}\\) for which the following equation has equal roots. <strong>[3]<\/strong><\/p>\n<div class=\"mathjax-overflow\">\\[\nx^{2}+4 k x+\\left(k^{2}-k+2\\right)=0\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\">\\[\nx^{2}+4 k x+\\left(k^{2}-k+2\\right)=0\n\\]<\/div>\n<p>Compare the equation to<\/p>\n<div class=\"mathjax-overflow\">\\[\na=1, b=4 k, c=k^{2}-k+2\n\\]<\/div>\n<p>Since, roots of the equation are equal<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\nb^{2}-4 a c &amp; =0 \\\\\n(4 k)^{2}-4 \\times 1 \\times\\left(k^{2}-k+2\\right) &amp; =0 \\\\\n16 k^{2}-4 k^{2}+4 k-8 &amp; =0 \\\\\n12 k^{2}+4 k-8 &amp; =0 \\\\\n3 k^{2}+k-2 &amp; =0 \\\\\n3 k^{2}+3 k-2 k-2 &amp; =0 \\\\\n3 k(k+1)-2(k+1) &amp; =0 \\\\\n(k+1)(3 k-2) &amp; =0 \\\\\nk &amp; =-1, k=2 \/ 3\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2018-q7-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>One map drawn to a scale of 1 : 50,000, a rectangular plot of land \\(\\boldsymbol{A} \\boldsymbol{B} \\boldsymbol{C} \\boldsymbol{D}\\) has the following dimensions. \\(\\boldsymbol{A} \\boldsymbol{B}=\\mathbf{6 ~ c m} ; \\boldsymbol{B} \\boldsymbol{C}=\\mathbf{8 ~ c m}\\) and all angles are right angles. Find :<\/p>\n<p>(i) the actual length of the diagonal distance \\(\\boldsymbol{A} \\boldsymbol{C}\\) of the plot in km.<\/p>\n<p>(ii) the actual area of the plot in sq km. <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) Since,<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; A B=6 \\mathrm{~cm} \\\\\n&amp; B C=8 \\mathrm{~cm}\n\\end{aligned}\n\\]<\/div>\n<p>By Pythagorean theorem<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\text { Diagonal } \\begin{aligned}\nA C &amp; =\\sqrt{\\mathrm{AB}^{2}+\\mathrm{BC}^{2}} \\\\\n&amp; =\\sqrt{(6)^{2}+(8)^{2}} \\\\\n&amp; =\\sqrt{100}=10 \\mathrm{~cm}\n\\end{aligned}\n\\]<\/div>\n<p>Since, it is given that<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{array}{rlrl}\n&amp; &amp; 1 \\mathrm{~cm} &amp; =50,000 \\mathrm{~cm} \\\\\n\\therefore &amp; 10 \\mathrm{~cm} &amp; =5,00,000 \\mathrm{~cm}\n\\end{array}\n\\]<\/div>\n<p>Diagonal \\(A C=5 \\mathrm{~km}\\)<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/12126fb9-3243-49d7-ba04-c01fd01e9dfc-5_306_425_2153_1203.jpg\" alt=\"ICSE Class 10 Mathematics 2018 Question 7(b) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>(ii)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\text { Area } &amp; =A B \\times B C \\\\\n&amp; =6 \\times 8\n\\end{aligned}\n\\]<\/div>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; =48 \\mathrm{~cm}^{2} \\\\\n&amp; =48 \\times(50,000 \\times 50,000) \\\\\n&amp; =1,20,00,00,00,000 \\mathrm{~cm}^{2} \\\\\n&amp; =12 \\mathrm{sq} . \\mathrm{km} \\\\\n{[1} &amp; \\left.\\mathrm{km}^{2}=10,00,00,00,000 \\mathrm{~cm}^{2}\\right]\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2018-q7-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>\\(\\boldsymbol{A}(\\mathbf{2}, \\mathbf{5}), \\boldsymbol{B}(-\\mathbf{1}, \\mathbf{2})\\) and \\(\\boldsymbol{C}(\\mathbf{5}, \\mathbf{8})\\) are the vertices of a triangle \\(\\boldsymbol{A} \\boldsymbol{B} \\boldsymbol{C}\\), &#8216; \\(\\boldsymbol{M}\\) &#8216; is a point on \\(\\boldsymbol{A} \\boldsymbol{B}\\) such that \\(A M: M B=1: 2\\). Find the co-ordinates of &#8216; \\(M\\) &#8216;. Hence find the equation of the line passing through the points \\(\\boldsymbol{C}\\) and \\(\\boldsymbol{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>Let \\(M\\left(x_{1}, y_{1}\\right)\\) be the point which divides the line segment \\(A B\\) in the ratio 1 : 2. Hence, Coordinate of \\(M\\) are<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\nM\\left(x_{1}, y_{1}\\right) &amp; =\\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}, \\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}} \\\\\nM\\left(x_{1}, y_{1}\\right) &amp; =\\left(\\frac{1 \\times(-1)+2 \\times 2}{1+2}, \\frac{1 \\times 2+2 \\times 5}{1+2}\\right) \\\\\n&amp; =\\left(\\frac{-1+4}{3}, \\frac{2+10}{3}\\right) \\\\\n&amp; =(1,4)\n\\end{aligned}\n\\]<\/div>\n<p>Coordinates of \\(M\\) are (1, 4). Now equation of line passing through \\(C(5,8)\\) and M(1, 4) is<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\ny-8 &amp; =\\frac{4-8}{1-5}(x-5) \\\\\ny-8 &amp; =\\frac{-4}{-4}(x-5) \\\\\ny-8 &amp; =x-5 \\\\\ny &amp; =x+3\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 2018<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2018-q8-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>\u20b9 7500 were divided equally among a certain number of children. Had there been 20 less children, each would have received \u20b9 100 more. Find the original number of children. <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 number of children be \\(x\\) and \u20b9 \\(y\\) is given to each children then<\/p>\n<div class=\"mathjax-overflow\">\\[\nx . y=7,500 \\qquad \\text{&#8230;(i)}\n\\]<\/div>\n<p>and<\/p>\n<div class=\"mathjax-overflow\">\\[\n(x-20)(y+100)=7,500\n\\]<\/div>\n\\(x y+100 x-20 y-2,000=7,500\\)\n\\(7,500+100 x-20 y-2,000=7,500\\)\n\\([\\because x y=7,500]\\)\n<div class=\"mathjax-overflow\">\\[\n\\begin{array}{r}\n100 x-20 y-2,000=0 \\\\\n100 x-20 \\times \\frac{7,500}{x}-2,000=0\n\\end{array}\n\\]<\/div>\n<div class=\"mathjax-overflow\">\\[\n\\text { [From (i) } y=\\frac{7,500}{x} \\text { ] }\n\\]<\/div>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n100 x^{2}-1,50,000-2,000 x &amp; =0 \\\\\nx^{2}-20 x-1,500 &amp; =0 \\\\\nx^{2}-50 x+30 x-1,500 &amp; =0 \\\\\nx(x-50)+30(x-50) &amp; =0 \\\\\n(x-50)(x+30) &amp; =0 \\\\\nx &amp; =50,-30\n\\end{aligned}\n\\]<\/div>\n<p>The number of children can not be &#8211; ve.<\/p>\n<div class=\"mathjax-overflow\">\\[\nx=50\n\\]<\/div>\n<p>\u2234 Number of children \\(=50\\)<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2018-q8-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>If the mean of the following distribution is 24, find the value of &#8216; \\(a\\) &#8216;. <strong>[3]<\/strong><\/p>\n<div class=\"math-responsive\"><table><thead><tr><th>Marks<\/th><th>0-10<\/th><th>10-20<\/th><th>20-30<\/th><th>30-40<\/th><th>40-50<\/th><\/tr><\/thead><tbody><tr><td>Number of students<\/td><td>7<\/td><td>\\(a\\)<\/td><td>8<\/td><td>10<\/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><div class=\"math-responsive\"><table><thead><tr><th>Marks<\/th><th>\\(\\operatorname{Mid} x_{i}\\)<\/th><th>\\(f_{i}\\)<\/th><th>\\(f_{i} x_{i}\\)<\/th><\/tr><\/thead><tbody><tr><td>0-10<\/td><td>5<\/td><td>7<\/td><td>35<\/td><\/tr><tr><td>10-20<\/td><td>15<\/td><td>\\(a\\)<\/td><td>\\(15 a\\)<\/td><\/tr><tr><td>20-30<\/td><td>25<\/td><td>8<\/td><td>200<\/td><\/tr><tr><td>30-40<\/td><td>35<\/td><td>10<\/td><td>350<\/td><\/tr><tr><td>40-50<\/td><td>45<\/td><td>5<\/td><td>225<\/td><\/tr><tr><td><\/td><td><\/td><td>\\(30+a\\)<\/td><td>\\(810+15 a\\)<\/td><\/tr><\/tbody><\/table><\/div>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\bar{x} &amp; =\\frac{\\Sigma f_{i} x_{i}}{\\Sigma f_{i}} \\\\\n24 &amp; =\\frac{810+15 a}{30+a} \\\\\n720+24 a &amp; =810+15 a \\\\\n24 a-15 a &amp; =810-720 \\\\\n9 a &amp; =90 \\\\\na &amp; =10\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2018-q8-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>Using ruler and compass only, construct a \\(\\triangle A B C\\) such that \\(B C=5 \\mathrm{~cm}\\) and \\(A B=6.5 \\mathrm{~cm}\\) and \\(\\angle A B C=\\mathbf{1 2 0}^{\\circ}\\).<\/p>\n<p>(i) Construct a circum-circle of \\(\\triangle A B C\\).<\/p>\n<p>(ii) Construct a cyclic quadrilateral \\(A B C D\\), such that \\(D\\) is equidistant from \\(A B\\) and \\(B C\\). <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>(a) Steps of construction :<\/p>\n<p>(i) Draw a line segment \\(B C=5 \\mathrm{~cm}\\).<\/p>\n<p>(ii) Construct \\(\\angle A B C=120^{\\circ}\\).<\/p>\n<p>(iii) Cut \\(B A=6.5 \\mathrm{~cm}\\)<\/p>\n<p>(iv) Join \\(A\\) to \\(C\\).<\/p>\n<p>(v) Construct perpendicular bisectors of \\(A B\\) and \\(B C\\), intersecting at \\(O\\). Join \\(A O\\).<\/p>\n<p>(vi) Taking \\(O\\) as centre, and \\(O A\\) as radius draw a circle, passing through \\(A, B\\) and \\(C\\).<\/p>\n<p>(b) (i) Draw the bisector of \\(\\angle A B C\\) such that it touches the circle at point \\(D\\).<\/p>\n<p>(ii) Join \\(A\\) to \\(D\\) and \\(C\\) to \\(D\\).<\/p>\n<p>(iii) \\(A B C D\\) is required cyclic quadrilateral.<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/12126fb9-3243-49d7-ba04-c01fd01e9dfc-6_410_436_2052_1198.jpg\" alt=\"ICSE Class 10 Mathematics 2018 Question 8(c) solution figure\" loading=\"lazy\" decoding=\"async\"><\/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 2018<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2018-q9-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>Priyanka has a recurring deposit account of \u20b9 1,000 per month at 10% per annum. If she gets \u20b9 5,550 as interest at the time of maturity, find the total time for which the account was held. <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>\\(P=\\text{\u20b9} 1,000, R=10 \\%\\), p.a., S.I. \\(=\\text{\u20b9} 5,550\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{gathered}\nn=? \\\\\n\\mathrm{SI}=\\frac{\\operatorname{Pn}(n+1)}{2} \\times \\frac{1}{12} \\times \\frac{R}{100} \\\\\n5,550=\\frac{1000 \\times n(n+1) \\times 10}{2 \\times 12 \\times 100} \\\\\nn(n+1)=1332 \\\\\nn^{2}+n-1332=0 \\\\\nn^{2}+37 n-36 n-1332=0 \\\\\nn(n+37)-36(n+37)=0 \\\\\n(n+37)(n-36)=0 \\\\\nn=36,-37\n\\end{gathered}\n\\]<\/div>\n<p>The value of time cannot be &#8211; ve. Hence<\/p>\n<div class=\"mathjax-overflow\">\\[\nn=36 \\text { months }\n\\]<\/div>\n<p>or \\(n=3\\) years<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2018-q9-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>In \\(\\triangle P Q R, M N\\) is parallel to \\(Q R\\) and<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\frac{P M}{M Q}=\\frac{2}{3}\n\\]<\/div>\n<p>(i) Find \\(\\frac{M N}{Q R}\\).<\/p>\n<p>(ii) Prove that \\(\\triangle O M N\\) and \\(\\triangle O R Q\\) are similar.<\/p>\n<p>(iii) Find, area of \\(\\triangle O M N\\) : Area of \\(\\triangle O R Q\\). <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\/12126fb9-3243-49d7-ba04-c01fd01e9dfc-7_409_647_816_258.jpg\" alt=\"ICSE Class 10 Mathematics 2018 Question 9(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>(i) Given \\(M N \\| Q R\\), then<\/p>\n<p>or \u2234<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\frac{P M}{P Q}=\\frac{M N}{Q R} \\text { (By Thales theorem) }\n\\]<\/div>\n<p>or<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\frac{P M}{P M+M Q}=\\frac{M N}{Q R}\n\\]<\/div>\n<p>or<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\frac{2}{2+3}=\\frac{M N}{Q R}\n\\]<\/div>\n<p>\u21d2<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\frac{M N}{Q R}=\\frac{2}{5}\n\\]<\/div>\n<p>(ii) \\(M N \\| Q R\\) (Given)<\/p>\n<p>\\(\\angle O M N=\\angle O R Q\\) (Alternate angle) \\(\\angle O N M=\\angle O Q R\\) (Alternate angle) \\(\\angle M O N=\\angle R O Q \\quad(\\) Vert. Opp. angles) \\(\\therefore \\triangle O M N \\sim \\triangle O R Q\\) (by AAA similarity)<\/p>\n<p>(iii) \u2235 \\(\\triangle O M N\\) and \\(\\triangle O R Q\\) are similar triangle.<\/p>\n<p>We know that ratio of the areas of two similar triangle is equal to ratio of the squares of the corres-ponding sides.<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\therefore \\quad \\frac{\\operatorname{ar}(\\triangle O M N)}{\\operatorname{ar}(\\triangle O R Q)} &amp; =\\frac{M N^{2}}{Q R^{2}} \\\\\n&amp; =\\left(\\frac{M N}{Q R}\\right)^{2}=\\left(\\frac{2}{5}\\right)^{2} \\\\\n&amp; =\\frac{4}{25}\n\\end{aligned}\n\\]<\/div>\n<p>Hence, \\(\\operatorname{ar}(\\triangle O M N): \\operatorname{ar}(\\triangle O R Q)=4: 25\\)<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2018-q9-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>The following figure represents a solid consisting of a right circular cylinder with a hemisphere at one end and a cone at the other. Their common radius is 7 cm. The height of the cylinder and cone are each of 4 cm. Find volume of the solid. <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\/12126fb9-3243-49d7-ba04-c01fd01e9dfc-7_479_377_1491_395.jpg\" alt=\"ICSE Class 10 Mathematics 2018 Question 9(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>Common radius \\((r)=7 \\mathrm{~cm}\\) Height of cylinder=Height of cone<\/p>\n<div class=\"mathjax-overflow\">\\[\n=h=4 \\mathrm{~cm}\n\\]<\/div>\n<p>Volume of solid =volume of hemisphere + volume of cylinder + volume of cone<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; =\\frac{2}{3} \\pi r^{3}+\\pi r^{2} h+\\frac{1}{3} \\pi r^{2} h \\\\\n&amp; =\\frac{2}{3} \\pi \\times 7 \\times 7 \\times 7+\\pi \\times 7 \\times 7 \\times 4+\\frac{1}{3} \\times \\pi \\times 7 \\times 7 \\times 4 \\\\\n&amp; =\\frac{686 \\pi}{3}+196 \\pi+\\frac{196 \\pi}{3} \\\\\n&amp; =\\frac{686 \\pi+588 \\pi+196 \\pi}{3} \\\\\n&amp; =\\frac{1,470 \\pi}{3}=490 \\pi \\mathrm{~cm}^{3} \\\\\n&amp; =\\frac{490 \\times 22}{7}=1540 \\mathrm{~cm}^{3}\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 2018<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2018-q10-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>Use Remainder theorem to factorize the following polynomial : <strong>[3]<\/strong><\/p>\n<div class=\"mathjax-overflow\">\\[\n2 x^{3}+3 x^{2}-9 x-10\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><p>Let<\/p>\n<div class=\"mathjax-overflow\">\\[\nf(x)=2 x^{3}+3 x^{2}-9 x-10\n\\]<\/div>\n<p>The factors of the constant terms are \u00b1 1, \u00b1 2, \u00b1 5 and \u00b1 10. We have<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; f(-1)=2(-1)^{3}+3(-1)^{2}-9(-1)-10 \\\\\n&amp; \\quad=-2+3+9-10=0\n\\end{aligned}\n\\]<\/div>\n<p>So, \\((x+1)\\) is a factor of \\(f(x)\\). We now divide \\(f(x)=2 x^{3}+3 x^{2}-9 x-10\\) by \\((x+1)\\) to get the other factor of \\(f(x)\\).<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; 2 x^{2}+x-10 \\\\\n&amp; x + 1 \\longdiv { 2 x ^ { 3 } + 3 x ^ { 2 } &#8211; 9 x &#8211; 1 0 } \\\\\n&amp; 2 x^{3}+2 x^{2} \\\\\n&amp; \\frac{-\\quad-}{x^{2}-9 x}-10 \\\\\n&amp; x^{2}+x \\\\\n&amp; \\frac{-\\quad-}{-10 x-10} \\\\\n&amp; -10 x-10 \\\\\n&amp; +\\quad+ \\\\\n&amp; 0\n\\end{aligned}\n\\]<\/div>\n<div class=\"mathjax-overflow\">\\[\n\\therefore \\quad \\begin{aligned}\n&amp; 2 x^{3}+3 x^{2}-9 x-10=(x+1)\\left(2 x^{2}+x-10\\right) \\\\\n&amp; =(x+1)\\left[2 x^{2}+5 x-4 x-10\\right] \\\\\n&amp; =(x+1)[x(2 x+5)-2(2 x+5)] \\\\\n&amp; =(x+1)(x-2)(2 x+5)\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2018-q10-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>In the figure given below &#8216; \\(O\\) &#8216; is the centre of the circle. If \\(Q R=O P\\) and \\(\\angle O R P=20^{\\circ}\\). Find the value of &#8216; \\(\\boldsymbol{x}\\) &#8216; giving reasons. <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\/12126fb9-3243-49d7-ba04-c01fd01e9dfc-8_311_533_270_318.jpg\" alt=\"ICSE Class 10 Mathematics 2018 Question 10(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>Given<\/p>\n<div class=\"mathjax-overflow\">\\[\nQ R=O P \\qquad \\text{&#8230;(i)}\n\\]<\/div>\n<p>and<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\angle O R P=20^{\\circ}\n\\]<\/div>\n<p>\u2235<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\angle O R Q=\\angle O R P=20^{\\circ}\n\\]<\/div>\n<p>\u2235<\/p>\n<div class=\"mathjax-overflow\">\\[\nO P=O Q\n\\]<\/div>\n<p>(Radius of the circle) \u2234<\/p>\n<div class=\"mathjax-overflow\">\\[\nQ R=O Q\n\\]<\/div>\n<p>(From (i)) In \\(\\triangle O Q R\\),<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\because \\quad O Q=Q R\n\\]<\/div>\n<p>\u2234<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\angle Q O R &amp; =\\angle O R Q \\\\\n&amp; =20^{\\circ}\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\/12126fb9-3243-49d7-ba04-c01fd01e9dfc-8_310_535_271_1150.jpg\" alt=\"ICSE Class 10 Mathematics 2018 Question 10(b) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>Now<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\angle O Q R+\\angle O R Q+\\angle Q O R=180^{\\circ}\n\\]<\/div>\n<p>(By angle sums property)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\angle O Q R+20^{\\circ}+20^{\\circ} &amp; =180^{\\circ} \\\\\n\\angle O Q R &amp; =180^{\\circ}-40^{\\circ} \\\\\n&amp; =140^{\\circ}\n\\end{aligned}\n\\]<\/div>\n<p>Now \\(\\angle O Q R+\\angle O Q P=180^{\\circ}\\) (By linear pair)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n140^{\\circ}+\\angle O Q P &amp; =180^{\\circ} \\\\\n\\angle O Q P &amp; =180^{\\circ}-140^{\\circ}=40^{\\circ}\n\\end{aligned}\n\\]<\/div>\n<p>In \\(\\triangle P O Q\\),<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\because &amp; &amp; O P=O Q \\\\\n\\therefore &amp; &amp; \\angle O P R=\\angle O Q P=40^{\\circ}\n\\end{aligned}\n\\]<\/div>\n<p>Now<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\angle P O Q+\\angle O P R+\\angle O Q P=180^{\\circ}\n\\]<\/div>\n<p>(By angle sum property)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\angle P O Q+40^{\\circ}+40^{\\circ} &amp; =180^{\\circ} \\\\\n\\angle P O Q &amp; =180^{\\circ}-80^{\\circ}=100^{\\circ}\n\\end{aligned}\n\\]<\/div>\n<p>Now<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\angle R O Q+\\angle P O Q+x &amp; =180^{\\circ} \\quad \\text { (By linear pair) } \\\\\n20^{\\circ}+100^{\\circ}+x &amp; =180^{\\circ} \\\\\nx &amp; =180^{\\circ}-120^{\\circ} \\\\\nx &amp; =60^{\\circ}\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2018-q10-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>The angle of elevation from a point \\(\\boldsymbol{P}\\) of the top of a tower \\(Q R, 50 \\mathrm{~m}\\) high is \\(60^{\\circ}\\) and that of the tower \\(P T\\) from a point \\(Q\\) is \\(30^{\\circ}\\). Find the height of the tower \\(\\boldsymbol{P} \\boldsymbol{T}\\), correct to the nearest metre. <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\/12126fb9-3243-49d7-ba04-c01fd01e9dfc-8_342_426_801_369.jpg\" alt=\"ICSE Class 10 Mathematics 2018 Question 10(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>Let the height of the tower \\(P T\\) be \\(h\\) and distance \\(P Q=x \\mathrm{~m}\\) In \\(\\triangle P R Q\\),<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\tan 60^{\\circ} &amp; =\\frac{R Q}{P Q} \\\\\n\\sqrt{3} &amp; =\\frac{50}{x}\n\\end{aligned}\n\\]<\/div>\n<p>or<\/p>\n<div class=\"mathjax-overflow\">\\[\nx=\\frac{50}{\\sqrt{3}}\n\\]<\/div>\n<p>In right \\(\\triangle P T Q\\),<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\tan 30^{\\circ} &amp; =\\frac{P T}{P Q}=\\frac{h}{x} \\\\\n\\frac{1}{\\sqrt{3}} &amp; =\\frac{h}{50 \/ \\sqrt{3}}\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\/12126fb9-3243-49d7-ba04-c01fd01e9dfc-8_347_429_2104_1202.jpg\" alt=\"ICSE Class 10 Mathematics 2018 Question 10(c) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n3 h &amp; =50 \\\\\nh &amp; =\\frac{50}{3}=16.66=17 \\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\">11<\/div><div><div class=\"xc-set-title\">Question 11<\/div><div class=\"xc-set-sub\">ICSE Class 10 Mathematics 2018<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2018-q11-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>The \\(\\mathbf{4}^{\\text {th }}\\) term of an A.P. is 22 and \\(\\mathbf{1 5}^{\\text {th }}\\) term is 66. Find the first term and the common difference. Hence find the sum of the series to 8 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 first term be a and common difference be \\(d\\).<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; T_{4}=a+3 d=22  \\qquad \\text{&#8230;(i)}\\\\\n&amp; T_{15}=a+14 d=66 \\qquad \\text{&#8230;(ii)}\n\\end{aligned}\n\\]<\/div>\n<p>Subtracting (ii) from (i)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\na+14 d-a-3 d &amp; =66-22 \\\\\n11 d &amp; =44 \\\\\nd &amp; =4\n\\end{aligned}\n\\]<\/div>\n<p>From (i)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\na+3 \\times 4 &amp; =22 \\\\\na &amp; =22-12 \\\\\n&amp; =10\n\\end{aligned}\n\\]<\/div>\n<p>Thus,<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\nS_{8} &amp; =\\frac{8}{2}[2 \\times 10+(8-1) 4] \\\\\n&amp; =4[20+28] \\\\\n&amp; =4 \\times 48=192\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2018-q11-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>Use Graph paper for this question. A survey regarding height (in cm) of 60 boys belonging to Class 10 of a school was conducted. The following data was recorded :<\/p>\n<div class=\"math-responsive\"><table><thead><tr><th>Height in cm<\/th><th>135140<\/th><th>140145<\/th><th>145150<\/th><th>150155<\/th><th>155160<\/th><th>160165<\/th><th>165-170<\/th><\/tr><\/thead><tbody><tr><td>No. of boys<\/td><td>4<\/td><td>8<\/td><td>20<\/td><td>14<\/td><td>7<\/td><td>6<\/td><td>1<\/td><\/tr><\/tbody><\/table><\/div>\n<p>Taking \\(\\mathbf{2 ~ c m}=\\) height of \\(\\mathbf{1 0 ~ c m}\\) along one axis and \\(\\mathbf{2} \\mathbf{~ c m}=\\mathbf{1 0}\\) boys along the other axis draw an ogive of the above distribution. Use the graph to estimate the following :<\/p>\n<p>(i) the median<\/p>\n<p>(ii) lower quartile<\/p>\n<p>(iii) if above 158 cm is considered as the tall boys of the class. Find the number of boys in the class who are tall. <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>Height (in cm)<\/th><th>Frequency \\(\\boldsymbol{(} \\boldsymbol{f} \\boldsymbol{)}\\)<\/th><th>Cumulative Frequency (c. f.)<\/th><\/tr><\/thead><tbody><tr><td>135-140<\/td><td>4<\/td><td>4<\/td><\/tr><tr><td>140-145<\/td><td>8<\/td><td>12<\/td><\/tr><tr><td>145-150<\/td><td>20<\/td><td>32<\/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\/12126fb9-3243-49d7-ba04-c01fd01e9dfc-9_246_641_237_1173.jpg\" alt=\"ICSE Class 10 Mathematics 2018 Question 11(b) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/12126fb9-3243-49d7-ba04-c01fd01e9dfc-9_993_756_496_1039.jpg\" alt=\"ICSE Class 10 Mathematics 2018 Question 11(b) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>Here, \\(n=60\\)<\/p>\n<p>(i) Median = \\(\\frac{60}{2}\\) th term = 30th term = 149.5<\/p>\n<p>(ii) Lower quartile \\(=\\frac{60}{4}\\) th term \\(=15\\) th term = 145.5<\/p>\n<p>(iii) Through mark of 158 on \\(x\\)-axis, draw a vertical line which meets the graph at a point on the ogive. Then through that point, draw a horizontal line which meets the \\(y\\)-axis at the mark of 48. Therefore, number of boys in the class who are tall \\(=60-48=12\\).<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/12126fb9-3243-49d7-ba04-c01fd01e9dfc-9_525_1175_2084_411.jpg\" alt=\"ICSE Class 10 Mathematics 2018 Question 11(b) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div><\/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\">34<\/span> <span class=\"rt-label rt-postfix\">minutes<\/span><\/span> ICSE Class 10 Mathematics PYQ Practice the complete ICSE Class 10 Mathematics 2018 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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