{"id":2332,"date":"2026-11-13T15:20:39","date_gmt":"2026-11-13T09:50:39","guid":{"rendered":"https:\/\/xamcontent.com\/cisce\/?p=2332"},"modified":"2026-11-13T15:20:40","modified_gmt":"2026-11-13T09:50:40","slug":"icse-class-10-mathematics-2023-question-paper-with-solutions","status":"publish","type":"post","link":"https:\/\/xamcontent.com\/cisce\/icse-class-10-mathematics-2023-question-paper-with-solutions\/","title":{"rendered":"ICSE Class 10 Mathematics 2023 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-2023-question-paper-with-solutions\/#ICSE_Mathematics_2023_%E2%80%94_Section_A\" title=\"ICSE Mathematics 2023 \u2014 Section A\">ICSE Mathematics 2023 \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-2023-question-paper-with-solutions\/#ICSE_Mathematics_2023_%E2%80%94_Section_B\" title=\"ICSE Mathematics 2023 \u2014 Section B\">ICSE Mathematics 2023 \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\">19<\/span> <span class=\"rt-label rt-postfix\">minutes<\/span><\/span>\n<div class=\"xc-nd xc-icse-physics-pyq 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 2023 Question Paper with Solutions<\/strong>, including MCQs, algebra, geometry, trigonometry, statistics, graphs and constructions. 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\">2023<\/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 2023<\/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 questions are given in brackets [ ].<\/p><p>Mathematical tables and graph papers 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_2023_%E2%80%94_Section_A\"><\/span>ICSE Mathematics 2023 \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\">Choose the correct answer from the given options.<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2023-q1-i\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">i<\/div><div class=\"xc-qtext\"><p>If \\(\\left[\\begin{array}{ll}2 &#038; 0 \\\\ 0 &#038; 4\\end{array}\\right]\\left[\\begin{array}{l}x \\\\ y\\end{array}\\right]=\\left[\\begin{array}{c}2 \\\\ -8\\end{array}\\right]\\), the value of \\(\\boldsymbol{x}\\) and \\(\\boldsymbol{y}\\) respectively are: <strong>[1]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">A<\/span><span>1, -2<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">B<\/span><span>-2, 1<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">C<\/span><span>1, 2<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">D<\/span><span>-2, -1<\/span><\/div>\n<\/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>Option (a) is correct.<\/p>\n<p><strong>Explanation:<\/strong> \\(\\left[\\begin{array}{ll}2 &#038; 0 \\\\ 0 &#038; 4\\end{array}\\right]\\left[\\begin{array}{c}x \\\\ y\\end{array}\\right]=\\left[\\begin{array}{c}2 \\\\ -8\\end{array}\\right]\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n{\\left[\\begin{array}{l}\n2 x \\\\\n4 y\n\\end{array}\\right] } &#038; =\\left[\\begin{array}{c}\n2 \\\\\n-8\n\\end{array}\\right] \\\\\n2 x &#038; =2, x=1 \\\\\n4 y &#038; =-8, y=-2\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2023-q1-ii\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">ii<\/div><div class=\"xc-qtext\"><p>If \\(\\boldsymbol{x}-\\mathbf{2}\\) is a factor of \\(\\boldsymbol{x}^{\\mathbf{3}}-\\boldsymbol{k} \\boldsymbol{x}-\\mathbf{1 2}\\), then the value of \\(\\boldsymbol{k}\\) is: <strong>[1]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">A<\/span><span>3<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">B<\/span><span>2<\/span><\/div>\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">C<\/span><span>-2<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">D<\/span><span>-3<\/span><\/div>\n<\/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>Option (c) is correct.<\/p>\n<p><strong>Explanation:<\/strong> \\((x-2)\\) is a factor of \\(f(x)\\) \u2234<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\text { Remainder } &#038; =f(2)=0 \\\\\nf(x) &#038; =x^{3}-k x-12 \\\\\nf(2) &#038; =(2)^{3}-k(2)-12 \\\\\n0 &#038; =8-2 k-12 \\\\\nk &#038; =-2\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2023-q1-iii\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">iii<\/div><div class=\"xc-qtext\"><p>In the given diagram RT is a tangent touching the circle at \\(\\mathbf{S}\\). If \\(\\angle \\mathbf{P S T}=\\mathbf{3 0}^{\\boldsymbol{\\circ}}\\) and \\(\\angle \\mathbf{S P Q}=\\mathbf{6 0}^{\\boldsymbol{\\circ}}\\) then \\(\\angle \\mathrm{PSQ}\\) is equal to: <strong>[1]<\/strong><\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/0c46b619-427e-45f1-9c68-599ad4677dfb-01_173_410_1854_447.jpg\" alt=\"ICSE Class 10 Mathematics 2023 Question 1(iii) figure\" loading=\"lazy\" decoding=\"async\"><\/div><\/div><\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">A<\/span><span>40\u00b0<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">B<\/span><span>30\u00b0<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">C<\/span><span>60\u00b0<\/span><\/div>\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">D<\/span><span>90\u00b0<\/span><\/div>\n<\/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>Option (d) is correct.<\/p>\n<p><strong>Explanation:<\/strong> <\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/0c46b619-427e-45f1-9c68-599ad4677dfb-02_176_410_1577_491.jpg\" alt=\"ICSE Class 10 Mathematics 2023 Question 1(iii) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<div class=\"mathjax-overflow\">\\[\n\\angle \\mathrm{PST}=\\angle \\mathrm{PQS}=30^{\\circ}\n\\]<\/div>\n<p>(alternate segment angle) In \\(\\triangle \\mathrm{PSQ}\\),<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\angle \\mathrm{PSQ}+\\angle \\mathrm{PQS}+\\angle \\mathrm{SPQ}=180^{\\circ}\n\\]<\/div>\n<p>(Sum of all angles of a triangle)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\angle \\mathrm{PSQ}+30^{\\circ}+60^{\\circ} &#038; =180^{\\circ} \\\\\n\\angle \\mathrm{PSQ} &#038; =90^{\\circ}\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2023-q1-iv\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">iv<\/div><div class=\"xc-qtext\"><p>A letter is chosen at random from all the letters of the English alphabets. The probability that the letter chosen is a vowel, is: <strong>[1]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">A<\/span><span>\\(\\frac{4}{26}\\)<\/span><\/div>\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">B<\/span><span>\\(\\frac{5}{26}\\)<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">C<\/span><span>\\(\\frac{21}{26}\\)<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">D<\/span><span>\\(\\frac{5}{24}\\)<\/span><\/div>\n<\/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>Option (b) is correct.<\/p>\n<p><strong>Explanation:<\/strong> Total no. of letters in English alphabets, \\(n(\\mathrm{~S})=26\\) Total no. of vowels in English alphabets, \\(n(\\mathrm{E})=05\\) \u2234 Required probability \\(\\mathrm{P}(\\mathrm{E})=\\frac{5}{26}\\)<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2023-q1-v\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">v<\/div><div class=\"xc-qtext\"><p>If 3 is a root of the quadratic equation \\(\\boldsymbol{x}^{\\mathbf{2}}-\\boldsymbol{p} \\boldsymbol{x}+\\) \\(\\boldsymbol{3}=\\mathbf{0}\\) then \\(\\boldsymbol{p}\\) is equal to: <strong>[1]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">A<\/span><span>4<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">B<\/span><span>3<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">C<\/span><span>5<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">D<\/span><span>2<\/span><\/div>\n<\/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>Option (a) is correct.<\/p>\n<p><strong>Explanation:<\/strong> \\(f(x)=x^{2}-p x+3=0\\) 3 is a root of \\(f(x)\\) \u2234<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{array}{r}\nf(3)=0 \\\\\n(3)^{2}-3 p+3=0\n\\end{array}\n\\]<\/div>\n<p>\u2234<\/p>\n<div class=\"mathjax-overflow\">\\[\np=4\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2023-q1-vi\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">vi<\/div><div class=\"xc-qtext\"><p>In the given figure \\(\\angle \\mathbf{B A P}=\\angle \\mathbf{D C P}=\\mathbf{7 0}{ }^{\\circ}, \\mathbf{P C}=\\) \\(\\mathbf{6 c m}\\) and \\(\\mathbf{C A}=\\mathbf{4 c m}\\), then PD : DB is: <strong>[1]<\/strong><\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/0c46b619-427e-45f1-9c68-599ad4677dfb-01_251_462_1273_1253.jpg\" alt=\"ICSE Class 10 Mathematics 2023 Question 1(vi) figure\" loading=\"lazy\" decoding=\"async\"><\/div><\/div><\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">A<\/span><span>5 : 3<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">B<\/span><span>3 : 5<\/span><\/div>\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">C<\/span><span>3 : 2<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">D<\/span><span>2 : 3<\/span><\/div>\n<\/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>Option (c) is correct.<\/p>\n<p><strong>Explanation:<\/strong> <\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/0c46b619-427e-45f1-9c68-599ad4677dfb-02_251_462_334_1325.jpg\" alt=\"ICSE Class 10 Mathematics 2023 Question 1(vi) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<div class=\"mathjax-overflow\">\\[\n\\angle A=\\angle C=70^{\\circ}\n\\]<\/div>\n<p>(Given) These are corresponding angles<\/p>\n\\(\\therefore D C \\| B A\\)\n<p>\u2234<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\frac{P D}{D B}=\\frac{P C}{A C}\n\\]<\/div>\n<p>(Basic proportional Theorem)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\frac{P D}{D B}=\\frac{6}{4}=\\frac{3}{2}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2023-q1-vii\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">vii<\/div><div class=\"xc-qtext\"><p>The printed price of an article is \u20b93080. If the rate of GST is 10% then the GST charged is: <strong>[1]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">A<\/span><span>\u20b9154<\/span><\/div>\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">B<\/span><span>\u20b9308<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">C<\/span><span>\u20b930.80<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">D<\/span><span>\u20b915.40<\/span><\/div>\n<\/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>Option (b) is correct.<\/p>\n<p><strong>Explanation:<\/strong> G.S.T. \\(=\\%\\) Rate \u00d7 printed price<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&#038; =\\frac{10}{100} \\times 3080 \\\\\n&#038; =\\text{\u20b9} 308\n\\end{aligned}\n\\]<\/div>\n<p>Therefore, the GST charged on the article is (rupees symbol) 308.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2023-q1-viii\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">viii<\/div><div class=\"xc-qtext\"><p>\\((\\mathbf{1}+\\boldsymbol{\\operatorname { s i n }} \\mathbf{A})(\\mathbf{1}-\\sin \\mathbf{A})\\) is equal to: <strong>[1]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">A<\/span><span>\\(\\operatorname{cosec}^{2} \\mathrm{~A}\\)<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">B<\/span><span>\\(\\sin ^{2} \\mathrm{~A}\\)<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">C<\/span><span>\\(\\sec ^{2} \\mathrm{~A}\\)<\/span><\/div>\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">D<\/span><span>\\(\\cos ^{2} \\mathrm{~A}\\)<\/span><\/div>\n<\/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>Option (d) is correct.<\/p>\n<p><strong>Explanation:<\/strong> \\((1+\\sin \\mathrm{A})(1-\\sin \\mathrm{A})\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&#038; =(1)^{2}-(\\sin \\mathrm{A})^{2} \\\\\n&#038; =1-\\sin ^{2} \\mathrm{~A} \\\\\n&#038; =\\sin ^{2} A+\\cos ^{2} A-\\sin ^{2} A \\\\\n&#038; =\\cos ^{2} A\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2023-q1-ix\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">ix<\/div><div class=\"xc-qtext\"><p>The coordinates of the vertices of \\(\\boldsymbol{\\triangle} \\mathbf{A B C}\\) are respectively (-4, -2), \\((6,2)\\) and \\((4,6)\\). The centroid G of \\(\\triangle \\mathrm{ABC}\\) is: <strong>[1]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">A<\/span><span>\\((2,2)\\)<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">B<\/span><span>\\((2,3)\\)<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">C<\/span><span>\\((3,3)\\)<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">D<\/span><span>(0, -1)<\/span><\/div>\n<\/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>Option (a) is correct.<\/p>\n<p><strong>Explanation:<\/strong> \\(G\\left(\\frac{x_{1}+x_{2}+x_{3}}{3}, \\frac{y_{1}+y_{2}+y_{3}}{3}\\right)\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&#038; =G\\left(\\frac{-4+6+4}{3}, \\frac{-2+2+6}{3}\\right) \\\\\n&#038; =G(2,2)\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2023-q1-x\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">x<\/div><div class=\"xc-qtext\"><p>The \\(\\mathbf{n}^{\\text {th }}\\) term an Arithmetic Progression (A.P.) is \\(\\mathbf{2} \\boldsymbol{n}+\\mathbf{5}\\). The \\(\\mathbf{1 0}^{\\text {th }}\\) term is: <strong>[1]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">A<\/span><span>7<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">B<\/span><span>15<\/span><\/div>\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">C<\/span><span>25<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">D<\/span><span>45<\/span><\/div>\n<\/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>Option (c) is correct.<\/p>\n<p><strong>Explanation:<\/strong> <\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\nt_{n} &#038; =2 n+5 \\\\\nt_{10} &#038; =2 \\times 10+5=25\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2023-q1-xi\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">xi<\/div><div class=\"xc-qtext\"><p>The mean proportional between 4 and 9 is: <strong>[1]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">A<\/span><span>4<\/span><\/div>\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">B<\/span><span>6<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">C<\/span><span>9<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">D<\/span><span>36<\/span><\/div>\n<\/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>Option (b) is correct.<\/p>\n<p><strong>Explanation:<\/strong> Mean proportional \\(=\\sqrt{4 \\times 9}=6\\)<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2023-q1-xii\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">xii<\/div><div class=\"xc-qtext\"><p>Which of the following cannot be determined graphically for a grouped frequency distribution? <strong>[1]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">A<\/span><span>Median<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">B<\/span><span>Mode<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">C<\/span><span>Quartiles<\/span><\/div>\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">D<\/span><span>Mean<\/span><\/div>\n<\/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>Option (d) is correct.<\/p>\n<p><strong>Explanation:<\/strong> Mean is a numerical measure that requires the individual data values, which cannot be determined graphically for a grouped frequency distribution.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2023-q1-xiii\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">xiii<\/div><div class=\"xc-qtext\"><p>Volume of a cylinder of height 3 cm is \\(\\mathbf{4 8} \\pi \\mathbf{~ c m}^{\\mathbf{3}}\\). Radius of the cylinder is: <strong>[1]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">A<\/span><span>48 cm<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">B<\/span><span>16 cm<\/span><\/div>\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">C<\/span><span>4 cm<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">D<\/span><span>24 cm<\/span><\/div>\n<\/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>Option (c) is correct.<\/p>\n<p><strong>Explanation:<\/strong> Vol. of cylinder \\(=48 \\pi \\mathrm{~cm}^{3}\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\pi r^{2} h &#038; =48 \\pi \\\\\nr^{2}(3) &#038; =48 \\\\\nr &#038; =4 \\mathrm{~cm}\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2023-q1-xiv\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">xiv<\/div><div class=\"xc-qtext\"><p>Naveen deposits \u20b9800 every month in a recurring deposit account for 6 months. If he receives \u20b94884 at the time of maturity, then the interest he earns is: <strong>[1]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">A<\/span><span>\u20b984<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">B<\/span><span>\u20b942<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">C<\/span><span>\u20b924<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">D<\/span><span>\u20b9284<\/span><\/div>\n<\/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>Option (a) is correct.<\/p>\n<p><strong>Explanation:<\/strong> Amount = Principle + Interest<\/p>\n<div class=\"mathjax-overflow\">\\[\n4884=800 \\times 6+\\text { Interest }\n\\]<\/div>\n<p>\u2234<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\text { Interest } &#038; =4884-4800 \\\\\n&#038; =\\text{\u20b9} 84\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2023-q1-xv\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">xv<\/div><div class=\"xc-qtext\"><p>The solution set for the inequation \\(\\mathbf{2} \\boldsymbol{x}+\\mathbf{4} \\leq \\mathbf{1 4}\\), \\(x \\varepsilon \\mathbf{W}\\) is: <strong>[1]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">A<\/span><span>\\{1, 2, 3, 4, 5\\}<\/span><\/div>\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">B<\/span><span>\\{0, 1, 2, 3, 4, 5\\}<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">C<\/span><span>\\{1, 2, 3, 4\\}<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">D<\/span><span>\\{0, 1, 2, 3, 4\\}<\/span><\/div>\n<\/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>Option (b) is correct.<\/p>\n<p><strong>Explanation:<\/strong> <\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n2 x+4 &#038; \\leq 14 \\\\\n2 x &#038; \\leq 14-4 \\\\\nx &#038; \\leq 5\n\\end{aligned}\n\\]<\/div>\n<p>\u2234 Solution set \\(\\{0,1,2,3,4,5\\}\\)<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"xc-set\">\n<div class=\"xc-set-header\"><div class=\"xc-set-icon\">2<\/div><div><div class=\"xc-set-title\">Question 2<\/div><div class=\"xc-set-sub\">ICSE Class 10 Mathematics 2023<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2023-q2-i\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">i<\/div><div class=\"xc-qtext\"><p>Find the value of &#8216; \\(\\boldsymbol{a}\\) &#8216; if \\(\\boldsymbol{x}-\\boldsymbol{a}\\) is a factor of the polynomial \\(3 x^{3}+x^{2}-a x-81\\). <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\">\\[\np(x)=3 x^{3}+x^{2}-a x-81\n\\]<\/div>\n<p>\\((x-a)\\) is a factor of \\(p(x)\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\np(a) &#038; =0 \\\\\n3(a)^{3}+(a)^{2}-a(a)-81 &#038; =0 \\\\\n3 a^{3}+a^{2}-a^{2}-81 &#038; =0 \\\\\na &#038; =3\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2023-q2-ii\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">ii<\/div><div class=\"xc-qtext\"><p>Salman deposits \u20b91000 every month in a recurring deposit account for 2 years. If he receives \u20b926000 on maturity, find:<\/p>\n<p>(a) the total interest Salman earns.<\/p>\n<p>(b) the rate of interest. <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>\\(\\mathrm{P}=\\text{\u20b9} 1000, n=2\\), years \\(=24\\) months, \\(r=\\) rate of interest<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\nI &#038; =P \\times \\frac{n(n+1)}{2} \\times \\frac{r}{100} \\times \\frac{1}{12} \\\\\nI &#038; =1000 \\times \\frac{24(24+1)}{2} \\times \\frac{r}{1200} \\\\\nI &#038; =250 r \\\\\nA &#038; =P+I \\\\\n26000 &#038; =1000 \\times 24+I \\\\\n2000 &#038; =I \\qquad \\text{&#8230;(i)}\n\\end{aligned}\n\\]<\/div>\n<p>&#8230;(ii)<\/p>\n<p>(a) Salman earns interest as \u20b92000<\/p>\n<p>From (i) and (ii)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n250 r &#038; =2000 \\\\\nr &#038; =8\n\\end{aligned}\n\\]<\/div>\n<p>\u2234 (b) rate of interest is 8%<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2023-q2-iii\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">iii<\/div><div class=\"xc-qtext\"><p>In the given figure O, is the centre of the circle. CE is a tangent to the circle at \\(\\mathbf{A}\\). If \\(\\angle \\mathbf{A B D}=\\mathbf{2 6}^{\\boldsymbol{\\circ}}\\), then find:<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/0c46b619-427e-45f1-9c68-599ad4677dfb-03_418_405_982_421.jpg\" alt=\"ICSE Class 10 Mathematics 2023 Question 2(iii) figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>(a) \u2220BDA<\/p>\n<p>(b) \u2220BAD<\/p>\n<p>(c) \u2220CAD<\/p>\n<p>(d) \u2220ODB <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)<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/0c46b619-427e-45f1-9c68-599ad4677dfb-03_305_300_291_1202.jpg\" alt=\"ICSE Class 10 Mathematics 2023 Question 2(iii) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<div class=\"mathjax-overflow\">\\[\n\\angle B D A=90^{\\circ} \\text { (Angle in a semicircle) }\n\\]<\/div>\n<p>(b) In \\(\\triangle B A D\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\angle B A D+\\angle A B D+\\angle A D B=180^{\\circ}\n\\]<\/div>\n<p>(Sum of all angles of \\(\\Delta\\) )<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&#038; &#038; \\angle B A D+26^{\\circ}+90^{\\circ} &#038; =180^{\\circ} \\\\\n&#038; \\therefore &#038; \\angle B A D &#038; =64^{\\circ}\n\\end{aligned}\n\\]<\/div>\n<p>(c)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&#038; \\angle C A D=\\angle A B D \\\\\n&#038; \\angle C A D=26^{\\circ}\n\\end{aligned}\n\\]<\/div>\n<p>(alternate segment angle) (d)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\nO D &#038; =O B &#038; &#038; \\text { (radius of circle) } \\\\\n\\therefore &#038; \\angle O D B &#038; =\\angle O B D &#038;\n\\end{aligned}\n\\]<\/div>\n<p>(Angles opposite to equal sides of a triangle are equal)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\therefore \\quad \\angle O D B=26^{\\circ}\n\\]<\/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 2023<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2023-q3-i\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">i<\/div><div class=\"xc-qtext\"><p>Solve the following quadratic equation:<\/p>\n<div class=\"mathjax-overflow\">\\[\nx^{2}+4 x-8=0\n\\]<\/div>\n<p>Give your answer correct to one decimal place. (Use mathematical tables if necessary.) <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>\\(x^{2}+4 x-8=0\\)\n<p>Compare the equation \\(a x^{2}+b x+c=0\\) \u2234<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\na &#038; =1, b=4 \\text { and } c=-8 \\\\\nD &#038; =b^{2}-4 a c=(4)^{2}-4(1)(-8) \\\\\n&#038; =16+32=48\n\\end{aligned}\n\\]<\/div>\n<p>\u2234 Roots are real and distinct<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\nx &#038; =\\frac{-b \\pm \\sqrt{D}}{2 a} \\\\\n&#038; =\\frac{-4 \\pm \\sqrt{48}}{2 \\times 1}=\\frac{-4 \\pm 4 \\sqrt{3}}{2}\n\\end{aligned}\n\\]<\/div>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&#038; x=-2 \\pm 2 \\sqrt{3} \\\\\n&#038; x=-2 \\pm 2 \\times 1.732 \\\\\n&#038; x=-2 \\pm 3.464 \\\\\n&#038; x=-2+3.464\n\\end{aligned}\n\\]<\/div>\n<p>and<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&#038; x=-2-3.464 \\\\\n&#038; x=1.464 \\text { and } x=-5.464 \\\\\n&#038; x=1.5 \\text { and } x=-5.5\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2023-q3-ii\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">ii<\/div><div class=\"xc-qtext\"><p>Prove the following identity: <strong>[4]<\/strong><\/p>\n<div class=\"mathjax-overflow\">\\[\n\\left(\\sin ^{2} \\theta-1\\right)\\left(\\tan ^{2} \\theta+1\\right)+1=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\">\\[\n\\begin{aligned}\n&#038; \\text { L.H.S. }=\\left(\\sin ^{2} \\theta-1\\right)\\left(\\tan ^{2} \\theta+1\\right)+1 \\\\\n&#038;=\\left(\\sin ^{2} \\theta-\\sin ^{2} \\theta-\\cos ^{2} \\theta\\right)\\left(\\sec ^{2} \\theta\\right)+1 \\\\\n&#038; \\begin{array}{r}\n1+\\tan ^{2} \\theta=\\sec ^{2} \\theta \\\\\n=\\left(-\\cos ^{2} \\theta\\right)\\left(\\frac{1}{\\cos ^{2} \\theta}\\right)+1 \\\\\n=-1+1 \\\\\n=0=\\text { R.H.S. }\n\\end{array} \\\\\n&#038; \\quad \\text { Hence Proved. }\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2023-q3-iii\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">iii<\/div><div class=\"xc-qtext\"><p>Use graph sheet to answer this question. Take 2 cm \\(=1\\) unit along both the axes.<\/p>\n<p>(a) Plot A, B, C where A(0, 4), B(1, 1) and C(4, 0)<\/p>\n<p>(b) Reflect A and B on the \\(\\boldsymbol{x}\\)-axis and name them as E and D respectively.<\/p>\n<p>(c) Reflect B through the origin and name if F. Write down the coordinates of F.<\/p>\n<p>(d) Reflect B and C on the \\(\\boldsymbol{y}\\)-axis and name them as H and G respectively.<\/p>\n<p>(e) Join points A, B, C, D, E, F, G, H and A in order and name the closed figure formed. <strong>[5]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/0c46b619-427e-45f1-9c68-599ad4677dfb-04_789_788_240_1022.jpg\" alt=\"ICSE Class 10 Mathematics 2023 Question 3(iii) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>(c) F(-1, -1)<\/p>\n<p>(e) Star<\/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_2023_%E2%80%94_Section_B\"><\/span>ICSE Mathematics 2023 \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\">4<\/div><div><div class=\"xc-set-title\">Question 4<\/div><div class=\"xc-set-sub\">ICSE Class 10 Mathematics 2023<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2023-q4-i\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">i<\/div><div class=\"xc-qtext\"><p>If \\(\\boldsymbol{A}=\\left[\\begin{array}{ll}1 &#038; 3 \\\\ 2 &#038; 4\\end{array}\\right], \\boldsymbol{B}=\\left[\\begin{array}{ll}1 &#038; 2 \\\\ 2 &#038; 4\\end{array}\\right], \\boldsymbol{C}=\\left[\\begin{array}{ll}4 &#038; 1 \\\\ 1 &#038; 5\\end{array}\\right]\\) and \\(\\boldsymbol{I}=\\) \\(\\left[\\begin{array}{ll}1 &#038; 0 \\\\ 0 &#038; 1\\end{array}\\right]\\). Find \\(\\boldsymbol{A}(\\boldsymbol{B}+\\boldsymbol{C})-\\mathbf{1 4} \\boldsymbol{I}\\) <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}\nA &#038; =\\left[\\begin{array}{ll}\n1 &#038; 3 \\\\\n2 &#038; 4\n\\end{array}\\right], B=\\left[\\begin{array}{ll}\n1 &#038; 2 \\\\\n2 &#038; 4\n\\end{array}\\right], C=\\left[\\begin{array}{ll}\n4 &#038; 1 \\\\\n1 &#038; 5\n\\end{array}\\right] \\\\\n&#038; A(B+C)-14 I \\\\\n&#038; I=\\left[\\begin{array}{ll}\n1 &#038; 0 \\\\\n0 &#038; 1\n\\end{array}\\right] \\\\\n&#038; =\\left[\\begin{array}{ll}\n1 &#038; 3 \\\\\n2 &#038; 4\n\\end{array}\\right]\\left\\{\\left[\\begin{array}{ll}\n1 &#038; 2 \\\\\n2 &#038; 4\n\\end{array}\\right]+\\left[\\begin{array}{ll}\n4 &#038; 1 \\\\\n1 &#038; 5\n\\end{array}\\right]\\right\\}-14\\left[\\begin{array}{ll}\n1 &#038; 0 \\\\\n0 &#038; 1\n\\end{array}\\right] \\\\\n&#038; =\\left[\\begin{array}{ll}\n1 &#038; 3 \\\\\n2 &#038; 4\n\\end{array}\\right]\\left[\\begin{array}{ll}\n5 &#038; 3 \\\\\n3 &#038; 9\n\\end{array}\\right]-\\left[\\begin{array}{cc}\n14 &#038; 0 \\\\\n0 &#038; 14\n\\end{array}\\right] \\\\\n&#038; =\\left[\\begin{array}{cc}\n5+9 &#038; 3+27 \\\\\n10+12 &#038; 6+36\n\\end{array}\\right]-\\left[\\begin{array}{cc}\n14 &#038; 0 \\\\\n0 &#038; 14\n\\end{array}\\right] \\\\\n&#038; =\\left[\\begin{array}{ll}\n14 &#038; 30 \\\\\n22 &#038; 42\n\\end{array}\\right]-\\left[\\begin{array}{cc}\n14 &#038; 0 \\\\\n0 &#038; 14\n\\end{array}\\right] \\\\\n&#038; =\\left[\\begin{array}{cc}\n0 &#038; 30 \\\\\n22 &#038; 28\n\\end{array}\\right]\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2023-q4-ii\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">ii<\/div><div class=\"xc-qtext\"><p>ABC is a triangle whose vertices are A(1, -1), B(0, 4) and C(-6, 4). D is the midpoint of BC. Find the:<\/p>\n<p>(a) coordinates of D.<\/p>\n<p>(b) equation of the median AD. <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=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/0c46b619-427e-45f1-9c68-599ad4677dfb-04_288_289_2070_1256.jpg\" alt=\"ICSE Class 10 Mathematics 2023 Question 4(ii) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>(a)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\mathrm{D} &#038; =\\left(\\frac{x_{1}+x_{2}}{2}, \\frac{y_{1}+y_{2}}{2}\\right) \\\\\n&#038; =\\left(\\frac{0-6}{2}, \\frac{4+4}{2}\\right)\n\\end{aligned}\n\\]<\/div>\n<p>therefore coordinates of D is (-3, 4)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\text { Slope of median } A D=\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\\frac{4+1}{-3-1}=\\frac{-5}{4}\n\\]<\/div>\n<p>Equation of the AD<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\ny-y_{1} &#038; =m\\left(x-x_{1}\\right) \\\\\ny+1 &#038; =\\frac{-5}{4}(x-1)\n\\end{aligned}\n\\]<\/div>\n<p>\u21d2<\/p>\n<div class=\"mathjax-overflow\">\\[\n5 x+4 y-1=0\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2023-q4-iii\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">iii<\/div><div class=\"xc-qtext\"><p>In the given figure, O is the centre of the circle. PQ is a tangent to the circle at T. Chord AB produced meets the tangent at P.<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/0c46b619-427e-45f1-9c68-599ad4677dfb-04_306_515_1879_369.jpg\" alt=\"ICSE Class 10 Mathematics 2023 Question 4(iii) figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\mathrm{AB} &#038; =9 \\mathrm{~cm}, \\mathrm{BP}=16 \\mathrm{~cm}, \\angle \\mathrm{PTB}=50^{\\circ} \\\\\n\\angle \\mathrm{OBA} &#038; =45^{\\circ}\n\\end{aligned}\n\\]<\/div>\n<p>Find:<\/p>\n<p>(a) length of PT<\/p>\n<p>(b) \u2220BAT<\/p>\n<p>(c) \u2220BOT<\/p>\n<p>(d) \u2220ABT <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)<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/0c46b619-427e-45f1-9c68-599ad4677dfb-05_305_518_612_372.jpg\" alt=\"ICSE Class 10 Mathematics 2023 Question 4(iii) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\nP T^{2} &#038; =P A \\times P B \\\\\n&#038; =(16+9) \\times 16 \\\\\nP T &#038; =\\sqrt{25 \\times 16}\n\\end{aligned}\n\\]<\/div>\n<p>\u2234<\/p>\n<div class=\"mathjax-overflow\">\\[\nP T=5 \\times 4=20 \\mathrm{~cm}\n\\]<\/div>\n<p>(b)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\angle \\mathrm{BAT}=\\angle \\mathrm{PTB}=50^{\\circ}\n\\]<\/div>\n<p>(alternate segment angle) (c)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\angle \\mathrm{BOT}=2 \\angle \\mathrm{BAT}\n\\]<\/div>\n<p>(Degree measure theorem)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\angle \\mathrm{BOT}=2 \\times 50=100^{\\circ}\n\\]<\/div>\n<p>(d) In \\(\\triangle \\mathrm{OBT}\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\angle \\mathrm{OBT}+\\angle \\mathrm{OTB}+\\angle \\mathrm{BOT}=180^{\\circ}\n\\]<\/div>\n<p>(Sum of all angles of \\(\\Delta\\) )<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\angle \\mathrm{OBT}+\\angle \\mathrm{OBT}+100^{\\circ} &#038; =180^{\\circ} \\\\\n\\angle \\mathrm{OBT} &#038; =\\angle \\mathrm{OTB}\n\\end{aligned}\n\\]<\/div>\n<p>(angles opposite to equal sides of a triangle are equal)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n2 \\angle \\mathrm{OBT} &#038; =180^{\\circ}-100^{\\circ}=80^{\\circ} \\\\\n\\angle \\mathrm{OBT} &#038; =\\frac{80^{\\circ}}{2}=40^{\\circ} \\\\\n\\angle \\mathrm{ABT} &#038; =\\angle \\mathrm{ABO}+\\angle \\mathrm{OBT} \\\\\n&#038; =45^{\\circ}+40^{\\circ}\n\\end{aligned}\n\\]<\/div>\n<p>\u2234<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\angle \\mathrm{ABT}=85^{\\circ}\n\\]<\/div><\/div>\n<\/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 2023<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2023-q5-i\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">i<\/div><div class=\"xc-qtext\"><p>Mrs. Arora bought the following articles from a departmental store:<\/p>\n<div class=\"math-responsive\"><table><thead><tr><th>S.No.<\/th><th>Item<\/th><th>Price<\/th><th>Rate of GST<\/th><th>Discount<\/th><\/tr><\/thead><tbody><tr><td>1.<\/td><td>Hair oil<\/td><td>\u20b91200<\/td><td>18%<\/td><td>\u20b9100<\/td><\/tr><tr><td>2.<\/td><td>Cashew nuts<\/td><td>\u20b9600<\/td><td>12%<\/td><td>&#8211;<\/td><\/tr><\/tbody><\/table><\/div>\n<p>Find the:<\/p>\n<p>(a) Total GST paid.<\/p>\n<p>(b) Total bill amount including GST. <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>Mrs Arora paid for<\/p>\n<p>1. Hair oil \\(\\text{\u20b9}(1200-100)=\\text{\u20b9} 1100\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\text { G.S.T. }=\\frac{18}{100} \\times 1100=\\text{\u20b9} 198\n\\]<\/div>\n<p>2. Cashew nuts<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\text { G.S.T. }=600 \\times \\frac{12}{100}=\\text{\u20b9} 72\n\\]<\/div>\n<p>(a) Total G.S.T. paid \\(=\\text{\u20b9} 198+72=\\text{\u20b9} 270\\)<\/p>\n<p>(b) Total Bill Amount including G.S.T.<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&#038; =\\text{\u20b9} 1100+600+270 \\\\\n&#038; =\\text{\u20b9} 1970\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2023-q5-ii\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">ii<\/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-5(x-9) \\geq 17-9 x>x+2, x \\in \\mathbf{R}\n\\]<\/div><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n-5(x-9) &#038; \\geq 17-9 x>x+2 \\\\\n-5(x-9) &#038; \\geq 17-9 x \\text { and } 17-9 x>x+2 \\\\\n-5 x+9 x &#038; \\geq 17-45 \\\\\n4 x &#038; \\geq-28 \\\\\nx &#038; \\geq-7\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\/0c46b619-427e-45f1-9c68-599ad4677dfb-05_88_343_1500_1244.jpg\" alt=\"ICSE Class 10 Mathematics 2023 Question 5(ii) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n17-9 x &#038; >x+2 \\\\\n-9 x-x &#038; >2-17 \\\\\n-10 x &#038; >-15 \\\\\nx &#038; <\\frac{-15}{-10} \\\\\nx &#038; <1.5\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\/0c46b619-427e-45f1-9c68-599ad4677dfb-05_91_342_1821_1245.jpg\" alt=\"ICSE Class 10 Mathematics 2023 Question 5(ii) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>Solution set<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/0c46b619-427e-45f1-9c68-599ad4677dfb-05_59_343_1964_1245.jpg\" alt=\"ICSE Class 10 Mathematics 2023 Question 5(ii) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<div class=\"mathjax-overflow\">\\[\n-7 \\leq x<1.5\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2023-q5-iii\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">iii<\/div><div class=\"xc-qtext\"><p>In the given figure, AC || DE || BF. If \\(\\mathbf{A C}=\\mathbf{2 4 ~ c m}, \\mathbf{E G}=\\mathbf{8 ~ c m}, \\mathbf{G B}=\\mathbf{1 6 ~ c m}, \\mathbf{B F}=\\mathbf{3 0 ~ c m}\\).<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/0c46b619-427e-45f1-9c68-599ad4677dfb-05_390_611_274_1152.jpg\" alt=\"ICSE Class 10 Mathematics 2023 Question 5(iii) figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>(a) Prove \\(\\Delta \\mathrm{GED}-\\triangle \\mathrm{GBF}\\)<\/p>\n<p>(b) Find DE<\/p>\n<p>(c) DB : AB <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)<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/0c46b619-427e-45f1-9c68-599ad4677dfb-05_389_612_2076_1203.jpg\" alt=\"ICSE Class 10 Mathematics 2023 Question 5(iii) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>In \\(\\triangle G E D\\) and \\(\\triangle G B F\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\angle E G D=\\angle B G F\n\\]<\/div>\n<p>(Vertically opposite angles)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\angle D E G=\\angle F B G\n\\]<\/div>\n<p>(Alternative angles \\(D E \\| B F\\) )<\/p>\n\\(\\therefore \\quad \\triangle G E D \\sim \\triangle G B F\\)\n<p>(AA Similarity Criterion) Hence Proved. (b)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\triangle G E D \\sim \\triangle G B F\n\\]<\/div>\n<p>(Proved)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\frac{G E}{G B}=\\frac{E D}{B F}=\\frac{G D}{G F} \\quad \\text { (by BPT) }\n\\]<\/div>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\frac{E D}{B F} &#038; =\\frac{G E}{G B} \\\\\n\\frac{E D}{30} &#038; =\\frac{8}{16} \\\\\nE D &#038; =\\frac{8}{16} \\times 30=15 \\mathrm{~cm} \\\\\nD E &#038; =15 \\mathrm{~cm}\n\\end{aligned}\n\\]<\/div>\n<p>(c) In \\(\\triangle A B C, D E \\| A C\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\angle B D E=\\angle B A C\n\\]<\/div>\n<p>and<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\angle B E D=\\angle B C A\n\\]<\/div>\n<p>by AA Similarity criterion, triangle \\(B D E \\sim\\) triangle \\(B A C\\).<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&#038; \\frac{D B}{A B}=\\frac{B E}{B C}=\\frac{D E}{A C} \\\\\n&#038; \\frac{D B}{A B}=\\frac{D E}{A C} \\\\\n&#038; \\frac{D B}{A B}=\\frac{15}{24}=\\frac{5}{8}\n\\end{aligned}\n\\]<\/div>\n<p>\u2234<\/p>\n<div class=\"mathjax-overflow\">\\[\nD B: A B=5: 8\n\\]<\/div><\/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 2023<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2023-q6-i\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">i<\/div><div class=\"xc-qtext\"><p>The following distribution gives the daily wages of 60 workers of a factory.<\/p>\n<div class=\"math-responsive\"><table><thead><tr><th>Daily income \u20b9<\/th><th>Number of worker (f)<\/th><\/tr><\/thead><tbody><tr><td>200-300<\/td><td>6<\/td><\/tr><tr><td>300-400<\/td><td>10<\/td><\/tr><tr><td>400-500<\/td><td>14<\/td><\/tr><tr><td>500-600<\/td><td>16<\/td><\/tr><tr><td>600-700<\/td><td>10<\/td><\/tr><tr><td>700-800<\/td><td>4<\/td><\/tr><\/tbody><\/table><\/div>\n<p>Use graph paper to answer this question. Take \\(2 \\mathrm{~cm}=\\text{\u20b9} 100\\) along one axis and \\(2 \\mathrm{~cm}=2\\) workers along the other axis. Draw a histogram and hence find the mode of the given distribution. <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=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/0c46b619-427e-45f1-9c68-599ad4677dfb-06_1001_752_1100_1042.jpg\" alt=\"ICSE Class 10 Mathematics 2023 Question 6(i) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2023-q6-ii\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">ii<\/div><div class=\"xc-qtext\"><p>The \\(\\mathbf{5}^{\\text {th }}\\) term and the \\(\\mathbf{9}^{\\text {th }}\\) term of an Arithmetic Progression are 4 and -12 respectively. Find:<\/p>\n<p>(a) the first term<\/p>\n<p>(b) common difference<\/p>\n<p>(c) sum of 16 terms of the AP. <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_{5}=4 \\text { and } t_{9}=-12\n\\]<\/div>\n<p>Let the first term be \\(a\\) and common difference be \\(d\\) respectively<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\na_{n} &#038; =a+(n-1) d \\\\\nt_{5} &#038; =4 \\\\\na+(5-1) d &#038; =4 \\\\\na+4 d &#038; =4 \\qquad \\text{&#8230;(i)}\n\\end{aligned}\n\\]<\/div>\n<p>Similarly, \\(a+8 d=-12\\) &#8230;(ii) from (i) and (ii)<\/p>\n<div class=\"mathjax-overflow\">\\[\nd=-4 \\text { and } a=20\n\\]<\/div>\n<p>(a) first term is 20<\/p>\n<p>(b) Common difference \\(=-4\\)<\/p>\n<p>(c)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\mathrm{S}_{n} &#038; =\\frac{n}{2}[2 a+(n+1) d] \\\\\n\\mathrm{S}_{16} &#038; =\\frac{16}{2}[2 \\times 20+(16-1)(-4)] \\\\\n&#038; =8[40-60)] \\\\\n&#038; =-160\n\\end{aligned}\n\\]<\/div>\n<p>\u2234 Sum of 16 terms is -160<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2023-q6-iii\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">iii<\/div><div class=\"xc-qtext\"><p>A and B are two points on the \\(\\boldsymbol{x}\\)-axis and \\(\\boldsymbol{y}\\)-axis respectively.<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/0c46b619-427e-45f1-9c68-599ad4677dfb-06_530_632_233_1141.jpg\" alt=\"ICSE Class 10 Mathematics 2023 Question 6(iii) figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>(a) Write down the coordinates of A and B.<\/p>\n<p>(b) P is a point on AB such that \\(\\mathbf{A P}: \\mathbf{P B}=\\mathbf{3}: \\mathbf{1}\\). Using section formula find the coordinates of point P.<\/p>\n<p>(c) Find the equation of a line passing through P and perpendicular to AB. <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)<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/0c46b619-427e-45f1-9c68-599ad4677dfb-07_489_586_823_368.jpg\" alt=\"ICSE Class 10 Mathematics 2023 Question 6(iii) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>Coordinates of A(4, 0) Coordinates of B(0, 4) (b)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&#038; P=\\left(\\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}}\\right) \\\\\n&#038; =\\left(\\frac{3 \\times 0+1 \\times 4}{3+1}, \\frac{3 \\times 4+1 \\times 0}{3+1}\\right) \\\\\n&#038; P=(1,3)\n\\end{aligned}\n\\]<\/div>\n<p>(c) Slope of line \\(A B=\\frac{4-0}{0-4}=-1\\)<\/p>\n<p>\u2234 Slope of perpendicular line \\(A B=1\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\ny-y_{1} &#038; =m\\left(x-x_{1}\\right) \\\\\ny-3 &#038; =1(x-1) \\\\\ny-3 &#038; =x-1 \\\\\nx-y+2 &#038; =0\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 2023<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2023-q7-i\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">i<\/div><div class=\"xc-qtext\"><p>A bag contains 25 cards, numbered through 1 to 25. A card is drawn at random. What is the probability that the number on the card drawn is:<\/p>\n<p>(a) multiple of 5<\/p>\n<p>(b) a perfect square<\/p>\n<p>(c) a prime 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><div class=\"mathjax-overflow\">\\[\n\\mathrm{S}=\\{1,2,3,4, \\ldots \\ldots \\ldots . . .25\\}\n\\]<\/div>\n<p>\u2234<\/p>\n<div class=\"mathjax-overflow\">\\[\nn(\\mathrm{~S})=25\n\\]<\/div>\n<p>(a) multiple of 5<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&#038; \\qquad \n\\mathrm{E} &#038; =\\{5,10,15,20,25\\} \\\\\nn(\\mathrm{E}) &#038; =5\n \\\\\n&#038; \\text { Probability of multiple of } 5=\\frac{n(\\mathrm{E})}{n(\\mathrm{~S})}=\\frac{5}{25} \\\\\n&#038; =\\frac{1}{5}\n\\end{aligned}\n\\]<\/div>\n<p>(b) Perfect square \\(=(1,4,9,16,25)\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\text { Required probability }=\\frac{5}{25}=\\frac{1}{5}\n\\]<\/div>\n<p>(c) a prime number \\{2, 3, 5, 7, 11, 13, 17, 19, 23\\}<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\text { Required probability }=\\frac{9}{25}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2023-q7-ii\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">ii<\/div><div class=\"xc-qtext\"><p>A man covers a distance of 100 km, travelling with a uniform speed of \\(\\boldsymbol{x}\\) km\/hr, had the speed been 5 km\/hr more it would have taken 1 hour less. Find \\(\\boldsymbol{x}\\) the original speed. <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 original speed be \\(x\\) km\/h<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\text { Time } &#038; =\\frac{\\text { Distance }}{\\text { Speed }} \\\\\nt_{1} &#038; =\\frac{100}{x} \\mathrm{~h} \\\\\nt_{2} &#038; =\\frac{100}{x+5} \\mathrm{~h} \\\\\nt_{1}-t_{2} &#038; =1 \\\\\n\\frac{100}{x}-\\frac{100}{x+5} &#038; =1 \\\\\n\\frac{100 x+500-100 x}{x(x+5)} &#038; =1 \\\\\n500 &#038; =x^{2}+5 x \\\\\nx^{2}+5 x-500 &#038; =0 \\\\\n(x+25)(x-20) &#038; =0\n\\end{aligned}\n\\]<\/div>\n<p>If \\(x+25=0 \\Rightarrow x=-25\\) it is not possible because speed can not be negative if \\(x-20=0\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\nx=20\n\\]<\/div>\n<p>\u2234 original speed be 20 km\/h<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2023-q7-iii\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">iii<\/div><div class=\"xc-qtext\"><p>A solid in the shape of a hemisphere of radius 7 cm, surmounted by a cone of height 4 cm. The solid is immersed completely in a cylindrical container filled with water to a certain height. If the radius of the cylinder is 14 cm, find the rise in the water level. <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\/0c46b619-427e-45f1-9c68-599ad4677dfb-07_305_257_387_1329.jpg\" alt=\"ICSE Class 10 Mathematics 2023 Question 7(iii) 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>Volume of solid = Vol. of hemisphere + Vol. of Cone.<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&#038; =\\frac{2}{3} \\pi r^{3}+\\frac{1}{3} \\pi r^{2} h=\\frac{\\pi r^{2}}{3}(2 r+h) \\\\\n&#038; =\\frac{\\pi \\times 7 \\times 7}{3}(2 \\times 7+4)=\\frac{49 \\pi}{3}(18) \\\\\n&#038; =294 \\pi \\mathrm{~cm}^{3}\n\\end{aligned}\n\\]<\/div>\n<p>Let the level of water rise in cylinder be \\(x \\mathrm{~cm}\\) \u2234 Rise in Vol. of water \\(=294 \\pi\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\pi R^{2} x &#038; =294 \\pi \\\\\n(14)^{2} x &#038; =294 \\\\\nx &#038; =\\frac{294}{14 \\times 14}=\\frac{21}{14} \\\\\nx &#038; =\\frac{3}{2} \\mathrm{~cm}=1.5 \\mathrm{~cm}\n\\end{aligned}\n\\]<\/div>\n<p>\u2234 level of water rise in cylinder \\(=1.5 \\mathrm{~cm}\\)<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"xc-set\">\n<div class=\"xc-set-header\"><div class=\"xc-set-icon\">8<\/div><div><div class=\"xc-set-title\">Question 8<\/div><div class=\"xc-set-sub\">ICSE Class 10 Mathematics 2023<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2023-q8-i\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">i<\/div><div class=\"xc-qtext\"><p>The following table gives the marks scored by a set of students in an examination. Calculate the mean of the distribution by using the short cut method. <strong>[3]<\/strong><\/p>\n<div class=\"math-responsive\"><table><thead><tr><th>Marks<\/th><th>Number of Students ( \\(\\boldsymbol{f}\\) )<\/th><\/tr><\/thead><tbody><tr><td>0-10<\/td><td>3<\/td><\/tr><tr><td>10-20<\/td><td>8<\/td><\/tr><tr><td>20-30<\/td><td>14<\/td><\/tr><tr><td>30-40<\/td><td>9<\/td><\/tr><tr><td>40-50<\/td><td>4<\/td><\/tr><tr><td>50-60<\/td><td>2<\/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>Number of Students (f)<\/th><th>\\(x\\)<\/th><th>\\(d=\\) \\(x-\\mathrm{A}\\)<\/th><th>\\(\\boldsymbol{f} \\boldsymbol{d}\\)<\/th><\/tr><\/thead><tbody><tr><td>0-10<\/td><td>3<\/td><td>5<\/td><td>-20<\/td><td>-60<\/td><\/tr><tr><td>10-20<\/td><td>8<\/td><td>15<\/td><td>-10<\/td><td>-80<\/td><\/tr><tr><td>20-30<\/td><td>14<\/td><td>25 = A<\/td><td>0<\/td><td>0<\/td><\/tr><tr><td>30-40<\/td><td>9<\/td><td>35<\/td><td>10<\/td><td>90<\/td><\/tr><tr><td>40-50<\/td><td>4<\/td><td>45<\/td><td>20<\/td><td>80<\/td><\/tr><tr><td>50-60<\/td><td>2<\/td><td>55<\/td><td>30<\/td><td>60<\/td><\/tr><tr><td><\/td><td>40<\/td><td><\/td><td><\/td><td>90<\/td><\/tr><\/tbody><\/table><\/div>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\bar{x} &#038; =A+\\frac{\\Sigma f d}{\\Sigma f} \\\\\n&#038; =25+\\frac{90}{40} \\\\\n&#038; =25+2.25 \\\\\n\\text { Mean } &#038; =27.25\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2023-q8-ii\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">ii<\/div><div class=\"xc-qtext\"><p>What number must be added to each of the numbers 4, 6, 8, 11 in order to get the four numbers in proportion? <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 \\(x\\) be added to get the four numbers in proportion<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&#038; \\therefore(4+x):(6+x)::(8+x):(11+x) \\\\\n&#038;(4+x)(11+x)=(6+x)(8+x) \\\\\n&#038; x^{2}+15 x+44=x^{2}+14 x+48 \\\\\n&#038; 15 x-14 x=48-44 \\\\\n&#038; x=4\n\\end{aligned}\n\\]<\/div>\n<p>\u2234 4 should be added to get the four numbers in proportion<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2023-q8-iii\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">iii<\/div><div class=\"xc-qtext\"><p>Using ruler and compass construct a triangle ABC in which \\(\\mathbf{A B}=\\mathbf{6 ~ c m}, \\angle \\mathbf{B A C}=\\mathbf{1 2 0}^{\\circ}\\) and \\(\\mathbf{A C}=\\mathbf{5}\\) cm. Construct a circle passing through A, B and C. Measure and write down the radius of the circle. <strong>[4]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/0c46b619-427e-45f1-9c68-599ad4677dfb-08_305_301_639_1141.jpg\" alt=\"ICSE Class 10 Mathematics 2023 Question 8(iii) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>Step of construction:<\/p>\n<p>(a) Draw \\(A B=6 \\mathrm{~cm}\\)<\/p>\n<p>(b) Make \\(\\angle B A D=120^{\\circ}\\)<\/p>\n<p>(c) Taking centre A and radius of \\(A C 5\\) cm draw an arc which intersect at the point \\(C\\)<\/p>\n<p>(d) join \\(B C\\). triangle \\(A B C\\) is required triangle<\/p>\n<p>(e) Draw perpendicular bisectors of side \\(A B\\) and \\(A C\\) respectively which intersect at the point \\(O\\)<\/p>\n<p>(f) Taking \\(O\\) as centre and \\(O A\\) as radius draw a circle which passes through the points \\(A, B\\) and C respectively.<\/p>\n<p>Now, measured the radius \\(O A=5.5 \\mathrm{~cm}\\)<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"xc-set\">\n<div class=\"xc-set-header\"><div class=\"xc-set-icon\">9<\/div><div><div class=\"xc-set-title\">Question 9<\/div><div class=\"xc-set-sub\">ICSE Class 10 Mathematics 2023<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2023-q9-i\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">i<\/div><div class=\"xc-qtext\"><p>Using Componendo and Dividendo solve for \\(\\boldsymbol{x}\\). <strong>[3]<\/strong><\/p>\n<div class=\"mathjax-overflow\">\\[\n\\frac{\\sqrt{2 x+2}+\\sqrt{2 x-1}}{\\sqrt{2 x+2}-\\sqrt{2 x-1}}=3\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{2 x+2}+\\sqrt{2 x-1}}{\\sqrt{2 x+2}-\\sqrt{2 x-1}}=3\n\\]<\/div>\n<p>Applying componendo and dividendo<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\frac{\\sqrt{2 x+2}+\\sqrt{2 x-1}+\\sqrt{2 x+2}-\\sqrt{2 x-1}}{\\sqrt{2 x+2}+\\sqrt{2 x-1}-\\sqrt{2 x+2}+\\sqrt{2 x-1}}=\\frac{3+1}{3-1}\n\\]<\/div>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&#038; \\frac{2 \\sqrt{2 x+2}}{2 \\sqrt{2 x-1}}=\\frac{4}{2}=2 \\\\\n&#038; \\frac{\\sqrt{2 x+2}}{\\sqrt{2 x-1}}=2\n\\end{aligned}\n\\]<\/div>\n<p>squaring both sides<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\frac{2 x+2}{2 x-1} &#038; =4 \\\\\n2 x+2 &#038; =8 x-4 \\\\\n2 x-8 x &#038; =-4-2 \\\\\n-6 x &#038; =-6 \\\\\nx &#038; =1\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2023-q9-ii\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">ii<\/div><div class=\"xc-qtext\"><p>Which term of the Arithmetic Progression (A.P.) 15, 30, 45, 60 &#8230; is 300? Hence find the sum of the terms of the Arithmetic Progression (A.P.) <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>15, 30, 45, 60, &#8230;&#8230;&#8230;&#8230;&#8230;.. 300<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\na &#038; =15, d=30-15=15 \\\\\na_{n} &#038; =300 \\\\\na+(n-1) d &#038; =300 \\\\\n15+(n-1)(15) &#038; =300 \\\\\n1+(n-1) &#038; =20 \\\\\nn &#038; =20\n\\end{aligned}\n\\]<\/div>\n<p>\\(20^{\\text {th }}\\) term is 300<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&#038; S_{n}=\\frac{n}{2}\\left[a+a_{n}\\right]=\\frac{20}{2}[15+300] \\\\\n&#038; S_{20}=3150\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2023-q9-iii\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">iii<\/div><div class=\"xc-qtext\"><p>From the top of a tower 100 m high a man observes the angles of depression of two ships A and B, on opposite sides of the lower as 45\u00b0 and 38\u00b0 respectively. If the foot of the tower and the ships are in the same horizontal line find the distance between the two ships A and B to the nearest metre. (Use Mathematical Tables for this question.) <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\/0c46b619-427e-45f1-9c68-599ad4677dfb-08_281_520_2203_1199.jpg\" alt=\"ICSE Class 10 Mathematics 2023 Question 9(iii) 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><div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/0c46b619-427e-45f1-9c68-599ad4677dfb-09_280_522_1704_309.jpg\" alt=\"ICSE Class 10 Mathematics 2023 Question 9(iii) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>In \\(\\triangle A C D\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\tan A &#038; =\\frac{C D}{A C} \\\\\n\\tan 45^{\\circ} &#038; =\\frac{100}{A C} \\\\\n1 &#038; =\\frac{100}{A C} \\\\\n\\therefore \\quad \\mathrm{AC} &#038; =100 \\mathrm{~m}\n\\end{aligned}\n\\]<\/div>\n<p>In \\(\\triangle B C D\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\tan B &#038; =\\frac{C D}{B C} \\\\\n\\tan 38^{\\circ} &#038; =\\frac{100}{B C} \\\\\n0.7813 &#038; =\\frac{100}{B C} \\\\\n\\therefore \\quad B C &#038; =\\frac{100}{0.7813} \\\\\n&#038; =127.99 \\mathrm{~m}\n\\end{aligned}\n\\]<\/div>\n<p>Distance between two ships \\(=A C+B C\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n= &#038; 100+127.99 \\\\\n227.99 \\mathrm{~m} &#038; =228 \\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\">10<\/div><div><div class=\"xc-set-title\">Question 10<\/div><div class=\"xc-set-sub\">ICSE Class 10 Mathematics 2023<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2023-q10-i\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">i<\/div><div class=\"xc-qtext\"><p>Factorize completely using factor theorem: <strong>[4]<\/strong><\/p>\n<div class=\"mathjax-overflow\">\\[\n2 x^{3}-x^{2}-13 x-6\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>\\(2 x^{3}-x^{2}-13 x-6\\)\n<p>Put<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\nx &#038; =-2 \\\\\n&#038; =2(-2)^{3}-(-2)^{2}-13(-2)-6 \\\\\n&#038; =-16-4+26-6 \\\\\n&#038; =0\n\\end{aligned}\n\\]<\/div>\n<p>\\(\\therefore(x+2)\\) is a factor of polynomial<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&#038; x+2) 2 x^{3}-x^{2}-13 x-6\\left(2 x^{2}-5 x-3\\right. \\\\\n&#038; 2 x^{3}+4 x^{2} \\\\\n&#038; \\frac{(-)(-)}{-5 x^{2}-13 x-6} \\\\\n&#038; -5 x^{2}-10 x \\\\\n&#038; \\begin{array}{c}\n(+) \\quad(+) \\\\\n\\hline-3 x-6 \\\\\n-3 x-6 \\\\\n(+)(+) \\\\\n\\hline \\times\n\\end{array} \\\\\n&#038; 2 x^{3}-x^{2}-13 x-6=(x+2)\\left(2 x^{2}-5 x-3\\right) \\\\\n&#038; =(x+2)\\left\\{2 x^{2}-6 x+x-3\\right\\} \\\\\n&#038; =(x+2)\\{2 x(x-3)+1(x-3)\\} \\\\\n&#038; =(x+2)(x-3)(2 x+1) \\\\\n&#038; \\therefore 2 x^{3}-x^{2}-13 x-6=(x-3)(2 x+1)(x+2)\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2023-q10-ii\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">ii<\/div><div class=\"xc-qtext\"><p>Use graph paper to answer this question. During a medical checkup of 60 students in a school, weights were recorded as follows:<\/p>\n<div class=\"math-responsive\"><table><thead><tr><th>Weight (in kg)<\/th><th>Number of Students<\/th><\/tr><\/thead><tbody><tr><td>28-30<\/td><td>2<\/td><\/tr><tr><td>30-32<\/td><td>4<\/td><\/tr><tr><td>32-34<\/td><td>10<\/td><\/tr><tr><td>34-36<\/td><td>13<\/td><\/tr><tr><td>36-38<\/td><td>15<\/td><\/tr><tr><td>38-40<\/td><td>9<\/td><\/tr><tr><td>40-42<\/td><td>5<\/td><\/tr><tr><td>42-44<\/td><td>2<\/td><\/tr><\/tbody><\/table><\/div>\n<p>Taking \\(\\mathbf{2} \\mathbf{~ c m}=\\mathbf{2 ~ k g}\\) along one axis and \\(\\mathbf{2 ~ c m}=\\mathbf{1 0}\\) students along the other axis draw an ogive. Use your graph to find the:<\/p>\n<p>(a) median<\/p>\n<p>(b) upper Quartile<\/p>\n<p>(c) number of students whose weight is above 37 kg <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>Weight (in kg)<\/th><th>Number of Students (f)<\/th><th>Cumulative frequency (cf)<\/th><\/tr><\/thead><tbody><tr><td>28-30<\/td><td>2<\/td><td>2<\/td><\/tr><tr><td>30-32<\/td><td>4<\/td><td>6<\/td><\/tr><tr><td>32-34<\/td><td>10<\/td><td>16<\/td><\/tr><tr><td>34-36<\/td><td>13<\/td><td>29<\/td><\/tr><tr><td>36-38<\/td><td>15<\/td><td>44<\/td><\/tr><tr><td>38-40<\/td><td>9<\/td><td>53<\/td><\/tr><tr><td>40-42<\/td><td>5<\/td><td>58<\/td><\/tr><tr><td>42-44<\/td><td>2<\/td><td>60<\/td><\/tr><\/tbody><\/table><\/div>\n<div class=\"math-responsive\"><table><thead><tr><th>Weight in kgs. (less than)<\/th><th>Number of Students<\/th><\/tr><\/thead><tbody><tr><td>28<\/td><td>0<\/td><\/tr><tr><td>30<\/td><td>2<\/td><\/tr><tr><td>32<\/td><td>6<\/td><\/tr><tr><td>34<\/td><td>16<\/td><\/tr><tr><td>36<\/td><td>29<\/td><\/tr><tr><td>38<\/td><td>44<\/td><\/tr><tr><td>40<\/td><td>53<\/td><\/tr><tr><td>42<\/td><td>58<\/td><\/tr><tr><td>44<\/td><td>60<\/td><\/tr><\/tbody><\/table><\/div>\n<p>(a) median \\(=36.2 \\mathrm{~kg}\\)<\/p>\n<p>(b) upper quartile \\(=38.2 \\mathrm{~kg}\\)<\/p>\n<p>(c) Number of students where weight is above \\(37 \\mathrm{~kg}=60-37=23\\) students<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/0c46b619-427e-45f1-9c68-599ad4677dfb-10_751_685_447_1142.jpg\" alt=\"ICSE Class 10 Mathematics 2023 Question 10(ii) 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\">19<\/span> <span class=\"rt-label rt-postfix\">minutes<\/span><\/span> ICSE Class 10 Mathematics PYQ Practice the complete ICSE Class 10 Mathematics 2023 Question Paper with Solutions, including MCQs, algebra, [&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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