{"id":2424,"date":"2026-11-13T18:03:08","date_gmt":"2026-11-13T12:33:08","guid":{"rendered":"https:\/\/xamcontent.com\/cisce\/?p=2424"},"modified":"2026-11-13T18:03:09","modified_gmt":"2026-11-13T12:33:09","slug":"icse-class-10-mathematics-2022-semester-2-question-paper-with-solutions","status":"publish","type":"post","link":"https:\/\/xamcontent.com\/cisce\/icse-class-10-mathematics-2022-semester-2-question-paper-with-solutions\/","title":{"rendered":"ICSE Class 10 Mathematics 2022 Semester 2 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-2022-semester-2-question-paper-with-solutions\/#Question_1_%E2%80%94_Multiple_Choice_Questions\" title=\"Question 1 \u2014 Multiple Choice Questions\">Question 1 \u2014 Multiple Choice Questions<\/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-2022-semester-2-question-paper-with-solutions\/#Descriptive_and_Numerical_Questions\" title=\"Descriptive and Numerical Questions\">Descriptive and Numerical Questions<\/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\">22<\/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 2022 Semester 2 Question Paper with Solutions<\/strong>, including MCQs, algebra, geometry, trigonometry, statistics, probability, graphs and mensuration. 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\">2022 Semester 2<\/div><\/div>\n<div class=\"xc-info-card\"><div class=\"xc-info-card-label\">Maximum Marks<\/div><div class=\"xc-info-card-value\">40<\/div><\/div>\n<div class=\"xc-info-card\"><div class=\"xc-info-card-label\">Time Allowed<\/div><div class=\"xc-info-card-value\">One hour 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 2022 Semester 2<\/div><\/div><\/div>\n<div class=\"xc-qtext\"><p>Attempt all questions from Section A and any three questions from Section B.<\/p><p>The marks intended for questions 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=\"Question_1_%E2%80%94_Multiple_Choice_Questions\"><\/span>Question 1 \u2014 Multiple Choice Questions<span class=\"ez-toc-section-end\"><\/span><\/h2><div class=\"xc-section-heading-sub\">10 marks \u2022 Attempt all questions.<\/div><\/div>\n<div class=\"xc-section-heading-stat\"><div class=\"xc-section-heading-stat-num\">10<\/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 answers from the given options.<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2022-sem2-q1-i\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">i<\/div><div class=\"xc-qtext\"><p>The probability of getting a number divisible by 3 in throwing a dice 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{1}{6}\\)<\/span><\/div>\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">B<\/span><span>\\(\\frac{1}{3}\\)<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">C<\/span><span>\\(\\frac{1}{2}\\)<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">D<\/span><span>\\(\\frac{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 (b) is correct.<\/p>\n<p><strong>Explanation:<\/strong> All possible out comes of a dice \\(=\\{1,2,3,4,5,6\\}\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\nn(\\mathrm{E})=6\n\\]<\/div>\n<p>Favourable outcomes (divisible by 3 )<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; =\\{3,6\\} \\\\\nn(\\mathrm{~F}) &amp; =2\n\\end{aligned}\n\\]<\/div>\n<p>Probability of getting a number divisible by 3<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; =\\frac{\\text { Favourable outcomes }}{\\text { Total no. of outcomes }} \\\\\n&amp; =\\frac{n(\\mathrm{~F})}{n(\\mathrm{E})}=\\frac{2}{6}=\\frac{1}{3}\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2022-sem2-q1-ii\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">ii<\/div><div class=\"xc-qtext\"><p>The volume of a conical tent is \\(\\mathbf{4 6 2} \\mathbf{~ m}^{\\mathbf{3}}\\) and the area of the base is \\(\\mathbf{1 5 4} \\mathbf{~ m}^{\\mathbf{2}}\\). The height of the cone 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>15 m<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">B<\/span><span>12 m<\/span><\/div>\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">C<\/span><span>9 m<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">D<\/span><span>24 m<\/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> Given, Volume of conical tent \\(=462 \\mathrm{~m}^{3}\\) and Area of the base \\(=154 \\mathrm{~m}^{2}\\) We know that<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\text { Volume of a conical tent }= &amp; \\frac{1}{3} \\pi r^{2} h \\\\\n= &amp; \\frac{1}{3} \\times \\text { Area of base } \\\\\n&amp; \\times \\text { height }\n\\end{aligned}\n\\]<\/div>\n<p>\u21d2<\/p>\n<div class=\"mathjax-overflow\">\\[\n462=\\frac{1}{3} \\times 154 \\times h\n\\]<\/div>\n<p>\u21d2<\/p>\n<div class=\"mathjax-overflow\">\\[\nh=\\frac{462 \\times 3}{154}=9 \\mathrm{~m}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2022-sem2-q1-iii\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">iii<\/div><div class=\"xc-qtext\"><p>The median class for the given distribution is: <strong>[1]<\/strong><\/p>\n<div class=\"math-responsive\"><table><thead><tr><th>Class Interval<\/th><th>0-10<\/th><th>10-20<\/th><th>20-30<\/th><th>30-40<\/th><\/tr><\/thead><tbody><tr><td>Frequency<\/td><td>2<\/td><td>4<\/td><td>3<\/td><td>5<\/td><\/tr><\/tbody><\/table><\/div><\/div><\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">A<\/span><span>0-10<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">B<\/span><span>10-20<\/span><\/div>\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">C<\/span><span>20-30<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">D<\/span><span>30-40<\/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 (c) is correct.<\/p>\n<p><strong>Explanation:<\/strong> <\/p>\n<div class=\"math-responsive\"><table><thead><tr><th>C. I.<\/th><th>Frequency<\/th><th>c. f.<\/th><\/tr><\/thead><tbody><tr><td>0-10<\/td><td>2<\/td><td>2<\/td><\/tr><tr><td>10-20<\/td><td>4<\/td><td>\\(2+4=6\\)<\/td><\/tr><tr><td>20-30<\/td><td>3 (f)<\/td><td>\\(6+3=9\\)<\/td><\/tr><tr><td>30-40<\/td><td>5<\/td><td>\\(9+5=14\\)<\/td><\/tr><tr><td><\/td><td>14<\/td><td><\/td><\/tr><\/tbody><\/table><\/div>\n<div class=\"mathjax-overflow\">\\[\nN=14 \\Rightarrow \\frac{N}{2}=7\n\\]<\/div>\n<p>We can see that cumulative frequency just greater than 7 is 9, which is lie in the class interval 20-30. So, the median class is 20-30.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2022-sem2-q1-iv\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">iv<\/div><div class=\"xc-qtext\"><p>If two lines are perpendicular to one another then the relation between their slopes \\(\\boldsymbol{m}_{\\mathbf{1}}\\) and \\(\\boldsymbol{m}_{\\mathbf{2}}\\) 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>\\(m_{1}=m_{2}\\)<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">B<\/span><span>\\(m_{1}=\\frac{1}{m_{2}}\\)<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">C<\/span><span>\\(m_{1}=-m_{2}\\)<\/span><\/div>\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">D<\/span><span>\\(m_{1 \\times} m_{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 (d) is correct.<\/p>\n<p><strong>Explanation:<\/strong> If two lines, having slopes \\(m_{1}\\) and \\(m_{2}\\), perpendicular to one another, then \\(m_{1}\\) \\(\\times m_{2}=-1\\)<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2022-sem2-q1-v\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">v<\/div><div class=\"xc-qtext\"><p>A lighthouse is 80 m high. The angle of elevation of its top from a point 80 m away from its foot along the same horizontal line 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>60\u00b0<\/span><\/div>\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">B<\/span><span>45\u00b0<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">C<\/span><span>30\u00b0<\/span><\/div>\n<div class=\"xc-opt\"><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 (b) is correct.<\/p>\n<p><strong>Explanation:<\/strong> Height of the light house \\(=80 \\mathrm{~m}\\) Distance of the point from the foot of the light house \\(=80 \\mathrm{~m}\\) Let the angle of elevation be \\(\\theta\\), then \\(\\tan \\theta\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n=\\frac{\\text { Perpendicular }}{\\text { Base }}\n\\]<\/div>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/83c2efd8-c79b-47f4-a50b-fab47900d3f3-2_421_522_2045_325.jpg\" alt=\"ICSE Class 10 Mathematics 2022 Semester 2 Question 1(v) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<div class=\"mathjax-overflow\">\\[\n\\tan \\theta=\\frac{\\mathrm{AB}}{\\mathrm{BC}}=\\frac{80}{80}=1\n\\]<\/div>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\tan \\theta= &amp; \\tan 45^{\\circ} \\\\\n&amp; \\quad\\left(\\because \\tan 45^{\\circ}=1\\right)\n\\end{aligned}\n\\]<\/div>\n<p>So, \\(\\theta=45^{\\circ}\\)<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2022-sem2-q1-vi\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">vi<\/div><div class=\"xc-qtext\"><p>The modal class of a given distribution always corresponds to the: <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>interval with highest frequency<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">B<\/span><span>interval with lowest frequency<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">C<\/span><span>the first interval<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">D<\/span><span>the last interval<\/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><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2022-sem2-q1-vii\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">vii<\/div><div class=\"xc-qtext\"><p>The coordinates of the point P(-3, 5) on reflecting on the X axis are: <strong>[1]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">A<\/span><span>\\((3,5)\\)<\/span><\/div>\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">B<\/span><span>(-3, -5)<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">C<\/span><span>(3, -5)<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">D<\/span><span>(-3, 5)<\/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> Given point (-3, 5) lies in II \\({ }^{\\text {nd }}\\) quadrant. So, after reflection sign of Y-coordinate will invert and sign of X-coordinate remain same. Reflection of this point will be in \\(I I \\mathrm{II}^{\\text {rd }}\\)-quadrant, i.e., (-3, -5).<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2022-sem2-q1-viii\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">viii<\/div><div class=\"xc-qtext\"><p>ABCD is a cyclic quadrilateral. If \\(\\angle \\mathbf{B A D}=\\) \\((\\mathbf{2} \\boldsymbol{x}+\\mathbf{5})^{\\circ}\\) and \\(\\angle \\mathbf{B C D}=(\\boldsymbol{x}+\\mathbf{1 0})^{\\circ}\\) then \\(\\boldsymbol{x}\\) 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\/83c2efd8-c79b-47f4-a50b-fab47900d3f3-1_318_287_1051_1273.jpg\" alt=\"ICSE Class 10 Mathematics 2022 Semester 2 Question 1(viii) 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>65\u00b0<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">B<\/span><span>45\u00b0<\/span><\/div>\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">C<\/span><span>55\u00b0<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">D<\/span><span>5\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 (c) is correct.<\/p>\n<p><strong>Explanation:<\/strong> Given, \\(\\angle \\mathrm{BAD}=(2 x+5)^{\\circ}\\) and<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\angle \\mathrm{BCD}=(x+10)^{\\circ}\n\\]<\/div>\n<p>So,<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\angle \\mathrm{BAD}+\\angle \\mathrm{BCD}=180^{\\circ}\n\\]<\/div>\n<p>(Sum of opposite angles of a cyclic quadrilateral are supplementary)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\Rightarrow \\quad(2 x+5)^{\\circ}+(x+10)^{\\circ} &amp; =180^{\\circ} \\\\\n(3 x+15)^{\\circ} &amp; =180^{\\circ} \\\\\n3 x &amp; =165^{\\circ} \\\\\nx &amp; =\\frac{165^{\\circ}}{3}=55^{\\circ}\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2022-sem2-q1-ix\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">ix<\/div><div class=\"xc-qtext\"><p>A(1, 4), B \\((4,1)\\) and C \\((x, 4)\\) are the vertices of \\(\\triangle \\mathrm{ABC}\\). If the centroid of the triangle is G \\((4,3)\\) then \\(x\\) 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>2<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">B<\/span><span>1<\/span><\/div>\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">C<\/span><span>7<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">D<\/span><span>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 (c) is correct.<\/p>\n<p><strong>Explanation:<\/strong> Coordinates of centroid of a triangle is given by<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\left(\\frac{x_{1}+x_{2}+x_{3}}{3}, \\frac{y_{1}+y_{2}+y_{3}}{3}\\right)\n\\]<\/div>\n<p>So, coordinates of the centroid of triangle ABC is given by \\(\\left(\\frac{1+4+x}{3}, \\frac{4+1+4}{3}\\right)\\) or \\(\\left(\\frac{5+x}{3}, 3\\right)\\)<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/83c2efd8-c79b-47f4-a50b-fab47900d3f3-2_357_456_1583_1190.jpg\" alt=\"ICSE Class 10 Mathematics 2022 Semester 2 Question 1(ix) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>Also, coordinates of centroid of \\(\\Delta \\mathrm{ABC}(4,3)\\) So, \\(4=\\frac{5+x}{3} \\Rightarrow 12=5+x \\Rightarrow x=7\\)<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2022-sem2-q1-x\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">x<\/div><div class=\"xc-qtext\"><p>The radius of a roller 100 cm long is 14 cm. The curved surface area of the roller is: (Take \\(\\pi=\\frac{22}{7}\\) ) <strong>[1]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">A<\/span><span>\\(13200 \\mathrm{~cm}^{2}\\)<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">B<\/span><span>\\(15400 \\mathrm{~cm}^{2}\\)<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">C<\/span><span>\\(4400 \\mathrm{~cm}^{2}\\)<\/span><\/div>\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">D<\/span><span>\\(8800 \\mathrm{~cm}^{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 (d) is correct.<\/p>\n<p><strong>Explanation:<\/strong> We have, radius of a roller \\(=14 \\mathrm{~cm}\\) height of the roller \\(=100 \\mathrm{~cm}\\) Now, C.S.A. of roller \\((\\) Cylinder \\()=2 \\pi r h\\) \\(=2 \\times \\frac{22}{7} \\times 14 \\times 100=8800 \\mathrm{~cm}^{2}\\) (Attempt any three questions from this Section.)<\/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=\"Descriptive_and_Numerical_Questions\"><\/span>Descriptive and Numerical Questions<span class=\"ez-toc-section-end\"><\/span><\/h2><div class=\"xc-section-heading-sub\">30 marks \u2022 Attempt any three questions from this Section.<\/div><\/div>\n<div class=\"xc-section-heading-stat\"><div class=\"xc-section-heading-stat-num\">30<\/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\">2<\/div><div><div class=\"xc-set-title\">Question 2<\/div><div class=\"xc-set-sub\">ICSE Class 10 Mathematics 2022 Semester 2<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2022-sem2-q2-i\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">i<\/div><div class=\"xc-qtext\"><p>Prove that <strong>[2]<\/strong><\/p>\n<div class=\"mathjax-overflow\">\\[\n\\frac{1}{1+\\sin \\theta}+\\frac{1}{1-\\sin \\theta}=2 \\sec ^{2} \\theta\n\\]<\/div><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>L.H.S.<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; =\\frac{1}{1+\\sin \\theta}+\\frac{1}{1-\\sin \\theta} \\\\\n&amp; =\\frac{(1-\\sin \\theta)+(1+\\sin \\theta)}{(1+\\sin \\theta)(1-\\sin \\theta)} \\\\\n&amp; =\\frac{2}{1-\\sin ^{2} \\theta}=\\frac{2}{\\cos ^{2} \\theta} \\quad\\left(\\because \\sin ^{2} \\theta+\\cos ^{2} \\theta=1\\right) \\\\\n&amp; =2 \\sec ^{2} \\theta=\\text { R.H.S. }\\left(\\because \\frac{1}{\\cos \\theta}=\\sec \\theta\\right)\n\\end{aligned}\n\\]<\/div>\n<p>Hence Proved.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2022-sem2-q2-ii\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">ii<\/div><div class=\"xc-qtext\"><p>Find &#8216; \\(a\\) &#8216; if \\(\\mathbf{A}(2 a+2,3), \\mathbf{B}(7,4)\\) and C ( \\(\\mathbf{2} \\boldsymbol{a}+\\mathbf{5}, \\mathbf{2}\\) ) are collinear. <strong>[2]<\/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 A \\((2 a+2,3), \\mathrm{B}(7,4)\\) and C \\((2 a+5,2)\\) are collinear. When three points are colliner, then<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\frac{1}{2}\\left[x_{1}\\left(y_{2}-y_{3}\\right)+x_{2}\\left(y_{3}-y_{1}\\right)+x_{3}\\left(y_{1}-y_{2}\\right)\\right]=0\n\\]<\/div>\n<p>or \\(x_{1}\\left(y_{2}-y_{3}\\right)+x_{2}\\left(y_{3}-y_{1}\\right)+x_{3}\\left(y_{1}-y_{2}\\right)=0\\) We have<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\Rightarrow \\quad \\begin{aligned}\nx_{1}=2 a+2, x_{2}=7, \\quad x_{3}=2 a+5 \\\\\ny_{1}=3, y_{2}=4, \\quad y_{3}=2 \\\\\n{[(2 a+2)(4-2)+7(2-3)+(2 a+5)} \\\\\n4 a+4-7-2 a-5=0 \\quad(3-4)]=0 \\\\\n2 a-8=0 \\\\\na=\\frac{8}{2}=4\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2022-sem2-q2-iii\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">iii<\/div><div class=\"xc-qtext\"><p>Calculate the mean of the following frequency distribution <strong>[3]<\/strong><\/p>\n<div class=\"math-responsive\"><table><thead><tr><th>Class Interval<\/th><th>5-15<\/th><th>15-25<\/th><th>25-35<\/th><th>35-45<\/th><th>45-55<\/th><\/tr><\/thead><tbody><tr><td>Frequency<\/td><td>2<\/td><td>6<\/td><td>4<\/td><td>8<\/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=\"math-responsive\"><table><thead><tr><th>C. I.<\/th><th>Class mark ( \\(\\boldsymbol{x}\\) )<\/th><th>Frequency (f)<\/th><th>\\(f \\times x\\)<\/th><\/tr><\/thead><tbody><tr><td>5-15<\/td><td>10<\/td><td>2<\/td><td>20<\/td><\/tr><tr><td>15-25<\/td><td>20<\/td><td>6<\/td><td>120<\/td><\/tr><tr><td>25-35<\/td><td>30<\/td><td>4<\/td><td>120<\/td><\/tr><tr><td>35-45<\/td><td>40<\/td><td>8<\/td><td>320<\/td><\/tr><tr><td>45-55<\/td><td>50<\/td><td>4<\/td><td>200<\/td><\/tr><tr><td>Total<\/td><td><\/td><td>24<\/td><td>780<\/td><\/tr><\/tbody><\/table><\/div>\n<p>We know, Mean \\((\\bar{x})=\\frac{\\sum f x}{\\sum f}\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\bar{x}=\\frac{780}{24}=32.5\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2022-sem2-q2-iv\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">iv<\/div><div class=\"xc-qtext\"><p>In the given figure O is the centre of the circle. \\(\\mathbf{P Q}\\) and \\(\\mathbf{P R}\\) are tangents and \\(\\angle \\mathbf{Q P R}=\\) 70\u00b0 Calculate:<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/83c2efd8-c79b-47f4-a50b-fab47900d3f3-3_329_342_1001_412.jpg\" alt=\"ICSE Class 10 Mathematics 2022 Semester 2 Question 2(iv) figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>(a) \\(\\angle \\mathrm{QOR}\\)<\/p>\n<p>(b) \\(\\angle \\mathrm{QSR}\\) <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>PQOR is a quadrilateral.<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/83c2efd8-c79b-47f4-a50b-fab47900d3f3-3_326_343_1006_1245.jpg\" alt=\"ICSE Class 10 Mathematics 2022 Semester 2 Question 2(iv) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>(a)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\angle \\mathrm{QPR}+\\angle \\mathrm{QOR}=180^{\\circ}\n\\]<\/div>\n<p>(supplementary angles)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n70^{\\circ}+\\angle \\mathrm{QOR} &amp; =180^{\\circ} \\\\\n\\angle \\mathrm{QOR} &amp; =110^{\\circ}\n\\end{aligned}\n\\]<\/div>\n<p>(b)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{array}{r}\n\\text { Reflex } \\\\\n\\qquad \\begin{array}{r}\n\\angle \\mathrm{QOR}=360^{\\circ}-110^{\\circ}=250^{\\circ} \\\\\n\\angle \\mathrm{QSR}\n\\end{array}=\\frac{1}{2} \\text { Reflex } \\angle \\mathrm{QOR}\n\\end{array}\n\\]<\/div>\n<p>( \u2235 The angle subtended by an arc at the centre is double the angle subtended by it on the remaining part of the circle.)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\angle \\mathrm{QSR}=\\frac{1}{2} \\times 250^{\\circ}=125^{\\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 2022 Semester 2<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2022-sem2-q3-i\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">i<\/div><div class=\"xc-qtext\"><p>A bag contains 5 white, 2 red and 3 black balls. A ball is drawn at random. What is the probability that the ball drawn is a red ball? <strong>[2]<\/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, No. of white balls= 5 No. of red balls \\(=2\\) No. of black balls \\(=3\\) Total no. of balls \\(=5+2+3\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n=10\n\\]<\/div>\n<p>Probability of getting a red ball<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; =\\frac{\\text { No. of favourable outcomes (red balls) }}{\\text { Total no. of balls }} \\\\\n&amp; =\\frac{2}{10}=\\frac{1}{5}\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2022-sem2-q3-ii\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">ii<\/div><div class=\"xc-qtext\"><p>A solid cone of radius 5 cm and height 9 cm is melted and made into small cylinders of radius of 0.5 cm and height 1.5 cm. Find the number of cylinders so formed. <strong>[2]<\/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, For cone, Radius \\((\\mathrm{R})=5 \\mathrm{~cm}\\), Height \\((\\mathrm{H})=9 \\mathrm{~cm}\\) For cylinder, Radius \\((r)=0.5 \\mathrm{~cm}\\), Height \\((h)=1.5 \\mathrm{~cm}\\) No. of cylinders \\(=\\frac{\\text { Volume of Cone }}{\\text { Volume of cylinder }}\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n=\\frac{\\frac{1}{3} \\pi \\mathrm{R}^{2} \\mathrm{H}}{\\pi r^{2} h}=\\frac{\\frac{1}{3} \\times(5)^{2} \\times 9}{(0.5)^{2} \\times 1.5}=\\frac{25 \\times 3}{0.25 \\times 1.5}=200\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2022-sem2-q3-iii\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">iii<\/div><div class=\"xc-qtext\"><p>Two lamp posts AB and CD each of height 100 m are on either side of the road. P is a point on the road between the two lamp posts. The angles of elevation of the top of the lamp posts from the point P are 60\u00b0and 30\u00b0. Find the distances PB and PD. <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\/83c2efd8-c79b-47f4-a50b-fab47900d3f3-4_306_618_265_312.jpg\" alt=\"ICSE Class 10 Mathematics 2022 Semester 2 Question 3(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>Given, Heights of the lamp posts \\(\\mathrm{AB}=\\mathrm{CD}=100 \\mathrm{~m}\\) and<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; \\angle \\mathrm{APB}=30^{\\circ} ; \\\\\n&amp; \\angle \\mathrm{CPD}=60^{\\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\/83c2efd8-c79b-47f4-a50b-fab47900d3f3-4_304_615_2080_369.jpg\" alt=\"ICSE Class 10 Mathematics 2022 Semester 2 Question 3(iii) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>Now, in \\(\\triangle \\mathrm{APR}\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\tan \\mathrm{P} &amp; =\\frac{\\text { Perpendicular }(\\mathrm{AB})}{\\text { Base }} \\\\\n\\tan 30^{\\circ} &amp; =\\frac{\\mathrm{AB}}{\\mathrm{BP}}=\\frac{100}{\\mathrm{BP}}\n\\end{aligned}\n\\]<\/div>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\frac{1}{\\sqrt{3}}=\\frac{100}{B P} &amp; \\Rightarrow B P=100 \\sqrt{3} \\mathrm{~m} \\\\\nB P &amp; =100 \\times 1.732=173.2 \\mathrm{~m}\n\\end{aligned}\n\\]<\/div>\n<p>Correction-In this question paper \\(\\angle \\mathrm{APB}=\\) 40\u00b0 is given, which should be 30\u00b0 to solve it. Now, In \\(\\triangle \\mathrm{PCD}\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\tan \\mathrm{P} &amp; =\\frac{\\mathrm{CD}}{\\mathrm{PD}}=\\frac{100}{\\mathrm{PD}} \\\\\n\\tan 60^{\\circ} &amp; =\\frac{100}{\\mathrm{PD}} \\\\\n\\sqrt{3} &amp; =\\frac{100}{\\mathrm{PD}}\n\\end{aligned}\n\\]<\/div>\n<p>\u21d2<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\mathrm{PD} &amp; =\\frac{100}{\\sqrt{3}} \\mathrm{~m} \\\\\n&amp; =\\frac{100 \\sqrt{3}}{3}=\\frac{173.2}{3} \\\\\n\\mathrm{PD} &amp; =57.73 \\mathrm{~m}\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2022-sem2-q3-iv\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">iv<\/div><div class=\"xc-qtext\"><p>Marks obtained by 100 students in an examination are given below.<\/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><th>50-60<\/th><\/tr><\/thead><tbody><tr><td>No of students<\/td><td>5<\/td><td>15<\/td><td>20<\/td><td>28<\/td><td>20<\/td><td>12<\/td><\/tr><\/tbody><\/table><\/div>\n<p>Draw a histogram for the given data using a graph paper and find the mode. Take \\(2 \\mathrm{~cm}=10\\) marks along one axis and \\(\\mathbf{2 ~ c m}=\\mathbf{1 0}\\) students along the other axis. <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>Mode = 35 Marks<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/83c2efd8-c79b-47f4-a50b-fab47900d3f3-4_579_797_1076_1018.jpg\" alt=\"ICSE Class 10 Mathematics 2022 Semester 2 Question 3(iv) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>To find mode, (i) First identify the rectangle with highest frequency (Modal class); here 30-40. (ii) Join the top corners of the modal rectangle with immediate next corners of the adjacent rectangles. (iii) Let the point where the joining lines cut each other (here A). Draw a perpendicular from A to \\(X\\)-axis. The point &#8216;P&#8217;, where the perpendicular meet the X-axis will give the mode.<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"xc-set\">\n<div class=\"xc-set-header\"><div class=\"xc-set-icon\">4<\/div><div><div class=\"xc-set-title\">Question 4<\/div><div class=\"xc-set-sub\">ICSE Class 10 Mathematics 2022 Semester 2<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2022-sem2-q4-i\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">i<\/div><div class=\"xc-qtext\"><p>Find a point P which divides internally the line segment joining the points A (-3, 9) and B (1, -3) in the ratio 1 : 3. <strong>[2]<\/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 points are<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\mathrm{A}\\left(x_{1}, y_{1}\\right)=(-3,9)\n\\]<\/div>\n<p>and<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\mathrm{B}\\left(x_{2}, y_{2}\\right)=(1,-3)\n\\]<\/div>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/83c2efd8-c79b-47f4-a50b-fab47900d3f3-5_103_527_1167_318.jpg\" alt=\"ICSE Class 10 Mathematics 2022 Semester 2 Question 4(i) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>Let \\(\\mathrm{P}(x, y)\\) divides the line segment AB in the ratio 1:3. Then by using section formula,<\/p>\n<div class=\"mathjax-overflow\">\\[\n=\\left(\\frac{1 \\times 1+3 \\times-3}{1+3}, \\frac{1 \\times-3+3 \\times 9}{1+3}\\right)\n\\]<\/div>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; \\mathrm{P}(x, y)=\\left(\\frac{-8}{4}, \\frac{24}{4}\\right) \\\\\n&amp; \\mathrm{P}(x, y)=(-2,6)\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2022-sem2-q4-ii\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">ii<\/div><div class=\"xc-qtext\"><p>A letter of the word &#8216;SECONDARY&#8217; is selected at random. What is the probability that the letter selected is not a vowel? <strong>[2]<\/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 word is &#8216;SECONDARY&#8217;<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\text { No. of vowels } &amp; =3 \\\\\n\\text { No. of consonants } &amp; =6 \\\\\n\\text { Total no. of letters } &amp; =9\n\\end{aligned}\n\\]<\/div>\n<p>Probability of not selecting a vowel = Probability of selecting a consonant<\/p>\n<div class=\"mathjax-overflow\">\\[\nP(\\text { not a vowel })=\\frac{6}{9}=\\frac{2}{3}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2022-sem2-q4-iii\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">iii<\/div><div class=\"xc-qtext\"><p>Use a graph paper for this question. Take 2 cm &#8211; 1 unit along both the axes.<\/p>\n<p>(a) Plot the points A (0, 4), B (2, 2), C(5, 2) and D \\((4,0), \\mathrm{E}(0,0)\\) is the origin.<\/p>\n<p>(b) Reflect B, C, D on the Y-axis and name them as B&#8217;, C&#8217; and D&#8217; respectively.<\/p>\n<p>(c) Join the points ABCDD&#8217;C&#8217;B&#8217; and A in order and give a geometrical name to the closed figure. <strong>[3]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>(a) See graph<\/p>\n<p>(b) Reflected points are : B&#8217; (-2, 2), C'(-5, 2) and D&#8217; (-4, 0)<\/p>\n<p>(c) On joining the points ABCDD&#8217;C&#8217;B&#8217;A, the shape obtain is not proper geometrical shape, but collection of a quadrilateral and triangle.<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/83c2efd8-c79b-47f4-a50b-fab47900d3f3-5_807_1074_1294_502.jpg\" alt=\"ICSE Class 10 Mathematics 2022 Semester 2 Question 4(iii) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2022-sem2-q4-iv\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">iv<\/div><div class=\"xc-qtext\"><p>A solid wooden cylinder is of radius 6 cm and height 16 cm. Two cones each of radius 2 cm and height 6 cm are drilled out of the cylinder. Find the volume of the remaining solid. <strong>[3]<\/strong><\/p>\n<div class=\"mathjax-overflow\">\\[\n\\left(\\text { Take } \\pi=\\frac{22}{7}\\right)\n\\]<\/div>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/83c2efd8-c79b-47f4-a50b-fab47900d3f3-5_368_331_656_418.jpg\" alt=\"ICSE Class 10 Mathematics 2022 Semester 2 Question 4(iv) 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: For Cylinder,<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\text { Radius }(\\mathrm{R}) &amp; =6 \\mathrm{~cm} \\\\\n\\text { Height }(\\mathrm{H}) &amp; =16 \\mathrm{~cm}\n\\end{aligned}\n\\]<\/div>\n<p>We know,<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\text { Volume } &amp; =\\pi \\mathrm{R}^{2} \\mathrm{H} \\\\\n&amp; =\\frac{22}{7} \\times(6)^{2} \\times(16) \\\\\n&amp; =\\frac{12672}{7} \\mathrm{~cm}^{3}\n\\end{aligned}\n\\]<\/div>\n<p>For cone,<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\text { Radius }(r) &amp; =2 \\mathrm{~cm} \\\\\n\\text { Height }(h) &amp; =6 \\mathrm{~cm}\n\\end{aligned}\n\\]<\/div>\n<p>We know,<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\text { Volume } &amp; =\\frac{1}{3} \\pi r^{2} h \\\\\n&amp; =\\frac{1}{3} \\times \\frac{22}{7} \\times(2)^{2} \\times 6 \\\\\n&amp; =\\frac{176}{7} \\mathrm{~cm}^{3}\n\\end{aligned}\n\\]<\/div>\n<p>Remaining Volume = Volume of cylinder &#8211; 2 \u00d7 Volume of a cone<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; =\\frac{12672}{7}-2 \\times \\frac{176}{7} \\\\\n&amp; =\\frac{12672}{7}-\\frac{352}{7} \\\\\n&amp; =\\frac{12672-352}{7} \\\\\n&amp; =\\frac{12320}{7} \\\\\n&amp; =1760 \\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\">5<\/div><div><div class=\"xc-set-title\">Question 5<\/div><div class=\"xc-set-sub\">ICSE Class 10 Mathematics 2022 Semester 2<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2022-sem2-q5-i\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">i<\/div><div class=\"xc-qtext\"><p>Two chords AB and CD of a circle intersect externally at \\(\\mathbf{E}\\). If \\(\\mathbf{E C}=\\mathbf{2 ~ c m}, \\mathbf{E A}=\\mathbf{3 ~ c m}\\) and \\(\\mathbf{A B}=\\mathbf{5 ~ c m}\\), find the length of CD. <strong>[2]<\/strong><\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/83c2efd8-c79b-47f4-a50b-fab47900d3f3-6_232_352_816_408.jpg\" alt=\"ICSE Class 10 Mathematics 2022 Semester 2 Question 5(i) 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>From the given figure,<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/83c2efd8-c79b-47f4-a50b-fab47900d3f3-6_234_351_2220_410.jpg\" alt=\"ICSE Class 10 Mathematics 2022 Semester 2 Question 5(i) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>We have \\(\\mathrm{EA}=3 \\mathrm{~cm}\\),<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; \\mathrm{AB}=5 \\mathrm{~cm} \\\\\n&amp; \\mathrm{EC}=2 \\mathrm{~cm}\n\\end{aligned}\n\\]<\/div>\n<p>When two chords intersect externally, then (from figure)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\mathrm{EA} \\times \\mathrm{EB}=\\mathrm{EC} \\times \\mathrm{ED}\n\\]<\/div>\n<p>or<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\mathrm{EA} \\times(\\mathrm{EA}+\\mathrm{AB})=\\mathrm{EC} \\times(\\mathrm{EC}+\\mathrm{CD})\n\\]<\/div>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\Rightarrow &amp; 3 \\times(3+5) \\\\\n=24 &amp; =4+2 \\mathrm{CD}\n\\end{aligned}\n\\]<\/div>\n<p>\u21d2<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n2 \\mathrm{CD} &amp; =20 \\\\\n\\mathrm{CD} &amp; =10 \\mathrm{~cm}\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2022-sem2-q5-ii\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">ii<\/div><div class=\"xc-qtext\"><p>Line AB is perpendicular to CD coordinates of B, C and D respectively (4, 0), (0, -1) and (4, 3).<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/83c2efd8-c79b-47f4-a50b-fab47900d3f3-6_313_388_1166_389.jpg\" alt=\"ICSE Class 10 Mathematics 2022 Semester 2 Question 5(ii) figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>Find<\/p>\n<p>(a) Slope of CD<\/p>\n<p>(b) Equation of AB <strong>[2]<\/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) Slope of line CD is given by<\/p>\n<div class=\"mathjax-overflow\">\\[\nm_{1}=\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\\frac{-1-3}{0-4}=1\n\\]<\/div>\n<p>(b) Slope of line AB, Which is perpendicular to CD given by<\/p>\n<div class=\"mathjax-overflow\">\\[\nm_{1}=\\frac{1}{-m_{2}}=-\\frac{1}{1}=-1\n\\]<\/div>\n<p>Now, equation of a line (AB), when slope and a coordinates is given<\/p>\n<div class=\"mathjax-overflow\">\\[\ny-y_{1}=m_{1}\\left(x-x_{1}\\right)\n\\]<\/div>\n<p>where<\/p>\n<div class=\"mathjax-overflow\">\\[\ny_{1}=0, m_{1}=-1, x_{1}=4\n\\]<\/div>\n<p>So,<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\ny-0 &amp; =-1(x-4) \\\\\ny &amp; =-x+4 \\\\\nx+y &amp; =4\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2022-sem2-q5-iii\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">iii<\/div><div class=\"xc-qtext\"><p>Prove that: <strong>[3]<\/strong><\/p>\n<div class=\"mathjax-overflow\">\\[\n\\frac{(1+\\sin \\theta)^{2}+(1-\\sin \\theta)^{2}}{2 \\cos ^{2} \\theta}=\\sec ^{2} \\theta+\\tan ^{2} \\theta\n\\]<\/div><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>L.H.S.<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; =\\frac{(1+\\sin \\theta)^{2}+(1-\\sin \\theta)^{2}}{2 \\cos ^{2} \\theta} \\\\\n&amp; =\\frac{1^{2}+\\sin ^{2} \\theta+2 \\sin \\theta+1^{2}+\\sin ^{2} \\theta-2 \\sin \\theta}{2 \\cos ^{2} \\theta} \\\\\n&amp; =\\frac{2+2 \\sin ^{2} \\theta}{2 \\cos ^{2} \\theta}=\\frac{1+\\sin ^{2} \\theta}{\\cos ^{2} \\theta} \\\\\n&amp; =\\frac{1}{\\cos ^{2} \\theta}+\\frac{\\sin ^{2} \\theta}{\\cos ^{2} \\theta} \\\\\n&amp; =\\sec ^{2} \\theta+\\tan ^{2} \\theta=\\text { R.H.S. Hence Proved. }\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2022-sem2-q5-iv\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">iv<\/div><div class=\"xc-qtext\"><p>The mean of the following distribution is 50. Find the unknown frequency. <strong>[3]<\/strong><\/p>\n<div class=\"math-responsive\"><table><thead><tr><th>Class Interval<\/th><th>Frequency<\/th><\/tr><\/thead><tbody><tr><td>0-20<\/td><td>6<\/td><\/tr><tr><td>20-40<\/td><td>\\(f\\)<\/td><\/tr><tr><td>40-60<\/td><td>8<\/td><\/tr><tr><td>60-80<\/td><td>12<\/td><\/tr><tr><td>80-100<\/td><td>8<\/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>C. I.<\/th><th>Class mark ( \\(\\boldsymbol{x}\\) )<\/th><th>Frequency (f)<\/th><th>\\(f \\times x\\)<\/th><\/tr><\/thead><tbody><tr><td>0-20<\/td><td>10<\/td><td>6<\/td><td>60<\/td><\/tr><tr><td>20-40<\/td><td>30<\/td><td>\\(f\\)<\/td><td>30f<\/td><\/tr><tr><td>40-60<\/td><td>50<\/td><td>8<\/td><td>400<\/td><\/tr><tr><td>60-80<\/td><td>70<\/td><td>12<\/td><td>840<\/td><\/tr><tr><td>80-100<\/td><td>90<\/td><td>8<\/td><td>720<\/td><\/tr><tr><td>Total<\/td><td><\/td><td>\\(34+f\\)<\/td><td>2020 + 30f<\/td><\/tr><\/tbody><\/table><\/div>\n<p>We know, Mean \\((\\bar{x})=\\frac{\\sum f x}{\\sum f}\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n50 &amp; =\\frac{2020+30 f}{34+f} \\\\\n1700+50 f &amp; =2020+30 f \\\\\n50 f-30 f &amp; =2020-1700 \\\\\n20 f &amp; =320 \\\\\nf &amp; =16\n\\end{aligned}\n\\]<\/div>\n<p>(given)<\/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 2022 Semester 2<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2022-sem2-q6-i\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">i<\/div><div class=\"xc-qtext\"><p>Prove that: <strong>[2]<\/strong><\/p>\n<div class=\"mathjax-overflow\">\\[\n1+\\frac{\\tan ^{2} \\theta}{1+\\sec \\theta}=\\sec \\theta\n\\]<\/div><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>L.H.S.<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; =1+\\frac{\\tan ^{2} \\theta}{1+\\sec \\theta} \\\\\n&amp; =1+\\frac{\\left(\\sec ^{2} \\theta-1\\right)}{1+\\sec \\theta} \\quad\\left(\\because \\sec ^{2} \\theta-\\tan ^{2} \\theta=1\\right) \\\\\n&amp; =1+\\frac{(\\sec \\theta+1)(\\sec \\theta-1)}{(1+\\sec \\theta)} \\\\\n&amp; =1+\\sec \\theta-1 \\\\\n&amp; =\\sec \\theta=\\text { R.H.S. }\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2022-sem2-q6-ii\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">ii<\/div><div class=\"xc-qtext\"><p>In the given figure A, B, C and D are points on the circle with centre O. Given \\(\\angle \\mathbf{A B C}=\\) 62\u00b0. Find:<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/83c2efd8-c79b-47f4-a50b-fab47900d3f3-7_315_290_578_438.jpg\" alt=\"ICSE Class 10 Mathematics 2022 Semester 2 Question 6(ii) figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>(a) \\(\\angle \\mathrm{ADC}\\)<\/p>\n<p>(b) \u2220CAB <strong>[2]<\/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) Given, \\(\\angle \\mathrm{ABC}=62^{\\circ}\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\angle \\mathrm{ABC}=\\angle \\mathrm{ADC}=62^{\\circ}\n\\]<\/div>\n<p>(\u2235 Angles in the same segment are equal)<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/83c2efd8-c79b-47f4-a50b-fab47900d3f3-7_313_288_2131_439.jpg\" alt=\"ICSE Class 10 Mathematics 2022 Semester 2 Question 6(ii) solution figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>(b) In \\(\\triangle \\mathrm{ABC}\\),<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; \\angle \\mathrm{ABC}=62^{\\circ} \\\\\n&amp; \\angle \\mathrm{ACB}=90^{\\circ}\n\\end{aligned}\n\\]<\/div>\n<p>(\u2235 Angles in a semicircle is a right angle) Now,<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\angle \\mathrm{ABC}+\\angle \\mathrm{ACB}+\\angle \\mathrm{CAB}=180^{\\circ}\n\\]<\/div>\n<p>(Angle sum property)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n62^{\\circ}+90^{\\circ}+\\angle \\mathrm{CAB} &amp; =180^{\\circ} \\\\\n\\angle \\mathrm{CAB} &amp; =180^{\\circ}-152^{\\circ} \\\\\n\\angle \\mathrm{CAB} &amp; =28^{\\circ}\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2022-sem2-q6-iii\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">iii<\/div><div class=\"xc-qtext\"><p>Find the equation of a line parallel to the line \\(2 \\boldsymbol{x}+\\boldsymbol{y}-\\mathbf{7}=\\mathbf{0}\\) and passing through the intersection of the lines \\(\\boldsymbol{x}+\\boldsymbol{y}-\\mathbf{4}=\\mathbf{0}\\) and \\(\\mathbf{2} \\boldsymbol{x}\\) \\(-y=8\\). <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\">\\[\nx+y-4=0 \\text { or } x+y=4\n\\]<\/div>\n<p>and<\/p>\n<div class=\"mathjax-overflow\">\\[\n2 x-y=8\n\\]<\/div>\n<p>On solving above two equations we get the intersection points as (4, 0). Now, equation of line parallel to \\(2 \\mathrm{x}+\\mathrm{y}-7=\\) 0 is given by<\/p>\n<div class=\"mathjax-overflow\">\\[\n2 x+y=\\lambda\n\\]<\/div>\n<p>which is passing through \\((4,0)\\) So,<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n2 \\times 4+0 &amp; =\\lambda \\\\\n\\lambda &amp; =8\n\\end{aligned}\n\\]<\/div>\n<p>So,the required equation of the line is 2x + y \\(=8\\) or,<\/p>\n<div class=\"mathjax-overflow\">\\[\n2 x+y-8=0\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-maths-2022-sem2-q6-iv\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">iv<\/div><div class=\"xc-qtext\"><p>Marks obtained by 40 students in an examination are given below.<\/p>\n<div class=\"math-responsive\"><table><thead><tr><th>Marks<\/th><th>10-20<\/th><th>20-30<\/th><th>30-40<\/th><th>40-50<\/th><th>50-60<\/th><th>60-70<\/th><\/tr><\/thead><tbody><tr><td>No of students<\/td><td>3<\/td><td>8<\/td><td>14<\/td><td>9<\/td><td>4<\/td><td>2<\/td><\/tr><\/tbody><\/table><\/div>\n<p>Using graph paper draw an ogive and estimate the median marks. Take \\(\\mathbf{2 ~ c m}=\\mathbf{1 0}\\) marks along one axis and \\(\\mathbf{2 c m}=\\mathbf{5}\\) students along the other axis. <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=\"math-responsive\"><table><thead><tr><th>Marks<\/th><th>No. of students<\/th><th>Marks less than<\/th><th>Cumulative frequency<\/th><\/tr><\/thead><tbody><tr><td>10-20<\/td><td>3<\/td><td>20<\/td><td>3<\/td><\/tr><tr><td>20-30<\/td><td>8<\/td><td>30<\/td><td>11<\/td><\/tr><tr><td>30-40<\/td><td>14<\/td><td>40<\/td><td>25<\/td><\/tr><tr><td>40-50<\/td><td>9<\/td><td>50<\/td><td>34<\/td><\/tr><tr><td>50-60<\/td><td>4<\/td><td>60<\/td><td>38<\/td><\/tr><tr><td>60-70<\/td><td>2<\/td><td>70<\/td><td>40<\/td><\/tr><\/tbody><\/table><\/div>\n<p>Other than the given class intervals, we assume a class interval 0-10 prior the first class with zero frequency. Now, plot the points (10, 0), (20, 3), (30, 11), (40, 25), (50, 34), (60, 38), \\((70,40)\\) on the graph paper. In order to obtain ogive, we draw a smooth curve passing through these points. In order to find the median, we first locate the point \\(\\frac{\\mathrm{N}}{2}=\\frac{40}{2}=20\\) on Y-axis. Let the point P from this draw a line parallel to X-axis cutting the curve at Q. From Q draw a line parallel to \\(Y\\)-axis meeting the X-axis at point M. The coordinate of M is 37.5. Hence, the median is 37.5. Now, plot the points (10, 0), (20, 3), (30, 11), (40, 25), (50, 34), (60, 38), \\((70,40)\\) on the graph paper. In order to obtain ogive, we draw a smooth curve passing through these, points.<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/83c2efd8-c79b-47f4-a50b-fab47900d3f3-8_994_996_318_578.jpg\" alt=\"ICSE Class 10 Mathematics 2022 Semester 2 Question 6(iv) 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\">22<\/span> <span class=\"rt-label rt-postfix\">minutes<\/span><\/span> ICSE Class 10 Mathematics PYQ Practice the complete ICSE Class 10 Mathematics 2022 Semester 2 Question Paper with Solutions, including [&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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