{"id":126486,"date":"2023-09-13T15:50:00","date_gmt":"2023-09-13T10:20:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=126486"},"modified":"2023-11-24T10:48:07","modified_gmt":"2023-11-24T05:18:07","slug":"rd-sharma-class-10-solutions-chapter-7-exercise-7-4","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-7-exercise-7-4\/","title":{"rendered":"RD Sharma Class 10 Solutions Chapter 7 Exercise 7.4 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-126501\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-7-Exercise-7.4.jpg\" alt=\"RD Sharma Class 10 Solutions Chapter 7 Exercise 7.4\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-7-Exercise-7.4.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-7-Exercise-7.4-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 7 Exercise 7.4:\u00a0<\/strong>The median is the point in a distribution that is in the middle. Students will experience finding the median of a discrete and grouped frequency distribution in this exercise. The <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-for-maths\/\"><strong>RD Sharma Class 10 Solutions<\/strong><\/a> is the best location to look for quick access to solutions. Download the <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-7-statistics\/\"><strong>RD Sharma Solutions for Class 10 Maths Chapter 7<\/strong><\/a> Exercise 7.4 PDF for further details on this exercise.<\/p>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_47_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=\"ez-toc-toggle-icon-1\"><label for=\"item-69d5684115201\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;direction:ltr;\"><svg style=\"fill: #000000;color:#000000\" 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: #000000;color:#000000\" 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><\/label><input  type=\"checkbox\" id=\"item-69d5684115201\"><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 eztoc-visibility-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:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-7-exercise-7-4\/#download-rd-sharma-class-10-solutions-chapter-7-exercise-74-free-pdf\" title=\"Download RD Sharma Class 10 Solutions Chapter 7 Exercise 7.4 Free PDF\">Download RD Sharma Class 10 Solutions Chapter 7 Exercise 7.4 Free PDF<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-7-exercise-7-4\/#access-answers-to-rd-sharma-solutions-class-10-maths-chapter-7-exercise-74-important-question-with-answers\" title=\"Access answers to RD Sharma Solutions Class 10 Maths Chapter 7 Exercise 7.4- Important Question with Answers\">Access answers to RD Sharma Solutions Class 10 Maths Chapter 7 Exercise 7.4- Important Question with Answers<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-7-exercise-7-4\/#faqs-on-rd-sharma-class-10-solutions-chapter-7-exercise-74\" title=\"FAQs on RD Sharma Class 10 Solutions Chapter 7 Exercise 7.4\">FAQs on RD Sharma Class 10 Solutions Chapter 7 Exercise 7.4<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-7-exercise-7-4\/#where-can-i-get-rd-sharma-class-10-solutions-chapter-7-exercise-74-free-pdf\" title=\"Where can I get RD Sharma Class 10 Solutions Chapter 7 Exercise 7.4 Free PDF?\">Where can I get RD Sharma Class 10 Solutions Chapter 7 Exercise 7.4 Free PDF?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-7-exercise-7-4\/#what-are-the-benefits-of-using-rd-sharma-class-10-solutions-chapter-7-exercise-74\" title=\"What are the benefits of using RD Sharma Class 10 Solutions Chapter 7 Exercise 7.4?\">What are the benefits of using RD Sharma Class 10 Solutions Chapter 7 Exercise 7.4?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-7-exercise-7-4\/#how-is-rd-sharma-class-10-solutions-chapter-7-exercise-74-helpful-for-board-exams\" title=\"How is RD Sharma Class 10 Solutions Chapter 7 Exercise 7.4 helpful for board exams?\">How is RD Sharma Class 10 Solutions Chapter 7 Exercise 7.4 helpful for board exams?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-10-solutions-chapter-7-exercise-74-free-pdf\"><\/span>Download RD Sharma Class 10 Solutions Chapter 7 Exercise 7.4 Free PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 800px;}<br \/>\n.pdfobject { border: 1px solid #666; }<br \/>\n<\/style><\/p>\n<p style=\"text-align: justify;\"><script src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br \/><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solution-Class-10-Chapter-7-Exercise-7.4.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solution-Class-10-Chapter-7-Exercise-7.4.pdf\">RD Sharma Class 10 Solutions Chapter 7 Exercise 7.4<\/a><\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-rd-sharma-solutions-class-10-maths-chapter-7-exercise-74-important-question-with-answers\"><\/span>Access answers to RD Sharma Solutions Class 10 Maths Chapter 7 Exercise 7.4- Important Question with Answers<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>1. Following are the lives in hours of 15 pieces of the components of aircraft engine. Find the median:\u00a0<\/strong><\/p>\n<p><strong>715, 724, 725, 710, 729, 745, 694, 699, 696, 712, 734, 728, 716, 705, 719.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Arranging the given data in ascending order, we have<\/p>\n<p>694, 696, 699, 705, 710, 712, 715, 716, 719, 721, 725, 728, 729, 734, 745<\/p>\n<p>As the number of terms is an old number, i.e., N = 15<\/p>\n<p>We use the following procedure to find the median.<\/p>\n<p>Median = (N + 1)\/2\u00a0<sup>th\u00a0<\/sup>term<\/p>\n<p>= (15 + 1)\/2\u00a0<sup>th\u00a0<\/sup>term<\/p>\n<p>= 8<sup>th<\/sup>\u00a0term<\/p>\n<p>So, the 8<sup>th<\/sup>\u00a0term in the arranged order of the given data should be the median.<\/p>\n<p>Therefore, 716 is the median of the data.<\/p>\n<p><strong>2. The following is the distribution of height of students of a certain class in a certain city:<\/strong><\/p>\n<table class=\"table table-bordered\">\n<tbody>\n<tr>\n<td><strong>Height (in cm):<\/strong><\/td>\n<td><strong>160 \u2013 162<\/strong><\/td>\n<td><strong>163 \u2013 165<\/strong><\/td>\n<td><strong>166 \u2013 168<\/strong><\/td>\n<td><strong>169 \u2013 171<\/strong><\/td>\n<td><strong>172 \u2013 174<\/strong><\/td>\n<\/tr>\n<tr>\n<td><strong>No of students:<\/strong><\/td>\n<td><strong>15<\/strong><\/td>\n<td><strong>118<\/strong><\/td>\n<td><strong>142<\/strong><\/td>\n<td><strong>127<\/strong><\/td>\n<td><strong>18<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong>Find the median height.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<table class=\"table table-bordered\">\n<tbody>\n<tr>\n<td>Class interval (exclusive)<\/td>\n<td>Class interval \u00a0(inclusive)<\/td>\n<td>Class interval frequency<\/td>\n<td>Cumulative frequency<\/td>\n<\/tr>\n<tr>\n<td>160 \u2013 162<\/td>\n<td>159.5 \u2013 162.5<\/td>\n<td>15<\/td>\n<td>15<\/td>\n<\/tr>\n<tr>\n<td>163 \u2013 165<\/td>\n<td>162.5 \u2013 165.5<\/td>\n<td>118<\/td>\n<td>133(F)<\/td>\n<\/tr>\n<tr>\n<td>166 \u2013 168<\/td>\n<td>165.5 \u2013 168.5<\/td>\n<td>142(f)<\/td>\n<td>275<\/td>\n<\/tr>\n<tr>\n<td>169 \u2013 171<\/td>\n<td>168.5 \u2013 171.5<\/td>\n<td>127<\/td>\n<td>402<\/td>\n<\/tr>\n<tr>\n<td>172 \u2013 174<\/td>\n<td>171.5 \u2013 174.5<\/td>\n<td>18<\/td>\n<td>420<\/td>\n<\/tr>\n<tr>\n<td>\u00a0<\/td>\n<td>\u00a0<\/td>\n<td>N = 420<\/td>\n<td>\u00a0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Here, we have N = 420,<\/p>\n<p>So, N\/2 = 420\/ 2 = 210<\/p>\n<p>The cumulative frequency just greater than N\/2 is 275 then 165.5 \u2013 168.5 is the median class such that<\/p>\n<p>L = 165.5, f = 142, F = 133 and h = (168.5 \u2013 165.5) = 3<\/p>\n<p><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 7 Statistics\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-7.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.4 - 1\" \/><\/p>\n<p>= 165.5 + 1.63<\/p>\n<p>= 167.13<\/p>\n<p><strong>3. Following is the distribution of I.Q. of 100 students. Find the median I.Q.<\/strong><\/p>\n<table class=\"table table-bordered\">\n<tbody>\n<tr>\n<td><strong>I.Q:<\/strong><\/td>\n<td><strong>55 \u2013 64<\/strong><\/td>\n<td><strong>65 \u2013 74<\/strong><\/td>\n<td><strong>75 \u2013 84<\/strong><\/td>\n<td><strong>85 \u2013 94<\/strong><\/td>\n<td><strong>95 \u2013 104<\/strong><\/td>\n<td><strong>105 \u2013 114<\/strong><\/td>\n<td><strong>115 \u2013 124<\/strong><\/td>\n<td><strong>125 \u2013 134<\/strong><\/td>\n<td><strong>135 \u2013 144<\/strong><\/td>\n<\/tr>\n<tr>\n<td><strong>No of students:<\/strong><\/td>\n<td><strong>1<\/strong><\/td>\n<td><strong>2<\/strong><\/td>\n<td><strong>9<\/strong><\/td>\n<td><strong>22<\/strong><\/td>\n<td><strong>33<\/strong><\/td>\n<td><strong>22<\/strong><\/td>\n<td><strong>8<\/strong><\/td>\n<td><strong>2<\/strong><\/td>\n<td><strong>1<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong>Solution:<\/strong><\/p>\n<table class=\"table table-bordered\">\n<tbody>\n<tr>\n<td>Class interval (exclusive)<\/td>\n<td>Class interval\u00a0 (inclusive)<\/td>\n<td>Class interval frequency<\/td>\n<td>Cumulative frequency<\/td>\n<\/tr>\n<tr>\n<td>55 \u2013 64<\/td>\n<td>54.5 \u2013 64-5<\/td>\n<td>1<\/td>\n<td>1<\/td>\n<\/tr>\n<tr>\n<td>65 \u2013 74<\/td>\n<td>64.5 \u2013 74.5<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<\/tr>\n<tr>\n<td>75 \u2013 84<\/td>\n<td>74.5 \u2013 84.5<\/td>\n<td>9<\/td>\n<td>12<\/td>\n<\/tr>\n<tr>\n<td>85 \u2013 94<\/td>\n<td>84.5 \u2013 94.5<\/td>\n<td>22<\/td>\n<td>34(F)<\/td>\n<\/tr>\n<tr>\n<td>95 \u2013 104<\/td>\n<td>94.5 \u2013 104.5<\/td>\n<td>33(f)<\/td>\n<td>67<\/td>\n<\/tr>\n<tr>\n<td>105 \u2013 114<\/td>\n<td>104.5 \u2013 114.5<\/td>\n<td>22<\/td>\n<td>89<\/td>\n<\/tr>\n<tr>\n<td>115 \u2013 124<\/td>\n<td>114.5 \u2013 124.5<\/td>\n<td>8<\/td>\n<td>97<\/td>\n<\/tr>\n<tr>\n<td>125 \u2013 134<\/td>\n<td>124.5 \u2013 134.5<\/td>\n<td>2<\/td>\n<td>98<\/td>\n<\/tr>\n<tr>\n<td>135 \u2013 144<\/td>\n<td>134.5 \u2013 144.5<\/td>\n<td>1<\/td>\n<td>100<\/td>\n<\/tr>\n<tr>\n<td>\u00a0<\/td>\n<td>\u00a0<\/td>\n<td>N = 100<\/td>\n<td>\u00a0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Here, we have N = 100,<\/p>\n<p>So, N\/2 = 100\/ 2 = 50<\/p>\n<p>The cumulative frequency just greater than N\/ 2 is 67 then the median class is (94.5 \u2013 104.5) such that L = 94.5, F = 33, h = (104.5 \u2013 94.5) = 10<\/p>\n<p><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 7 Statistics\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-7-1.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.4 - 2\" \/><\/p>\n<p>= 94.5 + 4.85<\/p>\n<p>= 99.35<\/p>\n<p><strong>4. Calculate the median from the following data:<\/strong><\/p>\n<table class=\"table table-bordered\">\n<tbody>\n<tr>\n<td><strong>Rent (in Rs):<\/strong><\/td>\n<td><strong>15 \u2013 25<\/strong><\/td>\n<td><strong>25 \u2013 35<\/strong><\/td>\n<td><strong>35 \u2013 45<\/strong><\/td>\n<td><strong>45 \u2013 55<\/strong><\/td>\n<td><strong>55 \u2013 65<\/strong><\/td>\n<td><strong>65 \u2013 75<\/strong><\/td>\n<td><strong>75 \u2013 85<\/strong><\/td>\n<td><strong>85 \u2013 95<\/strong><\/td>\n<\/tr>\n<tr>\n<td><strong>No of houses:<\/strong><\/td>\n<td><strong>8<\/strong><\/td>\n<td><strong>10<\/strong><\/td>\n<td><strong>15<\/strong><\/td>\n<td><strong>25<\/strong><\/td>\n<td><strong>40<\/strong><\/td>\n<td><strong>20<\/strong><\/td>\n<td><strong>15<\/strong><\/td>\n<td><strong>7<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong>Solution:<\/strong><\/p>\n<table class=\"table table-bordered\">\n<tbody>\n<tr>\n<td>Class interval<\/td>\n<td>Frequency<\/td>\n<td>Cumulative frequency<\/td>\n<\/tr>\n<tr>\n<td>15 \u2013 25<\/td>\n<td>8<\/td>\n<td>8<\/td>\n<\/tr>\n<tr>\n<td>25 \u2013 35<\/td>\n<td>10<\/td>\n<td>18<\/td>\n<\/tr>\n<tr>\n<td>35 \u2013 45<\/td>\n<td>15<\/td>\n<td>33<\/td>\n<\/tr>\n<tr>\n<td>45 \u2013 55<\/td>\n<td>25<\/td>\n<td>58(F)<\/td>\n<\/tr>\n<tr>\n<td>55 \u2013 65<\/td>\n<td>40(f)<\/td>\n<td>98<\/td>\n<\/tr>\n<tr>\n<td>65 \u2013 75<\/td>\n<td>20<\/td>\n<td>118<\/td>\n<\/tr>\n<tr>\n<td>75 \u2013 85<\/td>\n<td>15<\/td>\n<td>133<\/td>\n<\/tr>\n<tr>\n<td>85 \u2013 95<\/td>\n<td>7<\/td>\n<td>140<\/td>\n<\/tr>\n<tr>\n<td>\u00a0<\/td>\n<td>N = 140<\/td>\n<td>\u00a0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Here, we have N = 140,<\/p>\n<p>So, N\/2 = 140\/ 2 = 70<\/p>\n<p>The cumulative frequency just greater than N\/ 2 is 98 then the median class is 55 \u2013 65 such that L = 55, f = 40, F = 58, h = 65 \u2013 55 = 10<\/p>\n<p><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 7 Statistics\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-7-2.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.4 - 3\" \/><\/p>\n<p>= 55 + 3 = 58<\/p>\n<p><strong>5. Calculate the median from the following data:<\/strong><\/p>\n<table class=\"table table-bordered\">\n<tbody>\n<tr>\n<td><strong>Marks below:<\/strong><\/td>\n<td><strong>10 \u2013 20<\/strong><\/td>\n<td><strong>20 \u2013 30<\/strong><\/td>\n<td><strong>30 \u2013 40<\/strong><\/td>\n<td><strong>40 \u2013 50<\/strong><\/td>\n<td><strong>50 \u2013 60<\/strong><\/td>\n<td><strong>60 \u2013 70<\/strong><\/td>\n<td><strong>70 \u2013 80<\/strong><\/td>\n<td><strong>85 \u2013 95<\/strong><\/td>\n<\/tr>\n<tr>\n<td><strong>No of students:<\/strong><\/td>\n<td><strong>15<\/strong><\/td>\n<td><strong>35<\/strong><\/td>\n<td><strong>60<\/strong><\/td>\n<td><strong>84<\/strong><\/td>\n<td><strong>96<\/strong><\/td>\n<td><strong>127<\/strong><\/td>\n<td><strong>198<\/strong><\/td>\n<td><strong>250<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong>Solution:<\/strong><\/p>\n<table class=\"table table-bordered\">\n<tbody>\n<tr>\n<td>Marks below<\/td>\n<td>No. of students<\/td>\n<td>Class interval<\/td>\n<td>Frequency<\/td>\n<td>Cumulative frequency<\/td>\n<\/tr>\n<tr>\n<td>10<\/td>\n<td>15<\/td>\n<td>0 \u2013 10<\/td>\n<td>15<\/td>\n<td>15<\/td>\n<\/tr>\n<tr>\n<td>20<\/td>\n<td>35<\/td>\n<td>10 \u2013 20<\/td>\n<td>20<\/td>\n<td>35<\/td>\n<\/tr>\n<tr>\n<td>30<\/td>\n<td>60<\/td>\n<td>20 \u2013 30<\/td>\n<td>25<\/td>\n<td>60<\/td>\n<\/tr>\n<tr>\n<td>40<\/td>\n<td>84<\/td>\n<td>30 \u2013 40<\/td>\n<td>24<\/td>\n<td>84<\/td>\n<\/tr>\n<tr>\n<td>50<\/td>\n<td>96<\/td>\n<td>40 \u2013 50<\/td>\n<td>12<\/td>\n<td>96(F)<\/td>\n<\/tr>\n<tr>\n<td>60<\/td>\n<td>127<\/td>\n<td>50 \u2013 60<\/td>\n<td>31(f)<\/td>\n<td>127<\/td>\n<\/tr>\n<tr>\n<td>70<\/td>\n<td>198<\/td>\n<td>60 \u2013 70<\/td>\n<td>71<\/td>\n<td>198<\/td>\n<\/tr>\n<tr>\n<td>80<\/td>\n<td>250<\/td>\n<td>70 \u2013 80<\/td>\n<td>52<\/td>\n<td>250<\/td>\n<\/tr>\n<tr>\n<td>\u00a0<\/td>\n<td>\u00a0<\/td>\n<td>\u00a0<\/td>\n<td>N = 250<\/td>\n<td>\u00a0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Here, we have N = 250,<\/p>\n<p>So, N\/2 = 250\/ 2 = 125<\/p>\n<p>The cumulative frequency just greater than N\/ 2 is 127, then the median class is 50 \u2013 60 such that L = 50, f = 31, F = 96, h = 60 -50 = 10<\/p>\n<p><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 7 Statistics\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-7-3.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.4 - 4\" \/><\/p>\n<p>= 50 + 9.35<\/p>\n<p>= 59.35<\/p>\n<p><strong>6. Calculate the missing frequency from the following distribution, it being given that the median of the distribution is 24.<\/strong><\/p>\n<table class=\"table table-bordered\">\n<tbody>\n<tr>\n<td><strong>Age in years:<\/strong><\/td>\n<td><strong>0 \u2013 10<\/strong><\/td>\n<td><strong>10 \u2013 20<\/strong><\/td>\n<td><strong>20 \u2013 30<\/strong><\/td>\n<td><strong>30 \u2013 40<\/strong><\/td>\n<td><strong>40 \u2013 50<\/strong><\/td>\n<\/tr>\n<tr>\n<td><strong>No of persons:<\/strong><\/td>\n<td><strong>5<\/strong><\/td>\n<td><strong>25<\/strong><\/td>\n<td><strong>?<\/strong><\/td>\n<td><strong>18<\/strong><\/td>\n<td><strong>7<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let the unknown frequency be taken as x,<\/p>\n<table class=\"table table-bordered\">\n<tbody>\n<tr>\n<td>Class interval<\/td>\n<td>Frequency<\/td>\n<td>Cumulative frequency<\/td>\n<\/tr>\n<tr>\n<td>0 \u2013 10<\/td>\n<td>5<\/td>\n<td>5<\/td>\n<\/tr>\n<tr>\n<td>10 \u2013 20<\/td>\n<td>25<\/td>\n<td>30(F)<\/td>\n<\/tr>\n<tr>\n<td>20 \u2013 30<\/td>\n<td>x (f)<\/td>\n<td>30 + x<\/td>\n<\/tr>\n<tr>\n<td>30 \u2013 40<\/td>\n<td>18<\/td>\n<td>48 + x<\/td>\n<\/tr>\n<tr>\n<td>40 \u2013 50<\/td>\n<td>7<\/td>\n<td>55 + x<\/td>\n<\/tr>\n<tr>\n<td>\u00a0<\/td>\n<td>N = 170<\/td>\n<td>\u00a0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>It\u2019s given that<\/p>\n<p>Median = 24<\/p>\n<p>Then, median class = 20 \u2013 30; L = 20,\u00a0h = 30 -20 = 10, f = x,\u00a0F = 30<\/p>\n<p><strong><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 7 Statistics\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-7-4.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.4 - 5\" \/><\/strong><\/p>\n<p>4x = 275 + 5x \u2013 300<\/p>\n<p>4x \u2013 5x = \u2013 25<\/p>\n<p>\u2013 x = \u2013 25<\/p>\n<p>x = 25<\/p>\n<p>Therefore, the Missing frequency = 25<\/p>\n<p><strong>7. The following table gives the frequency distribution of married women by age at marriage.<\/strong><\/p>\n<table class=\"table table-bordered\">\n<tbody>\n<tr>\n<td><strong>Age (in years)<\/strong><\/td>\n<td><strong>Frequency<\/strong><\/td>\n<td><strong>Age (in years)<\/strong><\/td>\n<td><strong>Frequency<\/strong><\/td>\n<\/tr>\n<tr>\n<td><strong>15 \u2013 19<\/strong><\/td>\n<td><strong>53<\/strong><\/td>\n<td><strong>40 \u2013 44<\/strong><\/td>\n<td><strong>9<\/strong><\/td>\n<\/tr>\n<tr>\n<td><strong>20 \u2013 24<\/strong><\/td>\n<td><strong>140<\/strong><\/td>\n<td><strong>45 \u2013 49<\/strong><\/td>\n<td><strong>5<\/strong><\/td>\n<\/tr>\n<tr>\n<td><strong>25 \u2013 29<\/strong><\/td>\n<td><strong>98<\/strong><\/td>\n<td><strong>45 \u2013 49<\/strong><\/td>\n<td><strong>3<\/strong><\/td>\n<\/tr>\n<tr>\n<td><strong>30 \u2013 34<\/strong><\/td>\n<td><strong>32<\/strong><\/td>\n<td><strong>55 \u2013 59<\/strong><\/td>\n<td><strong>3<\/strong><\/td>\n<\/tr>\n<tr>\n<td><strong>35 \u2013 39<\/strong><\/td>\n<td><strong>12<\/strong><\/td>\n<td><strong>60 and above<\/strong><\/td>\n<td><strong>2<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong>Calculate the median and interpret the results.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<table class=\"table table-bordered\">\n<tbody>\n<tr>\n<td><strong>Class interval (exclusive)<\/strong><\/td>\n<td><strong>Class interval (inclusive)<\/strong><\/td>\n<td><strong>Frequency<\/strong><\/td>\n<td><strong>Cumulative frequency<\/strong><\/td>\n<\/tr>\n<tr>\n<td>15 \u2013 19<\/td>\n<td>14.5 \u2013 19.5<\/td>\n<td>53<\/td>\n<td>53 (F)<\/td>\n<\/tr>\n<tr>\n<td>20 \u2013 24<\/td>\n<td>19.5 \u2013 24.5<\/td>\n<td>140 (f)<\/td>\n<td>193<\/td>\n<\/tr>\n<tr>\n<td>25 \u2013 29<\/td>\n<td>24.5 \u2013 29.5<\/td>\n<td>98<\/td>\n<td>291<\/td>\n<\/tr>\n<tr>\n<td>30 \u2013 34<\/td>\n<td>29.5 \u2013 34.5<\/td>\n<td>32<\/td>\n<td>323<\/td>\n<\/tr>\n<tr>\n<td>35 \u2013 39<\/td>\n<td>34.5 \u2013 39.5<\/td>\n<td>12<\/td>\n<td>335<\/td>\n<\/tr>\n<tr>\n<td>40 \u2013 44<\/td>\n<td>39.5 \u2013 44.5<\/td>\n<td>9<\/td>\n<td>344<\/td>\n<\/tr>\n<tr>\n<td>45 \u2013 49<\/td>\n<td>44.5 \u2013 49.5<\/td>\n<td>5<\/td>\n<td>349<\/td>\n<\/tr>\n<tr>\n<td>50 \u2013 54<\/td>\n<td>49.5 \u2013 54.5<\/td>\n<td>3<\/td>\n<td>352<\/td>\n<\/tr>\n<tr>\n<td>55 \u2013 54<\/td>\n<td>54.5 \u2013 59.5<\/td>\n<td>3<\/td>\n<td>355<\/td>\n<\/tr>\n<tr>\n<td>60 and above<\/td>\n<td>59.5 and above<\/td>\n<td>2<\/td>\n<td>357<\/td>\n<\/tr>\n<tr>\n<td>\u00a0<\/td>\n<td>\u00a0<\/td>\n<td>N =357<\/td>\n<td>\u00a0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Here, we have N = 357,<\/p>\n<p>So, N\/2 = 357\/ 2 = 178.5<\/p>\n<p>The cumulative frequency just greater than N\/2 is 193, so then the median class is (19.5 \u2013 24.5) such that l = 19.5, f = 140, F = 53, h = 25.5 \u2013 19.5 = 5<\/p>\n<p><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 7 Statistics\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-7-5.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.4 - 6\" \/><\/p>\n<p>Median = 23.98<\/p>\n<p>This means nearly half the women were married between the ages of 15 and 25<\/p>\n<p><strong>8. The following table gives the distribution of the life time of 400 neon lamps:<\/strong><\/p>\n<table class=\"table table-bordered\">\n<tbody>\n<tr>\n<td><strong>Life time: (in hours)<\/strong><\/td>\n<td><strong>Number of lamps<\/strong><\/td>\n<\/tr>\n<tr>\n<td><strong>1500 \u2013 2000<\/strong><\/td>\n<td><strong>14<\/strong><\/td>\n<\/tr>\n<tr>\n<td><strong>2000 \u2013 2500<\/strong><\/td>\n<td><strong>56<\/strong><\/td>\n<\/tr>\n<tr>\n<td><strong>2500 \u2013 3000<\/strong><\/td>\n<td><strong>60<\/strong><\/td>\n<\/tr>\n<tr>\n<td><strong>3000 \u2013 3500<\/strong><\/td>\n<td><strong>86<\/strong><\/td>\n<\/tr>\n<tr>\n<td><strong>3500 \u2013 4000<\/strong><\/td>\n<td><strong>74<\/strong><\/td>\n<\/tr>\n<tr>\n<td><strong>4000 \u2013 4500<\/strong><\/td>\n<td><strong>62<\/strong><\/td>\n<\/tr>\n<tr>\n<td><strong>4500 \u2013 5000<\/strong><\/td>\n<td><strong>48<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong>Find the median life.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<table class=\"table table-bordered\">\n<tbody>\n<tr>\n<td>Life time<\/td>\n<td>Number of lamps fi<\/td>\n<td>Cumulative frequency (cf)<\/td>\n<\/tr>\n<tr>\n<td>1500 \u2013 2000<\/td>\n<td>14<\/td>\n<td>14<\/td>\n<\/tr>\n<tr>\n<td>2000 \u2013 2500<\/td>\n<td>56<\/td>\n<td>70<\/td>\n<\/tr>\n<tr>\n<td>2500 \u2013 3000<\/td>\n<td>60<\/td>\n<td>130(F)<\/td>\n<\/tr>\n<tr>\n<td>3000 \u2013 3500<\/td>\n<td>86(f)<\/td>\n<td>216<\/td>\n<\/tr>\n<tr>\n<td>3500 \u2013 4000<\/td>\n<td>74<\/td>\n<td>290<\/td>\n<\/tr>\n<tr>\n<td>4000 \u2013 4500<\/td>\n<td>62<\/td>\n<td>352<\/td>\n<\/tr>\n<tr>\n<td>4500 \u2013 5000<\/td>\n<td>48<\/td>\n<td>400<\/td>\n<\/tr>\n<tr>\n<td>\u00a0<\/td>\n<td>N = 400<\/td>\n<td>\u00a0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>It\u2019s seen that the cumulative frequency just greater than n\/2 (400\/2 = 200) is 216, and it belongs to the class interval 3000 \u2013 3500, which becomes the Median class = 3000 \u2013 3500<\/p>\n<p>Lower limits (l) of median class = 3000 and,<\/p>\n<p>Frequency (f) of median class = 86<\/p>\n<p>Cumulative frequency (cf) of class preceding median class = 130<\/p>\n<p>And the Class size (h) = 500<\/p>\n<p>Thus, by calculating the median by the formula, we get<\/p>\n<p><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 7 Statistics\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-7-6.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.4 - 7\" \/><\/p>\n<p>= 3000 + (35000\/86)<\/p>\n<p>= 3406.98<\/p>\n<p>Thus, the median lifetime of lamps is 3406.98 hours<\/p>\n<p><strong>9. The distribution below gives the weight of 30 students in a class. Find the median weight of students:<\/strong><\/p>\n<table class=\"table table-bordered\">\n<tbody>\n<tr>\n<td><strong>Weight (in kg):<\/strong><\/td>\n<td><strong>40 \u2013 45<\/strong><\/td>\n<td><strong>45 \u2013 50<\/strong><\/td>\n<td><strong>50 \u2013 55<\/strong><\/td>\n<td><strong>55 \u2013 60<\/strong><\/td>\n<td><strong>60 \u2013 65<\/strong><\/td>\n<td><strong>65 \u2013 70<\/strong><\/td>\n<td><strong>70 \u2013 75<\/strong><\/td>\n<\/tr>\n<tr>\n<td><strong>No of students:<\/strong><\/td>\n<td><strong>2<\/strong><\/td>\n<td><strong>3<\/strong><\/td>\n<td><strong>8<\/strong><\/td>\n<td><strong>6<\/strong><\/td>\n<td><strong>6<\/strong><\/td>\n<td><strong>3<\/strong><\/td>\n<td><strong>2<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong>Solution:<\/strong><\/p>\n<table class=\"table table-bordered\">\n<tbody>\n<tr>\n<td>Weight (in kg)<\/td>\n<td>Number of students fi<\/td>\n<td>Cumulative frequency (cf)<\/td>\n<\/tr>\n<tr>\n<td>40 \u2013 45<\/td>\n<td>2<\/td>\n<td>2<\/td>\n<\/tr>\n<tr>\n<td>45 \u2013 50<\/td>\n<td>3<\/td>\n<td>5<\/td>\n<\/tr>\n<tr>\n<td>50 \u2013 55<\/td>\n<td>8<\/td>\n<td>13<\/td>\n<\/tr>\n<tr>\n<td>55 \u2013 60<\/td>\n<td>6<\/td>\n<td>19<\/td>\n<\/tr>\n<tr>\n<td>60 \u2013 65<\/td>\n<td>6<\/td>\n<td>25<\/td>\n<\/tr>\n<tr>\n<td>65 \u2013 70<\/td>\n<td>3<\/td>\n<td>28<\/td>\n<\/tr>\n<tr>\n<td>70 \u2013 75<\/td>\n<td>2<\/td>\n<td>30<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>It\u2019s seen that the cumulative frequency just greater than n\/ 2 (i.e. 30\/ 2 = 15) is 19, belongs to class interval 55 \u2013 60.<\/p>\n<p>So, it\u2019s chosen that<\/p>\n<p>Median class = 55 \u2013 60<\/p>\n<p>Lower limit (l) of median class = 55<\/p>\n<p>Frequency (f) of median class = 6<\/p>\n<p>Cumulative frequency (cf) = 13<\/p>\n<p>And, Class size (h) = 5<\/p>\n<p>Thus, by calculating the median by the formula, we get<\/p>\n<p><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 7 Statistics\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-7-7.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.4 - 8\" \/><\/p>\n<p>= 55 + 10\/6 = 56.666<\/p>\n<p>So, the median weight is 56.67 kg.<\/p>\n<p><strong>10. Find the missing frequencies and the median for the following distribution if the mean is 1.46<\/strong><\/p>\n<table class=\"table table-bordered\">\n<tbody>\n<tr>\n<td><strong>No. of accidents:<\/strong><\/td>\n<td><strong>0<\/strong><\/td>\n<td><strong>1<\/strong><\/td>\n<td><strong>2<\/strong><\/td>\n<td><strong>3<\/strong><\/td>\n<td><strong>4<\/strong><\/td>\n<td><strong>5<\/strong><\/td>\n<td><strong>Total<\/strong><\/td>\n<\/tr>\n<tr>\n<td><strong>Frequencies (no. of days):<\/strong><\/td>\n<td><strong>46<\/strong><\/td>\n<td><strong>?<\/strong><\/td>\n<td><strong>?<\/strong><\/td>\n<td><strong>25<\/strong><\/td>\n<td><strong>10<\/strong><\/td>\n<td><strong>5<\/strong><\/td>\n<td><strong>200<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong>Solution:<\/strong><\/p>\n<table class=\"table table-bordered\">\n<tbody>\n<tr>\n<td>No. of accidents (x)<\/td>\n<td>No. of days (f)<\/td>\n<td>fx<\/td>\n<\/tr>\n<tr>\n<td>0<\/td>\n<td>46<\/td>\n<td>0<\/td>\n<\/tr>\n<tr>\n<td>1<\/td>\n<td>x<\/td>\n<td>x<\/td>\n<\/tr>\n<tr>\n<td>2<\/td>\n<td>y<\/td>\n<td>2y<\/td>\n<\/tr>\n<tr>\n<td>3<\/td>\n<td>25<\/td>\n<td>75<\/td>\n<\/tr>\n<tr>\n<td>4<\/td>\n<td>10<\/td>\n<td>40<\/td>\n<\/tr>\n<tr>\n<td>5<\/td>\n<td>5<\/td>\n<td>25<\/td>\n<\/tr>\n<tr>\n<td>\u00a0<\/td>\n<td>N = 200<\/td>\n<td>Sum = x + 2y + 140<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>It\u2019s given that N = 200<\/p>\n<p>\u21d2 46 + x + y + 25 + 10 + 5 = 200<\/p>\n<p>\u21d2 x + y = 200 \u2013 46 \u2013 25 \u2013 10 \u2013 5<\/p>\n<p>\u21d2 x + y = 114 \u2014- (i)<\/p>\n<p>And also given, Mean = 1.46<\/p>\n<p>\u21d2 Sum\/ N = 1.46<\/p>\n<p>\u21d2 (x + 2y + 140)\/ 200 = 1.46<\/p>\n<p>\u21d2 x + 2y = 292 \u2013 140<\/p>\n<p>\u21d2 x + 2y = 152 \u2014- (ii)<\/p>\n<p>Subtract equation (i) from equation (ii), and we get<\/p>\n<p>x + 2y \u2013 x \u2013 y = 152 \u2013 114<\/p>\n<p>\u21d2 y = 38<\/p>\n<p>Now, on putting the value of y in equation (i), we find x = 114 \u2013 38 = 76<\/p>\n<p>Thus, the table becomes:<\/p>\n<table class=\"table table-bordered\">\n<tbody>\n<tr>\n<td>No. of accidents (x)<\/td>\n<td>No. of days (f)<\/td>\n<td>Cumulative frequency<\/td>\n<\/tr>\n<tr>\n<td>0<\/td>\n<td>46<\/td>\n<td>46<\/td>\n<\/tr>\n<tr>\n<td>1<\/td>\n<td>76<\/td>\n<td>122<\/td>\n<\/tr>\n<tr>\n<td>2<\/td>\n<td>38<\/td>\n<td>160<\/td>\n<\/tr>\n<tr>\n<td>3<\/td>\n<td>25<\/td>\n<td>185<\/td>\n<\/tr>\n<tr>\n<td>4<\/td>\n<td>10<\/td>\n<td>195<\/td>\n<\/tr>\n<tr>\n<td>5<\/td>\n<td>5<\/td>\n<td>200<\/td>\n<\/tr>\n<tr>\n<td>\u00a0<\/td>\n<td>N = 200<\/td>\n<td>\u00a0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>It\u2019s seen that,<\/p>\n<p>N = 200 N\/2 = 200\/2 = 100<\/p>\n<p>So, the cumulative frequency just more than N\/2 is 122<\/p>\n<p>Therefore, the median is 1.<\/p>\n<p>We have provided complete details of RD Sharma Class 10 Solutions Chapter 7 Exercise 7.4. If you have any queries related to <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\"><strong>CBSE<\/strong><\/a>\u00a0Class 10, feel free to ask us in the comment section below.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-class-10-solutions-chapter-7-exercise-74\"><\/span>FAQs on RD Sharma Class 10 Solutions Chapter 7 Exercise 7.4<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1631268302453\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"where-can-i-get-rd-sharma-class-10-solutions-chapter-7-exercise-74-free-pdf\"><\/span>Where can I get RD Sharma Class 10 Solutions Chapter 7 Exercise 7.4 Free PDF?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can get RD Sharma Class 10 Solutions Chapter 7 Exercise 7.4 Free PDF from the above article.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631268445648\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"what-are-the-benefits-of-using-rd-sharma-class-10-solutions-chapter-7-exercise-74\"><\/span>What are the benefits of using RD Sharma Class 10 Solutions Chapter 7 Exercise 7.4?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>1. Correct answers according to the latest CBSE guidelines and syllabus.<br \/>2. The RD Sharma Class 10 Solutions Chapter 7 Exercise 7.4 is written in simple language to assist students in their board examination, &amp; competitive examination preparation.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631268589813\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-is-rd-sharma-class-10-solutions-chapter-7-exercise-74-helpful-for-board-exams\"><\/span>How is RD Sharma Class 10 Solutions Chapter 7 Exercise 7.4 helpful for board exams?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>For self-evaluation, RD Sharma Class 10 Solutions Chapter 7 Exercise 7.4 provides solutions with thorough descriptions as per term limits specified by the Board. Students will gain valuable experience solving these problems, allowing them to complete the assignment on time.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Class 10 Solutions Chapter 7 Exercise 7.4:\u00a0The median is the point in a distribution that is in the middle. Students will experience finding the median of a discrete and grouped frequency distribution in this exercise. The RD Sharma Class 10 Solutions is the best location to look for quick access to solutions. Download &#8230; <a title=\"RD Sharma Class 10 Solutions Chapter 7 Exercise 7.4 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-7-exercise-7-4\/\" aria-label=\"More on RD Sharma Class 10 Solutions Chapter 7 Exercise 7.4 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":126501,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411,2985,73410],"tags":[3243,9206,73520,4388],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126486"}],"collection":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/users\/238"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=126486"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126486\/revisions"}],"predecessor-version":[{"id":511889,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126486\/revisions\/511889"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/126501"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=126486"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=126486"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=126486"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}