{"id":14234,"date":"2020-09-09T17:56:02","date_gmt":"2020-09-09T12:26:02","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=14234"},"modified":"2021-07-02T15:01:54","modified_gmt":"2021-07-02T09:31:54","slug":"kseeb-solutions-for-class-7-maths-chapter-3-ex-3-1","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-for-class-7-maths-chapter-3-ex-3-1\/","title":{"rendered":"KSEEB Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.1"},"content":{"rendered":"

Students can Download Chapter 3 Data Handling Ex 3.1, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths<\/a>, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.<\/p>\n

Karnataka State Syllabus Class 7 Maths Chapter 3 Data Handling Ex 3.1<\/h2>\n

Question 1.
\nFind the range of heights of any ten students of your class.
\nSolution:
\nStudents can try themselves.
\n(This is practical question)<\/p>\n

Question 2.
\nOrganise the following marks ion a class assessment, in a tabular form.
\n4, 6, 7, 5, 3, 5, 4, 5, 2, 6, 2, 5, 1, 9, 6, 5, 8, 4, 6, 7
\n\"KSEEB
\ni) Which number is the highest ?
\nSolution:
\nHighest number is 9<\/p>\n

ii) Which number is the lowest ?
\nSolution:
\nLowest number is 1<\/p>\n

\"KSEEB<\/p>\n

iii) What is the range of the data ?
\nSolution:
\nRange of the given data = Highest number – Lowest number = 9 – 1 = 8<\/p>\n

iv) Find the arithmetic mean.
\nSolution:
\nArithmetic Mean =
\n\"KSEEB<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nFind the mean of the first five whole numbers.
\nSolution:
\nThe first whole numbers are 0, 1, 2, 3, 4
\nArithemetic mean of five whole numbers =
\n\"KSEEB<\/p>\n

Question 4.
\nA cricketer scores the following runs in eight innings :
\n58, 76, 40, 35, 46, 45, 0, 100. Find the mean score.
\nSolution:
\nThe scores of 8 innings = 58, 76, 40, 35, 46, 45, 0, 100
\n\"KSEEB<\/p>\n

\"KSEEB<\/p>\n

Question 5.
\nFollowing table shows the points of each player scored in four games :
\n\"KSEEB<\/p>\n

i) Find the mean to determine as average number of points scored per game.
\nSolution:
\nA\u2019s average number of points scored
\n\"KSEEB<\/p>\n

ii) To find the mean number of points per game for C, would you divide the total points by 3 or 4 ? Why ?
\nSolution:
\nTo find the mean number of points per game for C, we shall divide the total points by 3 only because he did not play the 3rd game.<\/p>\n

iii) B played in all the four games. How would you find the mean ?
\nSolution:
\n\u2018B\u2019 played all the four games. So we find
\n\"KSEEB<\/p>\n

\"KSEEB<\/p>\n

iv) Who is the best performer ?
\nSolution:
\nTo find out the best performer, we should know mean of all the 3 players. So we should calculate \u2018C\u2019s mean.
\n\"KSEEB
\n\u2234 The best performer is ‘A’<\/p>\n

Question 6.
\nThe marks (out of 100) obtained by a group of students in a science test are 85, 76, 90, 85, 39, 48, 56, 95, 81 and 75. Find the:
\nSolution:
\nMarks obtained by a group (in ascending order) = 39, 48, 56, 75, 76, 81, 85, 85, 90, 95
\ni) Highest and the lowest marks obtained by the students.
\nSolution:
\nHighest and the lowest marks obtained by the student 95 and 39 respectively<\/p>\n

ii) Range of the marks obtained.
\nRange = Highest score – Lowest score
\n= 95 -39 = 56<\/p>\n

iii) Mean marks obtained by the group.
\n\"KSEEB
\n= 39 + 48 + 56 + 75 + 76 + 81 + 85 + 85 + 90 + 95
\n\"KSEEB<\/p>\n

\"KSEEB<\/p>\n

Question 7.
\nThe enrollment in a school during six consecutive years was as follows :
\n1555, 1670, 1750, 2013, 2540, 2820. Find the mean enrollment of the school for this period.
\nSolution:
\nThe enrollment in a school for 6 successive years } = 1555, 1670, 1750, 2013, 2540, 2820.
\n\"KSEEB
\n\u2234 Mean enrollment of the school for 6 years is 2,058<\/p>\n

\"KSEEB<\/p>\n

Question 8.
\nThe rainfall (in mm) in a city on 7 days of a certain week was recorded as follows:
\n\"KSEEB
\ni) Find the range of rainfall in the above data.
\nSolution:
\nRange of the rainfall = Highest Rainfall – Lowest Rainfall
\n= 20.5 – 0.0 = 20.5<\/p>\n

ii) Find the mean rainfall for the week
\nSolution:
\n\"KSEEB<\/p>\n

iii) On how many days was the rainfall less than the mean rainfall.
\nSolution:
\nNo. of days of rainfall which was less than Mean rainfall } = 5 days<\/p>\n

\"KSEEB<\/p>\n

Question 9.
\nThe heights of 10 girls were measured in cms and the results are as follows: 135, 150, 139, 128, 151, 132, 146, 149, 143, 141.
\nSolution:
\nThe height of girls are arranged in ascending order} = 128, 132, 135, 139, 141, 143, 146, 149, 150, 151.<\/p>\n

i) What is the height of the tallest girl?
\nSolution:
\nThe height of the tallest girl is 151 cms<\/p>\n

ii) What is the height of the shortest girl?
\nSolution:
\nThe height of the shortest girl is 128cms<\/p>\n

iii) What is the range of the data ?
\nSolution:
\nThe range of the data = Highest height – lowest height
\n= 151 – 128 = 23<\/p>\n

iv) What is the mean height of the girls?
\nSolution:
\n\"KSEEB
\n= 128 + 132 + 135 + 139 + 141 + 143 + 146 + 149 + 150 + 151
\n\"KSEEB<\/p>\n

v) How many girls have heights more than the mean height.
\nSolution:
\nNo. of girls who were more than mean height = 5<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Chapter 3 Data Handling Ex 3.1, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations. Karnataka State Syllabus Class 7 Maths Chapter 3 Data Handling Ex 3.1 Question 1. Find the range of …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[84],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/14234"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=14234"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/14234\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=14234"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=14234"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=14234"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}