NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Exercise 3.1<\/h2>\n
Question 1.
\nFind the range of heights of any ten students of your class.
\nAnswer:
\nDo it yourself, one simple solution may be to consider the heights (in cm) of ten students as follows:
\n111, 118, 110, 120, 114, 113, 118, 117, 115, 119
\nAverage height
\n![\"NCERT](\"https:\/\/i2.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/06\/NCERT-Solutions-for-Class-7-Maths-Chapter-3-Data-Handling-Ex-3.1-1.png?resize=265%2C150&ssl=1\")
\n= 115.5 cm<\/p>\n
![\"NCERT](\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?resize=170%2C17&ssl=1\")
<\/p>\n
Question 2.
\nOrganise the following marks in 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(i) Which number is the highest?
\n(ii) Which number is the lowest?
\n(iii) What is the range of the data?
\n(iv) Find the arithmetic mean.
\nAnswer:
\n![\"NCERT](\"https:\/\/i2.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/06\/NCERT-Solutions-for-Class-7-Maths-Chapter-3-Data-Handling-Ex-3.1-2.png?resize=299%2C375&ssl=1\")
\n(i) The highest number = 9
\n(ii) The lowest number = 1
\n(iii) The range = Highest number – Lowest number = 9 – 1 = 8
\n(iv) Arithmetic Mean
\n= \\(\\frac{\\text { Sum of the marks }}{\\text { Number of students }}\\)
\n![\"NCERT](\"https:\/\/i2.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/06\/NCERT-Solutions-for-Class-7-Maths-Chapter-3-Data-Handling-Ex-3.1-3.png?resize=267%2C182&ssl=1\")
\n= \\(\\frac{100}{20}\\) = 5
\nThus, the mean marks = 5.<\/p>\n
Question 3.
\nFind the mean of the first five whole numbers.
\nAnswer:
\nFirst five whole numbers are 0, 1, 2, 3 and 4
\nSum 0 + 1 + 2 + 3 + 4 = 10
\nMean = \\(\\frac{\\text { Sum of the number }}{\\text { Number of whole numbers }}\\)
\n= \\(\\frac{10}{5}\\) = 2<\/p>\n
<\/p>\n
Question 4.
\nA cricketer scores the following runs in eight innings.
\n58, 76, 40, 35, 46, 45, 0, 100
\nFind the mean score.
\nAnswer:
\nMean score = \\(\\frac{\\text { Sum of the scores }}{\\text { Number of innings }}\\)
\n= \\(\\frac{58+76+40+35+46+45+0+100}{8}\\)
\n= \\(\\frac{400}{8}\\) = 50
\nMean score = 50<\/p>\n
Question 5.
\nFollowing table shows the points of each player scored in four games:
\n![\"NCERT](\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/06\/NCERT-Solutions-for-Class-7-Maths-Chapter-3-Data-Handling-Ex-3.1-4.png?resize=333%2C182&ssl=1\")
\nNow answer the following questions:
\n(i) Find the mean to determine A\u2019s average number of points scored per game.
\n(ii) To find the mean number of points per game for C, would you divide the total points by 3 or by 4?Why?
\n(iii) B played in all the four games. How would you find the mean?
\n(iv) Who is the best performer?
\nAnswer:
\n(i) Mean = \\(\\frac{\\text { Sum of the observations }}{\\text { Number of observations }}\\)
\n= \\(\\frac{14+16+10+10}{4}\\) = \\(\\frac{50}{4}\\) = 12.5
\nA\u2019s average score per game is 12.5<\/p>\n
(ii) Since C played only 3 games (He did not play the 3rd game)
\nTotal will be divided by 3<\/p>\n
(iii) Mean score of B
\n= \\(\\frac{\\text { Sum of the all observations }}{\\text { Number of observations }}\\)
\n= \\(\\frac{0+8+6+4}{4}\\) = \\(\\frac{18}{4}\\) = 4.5
\nThus, th average\u2019tnumber of points scored by is B is 4.5<\/p>\n
(iv) Mean score of C
\n= \\(\\frac{\\text { Sum of the all observations }}{\\text { Number of observations }}\\)
\n= \\(\\frac{8+11+13}{4}\\) = \\(\\frac{32}{3}\\) = 10.67
\nThus, mean number of points scored by C is 10.67
\nA\u2019s average score is = 12.5
\nB\u2019s average score is = 4.5
\nC\u2019s average score is = 10.67
\nThe best performer is A.<\/p>\n
<\/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:
\n(i) Highest and the lowest marks obtained by the students.
\n(ii) Range of the marks obtained.
\n(iii) Mean marks obtained by the group.
\nAnswer:
\nWrite the given marks in ascending order, we get 39, 48, 56, 75, 76, 81, 85, 85, 90, 95
\n(i) Highest marks = 95, Lowest marks = 39<\/p>\n
(ii) Range = Highest marks – Lowest marks = 95 – 39 = 56<\/p>\n
(iii) Mean = \\(\\frac{\\text { Sum of all the marks }}{\\text { Number of students }}\\)
\n![\"NCERT](\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/06\/NCERT-Solutions-for-Class-7-Maths-Chapter-3-Data-Handling-Ex-3.1-5.png?resize=257%2C77&ssl=1\")
\n= \\(\\frac{730}{10}\\) = 73
\nMean marks obtained by the group is 73.<\/p>\n
Question 7.
\nThe enrolment in a school during six consecutive years was as follows:
\n1555,1670,1750,2013,2540,2820 Find the mean enrolment of the school for this period.
\nAnswer:
\n![\"NCERT](\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/06\/NCERT-Solutions-for-Class-7-Maths-Chapter-3-Data-Handling-Ex-3.1-6.png?resize=313%2C140&ssl=1\")
\n\u2234 The mean enrolment is 2058 per year.<\/p>\n
Question 8.
\nThe rainfall (in mm) in a city on 7days of a certain week was recorded as follows:<\/p>\n
![\"NCERT](\"https:\/\/i2.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/06\/NCERT-Solutions-for-Class-7-Maths-Chapter-3-Data-Handling-Ex-3.1-7.png?resize=324%2C130&ssl=1\")
![\"NCERT](\"https:\/\/i2.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/06\/NCERT-Solutions-for-Class-7-Maths-Chapter-3-Data-Handling-Ex-3.1-8.png?resize=297%2C188&ssl=1\")