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 heights of any ten students of your class.
Solution:
Students can try themselves.
(This is practical question)
Question 2.
Organise the following marks ion a class assessment, in a tabular form.
4, 6, 7, 5, 3, 5, 4, 5, 2, 6, 2, 5, 1, 9, 6, 5, 8, 4, 6, 7
i) Which number is the highest ?
Solution:
Highest number is 9
ii) Which number is the lowest ?
Solution:
Lowest number is 1
iii) What is the range of the data ?
Solution:
Range of the given data = Highest number – Lowest number = 9 – 1 = 8
iv) Find the arithmetic mean.
Solution:
Arithmetic Mean =
Question 3.
Find the mean of the first five whole numbers.
Solution:
The first whole numbers are 0, 1, 2, 3, 4
Arithemetic mean of five whole numbers =
Question 4.
A cricketer scores the following runs in eight innings :
58, 76, 40, 35, 46, 45, 0, 100. Find the mean score.
Solution:
The scores of 8 innings = 58, 76, 40, 35, 46, 45, 0, 100
Question 5.
Following table shows the points of each player scored in four games :
i) Find the mean to determine as average number of points scored per game.
Solution:
A’s average number of points scored
ii) To find the mean number of points per game for C, would you divide the total points by 3 or 4 ? Why ?
Solution:
To 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.
iii) B played in all the four games. How would you find the mean ?
Solution:
‘B’ played all the four games. So we find
iv) Who is the best performer ?
Solution:
To find out the best performer, we should know mean of all the 3 players. So we should calculate ‘C’s mean.
∴ The best performer is ‘A’
Question 6.
The 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:
Solution:
Marks obtained by a group (in ascending order) = 39, 48, 56, 75, 76, 81, 85, 85, 90, 95
i) Highest and the lowest marks obtained by the students.
Solution:
Highest and the lowest marks obtained by the student 95 and 39 respectively
ii) Range of the marks obtained.
Range = Highest score – Lowest score
= 95 -39 = 56
iii) Mean marks obtained by the group.
= 39 + 48 + 56 + 75 + 76 + 81 + 85 + 85 + 90 + 95
Question 7.
The enrollment in a school during six consecutive years was as follows :
1555, 1670, 1750, 2013, 2540, 2820. Find the mean enrollment of the school for this period.
Solution:
The enrollment in a school for 6 successive years } = 1555, 1670, 1750, 2013, 2540, 2820.
∴ Mean enrollment of the school for 6 years is 2,058
Question 8.
The rainfall (in mm) in a city on 7 days of a certain week was recorded as follows:
i) Find the range of rainfall in the above data.
Solution:
Range of the rainfall = Highest Rainfall – Lowest Rainfall
= 20.5 – 0.0 = 20.5
ii) Find the mean rainfall for the week
Solution:
iii) On how many days was the rainfall less than the mean rainfall.
Solution:
No. of days of rainfall which was less than Mean rainfall } = 5 days
Question 9.
The heights of 10 girls were measured in cms and the results are as follows: 135, 150, 139, 128, 151, 132, 146, 149, 143, 141.
Solution:
The height of girls are arranged in ascending order} = 128, 132, 135, 139, 141, 143, 146, 149, 150, 151.
i) What is the height of the tallest girl?
Solution:
The height of the tallest girl is 151 cms
ii) What is the height of the shortest girl?
Solution:
The height of the shortest girl is 128cms
iii) What is the range of the data ?
Solution:
The range of the data = Highest height – lowest height
= 151 – 128 = 23
iv) What is the mean height of the girls?
Solution:
= 128 + 132 + 135 + 139 + 141 + 143 + 146 + 149 + 150 + 151
v) How many girls have heights more than the mean height.
Solution:
No. of girls who were more than mean height = 5