Excercise 3.1 Data Handling  NCERT Solutions Class 7
Exercise 3.1
Chapter 3 Ex.3.1 Question 1
Find the range of the heights of any \(10\) students of your class
Solution
What is known?
The heights of ten student.
What is unknown?
The range of heights.
Reasoning:
Range \(=\) Highest value \(\) Lowest value
Steps:
Serial no. 
Name of the students 
Height (in cm) 
\(1\) 
Yash 
\(126\) 
\(2\) 
Nidhi 
\(128\) 
\(3\) 
Toshi 
\(134\) 
\(4\) 
Abhishek 
\(148\) 
\(5\) 
Dimple 
\(146\) 
\(6\) 
Parul 
\(132\) 
\(7\) 
Neha 
\(129\) 
\(8\) 
Ananya 
\(137\) 
\(9\) 
Arya 
\(127\) 
\(10\) 
Dipti 
\(130\) 
\(\therefore\) Range \(=\) Highest value \(\) Lowest value
Highest value among this observation \(= 148\)
Lowest value among this observation \(= 126\)
\(\therefore\) Range \(= 148  126 = 22 \rm\,cm\)
Chapter 3 Ex.3.1 Question 2
Organize the following marks in 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?
(ii) Which number is the lowest?
(iii) What is the range of the data?
(iv) What is the arithmetic mean?
Solution
What is known?
Marks in class assessment.
What is unknown?
Highest marks,lowest marks,range,arithmetic mean.
Reasoning:
Range \(= \) Highest marks \(\) lowest marks
Steps:
S\(I\) NO 
Marks 
Tally marks 
No. of students (Frequency) 
\(1\) 
\(1\) 
I 
\(1\) 
\(2\) 
\(2\) 
II 
\(2\) 
\(3\) 
\(3\) 
I 
\(1\) 
\(4\) 
\(4\) 
III 
\(3\) 
\(5\) 
\(5\) 

\(5\) 
\(6\) 
\(6\) 
IIII 
\(4\) 
\(7\) 
\(7\) 
II 
\(2\) 
\(8\) 
\(8\) 
I 
\(1\) 
\(9\) 
\(9\) 
I 
\(1\) 
The highest number is \(9.\)
The lowest number is \(1.\)
The range of the data \(=\) Highest number \(\) Lowest number no.
\(= 9 \,– 1 = 8\)
\(\therefore\) Mean of first five whole numbers
\[= \frac{{{\text{Sum of all numbers}}}}{{{\text{Total numbers}}}}\]
Arithmetic mean
\[\begin{align}& =\frac{\left[ \begin{align}& 4+6+7+5+3+5+4 \\ & +5+2+6+2+5+1+ \\ & 9+6+5+8+4+6+7 \\ \end{align} \right]}{20} \\ & =\frac{100}{20} \\ & =5\\ \end{align}\]
Chapter 3 Ex.3.1 Question 3
Find the mean of the first \(5\) whole numbers.
Solution
What is known?
First 5 whole numbers
What is unknown?
Mean of first 5 whole numbers.
Reasoning:
Mean of first five whole numbers
\[\begin{align}=\frac{\text{Sum of all numbers}}{\text{Total numbers}}\end{align}\]
Steps:
The first five while numbers are \(0, 1, 2, 3, 4\)
\(\therefore\) Mean of first five whole numbers
\[\begin{align}=\frac{\text{Sum of all numbers}}{\text{Total numbers}}\end{align}\]
Arithmetic mean
\[\begin{align} &= \frac{{0 + 1 + 2 + 3 + 4}}{5}\\ &= \frac{{10}}{5}\\ &= 2\end{align}\]
Thus, the mean of first five whole numbers is \(5.\)
Chapter 3 Ex.3.1 Question 4
A cricketer scores the following runs in eight innings:
\(58, 76, 40, 35, 46, 45, 0, 100\)
Find the mean score.
Solution
What is known?
Runs in eight innings.
What is unknown?
Mean Score
Reasoning:
\(\begin{align}\text{Mean of score}\!=\!\frac{\text{Sum of all runs}} {\text{Number of innings }}\end{align}\)
Steps:
Total number of innings = \(8\)
Scores of cricketer in \(8\) innings
\[\begin{align} = 58,76,40,35,46,45,0,100\end{align}\]
Mean of score
\[\begin{align} &= \frac{{\left[ \begin{array}{l}58 + 76 + 40 + 35 +\\ 46 + 45 + 0 + 100\end{array} \right]}}{8}\\ &= \frac{{400}}{8}\\ &= 50\end{align}\]
Thus, the mean score of a cricketer in \(8\) innings is \(50.\)
Chapter 3 Ex.3.1 Question 5
Following table shows the points each player scored in four games: 
Player 
Game 1 
Game 2 
Game 3 
Game 4 
\(A\) 
\(14\) 
\(16\) 
\(10\) 
\(10\) 
\(B\) 
\(10\) 
\(8\) 
\(6\) 
\(4\) 
C 
\(8\) 
\(11\) 
Didn’t play 
\(13\) 
Now answer the following questions: 
(i) Find the mean to detetextine \(A’s\) average number of points scored per game.
(ii) To find the mean number of points per game for \(C,\) would you divide the points by \(3\) or \(4\)? Why?
(iii) \(B\) played in all the four games. How would you find the mean?
(iv) Who is the best performer?
Solution
Reasoning:
What is known?
Point of each player scored in four games.
What is unknown?
Average score, Mean
Reasoning:\( \begin{align}\text{Mean}=\frac{\text{Sum of score }}{\text{Number of games}}\end{align}\)
Steps:
Total number of games played by \(A = 4\)
Scores obtained by \(A = 14, 16, 10, 10\)
Mean of player \(A\)
\[\begin{align}&= \frac{{14 + 16 + 10 + 10}}{4}\\ &= \frac{{50}}{4}\\ &= 12.5\end{align}\]
ii) We should divide total points by \(3\) because player \(C\) played only three games.
iii) Total number of games played by \(B = 4\)
Scores obtained by \(B = 0, 8, 6, 4\)
Mean of player \(B \)
\[\begin{align}&= \frac{{{\text{Sum of scores by B}}}}{{{\text{No}}{\text{. of games played by B}}}}\\ &= \frac{{0 + 8 + 6 + 4}}{4}\\ &= \frac{{18}}{4}\\ &= 4.5\end{align}\]
iv) To find the best performer,
We should find the mean of all players.
Mean of player \(A = 12.5\)
Mean of player \(B = 4.5\)
Mean of player \(C\)
\[\begin{align} &= \frac{{{\text{sum of scores by C}}}}{{{\text{No}}{\text{. of games played by C}}}}\\ &=\frac{{8 + 11 + 13}}{3}\\ &= \frac{{32}}{3}\\ &= 10.67\end{align}\]
Therefore, on comparing means of all players, \(A\) is the best performer.
Chapter 3 Ex.3.1 Question 6
The marks (Out of \(100\)) obtained by a group of students in a science test are
\(85, 76, 90, 85, 39, 48, 56, 95, 81, 75.\) Find the: 
(i) Highest and lowest marks obtained by the students.
(ii) Range of the marks obtained.
(iii) Mean marks obtained by the group.
Solution
What is known?
Marks obtained by a group of students in a science test.
What is unknown?
Highest and lowest marks, Range of the marks, Mean marks
Reasoning:
Mean marks
\[\begin{align}=\frac{\text{Sum of marks}}{\text{Total no}\text{.of students}}\end{align}\]
Solution:
(i) Highest marks obtained by the student \(= 95\)
Lowest marks obtained by the student \(= 39\)
(ii) Range of the marks \(=\) Highest marks \(–\) Lowest marks
\(= 95\, – 39\\ = 56\)
Total no. of students \(= 10\)
Marks obtained by the group of students \(= 85, 76, 90, 85, 39, 48, 56, 95, 81, 75\)
Mean marks
\[\begin{align} &= \frac{{{\text{Sum of marks}}}}{{{\text{Total no }}{\text{.of students}}}}\\ &= \frac{{\left[ \begin{array}{l}85 + 76 + 90 + 85 + 39 +\\48 + 56 + 95 + 81 + 75\end{array} \right]}}{{10}}\\ &= \frac{{730}}{{10}}\\& = 73\end{align}\]
Thus, the mean marks obtained by a group of students are \(73.\)
Chapter 3 Ex.3.1 Question 7
The enrollment in a school during six consecutive years as follows: 
\(1555, 1670, 1750, 2013, 2540, 2820\)
Find the mean enrollment of the school for this period.
Solution
What is known?
Number of enrollments
What is unknown?
Mean
Reasoning:
Mean enrollment
\[\begin{align} = \frac{{\left[ \begin{align} {\text{sum of numbers}}\\ {\text{of enrollments}} \end{align} \right]}}{{{\text{Total numbers of years}}}}\end{align}\]
Steps:
Total no. of years \(= 6\)
The enrollments in a school during six consecutive years
\[\left[ \begin{align}{\rm{ = 1555, 1670, 1750, }}\\{\rm{ 2013, 2540, 2820}}\end{align} \right]\]
Mean enrollment
\[\begin{align} &= \frac{{\left[ \begin{array}{l}{\text{sum of numbers }}\\{\text{of enrollment}}\end{array} \right]}}{{\left[ \begin{array}{l}{\text{Total numbers }}\\{\text{of enrollment}}\end{array} \right]}}\\ &= \frac{{\left[ \begin{array}{l}1555 + 1670 + 1750 + \\2013 + 2540 + 2820\end{array} \right]}}{6}\\& = \frac{{12348}}{6}\\ &= 2058\end{align}\]
Thus, the mean enrollment of the school is \(2058.\)
Chapter 3 Ex.3.1 Question 8
The rainfall (in mm) in a city on \(7\) days of a certain week as recorded as follows:
Day 
Mon 
Tue 
Wed 
Thurs 
Fri 
Sat 
Sun 
Rain fall (in \(\rm mm\)) 
\(0.0\) 
\(12.2\) 
\(2.1\) 
\(0.0\) 
\(20.5\) 
\(5.5\) 
\(1.0\) 
(i) Find the range of the rainfall in the above data.
(ii) Find the mean rainfall of the week.
(iii) On how many days was the rainfall less than the mean rainfall?
Solution
What is known?
Rainfall in \(\rm mm\)
What is unknown?
Range of the rainfall and mean rainfall
Reasoning:
Mean of the rainfall in \((\text{mm})\)
\(\begin{align}=\frac{\text{sum of rain falls (in mm) }}{\text{Number of days}}\end{align}\)
Steps:
(i)Highest of the rainfall (in \(\rm m\rm m\)) \(= 20.5\)
Lowest of the rainfall (in \(\rm m\rm m\)) \(= 0.0\)
Range of the rainfall = Highest rainfall – Lowest rainfall
\(= 20.5 – 0.0\\ = 20. 5 \rm\,mm\)
(ii)Total number of days \(= 7\)
Record of the rainfall (in \(\rm mm\)) \(= 0.0, 12.2, 2.1, 0.0, 20.5, 5.5, 1.0\)
Mean of the rainfall in \((\text{mm})\)
\[\begin{align} & =\frac{\left[ \begin{align} & 0.0+12.2+2.1+0.0+ \\ & 20.5+5.5+1.0 \\ \end{align} \right]}{7} \\ & =\frac{41.3}{7} \\ & =5.9 \\ \end{align}\]
(iii)On \(5\) days rainfall was less than the mean rainfall i.e., Monday, Wednesday, Thursday, Saturday and Sunday.
Chapter 3 Ex.3.1 Question 9
The heights of the \(10\) girls were measured in cm and the results are as follows:
\(135, 150, 139, 128, 151, 132, 146, \\149, 143, 141.\)
(i) What is the height of the tallest girl?
(ii) What is the height of the shortest girl?
(iii) What is the range of the data?
(iv) What is the mean height of the girls?
(v) How many girls have height more than the mean height?
Solution
What is known?
Height of the \(10\) girls
What is unknown?
(i) Height of the tallest girl (ii) Height of the shortest girl (iii). Range of the data (iv) Mean height
Range \(=\) Height of tallest girl – Height of the shortest girl
\[\begin{align}\text{Mean}\!=\!\frac{\text{Sum of all heights}\left(\text{in mm}\right)}{\text{Number of girls}}\end{align}\]\(\)
Steps:
(i) The height of the tallest girl \(=151 \rm\,cm\)
(ii) The height of the shortest girl \(= 128 \rm \,cm\)
(iii) Range of the data \(= 151 – 128 =\) \(23\rm\, cm\)
Total number of girls \(= 10\)
Measurement of the heights of 10 girls
\[\begin{align}=& 135, 150, 139, 128, 151, 132, 146, \\&149, 143, 141.\end{align}\]
Mean height of the girls
\[\begin{align} & =\frac{\left[ \begin{align} & 135+150+139+128+151 \\& +132+146+149+143+141 \\\end{align} \right]}{10} \\& =\frac{1414}{10} \\& =14.14\ \text{cm} \\\end{align}\]
Five girls have heights more than the mean height i.e.: \(150, 151, 146, 149, 143.\)