NCERT Solutions Class 7 Maths Chapter 3 Data Handling Ex 3.1

NCERT Solutions Class 7 Mathematics 
Chapter – 3 (Data Handling)

The NCERT Solutions in English Language for Class 7 Mathematics Chapter – 3 Data Handling Exercise 3.1 has been provided here to help the students in solving the questions from this exercise. 

Chapter : 3 Data Handling

• NCERT Solution Class 7 Maths Exercise – 3.2
• NCERT Solution Class 7 Maths Exercise – 3.3
• NCERT Solution Class 7 Maths Exercise – 3.4

Exercise – 3.1 

1. Find the range of heights of any ten students of your class.

Solution –
Let us assume heights (in cm) of 10 students of our class.
= 130, 132, 135, 137, 139, 140, 142, 143, 145, 148
Here, minimum height = 130 cm
Maximum height = 148 cm
Then,
Range of Heights = Highest value – Lowest value
= 148 – 130
= 18 cm

2. Organise 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) Find the arithmetic mean.

Solution –
First, we have to arrange the given marks in ascending order.
= 1, 2, 2, 3, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 8, 9
Now, we will draw the frequency table of the given data.

Marks (xi)	Tally Marks	Frequency (fi)	fixi
1	|	1	1
2	||	2	4
3	|	1	3
4	|||	3	12
5		5	25
6	||||	4	24
7	||	2	14
8	|	1	8
9	|	1	9
		20	100

(i) Which number is the highest?
highest marks = 9

(ii) Which number is the lowest?
lowest marks = 1

(iii) What is the range of the data?
Range = Max. marks – Min. marks
= 9 – 1 = 8 

(iv) Find the arithmetic mean.
Arithmetic mean =

= = 5 

3. Find the mean of the first five whole numbers.

Solutions –
First 5 Whole numbers are 0, 1, 2, 3, and 4.
Mean = (Sum of first five whole numbers)/ (Total number of whole numbers)
Sum of five whole numbers = 0 + 1 + 2 + 3 +4 = 10
Total Number of whole numbers = 5
∴ Mean = = 2
∴ Mean of first five whole numbers is 2.

4. A cricketer scores the following runs in eight innings:
58, 76, 40, 35, 46, 45, 0, 100.
Find the mean score.

Solution –
Mean score = (Total runs scored by the cricketer in all innings)/ (Total number of innings Played by the cricketer)
Total runs scored by the cricketer in all innings = 58 + 76 + 40 + 35 + 46 + 45 + 0 + 100 = 400
Total number of innings = 8
∴  Mean = = 50
Mean score of the cricketer is 50.

5. Following table shows the points of each player scored in four games:

Now answer the following questions:
(i) Find the mean to determine 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 total points by 3 or by 4? Why?
(iii) B played in all the four games. How would you find the mean?
(iv) Who is the best performer?

Solution –

(i) A’s average number of points scored per game = Total points scored by A in 4 games/ Total number of games
= = = 12.5 points

(ii) Since, C did not play Game 3, he played only 3 games. So, the total will be divided by 3.

(iii) Number of points scored by B in all the games are Game 1 = 0, Game 2 = 8, Game 3 = 6, Game 4 = 4
Mean of B’s score = Total points scored by B in 4 games / Total number of games
=  = = 4.5 points

(vi) Now, we have to find the best performer among 3 players.
So, we have to find the average points of C = = = 10.67 points

By observing,
the average points scored A is 12.5 which is more than B and C.
Clearly, we can say that A is the best performer among three.

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:
(i) Highest and the lowest marks obtained by the students.
(ii) Range of the marks obtained.
(iii) Mean marks obtained by the group.

Solution –
Marks obtained are:
85, 76, 90, 85, 39, 48, 56, 95, 81 and 75

(i) Highest marks = 95
Lowest marks = 39

(ii) Range of the marks = Highest marks – Lowest marks
= 95 – 39 = 56

(iii) Mean marks = (Sum of all marks obtained by the group of students)/ (Total number of marks)

= = = 73

7. The enrolment in a school during six consecutive years was as follows:
1555, 1670, 1750, 2013, 2540, 2820.
Find the mean enrolment of the school for this period.

Solution –
Mean enrolment = (Sum of all observations) / (Number of observations)

=  = = 2058
∴ The mean enrolment of the school for this given period is 2058.

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.
(ii) Find the mean rainfall for the week.
(iii) On how many days was the rainfall less than the mean rainfall.

Solution –
(i) Range of rainfall = Highest rainfall – Lowest rainfall
= 20.5 – 0.0 = 20.5 mm

(ii) Mean of rainfall = (Sum of all observations)/ (Number of observation)

= = = 5.9 mm

(iii) Number of days on which the rainfall was less than the mean rainfall = Monday, Wednesday, Thursday, Saturday, Sunday = 5 days.

9. The heights of 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 heights more than the mean height.

Solution –
First we have to arrange the given data in an ascending order
= 128, 132, 135, 139, 141, 143, 146, 149, 150, 151

(i) The height of the tallest girl is 151 cm

(ii) The height of the shortest girl is 128 cm

(iii) Range of given data = Tallest height – Shortest height
= 151 – 128 = 23 cm

(iv) Mean height of the girls = Sum of height of all the girls/ Number of girls
= = = 141.4 cm

(v) Number of girls having more height than the mean height
= 150, 151, 146, 149 and 143
= 5 girls