NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.1

# NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.1

NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.1 are the part of NCERT Solutions for Class 7 Maths (Rationalised Contents). Here you can find the NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.1.

## NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.1 (Rationalised Contents)

### Ex 3.1 Class 7 Maths Question 1.

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

Solution:
Suppose the heights (in cm) of 10 students in your class be 150, 152, 151, 148, 149, 149, 150, 152, 155, 147.
Arranging the heights in ascending order, we have 147, 148, 149, 149, 150, 150, 151, 152, 152, 155.
Range of height of students = 155 – 147 = 8

### Ex 3.1 Class 7 Maths Question 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:
Arranging the marks in a class assessment in the tabular form:

(i) Highest number = 9

(ii) Lowest number = 1

(iii) Range = highest value – lowest value = 9 – 1 = 8

(iv) Arithmetic mean = Sum of marks ÷ Total number of data

= 100 ÷ 20 = 5

### Ex 3.1 Class 7 Maths Question 3.

Find the mean of the first five whole numbers.

Solution:
The first five whole numbers are 0, 1, 2, 3 and 4.
Their mean = (0 + 1 + 2 + 3 + 4) ÷ 5 = 10/5 = 2

### Ex 3.1 Class 7 Maths Question 4.

A cricketer scores the following runs in eight innings:

58, 76, 40, 35, 46, 45, 0, 100.

Find the mean score.

Solution:
Sum of the runs in all eight innings = 58 + 76 + 40 + 35 + 46 + 45 + 0 + 100 = 400
Number of innings = 8
Mean score = 400/8 = 50

### Ex 3.1 Class 7 Maths Question 5.

Following table shows the points of each player scored in four games:
(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 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 = (14 + 16 + 10 + 10)/4

50/4 = 12.5

(ii) C’s average points per game = (8 + 11 + 0 + 13)/4 = 32/4 = 8
Since we are comparing the performance, so we divide the total points by 4 to find the mean for C.

(iii) B’s average point per game = (0 + 8 + 6 + 4)/4 = 18/4 = 4.5
(To find B’s average, we find the sum of all observations and divide the sum by the number of observations.)

(iv) Since, 12.5 > 8 > 4.5
The best performer is A.

### Ex 3.1 Class 7 Maths 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:
(i) Highest and the lowest marks obtained by the students.
(ii) Range of the marks obtained.
(iii) Mean marks obtained by the group.

Solution:
Arranging the marks obtained by a group of students in ascending order, we get

39, 48, 56, 75, 76, 81, 85, 85, 90, 95
(i) The highest and the lowest marks obtained by the students are 95 and 39, respectively.
(ii) Range of the marks obtained = Highest marks – lowest marks

= 95 – 39 = 56
(iii) Mean marks

### Ex 3.1 Class 7 Maths Question 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:
Sum of the enrolment in the school during six consecutive years
= 1555 + 1670 + 1750 + 2013 + 2540 + 2820
= 12348
Mean enrolment = 12348/6 = 2058

### Ex 3.1 Class 7 Maths 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 the 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) Arranging the rainfall during the week in ascending order, we get
0.0, 0.0, 1.0, 2.1, 5.5, 12.2, 20.5
Range = Highest rainfall – lowest rainfall = 20.5 – 0.0 = 20.5

(ii) Sum of the rainfall during the week = 0.0 + 0.0 + 1.0 + 2.1 + 5.5 + 12.2 + 20.5 = 41.3
Mean rainfall = 41.3/7 = 5.9

(iii) The rainfall was less than the mean rainfall for 5 days.

### Ex 3.1 Class 7 Maths Question 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:
Arranging the heights (in cm) in the ascending order, we get

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 = tallest height – shortest height = 151 – 128 = 23 cm.

(v) Five girls have heights more than the mean height.