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

Rajasthan Board RBSE Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.1 Textbook Exercise Questions and Answers.

Rajasthan Board RBSE Solutions for Class 7 Maths in Hindi Medium & English Medium are part of RBSE Solutions for Class 7. Students can also read RBSE Class 7 Maths Important Questions for exam preparation. Students can also go through RBSE Class 7 Maths Notes to understand and remember the concepts easily. Students can access the data handling class 7 extra questions with answers and get deep explanations provided by our experts.

RBSE Class 7 Maths Solutions Chapter 3 Data Handling Ex 3.1

Question 1.
Find the range of heights of any 10 students of your class.
Answer:
Try it yourself.

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?
Answer:
RBSE Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.1 1
(i) 9 is the highest number.
(ii) 2 is the lowest number.
(iii) Range = highest value - lowest value = 9 - 1 = 8
(iv) Arithmetic mean = \(\frac{\text { Sum of observations }}{\text { Number of observations }}\)
Sum of observations = 4 + 6 + 7 + 5 + 3 + 5 + 4 + 5 + 2 + 6 + 2 + 5 + 1 + 9 + 6 + 5 + 8 + 4 + 6 + 7 = 100
Mean = \(\frac{100}{20}\) = 5

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

Question 3.
Find the mean of the first five whole numbers.
Answer:
First 5 whole numbers are 0, 1, 2, 3, and 4.
Mean = \(\frac{0+1+2+3+4}{5}\) = \(\frac{10}{5}\) = 2

Question 4.
A cricketer scores the following runs in eight innings :
58, 76, 40, 35, 46, 45, 0,100
Find the mean score.
Answer:
Mean Score = \(\frac{\text { Sum of all observations }}{\text { Number of observations }}\)
= \(\frac{58+76+40+35+46+45+0+100}{8}\)
= \(\frac{400}{8}\) = 50

Question 5.
Following table shows the points of each player scored in four games:
RBSE Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.1 2
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?
Answer:
(i) Mean = \(\frac{\text { Sum of observations }}{\text { Number of observations }}\)
= \(\frac{14+16+10+10}{4}\) = \(\frac{50}{4}\) = 12.5
So, A’s average number of points scored per game = 12.5

(ii) To find the mean number of points per game for C, we shall divide the total points by 3 because ‘C’ did not play game 3.

(iii) Mean score of 'B’
= \(\frac{\text { Sum of observations }}{\text { Number of observations }}\)
= \(\frac{0+8+6+4}{4}\) = \(\frac{18}{4}\) = 4.5

(iv) Mean score of 'C' = \(\frac{8+11+13}{3}\) = \(\frac{32}{3}\)
= 10.67 (As ‘C’ did not play game 3)
∴ The best performer is player A.

RBSE Solutions for Class 7 Maths Chapter 3 Data Handling 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 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.
Answer:
(i) Highest marks = 95
Lowest marks = 39

(ii) Range of marks = highest marks - lowest marks = 95 - 39 = 56

(iii) Mean marks
= \(\frac{\text { Sum of observations }}{\text { Number of observations }}\)
= \(\frac{48+56+95+81+75}{10}\)
= \(\frac{730}{10}\) = 73

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.
Answer:
Mean enrolment
= \(\frac{\text { Sum of observations }}{\text { Number of observations }}\)
= \(\frac{1555+1670+1750+2013+2540+2820}{6}\)
= \(\frac{12348}{6}\)
= 2058

Question 8.
The rainfall (in mm) in a city on 7 days of a certain week was recorded as follows:

Day

Rainfall (in mm)

Monday

0.0

Tuesday

12.2

Wednesday

2.1

Thursday

0.0

Friday

20.5

Saturday

5.5

Sunday

1.0

(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?
Answer:
(i) Maximum rainfall = 20.5 mm
Minimum rainfall = 0.0 mm
∴ Range of rainfall
= highest observation - lowest observation
= 20.5 mm - 0.0 mm = 20.5 mm

(ii) Mean rainfall for the week
= \(\frac{\text { Sum of observations }}{\text { Number of observations }}\)
= \(\frac{0.0+12.2+2.1+0.0+20.5+5.5+1.0}{7}\)
= \(\frac{41.3}{7}\) mm = 5.9 mm

(iii) On Monday, Wednesday, Thursday, Saturday and Sunday, i.e. 5 days the rainfall was less than the mean rainfall.

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

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?
Answer:
(i) Height of the tallest girl = 151 cm

(ii) Height of the shortest girl = 128 cm

(iii) Range = highest observation - lowest observation = 151 cm - 128 cm = 23 cm

(iv) Mean height
RBSE Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.1 3

Bhagya
Last Updated on June 4, 2022, 3:04 p.m.
Published June 3, 2022