RBSE Solutions for Class 7 Maths Chapter 17 Data Handling Ex 17.2

RBSE Solutions for Class 7 Maths Chapter 17 Data Handling Ex 17.2 is part of RBSE Solutions for Class 7 Maths. Here we have given Rajasthan Board RBSE Class 7 Maths Chapter 17 Data Handling Exercise 17.2.

Board	RBSE
Textbook	SIERT, Rajasthan
Class	Class 7
Subject	Maths
Chapter	Chapter 17
Chapter Name	Data Handling
Exercise	Ex 17.2
Number of Questions	10
Category	RBSE Solutions

Rajasthan Board RBSE Class 7 Maths Chapter 17 Data Handling Ex 17.2

Question 1
Following are the number of students (RBSESolutions.com) of a school from class 6 to 12.
78, 72, 67, 59, 54, 49, 48. Then find :
(i) In which class the number of students is maximum?
(ii) In which class the number of students is maximum?
(iii) What is the range of these data?
(iv) Find the mean of these data.
Solution:
(i) class 6 has 78 students which has maximum students.
(ii) In class 12 has 48 student which has minimum student.
(iii) Range of observation = class 6 students – class 12 students = 78 – 48 = 30, Range = 30
(iv) 78, 72, 67, 59, 54, 49, 48 are the number of students in class 6 – 12.

Question 2
Find the mean (RBSESolutions.com) of first 10 whole number.
Solution:
First 10 whole number 0, 1,2,3,4, 5,6, 7, 8,9
Mean

Question 3
A cricketer scored the runs in 6 innings as follows : 68, 03, 17, 78, 12, 104. Find the arithmetic mean of runs scored.
Solution:

Question 4
The number of passengers travelled in (RBSESolutions.com) the bus running from Bikaner to Udaipur from Monday to Friday are as follows :
45, 48, 32, 40, 30. What is the mean of passengers in each day?
Sol:
Mean
45 + 48 + 32 + 40 + 30 195

Question 5
Following crops were raised up to five years in a village. The (RBSESolutions.com) profit (in rupees) per acre on the crop were as follows :

Answer the following questions on the basis of the above table.
(i) Find the mean profit of each crop in five years.
(ii) Which crop should be (RBSESolutions.com) raised next year on the basis of above answer.
Solution:
(i) Millet crop mean of profit

(ii) we have observe that maximum profit in . growing the crop of Ground nut. So Ground nut crop should be grown.

Question 6
If arthemetic mean of digits 3, 4, 8, 5, x, 3 is 4 then (RBSESolutions.com) find the value of x.
Solution:
Mean = 4

⇒ [latex]\frac { 23+x }{ 6 }[/latex] = 4
⇒ 23 + x = 6 x 4 = 24
⇒ x = 24 – 13 = 1

Question 7
The number of books given to students in 10 days (RBSESolutions.com) from a library are as follows :
40, 57, 32, 59, 72, 66, 40, 62, 72, 60
Find the mean of books given to students every day.
Solution:

Question 8
The average of five numbers is 18. If four numbers are 22, 20, 14 and 13 then find (RBSESolutions.com) the fifth number.
Solution:
Average of 5 numbers = 18
Let the fifth number is x

Question 9
The temperature of a particular (RBSESolutions.com) week in a city is noted as below :

(i) Find the range of temperature by using the data.
(ii) Find the mean temperature of this week.
(iii) How many days did the temperature remain more than average?
Solution:
(i) According to observations maximum temperature was (RBSESolutions.com) on Wednesday and minimum temperature was on Saturday So, Range = 40°C – 35°C = – 5°C
(ii) Mean

(iii) Tuesday, Wednesday and Friday 3 days temperature was above the average temperature.

Question 10
In a singing competition in a school, three judges gave (RBSESolutions.com) marks out of 100 to four singer contestants as follows :

(1) What is the range of mark given by the judges?
(2) Find the mean of total marks.
(3) Point out the name of winner.
(4) What is the difference in the (RBSESolutions.com) means of winner and the contestant placed at the fourth place?
Solution:
(1) The judges has given minimum marks which are 49 and maximum marks 83
Range = 83 – 49 = 34

(2) Mean of marks

(3) By observing the marks we can see that judges (RBSESolutions.com) have given more marks to Suman so Suman is winner of competition

(4) Mean of winner = [latex]\frac { 82 + 75 + 83 }{ 3 }[/latex] = 80
Mean of contestant placed at the fourth place (Khushboo) = [latex]\frac { 49 + 56 + 51 }{ 3 }[/latex] = 52
So the difference of mean’s between winner and fourth position is = 80 – 52 = 28

We hope the RBSE Solutions for Class 7 Maths Chapter 17 Data Handling Ex 17.2 will help you. If you have any query regarding Rajasthan Board RBSE Class 7 Maths Chapter 17 Data Handling Exercise 17.2, drop a comment below and we will get back to you at the earliest.