RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Ex 17.1

RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Ex 17.1 is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 17 Measures of Central Tendency Exercise 17.1.

Board RBSE
Textbook SIERT, Rajasthan
Class Class 10
Subject Maths
Chapter Chapter 17
Chapter Name Measures of Central Tendency
Exercise Exercise 17.1
Number of Questions Solved 10
Category RBSE Solutions

Rajasthan Board RBSE Class 10 Maths Chapter 17 Measures of Central Tendency Ex 17.1

Ex 17.1 Class 10 RBSE Question 1.
If marks obtained by 10 students in (RBSESolutions.com) maths are 52, 75, 40, 70, 43, 40, 65, 35, 48 and 52 then find arithmetic mean.
Solution :
Given : Total students in class = 10
Ex 17.1 Class 10 RBSE Maths Chapter 17 Measures Of Central Tendency
Thus, arithmetic mean = 52

RBSE Class 10 Maths Chapter 17 Question 2.
The monthly salary (in ₹) of assistant workers in a (RBSESolutions.com) school are ₹1720, ₹1750, ₹1760 and ₹1710, then find arithmetic mean.
Solution :
RBSE Class 10 Maths Chapter 17 Ex 17.1 Measures Of Central Tendency
Thus, required arithmetic mean = ₹1735

Class 10 Maths RBSE Solution Chapter 17 Question 3.
If arithmetic mean (RBSESolutions.com) of digits 3, 4, 8, 5, x, 3, 2, 1 is 4, then find x.
Solution :
Given : A.M. = 4
Class 10 Maths RBSE Solution Chapter 17 Ex 17.1 Measures Of Central Tendency
⇒ 26 + x = 32
⇒ x = 32 – 26 = 6
Thus, x = 6

RBSE Solutions For Class 10 Maths Chapter 17 Question 4.
Runs scored in 10 innings by a cricketer (RBSESolutions.com) are 60, 62, 56, 64, 0, 57, 33, 27, 9 and 71. Find average run scored in these innings.
Solution:
We know that
RBSE Solutions For Class 10 Maths Chapter 17 Ex 17.1 Measures Of Central Tendency
Thus, average run scored = 43.9 runs.

Ch 17 Maths Class 10 RBSE Question 5.
Calculate arithmetic mean of marks obtained in English (RBSESolutions.com) in monthly test of 10 students
S.No. :                  1     2     3    4    5     6     7    8    9    10
Marks obtained : 30   28   32  12  18   20   25  15   26  14
Solution :
Ch 17 Maths Class 10 RBSE Ex 17.1 Measures Of Central Tendency
Thus, arithmetic mean = 22 marks

RBSE Solutions For Class 10 Maths Chapter 17.1 Question 6.
The number of books issued to students in 10 days (RBSESolutions.com) from library of a school are as follows :
300   405   455    489    375     280    418    502    300     476
Find the average books issued daily.
Solution :
Arithmetic mean
RBSE Solutions For Class 10 Maths Chapter 17 Ex 17.1 Measures Of Central Tendency
Thus, average books issued daily = 400 books.

RBSE Solution Class 10 Maths Chapter 17 Question 7.
The average weight of 25 students of group A of a (RBSESolutions.com) class is 51 kg. Whereas average weight of 35 students of group B is 54 kg. Find the average weight of 60 students of this class.
Solution :
Average weight of 25 students of group A = 51 kg
Thus, total of weight of 25 students
W1 = 51 x 25 = 1275 kg
Average weight of 35 students of group B = 54 kg
Thus, sum of weight of 35 students, W2 = 54 x 35 = 1890 kg
∴ Average weight of 60 students of class
= \(\frac { { W }_{ 1 }+{ W }_{ 2 } }{ 60 } \)
= \(\frac { 1275+1890 }{ 60 } =\frac { 3165 }{ 60 } \) = 52.75 kg
Thus, average wt. of 60 students = 52.75 kg

Class 10 Maths 17.1 Question 8.
The average of 5 numbers is 18, if one number is removed, then (RBSESolutions.com) average becomes 16. Find the removed number.
Solution :
Sum of 5 numbers = 18 × 5 = 90
Let removed number is x
Then, sum of four number = 16 × 4 = 64
⇒ Sum of four number + x = 90
⇒ 64 + x = 90
x = 90 – 64 = 26
Thus, fifth number = removed number = 26

Exercise 17.1 Class 10 RBSE Question 9.
The mean of 13 numbers is 24. If 3 is added to (RBSESolutions.com) each number, then what will the new mean?
Solution :
According to question,
Mean of 13 numbers = 24
Let 13 nos. are x1, x2, ….., x13
∵ Arithmetic mean
RBSE Solution Class 10 Maths Chapter 17 Ex 17.1 Measures Of Central Tendency
Thus, new mean will be (24 + 3) = 27.

Class 10 Maths Chapter 17.1 Question 10.
The average monthly salary of five employees of a (RBSESolutions.com) school is ₹3000. After retirement of one employee, average salary of remaining becomes ₹3200. What was the salary of fifth employee at the time of retirement ?
Solution :
Given : Average salary of 5 employees = ₹3000
Class 10 Maths Ex 17.1 Measures Of Central Tendency
⇒ Sum of salary of 5 employees = 3000 x 5 = ₹15000
Again, average salary of 4 employees = 3200
∵ Sum of salary of 4 employees = 3200 x 4 = ₹12800
Thus, salary of fifth employee = Sum of salary of 5 employees – Sum of salary of 4 employees
= ₹15000 – ₹12800 = ₹2200

We hope the given RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Ex 17.1 will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 17 Measures of Central Tendency Exercise 17.1, drop a comment below and we will get back to you at the earliest.