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RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Ex 17.5

March 25, 2019 by Veer Leave a Comment

RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Ex 17.5 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 17.5.

Board RBSE
Textbook SIERT, Rajasthan
Class Class 10
Subject Maths
Chapter Chapter 17
Chapter Name Measures of Central Tendency
Exercise Exercise 17.5
Number of Questions Solved 4
Category RBSE Solutions

Rajasthan Board RBSE Class 10 Maths Chapter 17 Measures of Central Tendency Ex 17.5

Question 1.
Astudent scored marks 46% in English, 67% ¡n Maths, 53% in Hindi, 72% in HLstory and 58% in Economics. As (RBSESolutions.com) compared to other subject weightage given to Mathematics, then find weightage mean marks of student.
Solution :
As per question
x1 = 46, x2 = 67, x3 = 53, x4 = 72, x5 = 58
Let weightage of Hindi, History and Economics = x
Then weightage of Maths = 2x
and of English = 3x
So, here w1 = 3x, w2 = 2x, w3 = x, w4 = x, w5 = x,
Then, weightage arithmetic mean
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Ex 17.5

RBSE Solutions

Question 2.
Two candidates A and B scored following marks in (RBSESolutions.com) different subject for entrance examination in Business School whose weightage are given alongwith:
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Ex 17.5
By finding weightage mean find from A and B which is more capable?
Solution :
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Ex 17.5
Mean weightage of A = \(\frac { 692 }{ 8 }\) = 86.5
Again
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Ex 17.5
Mean weightage of B = \(\frac { 698 }{ 8 }\) = 87.25
∵ Weightage mean of A is less than that of B, so B is more capable.

Question 3.
Marks obtained by a student in Mathematics in (RBSESolutions.com) three monthly tests are 85, 60 and 75 resp. And in annual exams begot 95 marks. Weightage of monthly tests are same where as weightage of annual exams is double that of monthly exams. Find the mean weightage of marks in Mathematics.
Solution :
According to question, x1 = 85, x2 = 60, x3 = 75, x4 = 95
and w1 = 1, w2 = 1, w3 = 1, w4 = 2,
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Ex 17.5

Question 4.
There are 45 students is a class in which 15 are girls. Average (RBSESolutions.com) weight of girls is 45 kg and of boys 52 kg. Find average weight of one student.
Solution :
According to question, average weight of 15 girls = 45 kg.
So, \(\bar { X }=\quad \frac { { \Sigma x }_{ i } }{ n } \)
⇒ 45 = \(\frac { { \Sigma x }_{ i } }{ 15 } \)
⇒ Σxi = 45 × 15 = 675
Total weight of 15 girls Σxi = 675 kg
Average weight of 30 boys = 52 kg

So, \(\bar { Y }=\quad \frac { { \Sigma y }_{ i } }{ 30 } \)
⇒ 52 = \(\frac { { \Sigma y }_{ i } }{ 30 } \)
⇒ Σyi = 52 × 30 = 1560
Total weight (RBSESolutions.com) of boys Σyi = 1560 kg
Average weight of 45 students
Σxi + Σyi = 675 + 1560 = 2235
Average weight of one student = 2235/45 = 49.67 kg.

RBSE Solutions

We hope the given RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Ex 17.5 will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 17 Measures of Central Tendency Exercise 17.5, drop a comment below and we will get back to you at the earliest.

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