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

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.

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

Ex 17.5 Class 10 RBSE 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

Exercise 17.5 Class 10 RBSE 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:

By finding weightage mean find from A and B which is more capable?
Solution :

Mean weightage of A = \(\frac { 692 }{ 8 }\) = 86.5
Again

Mean weightage of B = \(\frac { 698 }{ 8 }\) = 87.25
∵ Weightage mean of A is less than that of B, so B is more capable.

RBSE Class 10 Maths Chapter 17.5 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 17.5 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 = \(\frac { { \Sigma x }_{ i } }{ 15 } \)
⇒ Σx_i = 45 × 15 = 675
Total weight of 15 girls Σx_i = 675 kg
Average weight of 30 boys = 52 kg

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

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.