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

x_{1} = 46, x_{2} = 67, x_{3} = 53, x_{4} = 72, x_{5} = 58

Let weightage of Hindi, History and Economics = x

Then weightage of Maths = 2x

and of English = 3x

So, here w_{1} = 3x, w_{2} = 2x, w_{3} = x, w_{4} = x, w_{5} = x,

Then, weightage arithmetic mean

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 = [latex]\frac { 692 }{ 8 }[/latex] = 86.5

Again

Mean weightage of B = [latex]\frac { 698 }{ 8 }[/latex] = 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, x_{1} = 85, x_{2} = 60, x_{3} = 75, x_{4} = 95

and w_{1} = 1, w_{2} = 1, w_{3} = 1, w_{4} = 2,

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, [latex]\bar { X }=\quad \frac { { \Sigma x }_{ i } }{ n } [/latex]

⇒ 45 = [latex]\frac { { \Sigma x }_{ i } }{ 15 } [/latex]

⇒ Σx_{i} = 45 × 15 = 675

Total weight of 15 girls Σx_{i} = 675 kg

Average weight of 30 boys = 52 kg

So, [latex]\bar { Y }=\quad \frac { { \Sigma y }_{ i } }{ 30 } [/latex]

⇒ 52 = [latex]\frac { { \Sigma y }_{ i } }{ 30 } [/latex]

⇒ 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.

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