How to Find Class 10 CBSE Registration Number?

Board exams are perhaps the most important aspects of any budding student’s life. When we talk of Board exams, many worries, formalities, tasks, and problems naturally jump in – mostly due to the large scale applicability to every candidate who’s appearing in a particular year. Some of the most important details for any student are the registration number, roll number, unique ID/ registration number, examination centre, and so on. Knowing the course and syllabus is equally essential, but without a unique id and the CBSE registration number, no student can appear for their Boards in a particular year.

If you are someone appearing for class 10 CBSE board exams this year, you might be finding some difficulty in keeping it all together. More so, due to unprecedented situations. But worry no more, this article right here will guide you on everything you need to know about finding your CBSE registration number.

What is the CBSE registration number?

CBSE Registration Number as with other boards is an 8-digit number with the first four digits representing the centre code (usually your school code assigned by CBSE) and the last four numbers follow the roll number depending upon the number of candidates in a school. This number is assigned after the CBSE verification and registration in class 9 that every aspirant has to complete to be eligible for Boards.

This is usually the first step into confirming yourself as a CBSE candidate for a particular session, and naturally, completing it properly should be a priority. This is so because everything else directly or indirectly depends on whether you are correctly registered in the initial stage with the Board. The registration number can be found on the registration card that you get, and this is not available on the admit card for exams.

CBSE registrations for regular and private candidates

Usually, there are two kinds of candidates appearing for CBSE Board exams each year – Regular and Private candidates. The method for carrying out the registration process is different for both these cases.

Regular candidates 

A student enrolled in any CBSE affiliated school is known as a Regular Candidate.

The CBSE registration for Class 10 regular candidates is carried by their respective schools. Schools need to enter the number of students for whom registration is to be done. This must be decided before commencing the CBSE online registration. Each and every detail has to be very carefully entered with precision and verification as schools are not allowed to change the information later.

Private candidates

Students who do not attend the regular classes in schools and study on their own without any guidance from their school are registered as private candidates. The only difference is that their certificate will be different from the regular candidates. But the percentage of marks and marking scheme given will be the same as any regular student.

Compartment in Class 10 Board Exams

Students declared as ESSENTIAL REPEAT in their 2020 exams, candidates who have been placed in COMPARTMENT in the main exam, candidates with COMPARTMENT in September 2020, candidates declared FAIL in 2015, 2016, 2017, 2018, 2019, candidates of 2020 who wish to appear for improving their performance in any or all subjects, candidates of 2015 or after, willing to appear in an Additional subject, women candidates (bonafide resident of the National Capital Territory of Delhi and has attained the age for appearing in Class X), disabled candidates (bonafide resident of the National Capital Territory of Delhi and has attained the age for appearing in Class X) are all eligible to register themselves as private candidates.

The detailed steps of registration can be found for both Private and Regular candidates on the CBSE official website.

CBSE registrations 2021 – All you need to know

The essential details to be filled in the registration forms include:

  • Name of the candidate
  • Date of birth
  • Father name
  • Mother name
  • Gender
  • Caste category
  •  Email ID
  • Address
  • Photo
  • Signature
  • Mobile number

For CBSE Registration for the academic year 2020-21, The Central Board of Secondary Education has extended the registration dates for all private candidates. An official notification regarding the extension of registration dates was released by the Board on its official website – cbse.nic.in. According to the new update, those candidates who were not able to register can do so from 5th December 2020 to 9th December 2020.

All those candidates who are already registered can make necessary corrections on their registration form from 10th December 2020 to 14th December 2020. Candidates who wish to apply must visit the official CBSE website.

FAQ’s (Frequently Asked Questions) on finding class 10 CBSE registration number

Q 1. Is it necessary to pay a fee for registration?

Answer. Each candidate needs to pay a minimal fee to register.

Q 2. Is it possible to register past the last date?

Answer. No. Students must be very careful to complete the procedure by the last date.

Q 3. Can forms be corrected afterwards?

Answer. There is no provision for correction. Candidates and schools need to be very mindful.

MCQ Questions for Classes 6, 7, 8, 9, 10, 11, 12 | Multiple Choice Questions and Answers for All Subjects

If you are looking out for Multiple Choice Questions of Classes 6-12 across the web then you have landed on the right page. This article comprises various objective-type questions that make your learning process much simpler.  Practicing the MCQ Questions and Answers will give you an extra edge in your preparation as they are quite scoring. You can download the Objective Questions of Classes 6, 7, 8, 9, 10, 11, 12 available via quick links and practice offline too. Attempt History MCQ Questions Quiz and Computer MCQ Questions for Classes 1 to 12 available here and test your understanding of the topics.

Classwise MCQs for All Subjects with Answers

Understand all the topics thoroughly in NCERT Books by availing the Multiple Choice Question and Answers as they give you an overview of all topics. Try to Answer the Quiz Questions available here on your own and then verify with our answers to assess your preparation level. Allot time to the areas in which you are lagging and work accordingly. Practicing consistently will help you learn how to answer the questions in exams.

MCQ Questions for Class 10 with Answers

MCQ Questions for Class 9 with Answers

MCQ Questions for Class 8 with Answers

MCQ Questions for Class 7 with Answers

MCQ Questions for Class 6 with Answers

MCQ Questions for Class 12 with Answers

MCQ Questions for Class 11 with Answers

Download Objective Type Questions All Classes PDF Download

Solve the MCQ Questions of All Subjects on your own and test your understanding of important topics. All the MCQs are as per the Latest CBSE Syllabus Guidelines and you can score good grades in exams. You need not worry about the accuracy of the answers provided as they are given to you after extensive research by subject experts. Answer the Questions available here and remediate the knowledge gap accordingly and excel in your board exams or any other competitive exams.

Tips and Tricks to solve Multiple Choice Questions

Below are the simple tips and tricks to keep in mind while answering the MCQ Question and Answers. Use these simple hacks and answer like a pro in your CBSE Board Exams. They are in the following fashion

  • Try to read the Objective Questions carefully and answer them in your mind.
  • Eliminate the Wrong Answers and Choose the Best Answer.
  • Answer the known Multiple Choice Quiz Questions first so that you don’t run out of time in your exams.
  • Apply as many shortcuts as possible so that you can clear the exam with flying colors.
  • Take short breaks in between if possible.

FAQs on Multiple Choice Questions with Answers

  1. Where can I find Classwise MCQ Questions for all subjects?

You can get Class Wise Multiple Choice Questions for all subjects organized efficiently on our page.

  1. How to Learn MCQs Fastly?

You can learn MCQs fastly by simply practicing them as many times as possible as it is the only key to achieve success.

  1. Is there any Portal that Provides Subject Wise Objective Type Questions for Classes 6 to 12?

Aplustopper.com is a portal that provides the Multiple Choice Questions and Answers for Classes 6 to 12 all in one place.

  1. How do I Prepare for MCQs in CBSE Board Exams?

Reading alone will not help you reach your goals and you need to continuously practice the sample papers on a day-to-day basis so that it’s easy for you while solving them in actual CBSE Board Exams.

Conclusion

We as a team believe the knowledge shed above regarding the MCQ Questions with Answers for All Subjects has been useful in resolving your queries to the possible extent. If you have any doubts do contact us and we will get back to you. Keep in touch with our site to avail MCQ Questions, NCERT Books, NCERT Solutions, Previous Papers, etc.

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1

Are you looking for the best Maths NCERT Solutions Chapter 3 Ex 3.1 Class 10? Then, grab them from our page and ace up your preparation for CBSE Class 10 Exams.

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1 are part of NCERT Solutions for Class 10 Maths. Here are we have given Chapter 3 Pair of Linear Equations in Two Variables Class 10 NCERT Solutions Ex 3.1. 

NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables are provided here with simple step-by-step explanations. These solutions for Pair Of Linear Equations In Two Variables are extremely popular among class 1 0 students for Math Pair Of Linear Equations In Two Variables Solutions come handy for quickly completing your homework and preparing for exams.

BoardCBSE
TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 3
Chapter NamePair of Linear Equations in Two Variables
ExerciseEx 3.1
Number of Questions Solved3
CategoryNCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1

Page No: 44

Question 1. Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically.

Solution:
Let the present age of Aftab and his daughter be x and y respectively.
Seven years ago,
Age of Aftab = x – 7
Age of his daughter = y – 7
According to the given condition,
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 1
Three years hence,
Age of Aftab = x + 3
Age of his daughter = y + 3
According to the given condition,
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 2
Thus, the given conditions can be algebraically represented as:
x – 7y = -42
x – 3y = 6
x – 7y = -42 ⇒ × = -42 + 7y
Three solutions of this equation can be written in a table as follows:

x-707
y567

× -3y = 6 ⇒ × = 6 + 3y
Three solutions of this equation can be written in a table as follows:

x630
y0-1-2

The graphical representation is as follows:
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 3

Concept Insight: In order to represent a given situation mathematically, first see what we need to find out in the problem. Here, Aftab and his daughter’s present age needs to be found so, so the ages will be represented by variables x and y. The problem talks about their ages seven years ago and three years from now. Here, the words ‘seven years ago’ means we have to subtract 7 from their present ages, and ‘three years from now’ or ‘three years hence’ means we have to add 3 to their present ages. Remember in order to represent the algebraic equations graphically the solution set of equations must be taken as whole numbers only for the accuracy. Graph of the two linear equations will be represented by a straight line.

Question 2. The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys another bat and 3 more balls of the same kind for Rs 1300. Represent this situation algebraically and geometrically.

Solution:
Let the cost of a bat and a ball be Rs x and Rs y respectively.
The given conditions can be algebraically represented as:
3x + 6y = 3900
x + 2y = 1300
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 4
Three solutions of this equation can be written in a table as follows:

x300100-100
y500600700

x + 2y = 1300 ⇒ x = 1300 – 2y
Three solutions of this equation can be written in a table as follows: 

x300100-100
y500600700

The graphical representation is as follows: 
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 5

Concept Insight:Cost of bats and balls needs to be found so the cost of a ball and bat will be taken as the variables. Applying the conditions of total cost of bats and balls algebraic equations will be obtained. Then, in order to represent the obtained equations graphically take atleast three ordered pairs on both the equations in order to avoid any computational errors.

Question 3. The cost of 2 kg of apples and 1kg of grapes on a day was found to be Rs 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs 300. Represent the situation algebraically and geometrically.

Solution:
Let the cost of 1 kg of apples and 1 kg grapes be Rs. x and Rs. y.
The given condtions can be algebraically represented as:
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 6
Three solutions of this equation can be written in a table as follows:

x506070
y604020

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 7
Three solutions of this equation can be written in a table as follows: 

x708075
y10-100

The graphical representation is as follows: 
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 8

Concept Insight: cost of apples and grapes needs to be found so the cost of 1 kg apples and 1 kg grapes will be taken as the variables. From the given conditions of collective cost of apples and grapes, a pair of linear equations in two variables will be obtained. Then, in order to represent the obtained equations graphically, take the values of variables as whole numbers only. Since these values are large so take the suitable scale like 1 cm = 20.

We hope the NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1 help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 10 Maths

NCERT Class 10 Maths

NCERT Solutions for Class 10 Maths

NCERT Solutions for Class 10 Maths: As a student, it becomes difficult to solve some of the questions given in the textbook. There are solved examples at the start of the chapter that can you help understand the method to solve the question. But many times the practice questions can be very tiresome. There have been many cases where students have found it very difficult to solve the practice questions. One particular subject that has this problem is maths. Thus, we are providing you with NCERT solutions for class 10 Maths. This article will help you with all the chapters and their solutions so that you can understand them better and prepare well for the exams.

Teachers at APlusTopper.com solved  . Class 10 Maths NCERT Solutions are of great help in completing homework and clearing your doubts. All these from NCERT book For class 10 Maths have been solved keeping in mind the guidelines issued by the CBSE.

NCERT Solutions for Class 10 Maths

Class 10 is very important for students as this is the first time students will be appearing for a board exam. This also makes up for a whole new experience for the students. Thus, students become conscious and try and solve every question given in the textbook. So, a thorough knowledge of all the chapters in the textbook becomes a priority for everyone. This NCERT solution provided for class 10 maths can help you with this. It will also help you prepare the subject in very less time. We will try to go in detail for each and every chapter. Many of the topics for class 10 areas an extension to class 9. Thus, you don’t need to worry too much about the content. Rather just practice all the questions given in the textbook.

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

NCERT Solutions for Class 10 Maths Chapter 5 Aritthmetic Progressions

NCERT Solutions for Class 10 Maths Chapter 6 Triangles

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Tringonmetry

NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry

NCERT Solutions for Class 10 Maths Chapter 10 Circles

NCERT Solutions for Class 10 Maths Chapter 11 Constructions

NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes

NCERT Solutions for Class 10 Maths Chapter 14 Statistics

NCERT Solutions for Class 10 Maths Chapter 15 Probability

has also been done in accordance with the marking scheme usually followed by the CBSE Board. AplusToper.com sincerely hope that NCERT Class 10 Maths Solutions will be an effective tool in the hands of the students preparing for their  CBSE Board Exams.

Class 10 Maths Real Numbers Solutions

Most of this ncert maths chapter follows the content that you have learned in class 9. The important properties for this chapter are covered in section 1.2 and 1.3. These properties are the fundamental theorem of arithmetic and Euclid’s division algorithm. Besides this, you will relearn the irrational numbers in section 1.4. Also, decimal expressions of rational numbers are included in section 1.5. There are a plethora of applications that are associated with Euclid’s division algorithm. Mostly this algorithm deals with the divisibility of the integers. Also, it deals with the divisibility of the integers. This chapter 1 in NCERT will be helpful for calculating the HCF of two positive integers.

Class 10 Maths Chapter 2 Polynomials Solutions

There are a total of four exercises in polynomials. The last exercise of the four is optional. Every exercise given in this chapter is solved using different methods. You need to find the number of zeroes in p(x) polynomial in the first exercise. While there are two questions in the second exercise. As mentioned above, you need to find a relationship between the coefficient and zeroes in the first exercise. And in the second you have to find a quadratic polynomial. There are a total of five questions in the third exercise. In this, you need to divide polynomials and thereby obtain the number of zeroes in the polynomials. The optional exercise which is the last one also has five questions in it.

Class 10 Maths Chapter 3 Linear Equation in Two Variables Solutions

The basics of this topic were discussed in class 9. While in class 10, this topic goes into more complex scenarios. There are a total of seven exercises in this chapter for students to practice. Some of the important topics that are covered in this chapter are a pair of linear equations in 2 variables, graphical representations, solving a pair of linear equations using algebraic methods, and elimination method. Also, some of the topics in this chapter are substitution method, cross-multiplication method and graphical method for the solution of linear equations.

Class 10 Maths Chapter 4 Quadratic Equations Solutions

In this chapter, you will study quadratic equations in depth. Furthermore, you will study to find the roots of quadratic equations. There are also some of the daily life applications of quadratic equations in this chapter. A total of 4 exercises are there in this chapter. There are two questions in the first exercise where you need to find whether the equations are quadratic or not. In the second you will be given daily life examples that need to be converted into quadratic equations. Also, in the second exercise, you are asked to find the roots of quadratic equations using factorization. The main topics in this chapter are solutions by factorization and solutions of a quadratic equation using the square.

Class 10 Maths Chapter 5 Arithmetic Progression Solutions

This chapter has some daily life examples and then it gradually moves towards the complexities of the chapter. There are four exercises described in the chapter. These exercises are divided further into various questions where you need to find a term. In this chapter, you will learn how to find the nth term. Alongside you will also find the sum of n consecutive terms.

Class 10 Maths Chapter 6 Triangles Solutions

This is one of the biggest chapters in CBSE Syllabus. This is because it has six exercises on it. There are different questions in each exercise that are based on the properties of the triangle. It consists of 9 theorems in total. Thus, this is one of the important chapters in maths for class 10.

Class 10 Maths Chapter 7 Coordinate Geometry Solutions

The questions in this chapter are mostly based on finding the distances between two points. There are coordinates provided along with the triangle. The area of the triangle is found using these three points. This chapter has a total of four exercises in it.

Class 10 Maths Chapter 8 Trigonometry Solutions

This chapter is mainly on finding the trigonometric ratios for a side of the right angled triangle. These ratios are called trigonometric ratios. There are a total of 4 exercises in this chapter. The main topics covered in this chapter are trigonometric ratios, trigonometric identities, trigonometric ratios for complementary angles, and for specific angles.

Class 10 Maths Chapter 9 Applications of Trigonometry Solutions

There is only one exercise in this chapter and the questions are mostly related to the practical applications of trigonometry. You will also study interesting ways to use trigonometry in real life.

Class 10 Maths Chapter 10 Circles Solutions

There are only two exercises in a circle. First one has the basic questions related to circles. While in the second one you need to find various ways to prove the given equations.

Class 10 Maths Chapter 11 Constructions Solutions

In this chapter, you have found various constructions like constructions of tangents to a circle, division of line segment, etc. It has two exercises and the questions are related to them only.

Class 10 Maths Chapter 12 Areas of a Circle Solutions

In this chapter, you need to find the areas of a circle. Furthermore, you are also required to find the areas of ‘special’ parts of a circle. This chapter has three exercises.

Class 10 Maths Chapter 13 Surface Area and Volume Solutions

This chapter deals with problems of finding surface areas and volumes of various cubes, cylinder, and cuboid. There are a total of five exercises in this for students to practice.

Class 10 Maths Chapter 14 Statistics Solutions

This chapter mainly deals with finding the mean, median, and mode. The data given can be grouped or ungrouped. For your practice, there are a total of 4 exercises. There are also questions related to cumulative frequency in this chapter. Furthermore, this chapter covers topics like cumulative frequency distribution and drawing cumulative frequency curves.

Class 10 Maths Chapter 15 Probability Solutions

The chapter deals mainly with finding the probability of daily situations. This is mostly related to dices and coins.

We have tried to give you an overview of what NCERT chapters have. There are many exercises in the book for you to practice. You can follow the website for NCERT class 10 maths solutions and other updates. Hope given NCERT Solutions for Class 10 Maths helpful to finish your Home Work. If you find these solutions helpful, Please share with your friends.

RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C

RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions

Exercise 11C

Question 1.
Solution:
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 1.1
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 1.2
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 1.3
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 1.4

Question 2.
Solution:
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 2.1
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 2.2
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 2.3
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 2.4
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 2.5

Question 3.
Solution:
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 3.1

Question 4.
Solution:
Sn = 3n² + 6n
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 4.1

Question 5.
Solution:
Sn = 3n² – n
S1 = 3(1)² – 1 = 3 – 1 = 2
S2 = 3(2)² – 2 = 12 – 2 = 10
T2 = 10 – 2 = 8 and
T1 = 2
(i) First term = 2
(ii) Common difference = 8 – 2 = 6
Tn = a + (n – 1) d = 2 + (n – 1) x 6
= 2 + 6n – 6 = 6n – 4

Question 6.
Solution:
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 6.1
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 6.2
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 6.3

Question 7.
Solution:
Let a be first term and d be the common difference of an AP.
Since, we have,
am = a + (m – 1) d = \(\frac { 1 }{ n }\) …(i)
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 7.1
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 7.2

Question 8.
Solution:
AP is 21, 18, 15,…
Here, a = 21,
d = 18 – 21 = -3,
sum = 0
Let number of terms be n, then
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 8.1
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 8.2

Question 9.
Solution:
AP is 9, 17, 25,…
Here, a = 9, d = 17 – 9 = 8
Sum of terms = 636
Let number of terms be n, then
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 9.1
Which is not possible being negative and fraction.
n = 12
Number of terms = 12

Question 10.
Solution:
AP is 63, 60, 57, 54,…
Here, a = 63, d = 60 – 63 = -3 and sum = 693
Let number of terms be n, then
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 10.1
22th term is zero.
There will be no effect on the sum.
Number of terms are 21 or 22.

Question 11.
Solution:
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 11.1
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 11.2
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 11.3

Question 12.
Solution:
Odd numbers between 0 and 50 are 1, 3, 5, 7, 9, …, 49
Here, a = 1, d = 3 – 1 = 2, l = 49
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 12.1
= \(\frac { 25 }{ 2 }\) x 50 = 25 x 25 = 625

Question 13.
Solution:
Numbers between 200 and 400 which are divisible by 7 will be 203, 210, 217,…, 399
Here, a = 203, d = 7, l = 399
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 13.1

Question 14.
Solution:
First forty positive integers are 0, 1, 2, 3, 4, …
and numbers divisible by 6 will be 6, 12, 18, 24, … to 40 terms
Here, a = 6, d = 12 – 6 = 6, n = 40 .
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 14.1

Question 15.
Solution:
First 15 multiples of 8 are 8, 16, 24, 32, … to 15 terms
Here, a = 8, d = 16 – 8 = 8 , n = 15
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 15.1

Question 16.
Solution:
Multiples of 9 lying between 300 and 700 = 306, 315, 324, 333, …, 693
Here, a = 306, d = 9, l = 693
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 16.1

Question 17.
Solution:
Three digit numbers are 100, 101, …, 999
and numbers divisible by 13, will be 104, 117, 130, …, 988
Here, a = 104, d = 13, l = 988
Tn (l) = a + (n – 1) d
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 17.1

Question 18.
Solution:
Even natural numbers are 2, 4, 6, 8, 10, …
Even natural numbers which are divisible by 5 will be
10, 20, 30, 40, … to 100 terms
Here, a = 10, d = 20 – 10 = 10, n = 100
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 18.1

Question 19.
Solution:
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 19.1
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 19.2

Question 20.
Solution:
Let a be the first term and d be the common difference of the AP, then
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 20.1
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 20.2

Question 21.
Solution:
Let a be the first term and d be the common difference, then
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 21.1
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 21.2

Question 22.
Solution:
Let a be the first term and d be the common difference, then
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 22.1

Question 23.
Solution:
In an AP
a = 17, d = 9, l = 350
Let number of terms be n, then
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 23.1

Question 24.
Solution:
Let a be the first term, d be the common difference, then
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 24.1

Question 25.
Solution:
In an AP, let a be the first term and d be the common difference, then
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 25.1

Question 26.
Solution:
Let a be the first term and d be the common difference of an AP, then
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 26.1

Question 27.
Solution:
Let a be first term and d be the common difference, then
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 27.1
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 27.2

Question 28.
Solution:
Let first term be a and d be the common difference in an AP, then
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 28.1
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 28.2

Question 29.
Solution:
Let a be the first term and d be the common difference of an AP, then
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 29.1
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 29.2

Question 30.
Solution:
Let a1 and a2 be the first terms of the two APs and d be the common difference, then
a1 = 3, a2 = 8
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 30.1

Question 31.
Solution:
Let a be the first term and d be the common difference, then
S10 = -150
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 31.1
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 31.2

Question 32.
Solution:
Let a be the first term and d be the common difference, then
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 32.1
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 32.2

Question 33.
Solution:
Let a be the first term and d be the common difference of an AP, then
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 33.1
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 33.2

Question 34.
Solution:
(i) AP is 5, 12, 19,… to 50 terms
Here, a = 5, d = 12 – 5 = 7, n = 50
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 34.1
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 34.2

Question 35.
Solution:
Let a1, a2 be the first term and d1, d2 be common difference of the two AP’s respectively.
Given, ratio of sum of first n terms =
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 35.1
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 35.2

Question 36.
Solution:
Let a be the first term and d be the common difference of an AP.
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 36.1
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 36.2

Question 37.
Solution:
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 37.1

Question 38.
Solution:
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 38.1
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 38.2

Question 39.
Solution:
AP is -12, -9, -6,…, 21
Here, a = -12, d = -9 – (-12) = -9 + 12 = 3, l = 21
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 39.1

Question 40.
Solution:
S14 = 1505
Let a be the first term and d be the common difference, then
a = 10
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 40.1

Question 41.
Solution:
Let a be the first term and d be the common difference of an AP, then
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 41.1

Question 42.
Solution:
In the school, there are 12 classes, class 1 to 12 and each class has two sections.
Each class plants double of the class
i.e. class 1 plants two plant, class 2 plants 4 plants, class 3 plants 6 plants, and so on.
So, total plants will be for each class each sections = 2 + 4 + 6 + 8 … + 24
Here, a = 2, d = 2, l = 24, n = 12
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 42.1
Each class has. two sections.
Plants will also be doubt
i.e. Total plants = 156 x 2 = 312

Question 43.
Solution:
In a potato race,
Bucket is at 5 m from first potato and then the distance between the two potatoes is 3m.
There are 10 potatoes.
The player, pick potato and put it in the bucket one by one.
Total distance in m to be covered, for 1st, 2nd, 3rd, … potato.
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 43.1

Question 44.
Solution:
Total number of trees = 25
Distance between them = 5 m in a line.
There is a water tank which is 10 m from the first tree.
A gardener waters these plants separately.
Total distance for going and coming back
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 44.1
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 44.2

Question 45.
Solution:
Total sum = ₹ 700
Number of cash prizes = 7
Each prize in ₹ 20 less than its preceding prize.
Let first prize = ₹ x
Then second prize = ₹ (x – 20)
Third prize = ₹ (x – 40) and so on.
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 45.1

Question 46.
Solution:
Total savings = ₹ 33000
Total period = 10 months
Each month, a man saved ₹ 100 more than its preceding of month.
Let he saves ₹ x in the first month.
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 46.1

Question 47.
Solution:
Total debt to be paid = ₹ 36000
No. of monthly installments = 40
After paying 30 installments, \(\frac { 1 }{ 3 }\) of his debt left
i.e. ₹ 36000 x \(\frac { 1 }{ 3 }\) = ₹ 12000
and amount paid = ₹ 36000 – ₹ 12000 = ₹ 24000
Monthly installments are in AP.
Let first installment = x
and common difference = d
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 47.1
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 47.2

Question 48.
Solution:
A contractor will pay the penalty for not doing the work in time.
For the first day = ₹ 200
For second day = ₹ 250
For third day = ₹ 300 and so on
The work was delayed for 30 days
Total penalty, he paid
200 + 250 + 300 + ….. to 30 terms
Here, a = 200, d = 50, n = 30
Total sum = \(\frac { n }{ 2 }\) [2a + (n – 1) d]
= \(\frac { 30 }{ 2 }\) [2 x 200 + (30 – 1) x 50]
= 15 [400 + 29 x 50]
= 15 [400 + 1450]
= 15 x 1850 = ₹ 27750

Question 49.
Solution:
Child will put 5 rupee on 1st day, 10 rupee (2 x 5 rupee)
on 2nd day, 15 rupee (3 x 5 rupee) on 3rd day etc.
Total saving = 190 coins = 190 x 5 = 950 rupee
The above problem can be written as Arithmetic Progression series
5, 10, 15, 20, ……
with a = 5, d = 5, Sn = 950
Let n be the last day when piggy bank become full.
RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C 49.1
⇒ n (n + 20) – 19 (n + 20) = 0
⇒ (n + 20) (n – 19) = 0
⇒ n + 20 = 0 or n – 19 = 0
⇒ n = -20 or n = 19
cannot be negative, hence n = 19
She can put money for 19 days.
Total saving is 950 rupees.

Hope given RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions are helpful to complete your math homework.

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