NCERT Solutions - If you are a CBSE student of class 6, 7, 8, 9, 10, 11 or 12 looking to improve your marks, you can do so by referring to the solutions of NCERT questions of all exercises from the textbook. While you will be taught the chapter in school, how to use the concepts and learn the NCERT solutions will be a challenge, especially with some of the tricky questions. CBSE NCERT solutions will give you an idea of how to solve the questions and arrive at the answers for all the exercises that are given in the textbook. This means you need to just check the solutions of NCERT exercises for CBSE subjects and chapters to help you understand and help get more marks in your exam. The NCERT class wise solutions for class 6 to 12 are all listed on this page with links to each subject, chapter wise that will open up the answers to all questions asked in the textbook along with detailed NCERT chapter-wise solutions. Check it out by clicking the class you are in and then the subject followed by the chapter.
The stepping stone and foundation to higher studies in this class 12th. What makes it different from the other board exams is that you may need to plan for the entrance exams to pursue a degree, especially if you have taken up science or maths. Preparation includes board and entrance exams making it a stressful period. Now, with help in the form of CBSE NCERT solutions of class 12 for each exercise and chapter, learning will become easy more so when the solutions are by experts of the respective fields. The links are given below lead to the NCERT chapter-wise solutions for class 12 including the 13 chapters in maths, 15 chapters in Physics, 16 chapters each in chemistry and biology. Click on these links and know the solutions for the questions asked to prepare well for your board exams.
While NCERT class 11 maths is sort of unified, science is divided into physics, chemistry, and biology. The 16 chapters of maths are pretty scoring only if you know how to solve the problems. This is also the case with science. What better than expert solutions of NCERT class 11 to help out with these? The CBSE NCERT solutions for class 11 science include 15 chapters in physics,14 in chemistry and 22 in Biology respectively. They include detailed explanations of concepts, how to solve any problem and alternate ways to do so. Scoring in the exams shouldn't be a problem if you follow the given links and check the NCERT solutions given.
Every CBSE student of class 10 knows that there are 16 maths chapters to solve and 15 science chapters to study and understand. Having the expert solutions of NCERT class 10 textbook would be a boon to every student preparing for the board exams. The solutions for NCERT class 10 CBSE science and maths include the concept explanation, step marking and alternate ways of solving the problems. Just click on the given links to move and check out the NCERT chapter-wise solutions for class 10.
Entering this class becomes a crucial phase in life as this is where the actual learning for entrance exams like JEE Main, NEET for engineering and medical courses starts. This is the foundation phase for topics of science and maths for the next four years to come. It is at such a time that expert provided solutions for NCERT class 9 for Maths and science which are scoring subjects come in handy. These can be yours to use if you click on the links mentioned below. Use these to understand how to solve the questions asked and also get alternate solutions for NCERT class 9 maths and science. The NCERT solutions for class 9 science and maths chapters for each exercise and chapter are provided herewith.
Students studying in class 8 have to study 16 maths chapters and 18 science chapters. Class 8 NCERT solutions science and maths provided hereby subject experts can aid in solving the exercises faster and in an easy manner. To check them out, just click on the links below and the solutions of NCERT class 8 CBSE textbook will be displayed.
With a total of 15 chapters in class 7 maths and 18 chapters in science, it is important that you ace your game. Use the CBSE NCERT solutions provided here for these subjects by experts and start scoring well on your exams. Again, just click on the chapter you wish to check out and check the solutions to NCERT class 7 CBSE maths and science here.
If you are in class 6, this is very apt for you to score well. Maths and Science being very scoring subjects, class 6 NCERT chapter-wise solutions will be the right tool to help you to solve all problems. Class 6 maths has 14 chapters while class 6 science has 16 chapters and the CBSE NCERT subject wise solutions for these have been listed in detail below. All you need to do is to click on the link provided and that's it. The solutions to NCERT exercises for CBSE class 6 will be displayed according to the subject you choose and the chapter as well. check them out.
NCERT Solutions - If you are a CBSE student of class 6, 7, 8, 9, 10, 11 or 12 looking to improve your marks, you can do so by referring to the solutions of NCERT questions of all exercises from the textbook. While you will be taught the chapter in school, how to use the concepts and learn the NCERT solutions will be a challenge, especially with some of the tricky questions. CBSE NCERT solutions will give you an idea of how to solve the questions and arrive at the answers for all the exercises that are given in the textbook. This means you need to just check the solutions of NCERT exercises for CBSE subjects and chapters to help you understand and help get more marks in your exam. The NCERT class wise solutions for class 6 to 12 are all listed on this page with links to each subject, chapter wise that will open up the answers to all questions asked in the textbook along with detailed NCERT chapter-wise solutions. Check it out by clicking the class you are in and then the subject followed by the chapter.
The stepping stone and foundation to higher studies in this class 12th. What makes it different from the other board exams is that you may need to plan for the entrance exams to pursue a degree, especially if you have taken up science or maths. Preparation includes board and entrance exams making it a stressful period. Now, with help in the form of CBSE NCERT solutions of class 12 for each exercise and chapter, learning will become easy more so when the solutions are by experts of the respective fields. The links are given below lead to the NCERT chapter-wise solutions for class 12 including the 13 chapters in maths, 15 chapters in Physics, 16 chapters each in chemistry and biology. Click on these links and know the solutions for the questions asked to prepare well for your board exams.
While NCERT class 11 maths is sort of unified, science is divided into physics, chemistry, and biology. The 16 chapters of maths are pretty scoring only if you know how to solve the problems. This is also the case with science. What better than expert solutions of NCERT class 11 to help out with these? The CBSE NCERT solutions for class 11 science include 15 chapters in physics,14 in chemistry and 22 in Biology respectively. They include detailed explanations of concepts, how to solve any problem and alternate ways to do so. Scoring in the exams shouldn't be a problem if you follow the given links and check the NCERT solutions given.
Every CBSE student of class 10 knows that there are 16 maths chapters to solve and 15 science chapters to study and understand. Having the expert solutions of NCERT class 10 textbook would be a boon to every student preparing for the board exams. The solutions for NCERT class 10 CBSE science and maths include the concept explanation, step marking and alternate ways of solving the problems. Just click on the given links to move and check out the NCERT chapter-wise solutions for class 10.
Entering this class becomes a crucial phase in life as this is where the actual learning for entrance exams like JEE Main, NEET for engineering and medical courses starts. This is the foundation phase for topics of science and maths for the next four years to come. It is at such a time that expert provided solutions for NCERT class 9 for Maths and science which are scoring subjects come in handy. These can be yours to use if you click on the links mentioned below. Use these to understand how to solve the questions asked and also get alternate solutions for NCERT class 9 maths and science. The NCERT solutions for class 9 science and maths chapters for each exercise and chapter are provided herewith.
Students studying in class 8 have to study 16 maths chapters and 18 science chapters. Class 8 NCERT solutions science and maths provided hereby subject experts can aid in solving the exercises faster and in an easy manner. To check them out, just click on the links below and the solutions of NCERT class 8 CBSE textbook will be displayed.
With a total of 15 chapters in class 7 maths and 18 chapters in science, it is important that you ace your game. Use the CBSE NCERT solutions provided here for these subjects by experts and start scoring well on your exams. Again, just click on the chapter you wish to check out and check the solutions to NCERT class 7 CBSE maths and science here.
If you are in class 6, this is very apt for you to score well. Maths and Science being very scoring subjects, class 6 NCERT chapter-wise solutions will be the right tool to help you to solve all problems. Class 6 maths has 14 chapters while class 6 science has 16 chapters and the CBSE NCERT subject wise solutions for these have been listed in detail below. All you need to do is to click on the link provided and that's it. The solutions to NCERT exercises for CBSE class 6 will be displayed according to the subject you choose and the chapter as well. check them out.
NCERT Solutions for Class 10 - We all know that class 10 is a very important threshold for choosing the career path you may lead. Whether you will take up science, arts or commerce will all depend upon your class 10th performance. To get good scores for entry to some of the best schools or junior colleges, it is vital to perform well in class 10 and this is where the solutions of NCERT for class 10 come in hand. To score good marks, you need to know CBSE NCERT solutions for class 10 as these are the answers to the questions asked in the exam. NCERT solutions are good tools or keys to help unlock your exam problems. These contain the step by step process on how to arrive at the NCERT solutions for class 10.
Maths and science are two subjects that are scoring only if you know the answers or rather know how to arrive at them. If you have access to the solutions of NCERT for class 10 for these subjects, you will have enough help to understand the chapters and reach your goal of scoring well.
Click on the link given above for the subject you wish to check out.
Go to the chapter given in the subject and check the solutions given therein
NCERT solutions for class 10 mathematics
The subject has around 15 chapters with many exercises that test you in the concepts taught and formulae to be used. The solutions of NCERT for class 10 cover all this and more. You will not just find that maths is now easier but also find the interest to do better in the final board exams.
NCERT solutions for class 10 science
This subject covers physics, chemistry and biology topics in 16 chapters. Taking the help of solutions of NCERT for class 10 science will be a good step as you will be in a position to be well conversant with the science. A strong foundation will help you in class 11 and 12 as well.
Also Read -
Stuck in a question? Unable to understand how to solve a problem? No worries. Ask our experts here and you will receive prompt help with it.
Happy learning!!!
", "added_on": "2019-07-08 12:34:13.261789", "updated_on": "2020-02-24 10:37:23.562592", "featured_image": "img/featured/article-126282019-07-08-12-34-17.jpg", "stream_id": "18", "ec_article_id": "0", "ec_view_count": "0", "is_announcement": "0", "is_course": "0", "is_featured": "0", "is_news": "0", "exam_attribute_id": "0", "author_id": "46266", "editor_id": "0", "exam_id": "1388", "ec_data": "", "main_article": "0", "migrated": "0", "description": "NCERT Solutions for Class 10 - We all know that class 10 is a very important threshold for choosing the career path you may lead. Whether you will take up science, arts or commerce will all depend upon your class 10th performance. To get good scores for entry to some of the best schools or junior colleges, it is vital to perform well in class 10 and this is where the solutions of NCERT for class 10 come in hand. To score good marks, you need to know CBSE NCERT solutions for class 10 as these are the answers to the questions asked in the exam. NCERT solutions are good tools or keys to help unlock your exam problems. These contain the step by step process on how to arrive at the NCERT solutions for class 10.
Maths and science are two subjects that are scoring only if you know the answers or rather know how to arrive at them. If you have access to the solutions of NCERT for class 10 for these subjects, you will have enough help to understand the chapters and reach your goal of scoring well.
Click on the link given above for the subject you wish to check out.
Go to the chapter given in the subject and check the solutions given therein
NCERT solutions for class 10 mathematics
The subject has around 15 chapters with many exercises that test you in the concepts taught and formulae to be used. The solutions of NCERT for class 10 cover all this and more. You will not just find that maths is now easier but also find the interest to do better in the final board exams.
NCERT solutions for class 10 science
This subject covers physics, chemistry and biology topics in 16 chapters. Taking the help of solutions of NCERT for class 10 science will be a good step as you will be in a position to be well conversant with the science. A strong foundation will help you in class 11 and 12 as well.
Also Read -
Stuck in a question? Unable to understand how to solve a problem? No worries. Ask our experts here and you will receive prompt help with it.
Happy learning!!!
", "article_status": "3", "updated_by_id": "168560", "published_on": "2019-07-08 12:34:13.261789", "featured_image_text": "", "chapter_id": "0", "migrate_id": "0", "level_id": "10", "subject_id": "0", "is_published": "1", "paper_type_id": "0" }, { "id": "12646", "name": "NCERT Solutions for Class 10 Maths", "slug": "ncert-solutions-class-10-maths", "draft_description": "NCERT solutions for class 10 maths- Are you preparing for class 10 board exams and looking for solutions of NCERT, then you are at the right place. CBSE NCERT solutions class 10 maths are going to help you crack board exams and make learning easier. NCERT class 10 maths contain a total of 15 chapters and all the chapters are important to study. You should not ignore any single chapter because at least one question is sure from each chapter in your boardexams. It is necessary to solve all the exercises including the additional exercises to get a good score. NCERT solutions for class 10 maths will help you understand each question in detail. By solving the entire NCERT maths using CBSE NCERT solutions for class 10 maths and practicing the previous 5 years' questions you can easily score well in board exams. Geometry and trigonometry are one of the most important chapters in this class as most of the students feel troubled in geometry and trigonometry. For saving you from this trouble, here are the NCERT solutions for class 10 maths, you can click on the chapter you wish to study below. Also, in order to study other subjects or even higher classes, you can take help from NCERT solutions that are answered using step by step approach so that you do not miss any step while answering the problems in the examinations.
If we talk about futuristic importance, then most of the students will opt for a competitive examination after their school or graduation. If a student is good in class 10 maths then only a few additional things will be left to study. If you want to check the solutions for science or other subjects of class 10, then click on the link NCERT solutions for class 10.
Go to the chapter that you wish to check.
Click on the link given for that particular chapter of class 10 maths.
The chapter page with solutions of NCERT class 10 maths will open.
Check the questions and solutions given in the NCERT class 10 maths.
NCERT solutions for class 10 maths chapter 1 Real Numbers: In this chapter, you are going to learn about Euclid’s Division Lemma and this concept is used to find the HCF, LCM, the relation between HCF and LCM and fundamental theorem of arithmetic. Euclid’s Division Lemma will also be required if you want to prove that the given number is rational or irrational and the given rational number is terminating or non terminating repeating. There are many real-life examples of LCM and HCF mentioned in the NCERT solutions for class 10 maths. There are a total of 4 exercises with a total of 48 questions in this chapter.
Solutions of NCERT class 10 maths chapter 2 Polynomials: The chapter starts with an explanation of polynomials referring to the previous classes and then explains the geometrical interpretation of the zeroes of the polynomial. Then it gives the relation between zeroes and coefficient of a given polynomial and also explains how to carry out the polynomial divisions. You have to be good in class 9 maths chapter polynomial to beat this one. In NCERT solutions for class 10 maths chapter 2 polynomials, step by step explanation is given in the answers of all 4 exercises including the optional exercise.
CBSE NCERT solutions for class 10 maths chapter 3 Pair of Linear Equations in Two Variables: There are many real-life examples that can be formulated in linear equations in two variables. Let's take an example to understand: If a person buys some pens and pencils at a total of 41 Rs and the price of a pen and a pencil are 5 Rs and 4 Rs respectively. Then the statement can be represented as Many such problems are mentioned in the NCERT solutions for class 10 maths. The chapter covers various algebraic methods to solve a given pair of linear equations and a geometrical way of solving the pair of linear equations.
NCERT solutions for class 10 maths chapter 4 Quadratic Equations: We get a quadratic equation if a quadratic polynomial is equated to zero with the condition that the coefficient of the variable with power two is not equal to zero. The main content of the chapter includes the various methods of solving the quadratic equations and how to determine the consistency of quadratic equations from the value of discriminants. Every year around 8 to 10 marks questions come from this chapter only. In solutions of NCERT for class 10 maths chapter 4 quadratic equation, all the 4 exercises consisting of 46 questions are covered.
Solutions of NCERT class 10 maths chapter 5 Arithmetic Progressions: Any number series having the same difference between two immediate terms is called arithematic progression. For example- 2, 5, 8, 11 is an arithematic progression as the difference everywhere is 3. Every year around questions of 8 to 10 marks come from this chapter only. In this particular chapter, you will learn to solve questions related to nth /last term and questions related to sum of arithematic progression. Also, some questions are based on the usage of both the concepts together. There are some real-life situations mentioned in the NCERT solutions for class 10 maths. Studying a chapter by correlating to real-life examples will be interesting. In NCERT solutions for class 10 maths chapter 5 arithematic progression, we are covering all the 4 exercises including an optional exercise.
CBSE NCERT solutions for class 10 maths chapter 6 Triangles: A closed shape made using three straight lines is known as geometry. The NCERT Class 10 Chapter 6 starts with an explanation of the difference between the congruent figures and similar figures. In this particular chapter, we will learn similarity of triangles, criteria for similarity, area of triangles and understanding of Pythagoras theorem using the similarity of triangles. There are many interesting examples that can be solved using the similarity of triangles. In the solutions of this particular chapter, you will get solutions to all the 6 exercises including optional exercise consisting of 90 questions.
NCERT solutions for class 10 maths chapter 7 Coordinate Geometry: This is one of the chapters which you will use in almost all the areas your higher studies, whether it is science, commerce or arts. In class 10 maths, this chapter holds some specific kind of concepts and questions and if you do them well, then the things will become easy for you in the examination. The NCERT solutions for class 10 maths chapter 7 coordinate geometry include the distance formulae, section formulae and the concept of finding the area of triangles if the coordinate of the vertex is given. The solution to all the 4 exercises including an optional exercise consisting total of 45 questions are explained in a step by step manner.
Solutions of NCERT class 10 maths chapter 8 Introduction to Trigonometry: The chapter has originated from the right angle triangle. The properties, formulas, and ratios of trigonometry can be derived using length, breadth, and height of the right-angle triangle. In this chapter, there are a total of 4 exercises and questions of these exercises are mainly based on finding trigonometric ratios of the sides of a right-angled triangle. The NCERT Solutions for Class 10 Maths of this chapter covers important concepts to find trigonometric ratios, trigonometric ratios of some specific angles, trigonometric ratios of complementary angles and trigonometric identities.
CBSE NCERT solutions for class 10 maths chapter 9 Some Applications of Trigonometry: This chapter is just an extension of the concepts used in chapter 8. This is the application based chapter which tells us the usage of trigonometry in real life. "Trigonometry" unit holds 12 marks out of 80 marks in the CBSE final maths paper and we can expect one question of 3 or 4 marks from this chapter. In this, we will learn how trigonometry is helpful in finding the distance and height of different objects without actually measuring them. This chapter has only 1 exercise for which answers are there in solutions of NCERT for class 10 maths chapter 9 some applications of trigonometry.
NCERT solutions for class 10 maths chapter 10 Circles: A circle is nothing but a closed figure with a collection of points in a plane that are at a specific distance (radius) from a fixed point (center). This chapter is an important part of geometry along with triangles. This chapter in class 10 maths is an extension of class 9 maths chapter circles. In this chapter, we will discuss tangent to a circle and number of tangents from a point on the circle. The chapter holds good weightage in the board examinations. Every year approximately 10 to 12 marks questions come from this chapter only. The solution of NCERT class 10 maths covers all 2 exercises consisting of overall 26 questions.
Solutions of NCERT class 10 maths chapter 11 Constructions: This chapter, deals with the problem constructions. It is also known as a geometrical drawing which deals with constructing geometric figures using some specific geometric tools. It is a part of the geometry unit and geometry holds 15 marks in the final CBSE exams and construction is a scoring topic for students. In this chapter, there is a total of 2 exercises with 14 questions. In the first exercise, we will learn how to divide a line segment and in the second exercise, we will learn the construction of tangents to a circle.
CBSE NCERT solutions for class 10 maths chapter 12 Areas Related to Circles: In this chapter, we will deal with areas related to circles. This chapter is a part of the Mensuration unit, which holds 10 marks in the CBSE Class 10 final exam. In this chapter, there are 3 exercises with 35 questions in them. In this chapter, we will learn the perimeter and area of a circle, areas of sector and segment of a circle, areas of combinations of plane figures. In NCERT solutions for class 10 maths chapter 12 areas related to circles, detailed solutions to each question are given.
NCERT solutions for class 10 maths chapter 13 Surface Areas and Volumes: This comes under mensuration. We have already done some part of the mensuration in the class 9 maths. In this chapter, we will discuss how to find surface areas and volumes of combinations, which are formed by two or more than two of the basic solids such as cone, cuboid, cylinder, sphere, and hemisphere. Apart from all these shapes, a new shape frustum will be introduced here in this particular chapter. In this chapter, there are a total of 5 exercises with 38 questions in them. In all the exercises, we will deal with questions on finding surface areas and volumes of solids.
Solutions of NCERT class 10 maths chapter 14 Statistics: In the previous class, we have learned about data representation and have solved many calculation based questions on that. Statistics is also somehow related to data handling. In this chapter, the questions will be based on data and we will discuss how to find the measures of central tendency namely, mean, mode and median from the ungrouped data to that of grouped data. In this chapter, there are a total of 4 exercises with 25 questions. In NCERT Class 10 Maths Chapter 14 Statistics, you will also get to know about the concept of cumulative frequency, its distribution and how to draw cumulative frequency curves.
CBSE NCERT solutions for class 10 maths chapter 15 Probability: Probability is one of the newest concept you will learn here in class 10 mathematics. In the previous class, you have learned about empirical probabilities or experimental probabilities of events that are based on the results of actual experiments. In this NCERT class 10 maths, we will discuss the theoretical probability (also called classical probability) of an event, and solve simple questions based on this concept. In NCERT solutions for class 10 maths chapter 15 probability, solutions of 2 exercises including optional exercise consisting of overall 68 questions are covered.
Unit | Title |
1. | NCERT solutions for class 12 |
2. |
NCERT solutions for class 10 maths- Are you preparing for class 10 board exams and looking for solutions of NCERT, then you are at the right place. CBSE NCERT solutions class 10 maths are going to help you crack board exams and make learning easier. NCERT class 10 maths contain a total of 15 chapters and all the chapters are important to study. You should not ignore any single chapter because at least one question is sure from each chapter in your boardexams. It is necessary to solve all the exercises including the additional exercises to get a good score. NCERT solutions for class 10 maths will help you understand each question in detail. By solving the entire NCERT maths using CBSE NCERT solutions for class 10 maths and practicing the previous 5 years' questions you can easily score well in board exams. Geometry and trigonometry are one of the most important chapters in this class as most of the students feel troubled in geometry and trigonometry. For saving you from this trouble, here are the NCERT solutions for class 10 maths, you can click on the chapter you wish to study below. Also, in order to study other subjects or even higher classes, you can take help from NCERT solutions that are answered using step by step approach so that you do not miss any step while answering the problems in the examinations.
If we talk about futuristic importance, then most of the students will opt for a competitive examination after their school or graduation. If a student is good in class 10 maths then only a few additional things will be left to study. If you want to check the solutions for science or other subjects of class 10, then click on the link NCERT solutions for class 10.
Go to the chapter that you wish to check.
Click on the link given for that particular chapter of class 10 maths.
The chapter page with solutions of NCERT class 10 maths will open.
Check the questions and solutions given in the NCERT class 10 maths.
NCERT solutions for class 10 maths chapter 1 Real Numbers: In this chapter, you are going to learn about Euclid’s Division Lemma and this concept is used to find the HCF, LCM, the relation between HCF and LCM and fundamental theorem of arithmetic. Euclid’s Division Lemma will also be required if you want to prove that the given number is rational or irrational and the given rational number is terminating or non terminating repeating. There are many real-life examples of LCM and HCF mentioned in the NCERT solutions for class 10 maths. There are a total of 4 exercises with a total of 48 questions in this chapter.
Solutions of NCERT class 10 maths chapter 2 Polynomials: The chapter starts with an explanation of polynomials referring to the previous classes and then explains the geometrical interpretation of the zeroes of the polynomial. Then it gives the relation between zeroes and coefficient of a given polynomial and also explains how to carry out the polynomial divisions. You have to be good in class 9 maths chapter polynomial to beat this one. In NCERT solutions for class 10 maths chapter 2 polynomials, step by step explanation is given in the answers of all 4 exercises including the optional exercise.
CBSE NCERT solutions for class 10 maths chapter 3 Pair of Linear Equations in Two Variables: There are many real-life examples that can be formulated in linear equations in two variables. Let's take an example to understand: If a person buys some pens and pencils at a total of 41 Rs and the price of a pen and a pencil are 5 Rs and 4 Rs respectively. Then the statement can be represented as Many such problems are mentioned in the NCERT solutions for class 10 maths. The chapter covers various algebraic methods to solve a given pair of linear equations and a geometrical way of solving the pair of linear equations.
NCERT solutions for class 10 maths chapter 4 Quadratic Equations: We get a quadratic equation if a quadratic polynomial is equated to zero with the condition that the coefficient of the variable with power two is not equal to zero. The main content of the chapter includes the various methods of solving the quadratic equations and how to determine the consistency of quadratic equations from the value of discriminants. Every year around 8 to 10 marks questions come from this chapter only. In solutions of NCERT for class 10 maths chapter 4 quadratic equation, all the 4 exercises consisting of 46 questions are covered.
Solutions of NCERT class 10 maths chapter 5 Arithmetic Progressions: Any number series having the same difference between two immediate terms is called arithematic progression. For example- 2, 5, 8, 11 is an arithematic progression as the difference everywhere is 3. Every year around questions of 8 to 10 marks come from this chapter only. In this particular chapter, you will learn to solve questions related to nth /last term and questions related to sum of arithematic progression. Also, some questions are based on the usage of both the concepts together. There are some real-life situations mentioned in the NCERT solutions for class 10 maths. Studying a chapter by correlating to real-life examples will be interesting. In NCERT solutions for class 10 maths chapter 5 arithematic progression, we are covering all the 4 exercises including an optional exercise.
CBSE NCERT solutions for class 10 maths chapter 6 Triangles: A closed shape made using three straight lines is known as geometry. The NCERT Class 10 Chapter 6 starts with an explanation of the difference between the congruent figures and similar figures. In this particular chapter, we will learn similarity of triangles, criteria for similarity, area of triangles and understanding of Pythagoras theorem using the similarity of triangles. There are many interesting examples that can be solved using the similarity of triangles. In the solutions of this particular chapter, you will get solutions to all the 6 exercises including optional exercise consisting of 90 questions.
NCERT solutions for class 10 maths chapter 7 Coordinate Geometry: This is one of the chapters which you will use in almost all the areas your higher studies, whether it is science, commerce or arts. In class 10 maths, this chapter holds some specific kind of concepts and questions and if you do them well, then the things will become easy for you in the examination. The NCERT solutions for class 10 maths chapter 7 coordinate geometry include the distance formulae, section formulae and the concept of finding the area of triangles if the coordinate of the vertex is given. The solution to all the 4 exercises including an optional exercise consisting total of 45 questions are explained in a step by step manner.
Solutions of NCERT class 10 maths chapter 8 Introduction to Trigonometry: The chapter has originated from the right angle triangle. The properties, formulas, and ratios of trigonometry can be derived using length, breadth, and height of the right-angle triangle. In this chapter, there are a total of 4 exercises and questions of these exercises are mainly based on finding trigonometric ratios of the sides of a right-angled triangle. The NCERT Solutions for Class 10 Maths of this chapter covers important concepts to find trigonometric ratios, trigonometric ratios of some specific angles, trigonometric ratios of complementary angles and trigonometric identities.
CBSE NCERT solutions for class 10 maths chapter 9 Some Applications of Trigonometry: This chapter is just an extension of the concepts used in chapter 8. This is the application based chapter which tells us the usage of trigonometry in real life. "Trigonometry" unit holds 12 marks out of 80 marks in the CBSE final maths paper and we can expect one question of 3 or 4 marks from this chapter. In this, we will learn how trigonometry is helpful in finding the distance and height of different objects without actually measuring them. This chapter has only 1 exercise for which answers are there in solutions of NCERT for class 10 maths chapter 9 some applications of trigonometry.
NCERT solutions for class 10 maths chapter 10 Circles: A circle is nothing but a closed figure with a collection of points in a plane that are at a specific distance (radius) from a fixed point (center). This chapter is an important part of geometry along with triangles. This chapter in class 10 maths is an extension of class 9 maths chapter circles. In this chapter, we will discuss tangent to a circle and number of tangents from a point on the circle. The chapter holds good weightage in the board examinations. Every year approximately 10 to 12 marks questions come from this chapter only. The solution of NCERT class 10 maths covers all 2 exercises consisting of overall 26 questions.
Solutions of NCERT class 10 maths chapter 11 Constructions: This chapter, deals with the problem constructions. It is also known as a geometrical drawing which deals with constructing geometric figures using some specific geometric tools. It is a part of the geometry unit and geometry holds 15 marks in the final CBSE exams and construction is a scoring topic for students. In this chapter, there is a total of 2 exercises with 14 questions. In the first exercise, we will learn how to divide a line segment and in the second exercise, we will learn the construction of tangents to a circle.
CBSE NCERT solutions for class 10 maths chapter 12 Areas Related to Circles: In this chapter, we will deal with areas related to circles. This chapter is a part of the Mensuration unit, which holds 10 marks in the CBSE Class 10 final exam. In this chapter, there are 3 exercises with 35 questions in them. In this chapter, we will learn the perimeter and area of a circle, areas of sector and segment of a circle, areas of combinations of plane figures. In NCERT solutions for class 10 maths chapter 12 areas related to circles, detailed solutions to each question are given.
NCERT solutions for class 10 maths chapter 13 Surface Areas and Volumes: This comes under mensuration. We have already done some part of the mensuration in the class 9 maths. In this chapter, we will discuss how to find surface areas and volumes of combinations, which are formed by two or more than two of the basic solids such as cone, cuboid, cylinder, sphere, and hemisphere. Apart from all these shapes, a new shape frustum will be introduced here in this particular chapter. In this chapter, there are a total of 5 exercises with 38 questions in them. In all the exercises, we will deal with questions on finding surface areas and volumes of solids.
Solutions of NCERT class 10 maths chapter 14 Statistics: In the previous class, we have learned about data representation and have solved many calculation based questions on that. Statistics is also somehow related to data handling. In this chapter, the questions will be based on data and we will discuss how to find the measures of central tendency namely, mean, mode and median from the ungrouped data to that of grouped data. In this chapter, there are a total of 4 exercises with 25 questions. In NCERT Class 10 Maths Chapter 14 Statistics, you will also get to know about the concept of cumulative frequency, its distribution and how to draw cumulative frequency curves.
CBSE NCERT solutions for class 10 maths chapter 15 Probability: Probability is one of the newest concept you will learn here in class 10 mathematics. In the previous class, you have learned about empirical probabilities or experimental probabilities of events that are based on the results of actual experiments. In this NCERT class 10 maths, we will discuss the theoretical probability (also called classical probability) of an event, and solve simple questions based on this concept. In NCERT solutions for class 10 maths chapter 15 probability, solutions of 2 exercises including optional exercise consisting of overall 68 questions are covered.
Unit | Title |
1. | NCERT solutions for class 12 |
2. |
NCERT solutions for class 10 maths chapter 1 Real Numbers- In mathematics, there are two kinds of numbers- one is the real number and another one is the imaginary number. In this particular chapter, we are going to talk about real numbers. Solutions of NCERT class 10 maths chapter 1 real numbers carry detailed explanations for each and every question. CBSE NCERT solutions for class 10 maths chapter 1 real numbers can assist you while doing homework as well as for the board exam preparation. Real numbers are those numbers that can be represented in the number line. If anyone wants to go deeper into the number system then studying real numbers is the first step to master the number system. Real numbers include the in-depth classification of numbers, its applications, and properties related to various kinds of numbers. CBSE NCERT solutions for class 10 maths chapter 1 real numbers concepts are very important when you will be appearing in the competitive examinations. Apart from this, NCERT solutions for other subjects and classes can be downloaded by clicking on the link.
Three exercises of this chapter are explained below.
Types of questions asked from class 10 maths chapter 1 Real Numbers
CBSE Class 10 maths board exam will have the following types of questions from real numbers:
Euclid's division algorithm
Prime factorization
Highest common factor
Prime factorization
Prime & composite numbers
Rational & irrational numbers
Q1 (1) Use Euclid’s division algorithm to find the HCF of 135 and 225
Answer:
225 > 135. Applying Euclid's Division algorithm we get
since remainder 0 we again apply the algorithm
since remainder 0 we again apply the algorithm
since remainder = 0 we conclude the HCF of 135 and 225 is 45.
Q1 (2) Use Euclid’s division algorithm to find the HCF of 196 and 38220
Answer:
38220 > 196. Applying Euclid's Division algorithm we get
since remainder = 0 we conclude the HCF of 38220 and 196 is 196.
Q1 (3) Use Euclid’s division algorithm to find the HCF of 867 and 255
Answer:
867 > 225. Applying Euclid's Division algorithm we get
since remainder 0 we apply the algorithm again.
since 255 > 102
since remainder 0 we apply the algorithm again.
since 102 > 51
since remainder = 0 we conclude the HCF of 867 and 255 is 51.
Answer:
Let p be any positive integer. It can be expressed as
p = 6q + r
where and
but for r = 0, 2 or 4 p will be an even number therefore all odd positive integers can be written in the form 6q + 1, 6q + 3 or 6q + 5.
Answer:
The maximum number of columns in which they can march = HCF (32, 616)
Since 616 > 32, applying Euclid's Division Algorithm we have
Since remainder 0 we again apply Euclid's Division Algorithm
Since 32 > 8
Since remainder = 0 we conclude, 8 is the HCF of 616 and 32.
The maximum number of columns in which they can march is 8.
Answer:
Let x be any positive integer.
It can be written in the form 3q + r where and r = 0, 1 or 2
Case 1:
For r = 0 we have
x^{2} = (3q)^{2}
x^{2} = 9q^{2}
x^{2} = 3(3q^{2})
x^{2} = 3m
Case 2:
For r = 1 we have
x^{2} = (3q+1)^{2}
x^{2} = 9q^{2} + 6q +1
x^{2} = 3(3q^{2} + 2q) + 1
x^{2} = 3m + 1
Case 3:
For r = 2 we have
x^{2} = (3q+2)^{2}
x^{2} = 9q^{2} + 12q +4
x^{2} = 3(3q^{2} + 4q + 1) + 1
x^{2} = 3m + 1
Hence proved.
Answer:
Let x be any positive integer.
It can be written in the form 3q + r where and r = 0, 1 or 2
Case 1:
For r = 0 we have
x^{3} = (3q)^{3}
x^{3} = 27q^{3}
x^{3} = 9(3q^{3})
x^{3} = 9m
Case 2:
For r = 1 we have
x^{3} = (3q+1)^{3}
x^{3} = 27q^{3} + 27q^{2} + 9q + 1
x^{3} = 9(3q^{3} + 3q^{2} +q) + 1
x^{3} = 3m + 1
Case 3:
For r = 2 we have
x^{3} = (3q+2)^{3}
x^{3} = 27q^{3} + 54q^{2} + 36q + 8
x^{3} = 9(3q^{3} + 6q^{2} +4q) + 8
x^{3} = 3m + 8
Hence proved.
Answer:
The number can be as a product of its prime factors as follows
Q 1 (2) Express each number as a product of its prime factors: 156
Answer:
The given number can be expressed as follows
Q1 (3) Express each number as a product of its prime factors: 3825
Answer:
The number is expressed as the product of the prime factors as follows
Q1 (4) Express each number as a product of its prime factors: 5005
Answer:
The given number can be expressed as the product of its prime factors as follows.
Q1 (5) Express each number as a product of its prime factors: 7429
Answer:
The given number can be expressed as the product of their prime factors as follows
Answer:
26 = 2 x 13
91 = 7 x 13
HCF(26,91) = 13
LCM(26,91) = 2 x 7 x 13 = 182
HCF x LCM = 13 x 182 = 2366
26 x 91 = 2366
26 x 91 = HCF x LCM
Hence Verified
Answer:
The number can be expressed as the product of prime factors as
510 = 2 x 3 x 5 x 17
92 = 2^{2 }x 23
HCF(510,92) = 2
LCM(510,92) = 2^{2} x 3 x 5 x 17 x 23 = 23460
HCF x LCM = 2 x 23460 = 46920
510 x 92 = 46920
510 x 92 = HCF x LCM
Hence Verified
Answer:
336 is expressed as the product of its prime factor as
336 = 2^{4} x 3 x 7
54 is expressed as the product of its prime factor as
54 = 2 x 3^{3}
HCF(336,54) = 2 x 3 = 6
LCM(336,54) = 2^{4} x 3^{3 }x 7 = 3024
HCF x LCM = 6 x 3024 = 18144
336 x 54 = 18144
336 x 54 = HCF x LCM
Hence Verified
Answer:
The numbers can be written as the product of their prime factors as follows
12 = 2^{2} x 3
15 = 3 x 5
21 = 3 x 7
HCF = 3
LCM = 2^{2} x 3 x 5 x 7 = 420
Answer:
The given numbers are written as the product of their prime factors as follows
17 = 1 x 17
23 = 1 x 23
29 = 1 x 29
HCF = 1
LCM = 17 x 23 x 29 = 11339
Answer:
The given numbers are written as the product of their prime factors as follows
8 = 2^{3}
9 = 3^{2}
25 = 5^{2}
HCF = 1
LCM = 2^{3} x 3^{2} x 5^{2} = 1800
Q4 Given that HCF (306, 657) = 9, find LCM (306, 657).
Answer:
As we know the product of HCF and LCM of two numbers is equal to the product of the two numbers we have
HCF (306, 657) x LCM (306, 657) = 306 x 657
Q5 Check whether can end with the digit 0 for any natural number n.
Answer:
By prime factorizing we have
6^{n} = 2^{n} x 3^{n}
A number will end with 0 if it has at least 1 as the power of both 2 and 5 in its prime factorization. Since the power of 5 is 0 in the prime factorization of 6^{n} we can conclude that for no value of n 6^{n} will end with the digit 0.
Q6 Explain why 7 11 13 + 13 and 7 6 5 4 3 2 1 + 5 are composite numbers.
Answer:
7 x 11 x 13 + 13
= (7 x 11 + 1) x 13
= 78 x 13
= 2 x 3 x 13^{2}
7 x 6 x 5 x 4 x 3 x 2 x 1 + 5
= (7 x 6 x 4 x 3 x 2 x 1 + 1) x 5
= 5 x 1008
After Solving we observed that both the number are even numbers and the number rule says that we can take atleast two common out of two numbers. So that the number is composite number.
Answer:
The time after which they meet again at the starting point will be equal to the LCM of the times they individually take to complete one round.
Time taken by Sonia = 18 = 2 x 3^{2}
Time taken by Ravi = 12 = 2^{2} x 3
LCM(18,12) = 2^{2} x 3^{2} = 36
Therefore they would again meet at the starting point after 36 minutes.
Answer:
Let us assume is rational.
It means can be written in the form where p and q are co-primes and
Squaring both sides we obtain
From the above equation, we can see that p^{2} is divisible by 5, Therefore p will also be divisible by 5 as 5 is a prime number.
Therefore p can be written as 5r
p = 5r
p^{2} = (5r)^{2}
5q^{2} = 25r^{2}
q^{2} = 5r^{2}
From the above equation, we can see that q^{2} is divisible by 5, Therefore q will also be divisible by 5 as 5 is a prime number.
From (i) and (ii) we can see that both p and q are divisible by 5. This implies that p and q are not co-primes. This contradiction arises because our initial assumption that is rational was wrong. Hence proved that is irrational.
Answer:
Let us assume is rational.
This means can be wriiten in the form where p and q are co-prime integers.
As p and q are integers would be rational, this contradicts the fact that is irrational. This contradiction arises because our initial assumption that is rational was wrong. Therefore is irrational.
Q3 Prove that the following are irrationals :
Answer:
Let us assume is rational.
This means can be written in the form where p and q are co-prime integers.
Since p and q are co-prime integers will be rational, this contradicts the fact that is irrational. This contradiction arises because our initial assumption that is rational was wrong. Therefore is irrational.
Q3 (2) Prove that the following are irrationals :
Answer:
Let us assume is rational.
This means can be written in the form where p and q are co-prime integers.
As p and q are integers would be rational, this contradicts the fact that is irrational. This contradiction arises because our initial assumption that is rational was wrong. Therefore is irrational.
Q3 (3) Prove that the following are irrationals :
Answer:
Let us assume is rational.
This means can be written in the form where p and q are co-prime integers.
As p and q are integers would be rational, this contradicts the fact that is irrational. This contradiction arises because our initial assumption that is rational was wrong. Therefore is irrational.
NCERT solutions for class 10 maths chapter 1 Real Numbers Excercise: 1.4
Answer:
The denominator is of the form 2^{a} x 5^{b} where a = 0 and b = 5. Therefore the given rational number will have a terminating decimal expansion.
Answer:
The denominator is of the form 2^{a} x 5^{b} where a = 3 and b = 0. Therefore the given rational number will have a terminating decimal expansion.
Answer:
The denominator is not of the form 2^{a} x 5^{b}. Therefore the given rational number will have a non-terminating repeating decimal expansion.
Answer:
The denominator is of the form 2^{a} x 5^{b} where a = 6 and b = 1. Therefore the given rational number will have a terminating decimal expansion.
Answer:
The denominator is not of the form 2^{a} x 5^{b}. Therefore the given rational number will have a non-terminating repeating decimal expansion.
Answer:
The denominator is of the form 2^{a} x 5^{b} where a = 3 and b = 2. Therefore the given rational number will have a terminating decimal expansion.
Answer:
The denominator is not of the form 2^{a} x 5^{b}. Therefore the given rational number will have a non-terminating repeating decimal expansion.
Answer:
The denominator is of the form 2^{a} x 5^{b} where a = 0 and b = 1. Therefore the given rational number will have a terminating decimal expansion.
Answer:
The denominator is of the form 2^{a} x 5^{b} where a = 1 and b = 1. Therefore the given rational number will have a terminating decimal expansion.
Answer:
The denominator is not of the form 2^{a} x 5^{b}. Therefore the given rational number will have a non-terminating repeating decimal expansion.
Answer:
decimal expansions of rational numbers are
(i)
(ii)&nbsnbsp;
(iv)
(vI)
(viii)
(ix)
Answer:
The denominator is of the form 2^{a} x 5^{b} where a = 9 and b = 9. Therefore the given number is rational and has a terminating decimal expansion.
Answer:
Since the decimal part of the given number is non-terminating and non-repeating we can conclude that the given number is irrational and cannot be written in the form where p and q are integers.
Answer:
As the decimal part of the given number is non-terminating and repeating, the number is rational but its denominator will have factors other than 2 and 5.
Chapter No. | Chapter Name |
Chapter 1 | CBSE NCERT solutions for class 10 maths chapter 1 Real Numbers |
Chapter 2 | NCERT solutions for class 10 maths chapter 2 Polynomials |
Chapter 3 | Solutions of NCERT class 10 maths chapter 3 Pair of Linear Equations in Two Variables |
Chapter 4 | CBSE NCERT solutions for class 10 maths chapter 4 Quadratic Equations |
Chapter 5 | NCERT solutions for class 10 chapter 5 Arithmetic Progressions |
Chapter 6 | |
Chapter 7 | CBSE NCERT solutions for class 10 maths chapter 7 Coordinate Geometry |
Chapter 8 | NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry |
Chapter 9 | Solutions of NCERT class 10 maths chapter 9 Some Applications of Trigonometry |
Chapter 10 | |
Chapter 11 | |
Chapter 12 | Solutions of NCERT class 10 chapter maths chapter 12 Areas Related to Circles |
Chapter 13 | CBSE NCERT solutions class 10 maths chapter 13 Surface Areas and Volumes |
Chapter 14 | |
Chapter 15 |
After the completion of the NCERT syllabus, you can check past 5-year papers of board exams. Previous year papers will increase your dealing ability with a variety of questions.
NCERT syllabus coverage and previous year papers are enough tools to get a good score in the board examination. After covering NCERTs and the previous year papers, you can jump to the next chapters.
Keep working hard & happy learning!
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Three exercises of this chapter are explained below.
Types of questions asked from class 10 maths chapter 1 Real Numbers
CBSE Class 10 maths board exam will have the following types of questions from real numbers:
Euclid's division algorithm
Prime factorization
Highest common factor
Prime factorization
Prime & composite numbers
Rational & irrational numbers
Q1 (1) Use Euclid’s division algorithm to find the HCF of 135 and 225
Answer:
225 > 135. Applying Euclid's Division algorithm we get
since remainder 0 we again apply the algorithm
since remainder 0 we again apply the algorithm
since remainder = 0 we conclude the HCF of 135 and 225 is 45.
Q1 (2) Use Euclid’s division algorithm to find the HCF of 196 and 38220
Answer:
38220 > 196. Applying Euclid's Division algorithm we get
since remainder = 0 we conclude the HCF of 38220 and 196 is 196.
Q1 (3) Use Euclid’s division algorithm to find the HCF of 867 and 255
Answer:
867 > 225. Applying Euclid's Division algorithm we get
since remainder 0 we apply the algorithm again.
since 255 > 102
since remainder 0 we apply the algorithm again.
since 102 > 51
since remainder = 0 we conclude the HCF of 867 and 255 is 51.
Answer:
Let p be any positive integer. It can be expressed as
p = 6q + r
where and
but for r = 0, 2 or 4 p will be an even number therefore all odd positive integers can be written in the form 6q + 1, 6q + 3 or 6q + 5.
Answer:
The maximum number of columns in which they can march = HCF (32, 616)
Since 616 > 32, applying Euclid's Division Algorithm we have
Since remainder 0 we again apply Euclid's Division Algorithm
Since 32 > 8
Since remainder = 0 we conclude, 8 is the HCF of 616 and 32.
The maximum number of columns in which they can march is 8.
Answer:
Let x be any positive integer.
It can be written in the form 3q + r where and r = 0, 1 or 2
Case 1:
For r = 0 we have
x^{2} = (3q)^{2}
x^{2} = 9q^{2}
x^{2} = 3(3q^{2})
x^{2} = 3m
Case 2:
For r = 1 we have
x^{2} = (3q+1)^{2}
x^{2} = 9q^{2} + 6q +1
x^{2} = 3(3q^{2} + 2q) + 1
x^{2} = 3m + 1
Case 3:
For r = 2 we have
x^{2} = (3q+2)^{2}
x^{2} = 9q^{2} + 12q +4
x^{2} = 3(3q^{2} + 4q + 1) + 1
x^{2} = 3m + 1
Hence proved.
Answer:
Let x be any positive integer.
It can be written in the form 3q + r where and r = 0, 1 or 2
Case 1:
For r = 0 we have
x^{3} = (3q)^{3}
x^{3} = 27q^{3}
x^{3} = 9(3q^{3})
x^{3} = 9m
Case 2:
For r = 1 we have
x^{3} = (3q+1)^{3}
x^{3} = 27q^{3} + 27q^{2} + 9q + 1
x^{3} = 9(3q^{3} + 3q^{2} +q) + 1
x^{3} = 3m + 1
Case 3:
For r = 2 we have
x^{3} = (3q+2)^{3}
x^{3} = 27q^{3} + 54q^{2} + 36q + 8
x^{3} = 9(3q^{3} + 6q^{2} +4q) + 8
x^{3} = 3m + 8
Hence proved.
Answer:
The number can be as a product of its prime factors as follows
Q 1 (2) Express each number as a product of its prime factors: 156
Answer:
The given number can be expressed as follows
Q1 (3) Express each number as a product of its prime factors: 3825
Answer:
The number is expressed as the product of the prime factors as follows
Q1 (4) Express each number as a product of its prime factors: 5005
Answer:
The given number can be expressed as the product of its prime factors as follows.
Q1 (5) Express each number as a product of its prime factors: 7429
Answer:
The given number can be expressed as the product of their prime factors as follows
Answer:
26 = 2 x 13
91 = 7 x 13
HCF(26,91) = 13
LCM(26,91) = 2 x 7 x 13 = 182
HCF x LCM = 13 x 182 = 2366
26 x 91 = 2366
26 x 91 = HCF x LCM
Hence Verified
Answer:
The number can be expressed as the product of prime factors as
510 = 2 x 3 x 5 x 17
92 = 2^{2 }x 23
HCF(510,92) = 2
LCM(510,92) = 2^{2} x 3 x 5 x 17 x 23 = 23460
HCF x LCM = 2 x 23460 = 46920
510 x 92 = 46920
510 x 92 = HCF x LCM
Hence Verified
Answer:
336 is expressed as the product of its prime factor as
336 = 2^{4} x 3 x 7
54 is expressed as the product of its prime factor as
54 = 2 x 3^{3}
HCF(336,54) = 2 x 3 = 6
LCM(336,54) = 2^{4} x 3^{3 }x 7 = 3024
HCF x LCM = 6 x 3024 = 18144
336 x 54 = 18144
336 x 54 = HCF x LCM
Hence Verified
Answer:
The numbers can be written as the product of their prime factors as follows
12 = 2^{2} x 3
15 = 3 x 5
21 = 3 x 7
HCF = 3
LCM = 2^{2} x 3 x 5 x 7 = 420
Answer:
The given numbers are written as the product of their prime factors as follows
17 = 1 x 17
23 = 1 x 23
29 = 1 x 29
HCF = 1
LCM = 17 x 23 x 29 = 11339
Answer:
The given numbers are written as the product of their prime factors as follows
8 = 2^{3}
9 = 3^{2}
25 = 5^{2}
HCF = 1
LCM = 2^{3} x 3^{2} x 5^{2} = 1800
Q4 Given that HCF (306, 657) = 9, find LCM (306, 657).
Answer:
As we know the product of HCF and LCM of two numbers is equal to the product of the two numbers we have
HCF (306, 657) x LCM (306, 657) = 306 x 657
Q5 Check whether can end with the digit 0 for any natural number n.
Answer:
By prime factorizing we have
6^{n} = 2^{n} x 3^{n}
A number will end with 0 if it has at least 1 as the power of both 2 and 5 in its prime factorization. Since the power of 5 is 0 in the prime factorization of 6^{n} we can conclude that for no value of n 6^{n} will end with the digit 0.
Q6 Explain why 7 11 13 + 13 and 7 6 5 4 3 2 1 + 5 are composite numbers.
Answer:
7 x 11 x 13 + 13
= (7 x 11 + 1) x 13
= 78 x 13
= 2 x 3 x 13^{2}
7 x 6 x 5 x 4 x 3 x 2 x 1 + 5
= (7 x 6 x 4 x 3 x 2 x 1 + 1) x 5
= 5 x 1008
After Solving we observed that both the number are even numbers and the number rule says that we can take atleast two common out of two numbers. So that the number is composite number.
Answer:
The time after which they meet again at the starting point will be equal to the LCM of the times they individually take to complete one round.
Time taken by Sonia = 18 = 2 x 3^{2}
Time taken by Ravi = 12 = 2^{2} x 3
LCM(18,12) = 2^{2} x 3^{2} = 36
Therefore they would again meet at the starting point after 36 minutes.
Answer:
Let us assume is rational.
It means can be written in the form where p and q are co-primes and
Squaring both sides we obtain
From the above equation, we can see that p^{2} is divisible by 5, Therefore p will also be divisible by 5 as 5 is a prime number.
Therefore p can be written as 5r
p = 5r
p^{2} = (5r)^{2}
5q^{2} = 25r^{2}
q^{2} = 5r^{2}
From the above equation, we can see that q^{2} is divisible by 5, Therefore q will also be divisible by 5 as 5 is a prime number.
From (i) and (ii) we can see that both p and q are divisible by 5. This implies that p and q are not co-primes. This contradiction arises because our initial assumption that is rational was wrong. Hence proved that is irrational.
Answer:
Let us assume is rational.
This means can be wriiten in the form where p and q are co-prime integers.
As p and q are integers would be rational, this contradicts the fact that is irrational. This contradiction arises because our initial assumption that is rational was wrong. Therefore is irrational.
Q3 Prove that the following are irrationals :
Answer:
Let us assume is rational.
This means can be written in the form where p and q are co-prime integers.
Since p and q are co-prime integers will be rational, this contradicts the fact that is irrational. This contradiction arises because our initial assumption that is rational was wrong. Therefore is irrational.
Q3 (2) Prove that the following are irrationals :
Answer:
Let us assume is rational.
This means can be written in the form where p and q are co-prime integers.
As p and q are integers would be rational, this contradicts the fact that is irrational. This contradiction arises because our initial assumption that is rational was wrong. Therefore is irrational.
Q3 (3) Prove that the following are irrationals :
Answer:
Let us assume is rational.
This means can be written in the form where p and q are co-prime integers.
As p and q are integers would be rational, this contradicts the fact that is irrational. This contradiction arises because our initial assumption that is rational was wrong. Therefore is irrational.
NCERT solutions for class 10 maths chapter 1 Real Numbers Excercise: 1.4
Answer:
The denominator is of the form 2^{a} x 5^{b} where a = 0 and b = 5. Therefore the given rational number will have a terminating decimal expansion.
Answer:
The denominator is of the form 2^{a} x 5^{b} where a = 3 and b = 0. Therefore the given rational number will have a terminating decimal expansion.
Answer:
The denominator is not of the form 2^{a} x 5^{b}. Therefore the given rational number will have a non-terminating repeating decimal expansion.
Answer:
The denominator is of the form 2^{a} x 5^{b} where a = 6 and b = 1. Therefore the given rational number will have a terminating decimal expansion.
Answer:
The denominator is not of the form 2^{a} x 5^{b}. Therefore the given rational number will have a non-terminating repeating decimal expansion.
Answer:
The denominator is of the form 2^{a} x 5^{b} where a = 3 and b = 2. Therefore the given rational number will have a terminating decimal expansion.
Answer:
The denominator is not of the form 2^{a} x 5^{b}. Therefore the given rational number will have a non-terminating repeating decimal expansion.
Answer:
The denominator is of the form 2^{a} x 5^{b} where a = 0 and b = 1. Therefore the given rational number will have a terminating decimal expansion.
Answer:
The denominator is of the form 2^{a} x 5^{b} where a = 1 and b = 1. Therefore the given rational number will have a terminating decimal expansion.
Answer:
The denominator is not of the form 2^{a} x 5^{b}. Therefore the given rational number will have a non-terminating repeating decimal expansion.
Answer:
decimal expansions of rational numbers are
(i)
(ii)&nbsnbsp;
(iv)
(vI)
(viii)
(ix)
Answer:
The denominator is of the form 2^{a} x 5^{b} where a = 9 and b = 9. Therefore the given number is rational and has a terminating decimal expansion.
Answer:
Since the decimal part of the given number is non-terminating and non-repeating we can conclude that the given number is irrational and cannot be written in the form where p and q are integers.
Answer:
As the decimal part of the given number is non-terminating and repeating, the number is rational but its denominator will have factors other than 2 and 5.
Chapter No. | Chapter Name |
Chapter 1 | CBSE NCERT solutions for class 10 maths chapter 1 Real Numbers |
Chapter 2 | NCERT solutions for class 10 maths chapter 2 Polynomials |
Chapter 3 | Solutions of NCERT class 10 maths chapter 3 Pair of Linear Equations in Two Variables |
Chapter 4 | CBSE NCERT solutions for class 10 maths chapter 4 Quadratic Equations |
Chapter 5 | NCERT solutions for class 10 chapter 5 Arithmetic Progressions |
Chapter 6 | |
Chapter 7 | CBSE NCERT solutions for class 10 maths chapter 7 Coordinate Geometry |
Chapter 8 | NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry |
Chapter 9 | Solutions of NCERT class 10 maths chapter 9 Some Applications of Trigonometry |
Chapter 10 | |
Chapter 11 | |
Chapter 12 | Solutions of NCERT class 10 chapter maths chapter 12 Areas Related to Circles |
Chapter 13 | CBSE NCERT solutions class 10 maths chapter 13 Surface Areas and Volumes |
Chapter 14 | |
Chapter 15 |
After the completion of the NCERT syllabus, you can check past 5-year papers of board exams. Previous year papers will increase your dealing ability with a variety of questions.
NCERT syllabus coverage and previous year papers are enough tools to get a good score in the board examination. After covering NCERTs and the previous year papers, you can jump to the next chapters.
Keep working hard & happy learning!
", "article_status": "3", "updated_by_id": "624878", "published_on": "2019-07-16 08:58:35.069550", "featured_image_text": "", "chapter_id": "3485", "migrate_id": "0", "level_id": "10", "subject_id": "3", "is_published": "1", "paper_type_id": "0" }, { "id": "12640", "name": "NCERT Solutions for Class 10 Maths Chapter 10 Circles", "slug": "ncert-solutions-class-10-maths-chapter-10-circles", "draft_description": "NCERT Solutions for Class 10 Maths Chapter 10 Circles - In our previous classes, you have learnt that a circle is a closed shape with a collection of points in a plane that are at a specific distance ( called the radius) from a fixed point (called center). Solutions of NCERT class 10 maths chapter 10 Circles is covering the in-depth explanations to questions related to a circle. You have also studied important terms related to the circle like segment, arc, sector, chord, etc. In this chapter, there are two exercises with 17 questions in them. CBSE NCERT solutions for class 10 maths chapter 10 Circles are solved by subject experts to help students in their preparation keeping step by step marking in the mind.
This chapter introduces some complex and important terms like tangents, tangents to a circle, number of tangents from a point on a circle. In this chapter, we will study the different conditions that arise when a line and a circle are given in a plane. To solve these types of situations, we will learn the approach to apply the concept of the tangent to a circle in NCERT solutions for class 10 maths chapter 10 Circles. This chapter has fundamental concepts that are important for students in their future studies. Circles is a very interesting chapter due to the involvement of geometrical calculations and diagrams. These NCERT solutions can be used to study for competitive exams like JEE and NEET to build the basics. Two exercises of this chapter are explained below.
Q1 How many tangents can a circle have?
The lines that intersect the circle exactly at one single point are called tangents. In a circle, there can be infinitely many tangents.
(a) one
A tangent of a circle intersects the circle exactly in one single point.
(b) secant
It is a line that intersects the circle at two points.
(c) Two,
There can be only two parallel tangents to a circle.
(d) point of contact
The common point of a tangent and a circle.
(A) 12 cm
(B) 13 cm
(C) 8.5 cm
(D) cm.
The correct option is (d) = cm
It is given that the radius of the circle is 5 cm. OQ = 12 cm
According to question,
We know that
So, triangle OPQ is a right-angle triangle. By using Pythagoras theorem,
cm
AB is the given line and the line CD is the tangent to a circle at point M and parallels to the line AB. The line EF is a secant parallel to the AB
(A) 7 cm
(B) 12 cm
(C) 15 cm
(D) 24.5 cm
The correct option is (A) = 7 cm
Given that,
The length of the tangent (QT) is 24 cm and the length of OQ is 25 cm.
Suppose the length of the radius OT be cm.
We know that is a right angle triangle. So, by using Pythagoras theorem-
OT = 7 cm
(A)
(B)
(C)
(D)
The correct option is (b)
In figure,
Since POQT is quadrilateral. Therefore the sum of the opposite angles are 180
(A) 50°
(B) 60°
(C) 70°
(D) 80°
The correct option is (A)
It is given that, tangent PA and PB from point P inclined at
In triangle OAP and OBP
OA =OB (radii of the circle)
PA = PB (tangents of the circle)
Therefore, by SAS congruence
By CPCT,
Now, OPA = 80/4 = 40
In PAO,
= 50
Q4 Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Let line and line are two tangents of a circle and AB is the diameter of the circle.
OA and OB are perpendicular to the tangents and respectively.
therefore,
|| { 1 & 2 are alternate angles}
In the above figure, the line AXB is the tangent to a circle with center O. Here, OX is the perpendicular to the tangent AXB () at point of contact X.
Therefore, we have,
BXO + YXB =
OXY is a collinear
OX is passing through the center of the circle.
Given that,
the length of the tangent from the point A (AP) is 4 cm and the length of OA is 5 cm.
Since APO = 90^{0}
Therefore, APO is a right-angle triangle. By using Pythagoras theorem;
In the above figure, Pq is the chord to the larger circle, which is also tangent to a smaller circle at the point of contact R.
We have,
radius of the larger circle OP = OQ = 5 cm
radius of the small circle (OR) = 3 cm
OR PQ [since PQ is tangent to a smaller circle]
According to question,
In OPR and OQR
PRO = QRO {both }
OR = OR {common}
OP = OQ {both radii}
By RHS congruence OPR OQR
So, by CPCT
PR = RQ
Now, In OPR,
by using pythagoras theorem,
PR = 4 cm
Hence, PQ = 2.PR = 8 cm
To prove- AB + CD = AD + BC
Proof-
We have,
Since the length of the tangents drawn from an external point to a circle are equal
AP =AS .......(i)
BP = BQ.........(ii)
AS = AP...........(iii)
CR = CQ ...........(iv)
By adding all the equations, we get;
AP + BP +RD+ CR = AS +DS +BQ +CQ
(AP + BP) + (RD + CR) = (AS+DS)+(BQ + CQ)
AB + CD = AD + BC
Hence proved.
To prove- AOB =
Proof-
In AOP and AOC,
OA =OA [Common]
OP = OC [Both radii]
AP =AC [tangents from external point A]
Therefore by SSS congruence, AOP AOC
and by CPCT, PAO = OAC
..................(i)
Similarly, from OBC and OBQ, we get;
QBC = 2.OBC.............(ii)
Adding eq (1) and eq (2)
PAC + QBC = 180
2(OBC + OAC) = 180
(OBC + OAC) = 90
Now, in OAB,
Sum of interior angle is 180.
So, OBC + OAC + AOB = 180
AOB = 90
hence proved.
To prove -
Proof-
We have, PA and PB are two tangents, B and A are the point of contacts of the tangent to a circle. And , (since tangents and radius are perpendiculars)
According to question,
In quadrilateral PAOB,
OAP + APB +PBO +BOA =
90 + APB + 90 +BOA = 360
Hence proved.
Q11 Prove that the parallelogram circumscribing a circle is a rhombus.
To prove - the parallelogram circumscribing a circle is a rhombus
Proof-
ABCD is a parallelogram that circumscribes a circle with center O.
P, Q, R, S are the points of contacts on sides AB, BC, CD, and DA respectively
AB = CD .and AD = BC...........(i)
It is known that tangents drawn from an external point are equal in length.
RD = DS ...........(ii)
RC = QC...........(iii)
BP = BQ...........(iv)
AP = AS .............(v)
By adding eq (ii) to eq (v) we get;
(RD + RC) + (BP + AP) = (DS + AS) + (BQ + QC)
CD + AB = AD + BC
2AB = 2AD [from equation (i)]
AB = AD
Now, AB = AD and AB = CD
AB = AD = CD = BC
Hence ABCD is a rhombus.
Consider the above figure. Assume center O touches the sides AB and AC of the triangle at point E and F respectively.
Let the length of AE is x.
Now in ,
(tangents on the circle from point C)
(tangents on the circle from point B)
(tangents on the circle from point A)
Now AB = AE + EB
" ab="x" 8\\\\\\\\ bc="BD" dc\\\\\\\\=">" 14\\\\\\\\ ca="CF" fa\\\\\\\\=">" x\\\\\\\\\" src="\"https://entrancecorner.codecogs.com/gif.latex?%5C%5CAB%20%3D%20AE%20+%20EB%5C%5C%5C%5C%20%3D%3E%20AB%20%3D%20x%20+%208%5C%5C%5C%5C%20BC%20%3D%20BD%20+%20DC%5C%5C%5C%5C%20%3D%3E%20BC%20%3D%208+6%20%3D%2014%5C%5C%5C%5C%20CA%20%3D%20CF%20+%20FA%5C%5C%5C%5C%20%3D%3E%20CA%20%3D%206%20+%20x%5C%5C%5C%5C\"">
Now
" s="(x" 8 14 6 +x) 28) 14\" src="\"https://entrancecorner.codecogs.com/gif.latex?%5C%5Cs%20%3D%20%28AB%20+%20BC%20+%20CA%20%29/2%5C%5C%5C%5C%20%3D%3E%20s%20%3D%20%28x%20+%208%20+%2014%20+%206%20+x%29/2%5C%5C%5C%5C%20%3D%3E%20s%20%3D%20%282x%20+%2028%29/2%5C%5C%5C%5C%20%3D%3E%20s%20%3D%20x%20+%2014\"">
Area of triangle
Now the area of
Area of
Area of
Now Area of the = Area of + Area of + Area of
" 4x x) \\sqrt{3x(14+x)}]="14" x\" src="\"https://entrancecorner.codecogs.com/gif.latex?%5C%5C%3D%3E%204%5Csqrt%7B3x%2814+x%29%7D%3D%2028%20+%2012%20+%202x%20+%2016%20+%202x%20%5C%5C%5C%5C%3D%3E%204%5Csqrt%7B3x%2814+x%29%7D%20%3D%2056%20+%204x%20%5C%5C%5C%5C%3D%3E%204%5Csqrt%7B3x%2814+x%29%7D%20%3D%204%2814%20+%20x%29%20%5C%5C%5C%5C%3D%3E%20%5Csqrt%7B3x%2814+x%29%7D%5D%20%3D%2014%20+%20x\"">
On squaring both the side, we get
" 3x="14" x \\:\\:\\:\\:\\:\\: (14 => is\\: not\\: possible) \\\\\\\\=">" - 2x="14\\\\\\\\" src="\"https://entrancecorner.codecogs.com/gif.latex?%5C%5C3x%2814%20+%20x%29%20%3D%20%2814%20+%20x%29%5E2%5C%5C%5C%5C%20%3D%3E%203x%20%3D%2014%20+%20x%20%5C%3A%5C%3A%5C%3A%5C%3A%5C%3A%5C%3A%20%2814%20+%20x%20%3D%200%20%3D%3E%20x%20%3D%20-14%5C%3A%20is%5C%3A%20not%5C%3A%20possible%29%20%5C%5C%5C%5C%3D%3E%203x%20-%20x%20%3D%2014%5C%5C%5C%5C%20%3D%3E%202x%20%3D%2014%5C%5C%5C%5C%20%3D%3E%20x%20%3D%2014/2%5C%5C%5C%5C%20%3D%3E%20x%20%3D%207\"">
Hence
AB = x + 8
=> AB = 7+8
=> AB = 15
AC = 6 + x
=> AC = 6 + 7
=> AC = 13
Answer- AB = 15 and AC = 13
Given- ABCD is a quadrilateral circumscribing a circle. P, Q, R, S are the point of contact on sides AB, BC, CD, and DA respectively.
To prove-
Proof -
Join OP, OQ, OR and OS
In triangle DOS and DOR,
OD =OD [common]
OS = OR [radii of same circle]
DR = DS [length of tangents drawn from an external point are equal ]
By SSS congruency, DOS DOR,
and by CPCT, DOS = DOR
.............(i)
Similarily,
...............(2, 3, 4)
SImilarily,
Hence proved.
Chapter No. | Chapter Name |
Chapter 1 | CBSE NCERT solutions for class 10 maths chapter 1 Real Numbers |
Chapter 2 | |
Chapter 3 | Solutions of NCERT class 10 maths chapter 3 Pair of Linear Equations in Two Variables |
Chapter 4 | CBSE NCERT solutions for class 10 maths chapter 4 Quadratic Equations |
Chapter 5 | NCERT solutions for class 10 chapter 5 Arithmetic Progressions |
Chapter 6 | |
Chapter 7 | CBSE NCERT solutions for class 10 maths chapter 7 Coordinate Geometry |
Chapter 8 | NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry |
Chapter 9 | Solutions of NCERT class 10 maths chapter 9 Some Applications of Trigonometry |
Chapter 10 | CBSE NCERT solutions class 10 maths chapter 10 Circles |
Chapter 11 | |
Chapter 12 | Solutions of NCERT class 10 chapter maths chapter 12 Areas Related to Circles |
Chapter 13 | CBSE NCERT solutions class 10 maths chapter 13 Surface Areas and Volumes |
Chapter 14 | |
Chapter 15 |
It is the chapter the logic as well as memory, both are tested equally.
First, you should learn the theorems, terminologies, and concepts of circles.
Once you have done the theorems, go through some examples of the NCERT textbook.
When you have done the above-said points, then come to the practice exercises where your understanding of concepts will be tested.
While doing the practice exercises, you can take help of NCERT solutions for class 10 maths chapter 10 circles.
After doing all this, the last thing you should solve is past papers related to the particular chapter.
Keep working hard & happy learning!
", "added_on": "2019-07-14 18:16:27.575201", "updated_on": "2020-04-14 14:43:10.857407", "featured_image": "img/featured/article-126402019-07-14-18-16-28.jpg", "stream_id": "18", "ec_article_id": "0", "ec_view_count": "0", "is_announcement": "0", "is_course": "0", "is_featured": "0", "is_news": "0", "exam_attribute_id": "0", "author_id": "465402", "editor_id": "0", "exam_id": "1388", "ec_data": "", "main_article": "0", "migrated": "0", "description": "NCERT Solutions for Class 10 Maths Chapter 10 Circles - In our previous classes, you have learnt that a circle is a closed shape with a collection of points in a plane that are at a specific distance ( called the radius) from a fixed point (called center). Solutions of NCERT class 10 maths chapter 10 Circles is covering the in-depth explanations to questions related to a circle. You have also studied important terms related to the circle like segment, arc, sector, chord, etc. In this chapter, there are two exercises with 17 questions in them. CBSE NCERT solutions for class 10 maths chapter 10 Circles are solved by subject experts to help students in their preparation keeping step by step marking in the mind.
This chapter introduces some complex and important terms like tangents, tangents to a circle, number of tangents from a point on a circle. In this chapter, we will study the different conditions that arise when a line and a circle are given in a plane. To solve these types of situations, we will learn the approach to apply the concept of the tangent to a circle in NCERT solutions for class 10 maths chapter 10 Circles. This chapter has fundamental concepts that are important for students in their future studies. Circles is a very interesting chapter due to the involvement of geometrical calculations and diagrams. These NCERT solutions can be used to study for competitive exams like JEE and NEET to build the basics. Two exercises of this chapter are explained below.
Q1 How many tangents can a circle have?
The lines that intersect the circle exactly at one single point are called tangents. In a circle, there can be infinitely many tangents.
(a) one
A tangent of a circle intersects the circle exactly in one single point.
(b) secant
It is a line that intersects the circle at two points.
(c) Two,
There can be only two parallel tangents to a circle.
(d) point of contact
The common point of a tangent and a circle.
(A) 12 cm
(B) 13 cm
(C) 8.5 cm
(D) cm.
The correct option is (d) = cm
It is given that the radius of the circle is 5 cm. OQ = 12 cm
According to question,
We know that
So, triangle OPQ is a right-angle triangle. By using Pythagoras theorem,
cm
AB is the given line and the line CD is the tangent to a circle at point M and parallels to the line AB. The line EF is a secant parallel to the AB
(A) 7 cm
(B) 12 cm
(C) 15 cm
(D) 24.5 cm
The correct option is (A) = 7 cm
Given that,
The length of the tangent (QT) is 24 cm and the length of OQ is 25 cm.
Suppose the length of the radius OT be cm.
We know that is a right angle triangle. So, by using Pythagoras theorem-
OT = 7 cm
(A)
(B)
(C)
(D)
The correct option is (b)
In figure,
Since POQT is quadrilateral. Therefore the sum of the opposite angles are 180
(A) 50°
(B) 60°
(C) 70°
(D) 80°
The correct option is (A)
It is given that, tangent PA and PB from point P inclined at
In triangle OAP and OBP
OA =OB (radii of the circle)
PA = PB (tangents of the circle)
Therefore, by SAS congruence
By CPCT,
Now, OPA = 80/4 = 40
In PAO,
= 50
Q4 Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Let line and line are two tangents of a circle and AB is the diameter of the circle.
OA and OB are perpendicular to the tangents and respectively.
therefore,
|| { 1 & 2 are alternate angles}
In the above figure, the line AXB is the tangent to a circle with center O. Here, OX is the perpendicular to the tangent AXB () at point of contact X.
Therefore, we have,
BXO + YXB =
OXY is a collinear
OX is passing through the center of the circle.
Given that,
the length of the tangent from the point A (AP) is 4 cm and the length of OA is 5 cm.
Since APO = 90^{0}
Therefore, APO is a right-angle triangle. By using Pythagoras theorem;
In the above figure, Pq is the chord to the larger circle, which is also tangent to a smaller circle at the point of contact R.
We have,
radius of the larger circle OP = OQ = 5 cm
radius of the small circle (OR) = 3 cm
OR PQ [since PQ is tangent to a smaller circle]
According to question,
In OPR and OQR
PRO = QRO {both }
OR = OR {common}
OP = OQ {both radii}
By RHS congruence OPR OQR
So, by CPCT
PR = RQ
Now, In OPR,
by using pythagoras theorem,
PR = 4 cm
Hence, PQ = 2.PR = 8 cm
To prove- AB + CD = AD + BC
Proof-
We have,
Since the length of the tangents drawn from an external point to a circle are equal
AP =AS .......(i)
BP = BQ.........(ii)
AS = AP...........(iii)
CR = CQ ...........(iv)
By adding all the equations, we get;
AP + BP +RD+ CR = AS +DS +BQ +CQ
(AP + BP) + (RD + CR) = (AS+DS)+(BQ + CQ)
AB + CD = AD + BC
Hence proved.