cbse maths assignment for class 10

Class 10 Maths Chapter 1 Assignments

Class 10 Maths Chapter 1 Real Numbers Assignments and Worksheets with solutions and answers updated for academic session 2024-25. All the chapter is divided into 4 assignments taking easy, average, and difficult questions. These set of questions provide a complete revision for the preparation of CBSE and State board exams 2024-25.

CBSE NCERT Class 10 Maths Chapter 1 Real Numbers Assignments

• Class 10 Maths Chapter 1 Real Numbers Assignments 1
• Class 10 Maths Chapter 1 Real Numbers Assignments 1 Solutions
• Class 10 Maths Chapter 1 Real Numbers Assignments 2
• Class 10 Maths Chapter 1 Real Numbers Assignments 2 Solutions
• Class 10 Maths Chapter 1 Real Numbers Assignments 3
• Class 10 Maths Chapter 1 Real Numbers Assignments 3 Solutions
• Class 10 Maths Chapter 1 Real Numbers Assignments 4
• Class 10 Maths Chapter 1 Real Numbers Assignments 4 Solutions
• Class 10 Maths Chapter 1 Real Numbers Solution Page

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There are 4 assignments and worksheets. Assignments contains MCQ, Fill in the Blanks, and True false questions. We have covered every topic in the chapter 1 of class 10 Maths. Answers and solutions of each assignment is also given free to use. Students can download these assignments and solve themselves. After solving they can check the answers given on website.

There are some easy questions based on understanding only. Some questions contains tricky methods and 17 percent questions are little bit difficult one. Students need to put effort to explore these questions. These assignments are guarantee to get good marks in exams. The questions based on Case Study will also be included in later chapters.

Download links are given for each assignments. Download and print it and solve it. Match the answers and solutions given on website to know the accuracy. Assignments are deign level wise, so do the first assignment first and the fourth assignment in the last.

Please suggest us, if any, about these assignments, so that we can improve the quality of the contents. We are preparing assignments for all the chapters. Gradually it will be uploaded before the final exams. Students are advised to solve these assignments to prepare the chapter and get confidence in topics.

Are assignments helpful in final exams?

Assignments are helpful in the revision of chapter as well as preparation for exams.

How many assignment should do to prepare the chapter?

At least 3 or 4 assignments which cover the entire chapter will be sufficient to know the chapter well.

How do solve the assignment?

Do yourself each assignment. If a part of assignment is not cleared or creating confusion, discuss with your classmate or teachers and solve it again.

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• NCERT Solutions Class 10
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NCERT Solutions Class 10 - Free PDF Download

NCERT solutions of Class 10 are meant for helping the students in their preparation for the exams. NCERT solutions for Class 10 include all the questions and problems that are there in the NCERT textbooks with practical solutions. Every solution is explained in a practical and detailed manner in the NCERT solutions to help students get a clear knowledge of the subjects of Class 10. Mostly, the questions included in the NCERT solutions class 10 are likely to come in the exams, thus making the students more confident in getting good marks and aim to strengthen the core knowledge of the students by helping them in understanding the basics of every subject. Complications arise in every subject, but with the Class 10 NCERT solutions, students can face every complication.

Detailed Overview of NCERT Solutions for Class 10

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Subject wise NCERT Solutions for Class 10

Mentioned below in the table is the link to subject-wise NCERT solutions Class 10:

NCERT Solutions for Class 10 Maths

The NCERT Solutions for Class 10 Maths cover all the questions in the NCERT Maths textbook exercises. They're put together by our expert teachers to help you with your exam prep. Going through these solutions can really help you understand the topics in your NCERT Solutions for Class 10 Maths syllabus better. When you've finished your syllabus, it's a good idea to go through these solutions because they give you lots of practice questions for each topic. If you're stuck on any question, these solutions can clear things up for you. Plus, they make for great extra study material. They're written in a simple way that's easy to understand, just like your textbooks. Make sure to download and practice them regularly to do well in your board exams. Understanding each step in the solutions is key for effective prep.

NCERT Solutions for Class 10 Science 

The NCERT Solutions for Class 10 Science are a great resource for students to grasp tricky concepts and prepare for their Class 10 Board exams. By using these solutions, students can gauge their performance and identify their strengths and weaknesses. Understanding the topics thoroughly is crucial for exam success, and these solutions help achieve just that. Science requires a solid grasp of concepts, and these solutions simplify the learning process with their easy-to-understand language.

Our subject experts have prepared these solutions in line with the NCERT Solutions for Class 10 Science syllabus, covering Physics, Chemistry, and Biology comprehensively. They are particularly beneficial for students struggling with challenging problems. Regular practice with these solutions is recommended to excel in the Class 10 Board exams, making them an ideal study companion for efficient exam preparation.

NCERT Solutions for Class 10 Social Science

Similarly, for NCERT Solutions for Class 10 Social Science , the NCERT Textbook has been divided into History, Political Science, Geography, and Economics. Our team at Vedantu has provided separate answers for each book in the NCERT Solutions for Class 10 Social Science. These expertly crafted solutions are designed for easy comprehension, ensuring that students can access them effortlessly. Links to these solutions are provided below for easy access.

NCERT Solutions for Class 10 Hindi

The NCERT solutions for Class 10 Hindi offer detailed and expertly explained solutions for all the questions in the NCERT Hindi Books. They're a great help for students, especially when tackling homework or getting ready for exams. The questions and answers provided at the end of each chapter in NCERT Hindi Books are not just crucial for exams, but they also aid in grasping the concepts more effectively. That's why we suggest going through the NCERT Solutions for Class 10 Hindi thoroughly, taking notes, and jotting down solutions for each chapter to speed up your revision process. We believe that these solutions will be beneficial for you.

NCERT Solutions for Class 10 English

The NCERT solutions for class 10 English offer detailed, expert-level answers to all the questions found in the NCERT English Books. They're a real help for students, whether you're tackling homework or getting ready for exams. The questions and answers provided at the end of each chapter in the NCERT English Books aren't just crucial for exams—they're also really important for grasping the concepts more effectively. That's why we suggest going through the NCERT Solutions for class 10 English carefully and jotting down notes and solutions for each chapter. It'll make revising a whole lot quicker and easier.

NCERT Solutions for Class 10 in Hindi Medium

Unlocking Success: A Guide to Excelling in Class 10 with NCERT Solutions

As we all know, Class 10 is important in every student's career, so it is necessary to score good marks in Class 10. A student with good scoring in Class 10 can expect to have a good career in the future. Scoring good marks in Class 10 is not that easy because the subjects are complicated and difficult. Students facing trouble in Class 10 must download the NCERT Solutions Class 10 pdf to ease the burden and become confident of scoring good marks. NCERT solutions Class 10 pdf helps the students in making their preparation more efficient and productive. Students following this pdf can strengthen their core knowledge and achieve some expertise in every subject, even the difficult ones like Maths, Chemistry, physics, etc.

The questions in NCERT Class 10 solution pdf are solved in a practical way to explain to the students every aspect of a particular subject.

Foundation for Success: Mastering Class 10 Subjects with Comprehensive NCERT Solutions 

Class 10 being an important stage in every student's career; it is important that every student understands each subject of Class 10 in a detailed way. The learnings in Class 10 act as a basis for the future studies whatsoever the stream may be. Every stream has some subjects that are completely based on Class 10 subjects, so a student with good knowledge of Class 10 subjects will find higher studies understandable. A complete and detailed understanding of every subject in Class 10 is only possible through the  NCERT solution .

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NCERT solutions Class 10 are prepared by some expert teachers who have years of experience in this field, thus ensuring the students that the quality of the solutions is top class. As the solutions are prepared by experienced teachers, there are no mistakes committed in the solution; thus, the students can refer to it for their preparation confidently without any doubts. The teachers tried their best to explain each solution in a detailed and simpler manner to make students understand every aspect of a particular subject.

Practice Questions

Class 10th NCERT solutions have practical problems and exercises included in every chapter so as to refresh the knowledge a student has gained in a particular chapter. The exercise tries to cover the whole chapter for revision purposes. Through the NCERT Class 10th solution , students become capable of understanding every difficult area in a subject without causing much trouble. Students become confident and are not afraid of any surprises in the final exam - This confidence helps the students to score the highest possible marks in the exams.

Benefits of NCERT Solution of Class 10

NCERT solutions Class 10 is the key to achieve success for every student as it helps in strengthening the core knowledge of every student and make them experts in each and every subject. Some of the benefits of the NCERT Class 10th solution are:

The solutions are prepared according to the rules and format issued by the CBSE board, thus giving the students a preview of the pattern of questions they are going to face in the final exams.

The solutions are prepared by some expert teachers who have years of experience in this field, thus ensuring that there are no mistakes committed in these solutions. The students can be confident in the quality of the solutions.

The solutions are given to the questions which are most likely to come in the exams, thus making the preparation of the students efficient.

Important Related Links for NCERT Class 10

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Important Questions for CBSE Class 10

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CBSE Class 10 Maths Formulas

Vedantu's Class 10 NCERT solutions offers a comprehensive and user-friendly resource for students. The solutions, designed in simple language, help students understand complex concepts with ease. Vedantu's approach ensures that learning is engaging and effective. The solutions provided align with NCERT guidelines, making them a reliable companion for exam preparation. The platform's commitment to quality education is evident through its clear explanations and step-by-step solutions. Thus, Vedantu's NCERT Solutions for Class 10 serve as an invaluable tool for students, fostering a deeper understanding of subjects and boosting confidence in tackling examinations.

FAQs on NCERT Solutions Class 10

1. What is the Need for NCERT Solutions in Class 10?

In every student's life, Class 10 is an important stage as the marks scored in Class 10 can decide his/her future career. So it is important to score the highest possible marks in Class 10. It can be difficult because the subjects in Class 10 are complex and confusing, but with the help of NCERT solutions , students can easily understand the subjects of Class 10 without facing any trouble. NCERT solution can act as a saviour for students and can play an important role in strengthening the core knowledge of the students.

2. What are the Benefits of NCERT Solutions Class 10?

Some of the benefits of NCERT solutions Class 10 are discussed below:

The solutions are solved according to the format issued by the CBSE board, thus making it possible for the students to practice the format of the exam before appearing it.

The solutions here are for the questions that have the most possibility to come in the exams, thus making the preparation of the students more efficient and better.

The solutions are prepared by some teachers who are hired for their excellence in this field. These expert teachers help in improving the quality of the solutions. They also guarantee that no mistakes are committed in the solutions.

3. Which Chapters are covered in the NCERT Solutions for Class 10?

All the chapters from all the subjects for Class 10 are covered in the solutions pdf. Each pdf for the particular subject covers the chapters that are there for each subject prescribed by the CBSE. For example, Class 10 Science solutions are divided into Chemistry, Physics and Biology . Each chapter from each part is available for free download on the Vedantu website. All chapters are available for download online. Each chapter has a summary, solved NCERT textbook questions and extra questions for reference. 

4. How can I use Class 10 Maths NCERT Solutions to prepare for exams?

The NCERT Solutions are books developed by experts and experienced professionals to help students understand concepts better. The NCERT Solutions for Class 10 Maths has all the exercises and examples solved stepwise for the students to understand. Extra questions for additional practice are also given. The NCERT Solutions PDF is available for free download on the website. The students can practice thoroughly for the exam with the help of the PDF. Practising regularly can help you ace your exam. 

5. How many Chapters are there in Class 10 English?

Class 10 English has two textbooks prescribed by the NCERT in the syllabus. The two books are- First Flight and Footprints without Feet. The First Flight is the main literature textbook. It has 11 chapters. It includes stories touching on topics about life and learning. The other book has 10 chapters and all of them are prose or short poems. The NCERT Solution pdf provided for each of the books are concise and developed keeping in mind that all students will be able to understand them. They also have extra questions for the students to practice. 

6. How many Chapters are there in the NCERT Solutions for Class 10 Maths?

The NCERT Solutions for Class 10 Maths covers all 15 chapters from the NCERT textbook . The pdf has solved questions that are explained in a less complex manner step by step. It also has the list of formulas and concepts explained along with extra questions for the students to practice. Preparing from the NCERT textbook and the solutions PDF can assure you good marks in the final exam. 

7. Is Class 10 easy?

Class 10 is considered to be an important class in a student’s school life. It is the Class where you build your foundation for higher classes and decide what you want to do in Class 11 and 12 by choosing the appropriate subjects. It is also a class where you have to appear for board exams. The Class is not tough or daunting for anyone who has been thorough throughout the year. You have to be regular and practice what you learn in class everyday and you will be able to score more than 90% in your final exams. 

8. What are Class 10 NCERT Solutions?

Class 10 NCERT solutions are comprehensive guides that offer detailed explanations and answers to questions found in the Class 10 NCERT textbooks for subjects like Maths, Science, English, Hindi, and Social Science.

9. How can NCERT Solutions for Class 10 Maths benefit students?

NCERT solutions for Class 10 Maths provide step-by-step explanations to problems found in the Class 10 Maths textbook, aiding students in understanding concepts and preparing effectively for exams.

10. Where can I find NCERT Solutions for Class 10 Science?

You can find NCERT solutions for Class 10 Science online or in educational resources, offering solutions to questions from the Class 10 Science textbook, covering topics in Physics, Chemistry, and Biology.

11. Why are NCERT Solutions for Class 10 English important?

NCERT solutions for Class 10 English help students in comprehending literature, grammar, and language skills, enabling them to tackle questions from the Class 10 English textbook with confidence.

12. How do NCERT Solutions for Class 10 Hindi aid students in their studies?

NCERT solutions for Class 10 Hindi provide detailed explanations and interpretations of poems, stories, and grammar rules, assisting students in mastering the language and scoring well in exams.

13. What subjects are covered in NCERT Solutions for Class 10 Social Science?

NCERT solutions for Class 10 Social Science cover subjects like History, Geography, Political Science, and Economics, offering explanations and insights into key concepts and topics.

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• Chapter 3: Linear Equations In Two Variables

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

Ncert solutions for class 10 maths chapter 3 – cbse download free pdf.

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables  will help the students in understanding how the problems under this concept are solved. Maths is one subject that requires a lot of practice. The students appearing for the 10 th grade board examinations can turn to the NCERT Solutions Class 10 for reference. These solutions of the Chapter Pair of Linear Equations in Two Variables give step-wise answers to all the Maths problems in the NCERT textbook . An equation that can be of the form ax + by + c = 0, where a, b and c are real numbers, and a and b are not both zero, is called a linear equation in two variables x and y.

Download Exclusively Curated Chapter Notes for Class 10 Maths Chapter – 3 Pair of Linear Equations in Two Variables

Download most important questions for class 10 maths chapter – 3 pair of linear equations in two variables.

The NCERT Solutions for Class 10 Maths Chapter 3 also lets the students understand the fact that a solution of such an equation is a pair of values, one for x and the other for y, which makes the two sides of the equation equal. Students also learn that every solution of the equation is a point on the line representing it.

Students get to learn the problem-solving method of the questions present in Chapter 3 Pair of Linear Equations in Two Variables of CBSE Syllabus for 2023-24, revolving around the concepts mentioned above and more, by practising the NCERT Solutions provided below. These Solutions of NCERT are extremely beneficial from the CBSE examination perspective.

• Chapter 1 Real Numbers
• Chapter 2 Polynomials
• Chapter 3 Pair of Linear Equations in Two Variables
• Chapter 4 Quadratic Equations
• Chapter 5 Arithmetic Progressions
• Chapter 6 Triangles
• Chapter 7 Coordinate Geometry
• Chapter 8 Introduction to Trigonometry
• Chapter 9 Some Applications of Trigonometry
• Chapter 10 Circles
• Chapter 11 Constructions
• Chapter 12 Areas Related to Circles
• Chapter 13 Surface Areas and Volumes
• Chapter 14 Statistics
• Chapter 15 Probability
• Exercise 3.1 Chapter 3 Pair of Linear Equations in Two Variables
• Exercise 3.2 Chapter 3 Pair of Linear Equations in Two Variables
• Exercise 3.3 Chapter 3 Pair of Linear Equations in Two Variables
• Exercise 3.4 Chapter 3 Pair of Linear Equations in Two Variables
• Exercise 3.5 Chapter 3 Pair of Linear Equations in Two Variables
• Exercise 3.6 Chapter 3 Pair of Linear Equations in Two Variables
• Exercise 3.7 Chapter 3 Pair of Linear Equations in Two Variables

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Access Answers to NCERT Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variables

Exercise 3.1 page: 44.

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.

Solutions: Let the present age of Aftab be ‘x’.

And, the present age of his daughter be ‘y’.

Now, we can write, seven years ago,

Age of Aftab = x-7

Age of his daughter = y-7

According to the question,

x−7 = 7(y−7)

⇒x−7 = 7y−49

⇒x−7y = −42         ………………………(i)

Also, three years from now or after three years,

Age of Aftab will become = x+3.

Age of his daughter will become = y+3

According to the situation given,

x+3 = 3(y+3)

⇒x+3 = 3y+9

⇒x−3y = 6       …………..…………………(ii)

Subtracting equation (i) from equation (ii) we have

(x−3y)−(x−7y) = 6−(−42)

⇒−3y+7y = 6+42

The algebraic equation is represented by

For, x−7y = −42 or x = −42+7y

The solution table is

For,  x−3y = 6   or     x = 6+3y

The graphical representation is:

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.

Solutions: Let us assume that the cost of a bat be ‘Rs x’

And,the cost of a ball be ‘Rs y’

According to the question, the algebraic representation is

3x+6y = 3900

And x+3y = 1300

For, 3x+6y = 3900

Or x = (3900-6y)/3

For, x+3y = 1300

Or x = 1300-3y

The graphical representation is as follows.

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.

Solutions: Let the cost of 1 kg of apples be ‘Rs. x’

And, cost of 1 kg of grapes be ‘Rs. y’

And 4x+2y = 300

For, 2x+y = 160 or y = 160−2x, the solution table is;

For 4x+2y = 300 or y = (300-4x)/2, the solution table is;

The graphical representation is as follows;

Exercise 3.2 Page: 49

1. Form the pair of linear equations in the following problems, and find their solutions graphically.

(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.

(ii) 5 pencils and 7 pens together cost 50, whereas 7 pencils and 5 pens together cost 46. Find the cost of one pencil and that of one pen.

(i)Let there are x number of girls and y number of boys. As per the given question, the algebraic expression can be represented as follows.

Now, for x+y = 10 or x = 10−y, the solutions are;

For x – y = 4 or x = 4 + y, the solutions are;

From the graph, it can be seen that the given lines cross each other at point (7, 3). Therefore, there are 7 girls and 3 boys in the class.

(ii) Let 1 pencil costs Rs.x and 1 pen costs Rs.y.

According to the question, the algebraic expression cab be represented as;

5x + 7y = 50

7x + 5y = 46

For, 5x + 7y = 50 or  x = (50-7y)/5, the solutions are;

For 7x + 5y = 46 or x = (46-5y)/7, the solutions are;

Hence, the graphical representation is as follows;

From the graph, it is can be seen that the given lines cross each other at point (3, 5).

So, the cost of a pencil is 3/- and cost of a pen is 5/-.

2. On comparing the ratios a1/a2 , b1/b2 , c1/c2 find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:

(i) 5x – 4y + 8 = 0

7x + 6y – 9 = 0

(ii) 9x + 3y + 12 = 0

18x + 6y + 24 = 0

(iii) 6x – 3y + 10 = 0

2x – y + 9 = 0

(i) Given expressions;

5x−4y+8 = 0

7x+6y−9 = 0

Comparing these equations with a1x+b1y+c1 = 0

And a2x+b2y+c2 = 0

a1 = 5, b1 = -4, c1 = 8

a2 = 7, b2 = 6, c2 = -9

(a1/a2) = 5/7

(b1/b2) = -4/6 = -2/3

(c1/c2) = 8/-9

Since, (a1/a2) ≠ (b1/b2)

So, the pairs of equations given in the question have a unique solution and the lines cross each other at exactly one point.

(ii) Given expressions;

9x + 3y + 12 = 0

a1 = 9, b1 = 3, c1 = 12

a2 = 18, b2 = 6, c2 = 24

(a1/a2) = 9/18 = 1/2

(b1/b2) = 3/6 = 1/2

(c1/c2) = 12/24 = 1/2

Since (a1/a2) = (b1/b2) = (c1/c2)

So, the pairs of equations given in the question have infinite possible solutions and the lines are coincident.

(iii) Given Expressions;

6x – 3y + 10 = 0

a1 = 6, b1 = -3, c1 = 10

a2 = 2, b2 = -1, c2 = 9

(a1/a2) = 6/2 = 3/1

(b1/b2) = -3/-1 = 3/1

(c1/c2) = 10/9

Since (a1/a2) = (b1/b2) ≠ (c1/c2)

So, the pairs of equations given in the question are parallel to each other and the lines never intersect each other at any point and there is no possible solution for the given pair of equations.

3. On comparing the ratio, (a1/a2) , (b1/b2) , (c1/c2) find out whether the following pair of linear equations are consistent, or inconsistent.

(i) 3x + 2y = 5 ; 2x – 3y = 7

(ii) 2x – 3y = 8 ; 4x – 6y = 9

(iii)(3/2)x+(5/3)y = 7; 9x – 10y = 14

(iv) 5x – 3y = 11 ; – 10x + 6y = –22

(v)(4/3)x+2y = 8 ; 2x + 3y = 12

(i) Given : 3x + 2y = 5 or 3x + 2y -5 = 0

and 2x – 3y = 7 or 2x – 3y -7 = 0

Comparing these equations with a1x+b1y+c1 = 0

a1 = 3, b1 = 2, c1 = -5

a2 = 2, b2 = -3, c2 = -7

(a1/a2) = 3/2

(b1/b2) = 2/-3

(c1/c2) = -5/-7 = 5/7

So, the given equations intersect each other at one point and they have only one possible solution. The equations are consistent.

(ii) Given 2x – 3y = 8 and 4x – 6y = 9

a1 = 2, b1 = -3, c1 = -8

a2 = 4, b2 = -6, c2 = -9

(a1/a2) = 2/4 = 1/2

(b1/b2) = -3/-6 = 1/2

(c1/c2) = -8/-9 = 8/9

Since , (a1/a2) = (b1/b2) ≠ (c1/c2)

So, the equations are parallel to each other and they have no possible solution. Hence, the equations are inconsistent.

(iii)Given (3/2)x + (5/3)y = 7 and 9x – 10y = 14

a1 = 3/2, b1 = 5/3, c1 = -7

a2 = 9, b2 = -10, c2 = -14

(a1/a2) = 3/(2×9) = 1/6

(b1/b2) = 5/(3× -10)= -1/6

(c1/c2) = -7/-14 = 1/2

So, the equations are intersecting  each other at one point and they have only one possible solution. Hence, the equations are consistent.

(iv) Given, 5x – 3y = 11 and – 10x + 6y = –22

a1 = 5, b1 = -3, c1 = -11

a2 = -10, b2 = 6, c2 = 22

(a1/a2) = 5/(-10) = -5/10 = -1/2

(b1/b2) = -3/6 = -1/2

(c1/c2) = -11/22 = -1/2

These linear equations are coincident lines and have infinite number of possible solutions. Hence, the equations are consistent.

(v)Given, (4/3)x +2y = 8 and 2x + 3y = 12

a1 = 4/3 , b1= 2 , c1 = -8

a2 = 2, b2 = 3 , c2 = -12

(a1/a2) = 4/(3×2)= 4/6 = 2/3

(b1/b2) = 2/3

(c1/c2) = -8/-12 = 2/3

4. Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:

(i) x + y = 5, 2x + 2y = 10

(ii) x – y = 8, 3x – 3y = 16

(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0

(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0

(i)Given, x + y = 5 and 2x + 2y = 10

(a1/a2) = 1/2

(b1/b2) = 1/2

(c1/c2) = 1/2

∴The equations are coincident and they have infinite number of possible solutions.

So, the equations are consistent.

For, x + y = 5 or x = 5 – y

For 2x + 2y = 10 or x = (10-2y)/2

So, the equations are represented in graphs as follows:

From the figure, we can see, that the lines are overlapping each other.

Therefore, the equations have infinite possible solutions.

(ii) Given, x – y = 8 and 3x – 3y = 16

(a1/a2) = 1/3

(b1/b2) = -1/-3 = 1/3

(c1/c2) = 8/16 = 1/2

Since, (a1/a2) = (b1/b2) ≠ (c1/c2)

The equations are parallel to each other and have no solutions. Hence, the pair of linear equations is inconsistent.

(iii) Given, 2x + y – 6 = 0 and 4x – 2y – 4 = 0

(a1/a2) = 2/4 = ½

(b1/b2) = 1/-2

(c1/c2) = -6/-4 = 3/2

The given linear equations are intersecting each other at one point and have only one solution. Hence, the pair of linear equations is consistent.

Now, for 2x + y – 6 = 0 or y = 6 – 2x

And for 4x – 2y – 4 = 0 or y = (4x-4)/2

From the graph, it can be seen that these lines are intersecting each other at only one point,(2,2).

(iv) Given, 2x – 2y – 2 = 0 and 4x – 4y – 5 = 0

(b1/b2) = -2/-4 = 1/2

(c1/c2) = 2/5

Since, a1/a2 = b1/b2 ≠ c1/c2

Thus, these linear equations have parallel and have no possible solutions. Hence, the pair of linear equations are inconsistent.

5. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

Solutions: Let us consider.

The width of the garden is x and length is y.

Now, according to the question, we can express the given condition as;

Now, taking y – x = 4 or y = x + 4

For y + x = 36, y = 36 – x

The graphical representation of both the equation is as follows;

From the graph you can see, the lines intersects each other at a point(16, 20). Hence, the width of the garden is 16 and length is 20.

6. Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:

(i) Intersecting lines

(ii) Parallel lines

(iii) Coincident lines

(i) Given the linear equation 2x + 3y – 8 = 0.

To find another linear equation in two variables such that the geometrical representation of the pair so formed is intersecting lines, it should satisfy below condition;

(a1/a2) ≠ (b1/b2)

Thus, another equation could be 2x – 7y + 9 = 0, such that;

(a1/a2) = 2/2 = 1 and (b1/b2) = 3/-7

Clearly, you can see another equation satisfies the condition.

(ii) Given the linear equation 2x + 3y – 8 = 0.

To find another linear equation in two variables such that the geometrical representation of the pair so formed is parallel lines, it should satisfy below condition;

(a1/a2) = (b1/b2) ≠ (c1/c2)

Thus, another equation could be 6x + 9y + 9 = 0, such that;

(a1/a2) = 2/6 = 1/3

(b1/b2) = 3/9= 1/3

(c1/c2) = -8/9

(iii) Given the linear equation 2x + 3y – 8 = 0.

To find another linear equation in two variables such that the geometrical representation of the pair so formed is coincident lines, it should satisfy below condition;

(a1/a2) = (b1/b2) = (c1/c2)

Thus, another equation could be 4x + 6y – 16 = 0, such that;

(a1/a2) = 2/4 = 1/2 ,(b1/b2) = 3/6 = 1/2, (c1/c2) = -8/-16 = 1/2

7. Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.

Solution: Given, the equations for graphs are x – y + 1 = 0 and 3x + 2y – 12 = 0.

For, x – y + 1 = 0 or x = -1+y

For, 3x + 2y – 12 = 0 or x = (12-2y)/3

Hence, the graphical representation of these equations is as follows;

From the figure, it can be seen that these lines are intersecting each other at point (2, 3) and x-axis at (−1, 0) and (4, 0). Therefore, the vertices of the triangle are (2, 3), (−1, 0), and (4, 0).

Exercise 3.3 Page: 53

1. Solve the following pair of linear equations by the substitution method

(i) x + y = 14

(ii) s – t = 3

(s/3) + (t/2) = 6

(iii) 3x – y = 3

9x – 3y = 9

(iv) 0.2x + 0.3y = 1.3

0.4x + 0.5y = 2.3

(v) √2 x+√3 y = 0

√3 x-√8 y = 0

(vi) (3x/2) – (5y/3) = -2

(x/3) + (y/2) = (13/6)

x + y = 14 and x – y = 4 are the two equations.

From 1 st equation, we get,

Now, substitute the value of x in second equation to get,

(14 – y) – y = 4

14 – 2y = 4

By the value of y, we can now find the exact value of x;

∵ x = 14 – y

∴ x = 14 – 5

Hence, x = 9 and y = 5.

(ii) Given,

s – t = 3 and (s/3) + (t/2) = 6 are the two equations.

s = 3 + t ________________(1)

Now, substitute the value of s in second equation to get,

(3+t)/3 + (t/2) = 6

⇒ (2(3+t) + 3t )/6 = 6

⇒ (6+2t+3t)/6 = 6

⇒ (6+5t) = 36

Now, substitute the value of t in equation (1)

s = 3 + 6 = 9

Therefore, s = 9 and t = 6.

(iii) Given,

3x – y = 3 and 9x – 3y = 9 are the two equations.

x = (3+y)/3

Now, substitute the value of x in the given second equation to get,

9(3+y)/3 – 3y = 9

⇒9 +3y -3y = 9

Therefore, y has infinite values and since, x = (3+y) /3, so x also has infinite values.

(iv) Given,

0.2x + 0.3y = 1.3 and 0.4x + 0.5y = 2.3are the two equations.

x = (1.3- 0.3y)/0.2 _________________(1)

0.4(1.3-0.3y)/0.2 + 0.5y = 2.3

⇒2(1.3 – 0.3y) + 0.5y = 2.3

⇒ 2.6 – 0.6y + 0.5y = 2.3

⇒ 2.6 – 0.1 y = 2.3

⇒ 0.1 y = 0.3

Now, substitute the value of y in equation (1), we get,

x = (1.3-0.3(3))/0.2 = (1.3-0.9)/0.2 = 0.4/0.2 = 2

Therefore, x = 2 and y = 3.

√2 x + √3 y = 0 and √3 x – √8 y = 0

are the two equations.

x = – (√3/√2)y __________________(1)

Putting the value of x in the given second equation to get,

√3(-√3/√2)y – √8y = 0 ⇒ (-3/√2)y- √8 y = 0

Therefore, x = 0 and y = 0.

(vi) Given,

(3x/2)-(5y/3) = -2 and (x/3) + (y/2) = 13/6 are the two equations.

(3/2)x = -2 + (5y/3)

⇒ x = 2(-6+5y)/9 = (-12+10y)/9 ………………………(1)

((-12+10y)/9)/3 + y/2 = 13/6

⇒y/2 = 13/6 –( (-12+10y)/27 ) + y/2 = 13/6

(3x/2) – 5(3)/3 = -2

⇒ (3x/2) – 5 = -2

2. Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.

2x + 3y = 11…………………………..(I)

2x – 4y = -24………………………… (II)

From equation (II), we get

x = (11-3y)/2 ………………….(III)

Substituting the value of x in equation (II), we get

2(11-3y)/2 – 4y = 24

11 – 3y – 4y = -24

y = 5……………………………………..(IV)

Putting the value of y in equation (III), we get

x = (11-3×5)/2 = -4/2 = -2

Hence, x = -2, y = 5

Therefore the value of m is -1.

3. Form the pair of linear equations for the following problems and find their solution by substitution method.

(i) The difference between two numbers is 26 and one number is three times the other. Find them.

Let the two numbers be x and y respectively, such that y > x.

y = 3x ……………… (1)

y – x = 26 …………..(2)

Substituting the value of (1) into (2), we get

3x – x = 26

x = 13 ……………. (3)

Substituting (3) in (1), we get y = 39

Hence, the numbers are 13 and 39.

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

Let the larger angle by x o  and smaller angle be y o .

We know that the sum of two supplementary pair of angles is always 180 o .

x + y = 180 o ……………. (1)

x – y = 18 o  ……………..(2)

From (1), we get x = 180 o  – y …………. (3)

Substituting (3) in (2), we get

180 o  – y – y =18 o

162 o  = 2y

y = 81 o  ………….. (4)

Using the value of y in (3), we get

x = 180 o  – 81 o

Hence, the angles are 99 o  and 81 o .

(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs.3800. Later, she buys 3 bats and 5 balls for Rs.1750. Find the cost of each bat and each ball.

Let the cost a bat be x and cost of a ball be y.

7x + 6y = 3800 ………………. (I)

3x + 5y = 1750 ………………. (II)

From (I), we get

y = (3800-7x)/6………………..(III)

Substituting (III) in (II). we get,

3x+5(3800-7x)/6 =1750

⇒3x+ 9500/3 – 35x/6 = 1750

⇒3x- 35x/6 = 1750 – 9500/3

⇒(18x-35x)/6 = (5250 – 9500)/3

⇒-17x/6 = -4250/3

⇒-17x = -8500

x = 500 ……………………….. (IV)

Substituting the value of x in (III), we get

y = (3800-7 ×500)/6 = 300/6 = 50

Hence, the cost of a bat is Rs 500 and cost of a ball is Rs 50.

(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?

Let the fixed charge be Rs x and per km charge be Rs y.

x + 10y = 105 …………….. (1)

x + 15y = 155 …………….. (2)

From (1), we get x = 105 – 10y ………………. (3)

Substituting the value of x in (2), we get

105 – 10y + 15y = 155

y = 10 …………….. (4)

Putting the value of y in (3), we get

x = 105 – 10 × 10 = 5

Hence, fixed charge is Rs 5 and per km charge = Rs 10

Charge for 25 km = x + 25y = 5 + 250 = Rs 255

(v) A fraction becomes 9/11 , if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction.

Let the fraction be x/y.

(x+2) /(y+2) = 9/11

11x + 22 = 9y + 18

11x – 9y = -4 …………….. (1)

(x+3) /(y+3) = 5/6

6x + 18 = 5y +15

6x – 5y = -3 ………………. (2)

From (1), we get x = (-4+9y)/11 …………….. (3)

6(-4+9y)/11 -5y = -3

-24 + 54y – 55y = -33

y = 9 ………………… (4)

Substituting the value of y in (3), we get

x = (-4+9×9 )/11 = 7

Hence the fraction is 7/9.

(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

Let the age of Jacob and his son be x and y respectively.

(x + 5) = 3(y + 5)

x – 3y = 10 …………………………………….. (1)

(x – 5) = 7(y – 5)

x – 7y = -30 ………………………………………. (2)

From (1), we get x = 3y + 10 ……………………. (3)

3y + 10 – 7y = -30

y = 10 ………………… (4)

x = 3 x 10 + 10 = 40

Hence, the present age of Jacob’s and his son is 40 years and 10 years respectively.

Exercise 3.4 Page: 56

1. Solve the following pair of linear equations by the elimination method and the substitution method:

(i) x + y = 5 and 2x – 3y = 4

(ii) 3x + 4y = 10 and 2x – 2y = 2

(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7

(iv) x/2+ 2y/3 = -1 and x-y/3 = 3

By the method of elimination.

x + y = 5 ……………………………….. (i)

2x – 3y = 4 ……………………………..(ii)

When the equation (i) is multiplied by 2, we get

2x + 2y = 10 ……………………………(iii)

When the equation (ii) is subtracted from (iii) we get,

y = 6/5 ………………………………………(iv)

Substituting the value of y in eq. (i) we get,

x=5−6/5 = 19/5

∴x = 19/5 , y = 6/5

By the method of substitution.

From the equation (i), we get:

x = 5 – y………………………………….. (v)

When the value is put in equation (ii) we get,

2(5 – y) – 3y = 4

When the values are substituted in equation (v), we get:

x =5− 6/5 = 19/5

∴x = 19/5 ,y = 6/5

3x + 4y = 10……………………….(i)

2x – 2y = 2 ………………………. (ii)

When the equation (i) and (ii) is multiplied by 2, we get:

4x – 4y = 4 ………………………..(iii)

When the Equation (i) and (iii) are added, we get:

x = 2 ……………………………….(iv)

Substituting equation (iv) in (i) we get,

6 + 4y = 10

Hence, x = 2 and y = 1

By the method of Substitution

From equation (ii) we get,

x = 1 + y……………………………… (v)

Substituting equation (v) in equation (i) we get,

3(1 + y) + 4y = 10

When y = 1 is substituted in equation (v) we get,

A = 1 + 1 = 2

Therefore, A = 2 and B = 1

By the method of elimination:

3x – 5y – 4 = 0 ………………………………… (i)

9x = 2y + 7

9x – 2y – 7 = 0 …………………………………(ii)

When the equation (i) and (iii) is multiplied we get,

9x – 15y – 12 = 0 ………………………………(iii)

When the equation (iii) is subtracted from equation (ii) we get,

y = -5/13 ………………………………………….(iv)

When equation (iv) is substituted in equation (i) we get,

3x +25/13 −4=0

∴x = 9/13 and y = -5/13 

By the method of Substitution:

From the equation (i) we get,

x = (5y+4)/3 …………………………………………… (v)

Putting the value (v) in equation (ii) we get,

9(5y+4)/3 −2y −7=0

Substituting this value in equation (v) we get,

x = (5(-5/13)+4)/3

∴x = 9/13 , y = -5/13

(iv) x/2 + 2y/3 = -1 and x-y/3 = 3

By the method of Elimination.

3x + 4y = -6 …………………………. (i)

3x – y = 9 ……………………………. (ii)

When the equation (ii) is subtracted from equation (i) we get,

y = -3 ………………………………….(iii)

When the equation (iii) is substituted in (i) we get,

3x – 12 = -6

Hence, x = 2 , y = -3

From the equation (ii) we get,

x = (y+9)/3…………………………………(v)

Putting the value obtained from equation (v) in equation (i) we get,

3(y+9)/3 +4y =−6

When y = -3 is substituted in equation (v) we get,

x = (-3+9)/3 = 2

Therefore, x = 2 and y = -3

2. Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:

(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes if we only add 1 to the denominator. What is the fraction?

Let the fraction be a/b

According to the given information,

(a+1)/(b-1) = 1

=> a – b = -2 ………………………………..(i)

a/(b+1) = 1/2

=> 2a-b = 1…………………………………(ii)

When equation (i) is subtracted from equation (ii) we get,

a = 3 …………………………………………………..(iii)

When a = 3 is substituted in equation (i) we get,

Hence, the fraction is 3/5.

(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

Let us assume, present age of Nuri is x

And present age of Sonu is y.

According to the given condition, we can write as;

x – 5 = 3(y – 5)

x – 3y = -10…………………………………..(1)

x + 10 = 2(y +10)

x – 2y = 10…………………………………….(2)

Subtract eq. 1 from 2, to get,

y = 20 ………………………………………….(3)

Substituting the value of y in eq.1, we get,

x – 3.20 = -10

x – 60 = -10

Age of Nuri is 50 years

Age of Sonu is 20 years.

(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

Let the unit digit and tens digit of a number be x and y respectively.

Then, Number (n) = 10B + A

N after reversing order of the digits = 10A + B

According to the given information, A + B = 9…………………….(i)

9(10B + A) = 2(10A + B)

88 B – 11 A = 0

-A + 8B = 0 ………………………………………………………….. (ii)

Adding the equations (i) and (ii) we get,

B = 1……………………………………………………………………….(3)

Substituting this value of B, in the equation (i) we get A= 8

Hence the number (N) is 10B + A = 10 x 1 +8 = 18

(iv) Meena went to a bank to withdraw Rs.2000. She asked the cashier to give her Rs.50 and Rs.100 notes only. Meena got 25 notes in all. Find how many notes of Rs.50 and Rs.100 she received.

Let the number of Rs.50 notes be A and the number of Rs.100 notes be B

A + B = 25 ……………………………………………………………………….. (i)

50A + 100B = 2000 ………………………………………………………………(ii)

When equation (i) is multiplied with (ii) we get,

50A + 50B = 1250 …………………………………………………………………..(iii)

Subtracting the equation (iii) from the equation (ii) we get,

Substituting in the equation (i) we get,

Hence, Meena has 10 notes of Rs.50 and 15 notes of Rs.100.

(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs.27 for a book kept for seven days, while Susy paid Rs.21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Let the fixed charge for the first three days be Rs.A and the charge for each day extra be Rs.B.

According to the information given,

A + 4B = 27 …………………………………….…………………………. (i)

A + 2B = 21 ……………………………………………………………….. (ii)

When equation (ii) is subtracted from equation (i) we get,

B = 3 …………………………………………………………………………(iii)

Substituting B = 3 in equation (i) we get,

A + 12 = 27

Hence, the fixed charge is Rs.15

And the Charge per day is Rs.3

Exercise 3.5 Page: 62

1. Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method.

(i) x – 3y – 3 = 0 and 3x – 9y – 2 = 0 (ii) 2x + y = 5 and 3x + 2y = 8

(iii) 3x – 5y = 20 and 6x – 10y = 40 (iv) x – 3y – 7 = 0 and 3x – 3y – 15 = 0

(i) Given, x – 3y – 3 =0 and 3x – 9y -2 =0

a 1 /a 2 =1/3 ,         b 1 /b 2 = -3/-9 =1/3,     c 1 /c 2 =-3/-2 = 3/2

(a 1 /a 2 ) = (b 1 /b 2 ) ≠ (c 1 /c 2 )

Since, the given set of lines are parallel to each other they will not intersect each other and therefore there is no solution for these equations.

(ii) Given, 2x + y = 5 and 3x +2y = 8

a 1 /a 2 = 2/3 , b 1 /b 2 = 1/2 , c 1 /c 2 = -5/-8

(a 1 /a 2 ) ≠ (b 1 /b 2 )

Since they intersect at a unique point these equations will have a unique solution by cross multiplication method:

x/(b 1 c 2 -c 1 b 2 ) = y/(c 1 a 2 – c 2 a1) = 1/(a 1 b 2 -a 2 b 1 )

x/(-8-(-10)) = y/(-15-(-16)) = 1/(4-3)

x/2 = y/1 = 1

∴ x = 2 and y =1

(iii) Given, 3x – 5y = 20 and 6x – 10y = 40

(a 1 /a 2 ) = 3/6 = 1/2

(b 1 /b 2 ) = -5/-10 = 1/2

(c 1 /c 2 ) = 20/40 = 1/2

a 1 /a 2 = b 1 /b 2 = c 1 /c 2

Since the given sets of lines are overlapping each other there will be infinite number of solutions for this pair of equation.

(iv) Given, x – 3y – 7 = 0 and 3x – 3y – 15 = 0

(a 1 /a 2 ) = 1/3

(b 1 /b 2 ) = -3/-3 = 1

(c 1 /c 2 ) = -7/-15

a 1 /a 2 ≠ b 1 /b 2

Since this pair of lines are intersecting each other at a unique point, there will be a unique solution.

By cross multiplication,

x/(45-21) = y/(-21+15) = 1/(-3+9)

x/24 = y/ -6 = 1/6

x/24 = 1/6 and y/-6 = 1/6

∴ x = 4 and y = 1.

2. (i) For which values of a and b does the following pair of linear equations have an infinite number of solutions?

2x + 3y = 7

(a – b) x + (a + b) y = 3a + b – 2

(ii) For which value of k will the following pair of linear equations have no solution?

(2k – 1) x + (k – 1) y = 2k + 1

(i) 3y + 2x -7 =0

(a + b)y + (a-b)y – (3a + b -2) = 0

a 1 /a 2 = 2/(a-b) ,               b 1 /b 2 = 3/(a+b) ,               c 1 /c 2 = -7/-(3a + b -2)

For infinitely many solutions,

Thus 2/(a-b) = 7/(3a+b– 2)

6a + 2b – 4 = 7a – 7b

a – 9b = -4  ……………………………….(i)

2/(a-b) = 3/(a+b)

2a + 2b = 3a – 3b

a – 5b = 0 ……………………………….….(ii)

Subtracting (i) from (ii), we get

Substituting this eq. in (ii), we get

a -5 x 1= 0

Thus at a = 5 and b = 1 the given equations will have infinite solutions.

(ii) 3x + y -1 = 0

(2k -1)x  +  (k-1)y – 2k -1 = 0

a 1 /a 2 = 3/(2k -1) ,           b 1 /b 2 = 1/(k-1), c 1 /c 2  = -1/(-2k -1) = 1/( 2k +1)

For no solutions

a 1 /a 2 = b 1 /b 2 ≠ c 1 /c 2

3/(2k-1) = 1/(k -1)     ≠ 1/(2k +1)

3/(2k –1) = 1/(k -1)

3k -3 = 2k -1

Therefore, for k = 2 the given pair of linear equations will have no solution.

3. Solve the following pair of linear equations by the substitution and cross-multiplication methods:

8x + 5y = 9

3x + 2y = 4

8x + 5y = 9 …………………..(1)

3x + 2y = 4 ……………….….(2)

From equation (2) we get

x = (4 – 2y )/ 3  ……………………. (3)

Using this value in equation 1, we get

8(4-2y)/3 + 5y = 9

32 – 16y +15y = 27

y = 5 ……………………………….(4)

Using this value in equation (2), we get

3x + 10 = 4

Thus, x = -2 and y = 5.

Now, Using Cross Multiplication method:

8x +5y – 9 = 0

3x + 2y – 4 = 0

x/(-20+18) = y/(-27 + 32 ) = 1/(16-15)

-x/2 = y/5 =1/1

∴ x = -2 and y =5.

4. Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method:

(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay Rs.1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs.1180 as hostel charges. Find the fixed charges and the cost of food per day.

(ii) A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 when 8 is added to its denominator. Find the fraction.

(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?

(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?

(v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

(i) Let x be the fixed charge and y be the charge of food per day.

x + 20y = 1000……………….. (i)

x + 26y = 1180………………..(ii)

Subtracting (i) from  (ii) we get

Using this value in equation (ii) we get

x = 1180 -26 x 30

Therefore, fixed charges is Rs.400 and charge per day is Rs.30.

(ii)  L et the fraction be x/y.

So, as per the question given,

(x -1)/y = 1/3 => 3x – y = 3…………………(1)

x/(y + 8) = 1/4  => 4x –y =8 ………………..(2)

Subtracting equation (1) from (2) , we get

x = 5 ………………………………………….(3)

Using this value in equation (2), we get,

(4×5)– y = 8

Therefore, the fraction is 5/12.

(iii) Let the number of right answers is x and number of wrong answers be y

According to the given question;

3x−y=40……..(1)

⇒2x−y=25…….(2)

Subtracting equation (2) from equation (1), we get;

x = 15 ….….(3)

Putting this in equation (2), we obtain;

30 – y = 25

Therefore, number of right answers = 15 and number of wrong answers = 5

Hence, total number of questions = 20

(iv)  Let x km/h be the speed of car from point A and y km/h be the speed of car from point B.

If the car travels in the same direction,

5x – 5y = 100

x – y = 20 …………………………………(i)

If the car travels in the opposite direction,

x + y = 100………………………………(ii)

Solving equation (i) and (ii), we get

x = 60 km/h………………………………………(iii)

Using this in equation (i), we get,

60 – y = 20

y = 40 km/h

Therefore, the speed of car from point A = 60 km/h

Speed of car from point B = 40 km/h.

The length of rectangle = x unit

And breadth of the rectangle = y unit

Now, as per the question given,

(x – 5) (y + 3) = xy -9

3x – 5y – 6 = 0……………………………(1)

(x + 3) (y + 2) = xy + 67

2x + 3y – 61 = 0…………………………..(2)

Using cross multiplication method, we get,

x/(305 +18) = y/(-12+183) = 1/(9+10)

x/323 = y/171 = 1/19

Therefore, x = 17 and y = 9.

Hence, the length of rectangle = 17 units

And breadth of the rectangle = 9 units

Exercise 3.6 Page: 67

1. Solve the following pairs of equations by reducing them to a pair of linear equations:

(i) 1/2x + 1/3y = 2

1/3x + 1/2y = 13/6

Let us assume 1/x = m and 1/y = n  , then the equation will change as follows.

m/2 + n/3 = 2

⇒ 3m+2n-12 = 0…………………….(1)

m/3 + n/2 = 13/6

⇒ 2m+3n-13 = 0……………………….(2)

Now, using cross-multiplication method, we get,

m/(-26-(-36) ) = n/(-24-(-39)) = 1/(9-4)

m/10 = n/15 = 1/5

m/10 = 1/5 and n/15 = 1/5

So, m = 2 and n = 3

1/x = 2 and 1/y = 3

x = 1/2 and y = 1/3

(ii) 2/√x + 3/√y = 2

4/√x + 9/√y = -1

Substituting 1/√x = m and 1/√y = n in the given equations, we get

2m + 3n = 2 ………………………..(i)

4m – 9n = -1 ………………………(ii)

Multiplying equation (i) by 3, we get

6m + 9n = 6 ………………….…..(iii)

Adding equation (ii) and (iii), we get

m = 1/2…………………………….…(iv)

Now by putting the value of ‘m’ in equation (i), we get

2×1/2 + 3n = 2

Hence, x = 4 and y = 9

(iii) 4/x + 3y = 14

3/x -4y = 23

Putting in the given equation we get,

So, 4m + 3y = 14     => 4m + 3y – 14 = 0  ……………..…..(1)

3m – 4y = 23     => 3m – 4y – 23 = 0  ……………………….(2)

By cross-multiplication, we get,

m/(-69-56) = y/(-42-(-92)) = 1/(-16-9)

-m/125 = y/50 = -1/ 25

-m/125 = -1/25 and y/50 = -1/25

m = 5 and b = -2

m = 1/x = 5

So , x = 1/5

(iv) 5/(x-1) + 1/(y-2) = 2

6/(x-1) – 3/(y-2) = 1

Substituting 1/(x-1) = m and 1/(y-2) = n  in the given equations, we get,

5m + n = 2 …………………………(i)

6m – 3n = 1 ……………………….(ii)

15m + 3n = 6 …………………….(iii)

Adding (ii) and (iii) we get

Putting this value in equation (i), we get

5×1/3 + n = 2

n = 2- 5/3 = 1/3

m = 1/ (x-1)

⇒ 1/3 = 1/(x-1)

n = 1/(y-2)

⇒ 1/3 = 1/(y-2)

Hence, x = 4 and y = 5

(v) (7x-2y)/ xy = 5

(8x + 7y)/xy = 15

(7x-2y)/ xy = 5

7/y – 2/x = 5…………………………..(i)

8/y + 7/x = 15…………………………(ii)

Substituting 1/x =m in the given equation we get,

– 2m + 7n = 5     => -2 + 7n – 5 = 0  ……..(iii)

7m + 8n = 15     => 7m + 8n – 15 = 0 ……(iv)

By cross-multiplication method, we get,

m/(-105-(-40)) = n/(-35-30) = 1/(-16-49)

m/(-65) = n/(-65) = 1/(-65)

m/-65 = 1/-65

n/(-65) = 1/(-65)

m = 1 and n = 1

m = 1/x = 1        n = 1/x = 1

Therefore, x = 1 and y = 1

(vi) 6x + 3y = 6xy

2x + 4y = 5xy

6x + 3y = 6xy

6/y + 3/x = 6

Let 1/x = m and 1/y = n

=> 6n +3m = 6

=>3m + 6n-6 = 0…………………….(i)

=> 2/y + 4/x = 5

=> 2n +4m = 5

=> 4m+2n-5 = 0……………………..(ii)

3m + 6n – 6 = 0

4m + 2n – 5 = 0

By cross-multiplication method, we get

m/(-30 –(-12)) = n/(-24-(-15)) = 1/(6-24)

m/-18 = n/-9 = 1/-18

m/-18 = 1/-18

n/-9 = 1/-18

m = 1 and n = 1/2

m = 1/x = 1 and n = 1/y = 1/2

x = 1 and y = 2

Hence, x = 1 and y = 2

(vii) 10/(x+y) + 2/(x-y) = 4

15/(x+y) – 5/(x-y) = -2

Substituting 1/x+y = m and 1/x-y = n in the given equations, we get,

10m + 2n = 4      =>  10m + 2n – 4 = 0      ………………..…..(i)

15m – 5n = -2     =>   15m – 5n + 2 = 0    ……………………..(ii)

Using cross-multiplication method, we get,

m/(4-20) = n/(-60-(20)) = 1/(-50 -30)

m/-16 = n/-80 = 1/-80

m/-16 = 1/-80 and n/-80 = 1/-80

m = 1/5 and n = 1

m = 1/(x+y) = 1/5

x+y = 5 …………………………………………(iii)

n = 1/(x-y) = 1

x-y = 1……………………………………………(iv)

Adding equation (iii) and (iv), we get

2x = 6   => x = 3 …….(v)

Putting the value of x = 3 in equation (3), we get

Hence, x = 3 and y = 2

(viii) 1/(3x+y) + 1/(3x-y) = 3/4

1/2(3x+y) – 1/2(3x-y) = -1/8

Substituting 1/(3x+y) = m and 1/(3x-y) = n in the given equations, we get,

m + n = 3/4 …………………………….…… (1)

m/2 – n/2 = -1/8

m – n = -1/4  …………………………..…(2)

Adding (1) and (2), we get

2m = 3/4 – 1/4

Putting in (2), we get

1/4 – n = -1/4

n = 1/4 + 1/4 = 1/2

m = 1/(3x+y) = 1/4

3x + y = 4  …………………………………(3)

n = 1/( 3x-y) = 1/2

3x – y = 2 ………………………………(4)

Adding equations (3) and (4), we get

x = 1 ……………………………….(5)

Putting in (3), we get

3(1) + y = 4

Hence, x = 1 and y = 1

2. Formulate the following problems as a pair of equations, and hence find their solutions:

(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.

(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.

(iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

(i) Let us consider,

Speed of Ritu in still water = x km/hr

Speed of Stream = y km/hr

Now, speed of Ritu during,

Downstream = x + y km/h

Upstream = x – y km/h

As per the question given,

2(x+y) = 20

Or x + y = 10……………………….(1)

And, 2(x-y) = 4

Or x – y = 2………………………(2)

Adding both the eq.1 and 2, we get,

Putting the value of x in eq.1, we get,

Speed of Ritu rowing in still water = 6 km/hr

Speed of Stream = 4 km/hr

(ii) Let us consider,

Number of days taken by women to finish the work = x

Number of days taken by men to finish the work = y

Work done by women in one day = 1/x

Work done by women in one day = 1/y

4(2/x + 5/y) = 1

(2/x + 5/y) = 1/4

And, 3(3/x + 6/y) = 1

(3/x + 6/y) = 1/3

Now, put 1/x=m and 1/y=n, we get,

2m + 5n = 1/4 => 8m + 20n = 1…………………(1)

3m + 6n =1/3 => 9m + 18n = 1………………….(2)

Now, by cross multiplication method, we get here,

m/(20-18) = n/(9-8) = 1/ (180-144)

m/2 = n/1 = 1/36

m = 1/x = 1/18

n = 1/y = 1/36

Number of days taken by women to finish the work = 18

Number of days taken by men to finish the work = 36.

(iii) Let us consider,

Speed of the train = x km/h

Speed of the bus = y km/h

According to the given question,

60/x + 240/y = 4 …………………(1)

100/x + 200/y = 25/6 …………….(2)

Put 1/x=m and 1/y=n, in the above two equations;

60m + 240n = 4……………………..(3)

100m + 200n = 25/6

600m + 1200n = 25 ………………….(4)

Multiply eq.3 by 10, to get,

600m + 2400n = 40 ……………………(5)

Now, subtract eq.4 from 5, to get,

n = 15/1200 = 1/80

Substitute the value of n in eq. 3, to get,

60m + 3 = 4

m = 1/x = 1/60

And y = 1/n

Speed of the train = 60 km/h

Speed of the bus = 80 km/h

Exercise 3.7 Page: 68

1. The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.

The age difference between Ani and Biju is 3 yrs.

Either Biju is 3 years older than Ani or Ani is 3 years older than Biju. 

From both cases, we find out that Ani’s father’s age is 30 yrs more than that of Cathy’s age.

Let the ages of Ani and Biju be A and B, respectively.

Therefore, the age of Dharam = 2 x A = 2A yrs.

And the age of Biju’s sister Cathy is B/2 yrs.

By using the information that is given,

When Ani is older than Biju by 3 yrs, then A – B = 3 …..(1)

2A − B/2 = 30

4A – B = 60 ….(2)

By subtracting the equation (1) from (2), we get;

3A = 60 – 3 = 57

A = 57/3 = 19

Therefore, the age of Ani = 19 yrs

And the age of Biju is 19 – 3 = 16 yrs.

When Biju is older than Ani,

B – A = 3 ….(1)

Adding the equations (1) and (2), we get;

Therefore, the age of Ani is 21 yrs

And the age of Biju is 21 + 3 = 24 yrs.

2. One says, “Give me a hundred, friend! I shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount of their (respective) capital? [From the Bijaganita of Bhaskara II] [Hint : x + 100 = 2(y – 100), y + 10 = 6(x – 10)].

Let the capital amount with two friends be Rs. x and Rs. y, respectively. 

As per the given,

x + 100 = 2(y − 100)…..(i)

6(x − 10) = (y + 10)….(ii)

Consider the equation (i),

x + 100 = 2(y − 100)

x + 100 = 2y − 200

x − 2y = −300…..(iii)

From equation (ii),

6x − 60 = y + 10

6x − y = 70…..(iv)

(iv) × 2 – (iii)

12x – 2y – (x – 2y) = 140 – (-300)

Substituting x = 40 in equation (iii), we get;

40 – 2y = -300

Therefore, the two friends had Rs. 40 and Rs. 170 with them.

3. A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.

Let the speed of the train be x km/hr and the time taken by the train to travel a distance be t hours, and the d km be the distance.

Speed of the train = Distance travelled by train / Time taken to travel that distance

d = xt …..(i)

Case 1: When the speed of the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time.

(x + 10) = d/(t – 2)

(x + 10)(t – 2) = d

xt + 10t – 2x – 20 = d

d + 10t – 2x = 20 + d [From (i)]

10t – 2x = 20…..(ii)

Case 2: When the train was slower by 10 km/h, it would have taken 3 hours more than the scheduled time.

So, (x – 10) = d/(t + 3)

(x – 10)(t + 3) = d

xt – 10t + 3x – 30 = d

d – 10t + 3x = 30 + d [From (i)]

-10t + 3x = 30…..(iii)

Adding (ii) and (iii), we get;

Thus, the speed of the train is 50 km/h.

Substituting x = 50 in equation (ii), we get;

10t – 100 = 20

t = 12 hours

Distance travelled by train, d = xt

Hence, the distance covered by the train is 600 km.

4. The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.

Let x be the number of rows and y be the number of students in a row.

Total students in the class = Number of rows × Number of students in a row

Total number of students = (x − 1) (y + 3)

xy = (x − 1) (y + 3) 

xy = xy − y + 3x − 3

3x − y − 3 = 0

3x − y = 3…..(i)

Total number of students = (x + 2) (y − 3)

xy = xy + 2y − 3x − 6

3x − 2y = −6…..(ii)

Subtracting equation (ii) from (i), we get;

(3x − y) − (3x − 2y) = 3 − (−6)

− y + 2y = 9

Substituting y = 9 in equation (i), we get;

Therefore, the total number of students in a class = xy = 4 × 9 = 36

5. In a ∆ABC, ∠ C = 3 ∠ B = 2 (∠ A + ∠ B). Find the three angles.

∠C = 3 ∠B = 2(∠B + ∠A)

3∠B = 2 ∠A + 2 ∠B

2∠A – ∠B= 0- – – – – – – – – – – – (i)

We know that the sum of a triangle’s interior angles is 180°.

Thus, ∠ A +∠B + ∠C = 180°

∠A + ∠B +3 ∠B = 180°

∠A + 4 ∠B = 180°– – – – – – – – – – – – – – -(ii)

Multiplying equation (i) by 4, we get;

8 ∠A – 4 ∠B = 0- – – – – – – – – – – – (iii)

Adding equations (iii) and (ii), we get;

9 ∠A = 180°

Using this in equation (ii), we get;

20° + 4∠B = 180°

∠C = 3∠B = 3 x 40 = 120°

Therefore, ∠A = 20°, ∠B = 40°, and ∠C = 120°.

6. Draw the graphs of the equations 5x – y = 5 and 3x – y = 3. Determine the co-ordinates of the vertices of the triangle formed by these lines and the y axis.

⇒ y = 5x – 5

Its solution table will be.

Also given,3x – y = 3

The graphical representation of these lines will be as follows:

From the above graph, we can see that the coordinates of the vertices of the triangle formed by the lines and the y-axis are (1, 0), (0, -5) and (0, -3).

7. Solve the following pair of linear equations:

(i) px + qy = p – q

qx – py = p + q

(ii) ax + by = c

bx + ay = 1 + c

(iii) x/a – y/b = 0

ax + by = a 2 + b 2

(iv) (a – b)x + (a + b) y = a 2 – 2ab – b 2

(a + b)(x + y) = a 2 + b 2

(v) 152x – 378y = – 74

–378x + 152y = – 604

(i) px + qy = p – q……………(i)

qx – py = p + q……………….(ii)

Multiplying equation (i) by p and equation (ii) by q, we get;

p 2 x + pqy = p 2  − pq ………… (iii)

q 2 x − pqy = pq + q 2  ………… (iv)

Adding equations (iii) and (iv), we get;

p 2 x + q 2 x = p 2 + q 2

(p 2 + q 2 ) x = p 2  + q 2

x = (p 2 + q 2 )/ (p 2 + q 2 ) = 1

Substituting x = 1 in equation (i), we have;

p(1) + qy = p – q

qy = p – q – p

(ii) ax + by= c…………………(i)

bx + ay = 1+ c………… ..(ii)

Multiplying equation (i) by a and equation (ii) by b, we get;

a 2 x + aby = ac ………………… (iii)

b 2 x + aby = b + bc…………… (iv)

Subtracting equation (iv) from equation (iii),

(a 2 – b 2 ) x = ac − bc– b

x = (ac − bc – b)/ (a 2 – b 2 )

x = c(a – b) – b / (a 2  – b 2 )

From equation (i), we obtain

ax + by = c

a{c(a − b) − b)/ (a 2 – b 2 )} + by = c

{[ac(a−b)−ab]/ (a 2 – b 2 )} + by = c

by = c – {[ac(a − b) − ab]/(a 2  – b 2 )}

by = (a 2 c – b 2 c – a 2 c + abc + ab)/ (a 2 – b 2 )

by = [abc – b 2 c + ab]/ (a 2 – b 2 )

by = b(ac – bc + a)/(a 2 – b 2 )

y = [c(a – b) + a]/(a 2 – b 2 )

x/a – y/b = 0

⇒ bx − ay = 0 ……. (i)

ax + by = a 2 + b 2 …….. (ii)

Multiplying equations (i) and (ii) by b and a, respectively, we get;

b 2 x − aby = 0 …………… (iii)

a 2 x + aby = a 3 + ab 2  …… (iv)

b 2 x + a 2 x = a 3 + ab 2

x(b 2 + a 2 ) = a(a 2  + b 2 ) 

Substituting x = 1 in equation (i), we get;

b(a) − ay = 0

ab − ay = 0

(a + b) y + (a – b) x = a 2 − 2ab − b 2  …………… (i)

(x + y)(a + b) = a 2 + b 2

(a + b) y + (a + b) x = a 2 + b 2  ………………… (ii)

Subtracting equation (ii) from equation (i), we get;

(a − b) x − (a + b) x = (a 2 − 2ab − b 2 ) − (a 2 + b 2 )

x(a − b − a − b) = − 2ab − 2b 2

− 2bx = − 2b (a + b)

Substituting x = a + b in equation (i), we get;

y (a + b) + (a + b)(a − b) = a 2 − 2ab – b 2

a 2 − b 2  + y(a + b) = a 2 − 2ab – b 2

(a + b)y = −2ab

y = -2ab/(a + b)

(v) 152x – 378y = – 74 

152x – 378y = – 74 ….(i)

–378x + 152y = – 604….(ii)

From equation (i),

152x + 74 = 378y

y = (152x + 74)/378

y = (76x + 37)/189…..(iii)

Substituting the value of y in equation (ii), we get;

-378x + 152[(76x + 37)/189] = -604

(-378x)189  + [152(76x) + 152(37)] = (-604)(189)

-71442x + 11552x + 5624 = -114156

-59890x = -114156 – 5624 = -119780

x = -119780/-59890

Substituting x = 2 in equation (iii), we get;

y = [76(2) + 37]/189

= (152 + 37)/189

Therefore, x = 2 and y = 1

8. ABCD is a cyclic quadrilateral (see Fig. 3.7). Find the angles of the cyclic quadrilateral.

Given that ABCD is a cyclic quadrilateral.

As we know, the opposite angles of a cyclic quadrilateral are supplementary.

∠A + ∠C = 180

4y + 20 + (-4x) = 180

-4x + 4y = 160

⇒ -x + y = 40….(i)

∠B + ∠D = 180

3y – 5 + (-7x + 5) = 180

⇒ -7x + 3y = 180…..(ii)

Equation (ii) – 3 × (i),

-7x + 3y – (-3x + 3y) = 180 – 120

Substituting x = -15 in equation (i), we get;

-(-15) + y = 40

y = 40 – 15 = 25

Therefore, x = -15 and y = 25.

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

Chapter 3 – Pair of Linear Equations in Two Variables holds a weightage of 11 marks in the examinations. This chapter gives an account of the various topics related to the linear equations in two variables . The topics discussed in the chapter are mentioned below:

3.1 Introduction In earlier classes, you have studied Linear Equations in Two Variables. You have also studied that a Linear Equation in Two Variables has infinitely many solutions. In this chapter, the knowledge of Linear Equations in Two Variables shall be recalled and extended to that of Pair of Linear Equations In Two Variables. 

3.2 Pair of Linear Equations in Two Variables An equation which can be put in the form ax + by + c = 0, where a, b and c are real numbers, and a and b are not zeros, is called a linear equation in two variables x and y. The solution of such a problem is a pair of values, one for x and the other for y, which makes the two sides of the equation equal. The topic also discusses the geometrical representation of Pair of Linear Equations in Two Variables along with suitable examples.

3.3 Graphical Method of Solution of a Pair of Linear Equations In the previous section, you have seen how we can graphically represent a pair of linear equations as two lines. You have also seen that the lines may intersect, or may be parallel, or may coincide. In this section, you will know how to solve it in each case from the geometrical point of view.

3.4 Algebraic Methods of Solving a Pair of Linear Equations In the previous section, we discussed how to solve a pair of linear equations graphically. In some of the cases, the graphical method is not convenient. In this topic, we shall discuss various algebraic methods such as the Substitution Method, Elimination Method, and Cross – Multiplication Method. Each subtopic is explained elaborately with suitable examples, for better understanding.

3.5 Equations Reducible to a Pair of Linear Equation in Two Variables In this section, we shall discuss the solution of such pairs of equations which are not linear but can be reduced to linear form by making some suitable substitutions. The process is explained through some examples related to the subtopic.

3.6 Summary The Summary section consists of the overall important points that need to be memorized while solving the exercise questions of the chapter Pair of Linear Equations in Two Variables. The points mentioned in this section will help you to revise all the concepts mentioned in the chapter.

Two linear equations in the same two variables are called pair of linear equations in two variables. The pair of linear equations in two variables can be represented graphically and algebraically. The graph can be represented by two lines:

• If the lines intersect at a point, the pair of equations is said to be consistent.
• If the lines coincide, the pair of equations is dependent.
• If the lines are parallel, the pair of equations is inconsistent.

Algebraically, the following methods can be used to solve the pair of linear equations in two variables:

• Substitution method
• Elimination method
• Cross-multiplication method

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

• NCERT Solutions are created by subject experts.
• The answers are provided after a lot of brainstorming and are accurate.
• These contain questions related to all the important topics.
• NCERT Solutions Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variables includes the solutions to the exercises given in the textbook as well.

The faculty have curated the NCERT Class 10 Maths Solutions in a lucid manner to improve the problem-solving abilities of the students. For a more clear idea about Pair of Linear Equations in Two Variables, students can refer to the study materials available at BYJU’S.

• RD Sharma Solutions for Class 10 Maths Pair of Linear Equations in Two Variables

Disclaimer – 

Dropped Topics – 

3.2 Pair of linear equations in two variables 3.3 Graphical method of solution of a pair of linear equations

3.4.3 Cross-multiplication method 3.5 Equation reducible to a pair of linear equations in two variables

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We have provided below free printable Class 10 Mathematics Number System Assignments for Download in PDF. The Assignments have been designed based on the latest NCERT Book for Class 10 Mathematics Number System . These Assignments for Grade 10 Mathematics Number System cover all important topics which can come in your standard 10 tests and examinations. Free printable Assignments for CBSE Class 10 Mathematics Number System , school and class assignments, and practice test papers have been designed by our highly experienced class 10 faculty. You can free download CBSE NCERT printable Assignments for Mathematics Number System Class 10 with solutions and answers. All Assignments and test sheets have been prepared by expert teachers as per the latest Syllabus in Mathematics Number System Class 10. Students can click on the links below and download all Pdf Assignments for Mathematics Number System class 10 for free. All latest Kendriya Vidyalaya Class 10 Mathematics Number System Assignments with Answers and test papers are given below.

Mathematics Number System Class 10 Assignments Pdf Download

We have provided below the biggest collection of free CBSE NCERT KVS Assignments for Class 10 Mathematics Number System . Students and teachers can download and save all free Mathematics Number System assignments in Pdf for grade 10th. Our expert faculty have covered Class 10 important questions and answers for Mathematics Number System as per the latest syllabus for the current academic year. All test papers and question banks for Class 10 Mathematics Number System and CBSE Assignments for Mathematics Number System Class 10 will be really helpful for standard 10th students to prepare for the class tests and school examinations. Class 10th students can easily free download in Pdf all printable practice worksheets given below.

Topicwise Assignments for Class 10 Mathematics Number System Download in Pdf

Advantages of Class 10 Mathematics Number System Assignments

• As we have the best and largest collection of Mathematics Number System assignments for Grade 10, you will be able to easily get full list of solved important questions which can come in your examinations.
• Students will be able to go through all important and critical topics given in your CBSE Mathematics Number System textbooks for Class 10 .
• All Mathematics Number System assignments for Class 10 have been designed with answers. Students should solve them yourself and then compare with the solutions provided by us.
• Class 10 Students studying in per CBSE, NCERT and KVS schools will be able to free download all Mathematics Number System chapter wise worksheets and assignments for free in Pdf
• Class 10 Mathematics Number System question bank will help to improve subject understanding which will help to get better rank in exams

Frequently Asked Questions by Class 10 Mathematics Number System students

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We provide here Standard 10 Mathematics Number System chapter-wise assignments which can be easily downloaded in Pdf format for free.

You can click on the links above and get assignments for Mathematics Number System in Grade 10, all topic-wise question banks with solutions have been provided here. You can click on the links to download in Pdf.

We have provided here topic-wise Mathematics Number System Grade 10 question banks, revision notes and questions for all difficult topics, and other study material.

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