case study class 10 maths linear equations

Class 10 Maths Case Study Questions Chapter 3 Pair of Linear Equations in Two Variables

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Case study Questions in the Class 10 Mathematics Chapter 3  are very important to solve for your exam. Class 10 Maths Chapter 3 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving case study-based   questions for Class 10 Maths Chapter 3  Pair of Linear Equations in Two Variables

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In CBSE Class 10 Maths Paper, Students will have to answer some questions based on Assertion and Reason. There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Pair of Linear Equations in Two Variables Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 10 Maths  Chapter 3 Pair of Linear Equations in Two Variables

Case Study/Passage-Based Questions

(i) 1 st  situation can be represented algebraically as

Answer: (d) 2x+3y=46

(ii) 2 nd  situation can be represented algebraically as

Answer: (c) 3x + 5y = 74

(iii), Fare from Ben~aluru to Malleswaram is

Answer: (b) Rs 8

(iv) Fare from Bengaluru to Yeswanthpur is

Answer: (a) Rs 10

(v) The system oflinear equations represented by both situations has

Answer: (c) unique solution

Case Study 2: The scissors which are so common in our daily life use, its blades represent the graph of linear equations.

Let the blades of a scissor are represented by the system of linear equations:

x + 3y = 6 and 2x – 3y = 12

(i) The pivot point (point of intersection) of the blades represented by the linear equation x + 3y = 6 and 2x – 3y = 12 of the scissor is (a) (2, 3) (b) (6, 0) (c) (3, 2) (d) (2, 6)

Answer: (b) (6, 0)

(ii) The points at which linear equations x + 3y = 6 and 2x – 3y = 12 intersect y – axis respectively are (a) (0, 2) and (0, 6) (b) (0, 2) and (6, 0) (c) (0, 2) and (0, –4) (d) (2, 0) and (0, –4)

Answer: (c) (0, 2) and (0, –4)

(iii) The number of solution of the system of linear equations x + 2y – 8 = 0 and 2x + 4y = 16 is (a) 0 (b) 1 (c) 2 (d) infinitely many

Answer: (d) infinitely many

(iv) If (1, 2) is the solution of linear equations ax + y = 3 and 2x + by = 12, then values of a and b are respectively (a) 1, 5 (b) 2, 3 (c) –1, 5 (d) 3, 5

Answer: (a) 1, 5

(v) If a pair of linear equations in two variables is consistent, then the lines represented by two equations are (a) intersecting (b) parallel (c) always coincident (d) intersecting or coincident

Answer: (d) intersecting or coincident

Hope the information shed above regarding Case Study and Passage Based Questions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 10 Maths Pair of Linear Equations in Two Variables Case Study and Passage Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible By Team Study Rate

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Class 10 maths case study based questions chapter 3 pair of linear equations cbse board term 1 with answer key.

          Hello students, Welcome to Maths Easy Institute.  

CASE STUDY 1:   

A library is a collection of materials, books, and media that are easily accessible to everyone. Here, you can find books from different genres such that as science fiction, fiction, and many research papers. It is also a great place to socialize with your community members, which will help you build relationships with people of similar interests.

One day, two friends Sarita and Babita go to a library for some books. The library has fixed charges for the first 3 days and an additional charge for each day thereafter. If anyone takes the book for 10 days then he/she has to pay fixed charges for 3 days and additional charges for 7 days. 

Sarita take a book from Library for 7 days and paid 27 rs and Babita  take a book from Library for 5 days and paid 21 rs .

(a) If fixed charges for the library for the first 3 days is x Rs and additional charges for each day is y Rs/day then pair of linear equation satisfying the  Sarita case is:

(b)  If fixed charges for the library for first 3 days is x Rs and additional charges for each is y Rs/day then pair of linear equation satisfying Babita case is:

(c) Fixed charges and additional charges of the library is:

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CBSE Class 10 Maths Case Study Questions PDF

Download Case Study Questions for Class 10 Mathematics to prepare for the upcoming CBSE Class 10 Final Exam. These Case Study and Passage Based questions are published by the experts of CBSE Experts for the students of CBSE Class 10 so that they can score 100% on Boards.

CBSE Class 10 Mathematics Exam 2024  will have a set of questions based on case studies in the form of MCQs. The CBSE Class 10 Mathematics Question Bank on Case Studies, provided in this article, can be very helpful to understand the new format of questions. Share this link with your friends.

Table of Contents

Chapterwise Case Study Questions for Class 10 Mathematics

Inboard exams, students will find the questions based on assertion and reasoning. Also, there will be a few questions based on case studies. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

The above  Case studies for Class 10 Maths will help you to boost your scores as Case Study questions have been coming in your examinations. These CBSE Class 10 Mathematics Case Studies have been developed by experienced teachers of cbseexpert.com for the benefit of Class 10 students.

• Class 10th Science Case Study Questions
• Assertion and Reason Questions of Class 10th Science
• Assertion and Reason Questions of Class 10th Social Science

Class 10 Maths Syllabus 2024

Chapter-1  real numbers.

Starting with an introduction to real numbers, properties of real numbers, Euclid’s division lemma, fundamentals of arithmetic, Euclid’s division algorithm, revisiting irrational numbers, revisiting rational numbers and their decimal expansions followed by a bunch of problems for a thorough and better understanding.

Chapter-2  Polynomials

This chapter is quite important and marks securing topics in the syllabus. As this chapter is repeated almost every year, students find this a very easy and simple subject to understand. Topics like the geometrical meaning of the zeroes of a polynomial, the relationship between zeroes and coefficients of a polynomial, division algorithm for polynomials followed with exercises and solved examples for thorough understanding.

Chapter-3  Pair of Linear Equations in Two Variables

This chapter is very intriguing and the topics covered here are explained very clearly and perfectly using examples and exercises for each topic. Starting with the introduction, pair of linear equations in two variables, graphical method of solution of a pair of linear equations, algebraic methods of solving a pair of linear equations, substitution method, elimination method, cross-multiplication method, equations reducible to a pair of linear equations in two variables, etc are a few topics that are discussed in this chapter.

Chapter-4  Quadratic Equations

The Quadratic Equations chapter is a very important and high priority subject in terms of examination, and securing as well as the problems are very simple and easy. Problems like finding the value of X from a given equation, comparing and solving two equations to find X, Y values, proving the given equation is quadratic or not by knowing the highest power, from the given statement deriving the required quadratic equation, etc are few topics covered in this chapter and also an ample set of problems are provided for better practice purposes.

Chapter-5  Arithmetic Progressions

This chapter is another interesting and simpler topic where the problems here are mostly based on a single formula and the rest are derivations of the original one. Beginning with a basic brief introduction, definitions of arithmetic progressions, nth term of an AP, the sum of first n terms of an AP are a few important and priority topics covered under this chapter. Apart from that, there are many problems and exercises followed with each topic for good understanding.

Chapter-6  Triangles

This chapter Triangle is an interesting and easy chapter and students often like this very much and a securing unit as well. Here beginning with the introduction to triangles followed by other topics like similar figures, the similarity of triangles, criteria for similarity of triangles, areas of similar triangles, Pythagoras theorem, along with a page summary for revision purposes are discussed in this chapter with examples and exercises for practice purposes.

Chapter-7  Coordinate Geometry

Here starting with a general introduction, distance formula, section formula, area of the triangle are a few topics covered in this chapter followed with examples and exercises for better and thorough practice purposes.

Chapter-8  Introduction to Trigonometry

As trigonometry is a very important and vast subject, this topic is divided into two parts where one chapter is Introduction to Trigonometry and another part is Applications of Trigonometry. This Introduction to Trigonometry chapter is started with a general introduction, trigonometric ratios, trigonometric ratios of some specific angles, trigonometric ratios of complementary angles, trigonometric identities, etc are a few important topics covered in this chapter.

Chapter-9  Applications of Trigonometry

This chapter is the continuation of the previous chapter, where the various modeled applications are discussed here with examples and exercises for better understanding. Topics like heights and distances are covered here and at the end, a summary is provided with all the important and frequently used formulas used in this chapter for solving the problems.

Chapter-10  Circle

Beginning with the introduction to circles, tangent to a circle, several tangents from a point on a circle are some of the important topics covered in this chapter. This chapter being practical, there are an ample number of problems and solved examples for better understanding and practice purposes.

Chapter-11  Constructions

This chapter has more practical problems than theory-based definitions. Beginning with a general introduction to constructions, tools used, etc, the topics like division of a line segment, construction of tangents to a circle, and followed with few solved examples that help in solving the exercises provided after each topic.

Chapter-12  Areas related to Circles

This chapter problem is exclusively formula based wherein topics like perimeter and area of a circle- A Review, areas of sector and segment of a circle, areas of combinations of plane figures, and a page summary is provided just as a revision of the topics and formulas covered in the entire chapter and also there are many exercises and solved examples for practice purposes.

Chapter-13  Surface Areas and Volumes

Starting with the introduction, the surface area of a combination of solids, the volume of a combination of solids, conversion of solid from one shape to another, frustum of a cone, etc are to name a few topics explained in detail provided with a set of examples for a better comprehension of the concepts.

Chapter-14  Statistics

In this chapter starting with an introduction, topics like mean of grouped data, mode of grouped data, a median of grouped, graphical representation of cumulative frequency distribution are explained in detail with exercises for practice purposes. This chapter being a simple and easy subject, securing the marks is not difficult for students.

Chapter-15  Probability

Probability is another simple and important chapter in examination point of view and as seeking knowledge purposes as well. Beginning with an introduction to probability, an important topic called A theoretical approach is explained here. Since this chapter is one of the smallest in the syllabus and problems are also quite easy, students often like this chapter

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CBSE Case Study Questions for Class 10 Maths Linear Equations in Two Variables Free PDF

Mere Bacchon, you must practice the CBSE Case Study Questions Class 10 Maths Linear Equations in Two Variables  in order to fully complete your preparation . They are very very important from exam point of view. These tricky Case Study Based Questions can act as a villain in your heroic exams!

I have made sure the questions (along with the solutions) prepare you fully for the upcoming exams. To download the latest CBSE Case Study Questions , just click ‘ Download PDF ’.

CBSE Case Study Questions for Class 10 Maths Linear Equations in Two Variables PDF

Mcq set 1 -, mcq set 2 -, checkout our case study questions for other chapters.

• Chapter 1: Real Numbers Case Study Questions
• Chapter 2: Polynomials Case Study Questions
• Chapter 4: Quadratic Equation Case Study Questions
• Chapter 5: Arithmetic Progressions Case Study Questions

How should I study for my upcoming exams?

First, learn to sit for at least 2 hours at a stretch

Solve every question of NCERT by hand, without looking at the solution.

Solve NCERT Exemplar (if available)

Sit through chapter wise FULLY INVIGILATED TESTS

Practice MCQ Questions (Very Important)

Practice Assertion Reason & Case Study Based Questions

Sit through FULLY INVIGILATED TESTS involving MCQs. Assertion reason & Case Study Based Questions

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Case Based Questions (MCQ)

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Question 5 - Case Based Questions (MCQ) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables

Last updated at April 16, 2024 by Teachoo

John and Jivanti are playing with the marbles in  the playground. They together have 45 marbles  and John has 15 marbles more than Jivanti.

This question is exactly same as Example 1 (i) - Chapter 4 Class 10 - Quadratic Equations

The number of marbles Jivanti had:

(a) 15  , (c) 40  .

The number of marbles John had:

(a) 40  , (c) 15  .

If 45 is replaced by 55 in the above case discussed  in the question, then the number of marbles Jivanti  have:

(c) 20  .

According to the question 3, the number of marbles  John have:

(a) 30  , (c) 45  .

The given problem is based on which mathematical  concept?

(a) pair of linear equations   , (b) quadratic equations, (c) polynomials   , (d) none of the above.

Question John and Jivanti are playing with the marbles in the playground. They together have 45 marbles and John has 15 marbles more than Jivanti. Question 1 The number of marbles Jivanti had: (a) 15 (b) 30 (c) 40 (d) 5 Let Number of marbles with John = x Number of marbles with Jivanti = y Given that They together have 45 marbles x + y = 45 And, John has 15 marbles more than Jivanti x = y + 15 x – y = 15 Adding (1) and (2) (x + y) + (x − y) = 45 + 15 2x = 60 x = 60/2 x = 30 Putting x = 30 in (1) x + y = 45 30 + y = 45 y = 45 − 30 y = 15 So, x = 30 and y = 15 Thus, Number of marbles with John = x = 30 Number of marbles with Jivanti = y = 15 So, the correct answer is (A) Question 2 The number of marbles John had: (a) 40 (b) 30 (c) 15 (d) 20 Number of marbles with John = 30 So, the correct answer is (B) Question 3 If 45 is replaced by 55 in the above case discussed in the question, then the number of marbles Jivanti have: (a) 15 (b) 30 (c) 20 (d) 35 If 45 is replaced by 55, our equations will be x + y = 55 ...(1) x – y = 15 ...(2) Adding (1) and (2) (x + y) + (x − y) = 55 + 15 2x = 70 x = 70/2 x = 35 Putting x = 35 in (1) x + y = 55 35 + y = 55 y = 55 − 35 y = 20 Thus, Number of marbles with John = x = 35 Number of marbles with Jivanti = y = 20 So, the correct answer is (C) Question 4 According to the question 3, the number of marbles John have: (a) 30 (b) 40 (c) 45 (d) 35 From above question 3, Number of marbles with John = x = 35 So, the correct answer is (D) Question 5 The given problem is based on which mathematical concept? (a) pair of linear equations (b) Quadratic equations (c) Polynomials (d) None of the above The given Problem is based on Pair of linear equations. So, the correct answer is (A)

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Case Study on Pair of Equations in Two Variables Class 10 Maths PDF

The passage-based questions are commonly known as case study questions. Students looking for Case Study on Pair of Equations in Two Variables Class 10 Maths can use this page to download the PDF file. 

The case study questions on Pair of Equations in Two Variables are based on the CBSE Class 10 Maths Syllabus, and therefore, referring to the Pair of Equations in Two Variables case study questions enable students to gain the appropriate knowledge and prepare better for the Class 10 Maths board examination. Continue reading to know how should students answer it and why it is essential to solve it, etc.

Case Study on Pair of Equations in Two Variables Class 10 Maths with Solutions in PDF

Our experts have also kept in mind the challenges students may face while solving the case study on Pair of Equations in Two Variables, therefore, they prepared a set of solutions along with the case study questions on Pair of Equations in Two Variables.

The case study on Pair of Equations in Two Variables Class 10 Maths with solutions in PDF helps students tackle questions that appear confusing or difficult to answer. The answers to the Pair of Equations in Two Variables case study questions are very easy to grasp from the PDF - download links are given on this page.

Why Solve Pair of Equations in Two Variables Case Study Questions on Class 10 Maths?

There are three major reasons why one should solve Pair of Equations in Two Variables case study questions on Class 10 Maths - all those major reasons are discussed below:

• To Prepare for the Board Examination: For many years CBSE board is asking case-based questions to the Class 10 Maths students, therefore, it is important to solve Pair of Equations in Two Variables Case study questions as it will help better prepare for the Class 10 board exam preparation.
• Develop Problem-Solving Skills: Class 10 Maths Pair of Equations in Two Variables case study questions require students to analyze a given situation, identify the key issues, and apply relevant concepts to find out a solution. This can help CBSE Class 10 students develop their problem-solving skills, which are essential for success in any profession rather than Class 10 board exam preparation.
• Understand Real-Life Applications: Several Pair of Equations in Two Variables Class 10 Maths Case Study questions are linked with real-life applications, therefore, solving them enables students to gain the theoretical knowledge of Pair of Equations in Two Variables as well as real-life implications of those learnings too.

How to Answer Case Study Questions on Pair of Equations in Two Variables?

Students can choose their own way to answer Case Study on Pair of Equations in Two Variables Class 10 Maths, however, we believe following these three steps would help a lot in answering Class 10 Maths Pair of Equations in Two Variables Case Study questions.

• Read Question Properly: Many make mistakes in the first step which is not reading the questions properly, therefore, it is important to read the question properly and answer questions accordingly.
• Highlight Important Points Discussed in the Clause: While reading the paragraph, highlight the important points discussed as it will help you save your time and answer Pair of Equations in Two Variables questions quickly.
• Go Through Each Question One-By-One: Ideally, going through each question gradually is advised so, that a sync between each question and the answer can be maintained. When you are solving Pair of Equations in Two Variables Class 10 Maths case study questions make sure you are approaching each question in a step-wise manner.

What to Know to Solve Case Study Questions on Class 10 Pair of Equations in Two Variables?

 A few essential things to know to solve Case Study Questions on Class 10 Pair of Equations in Two Variables are -

• Basic Formulas of Pair of Equations in Two Variables: One of the most important things to know to solve Case Study Questions on Class 10 Pair of Equations in Two Variables is to learn about the basic formulas or revise them before solving the case-based questions on Pair of Equations in Two Variables.
• To Think Analytically: Analytical thinkers have the ability to detect patterns and that is why it is an essential skill to learn to solve the CBSE Class 10 Maths Pair of Equations in Two Variables case study questions.
• Strong Command of Calculations: Another important thing to do is to build a strong command of calculations especially, mental Maths calculations.

Where to Find Case Study on Pair of Equations in Two Variables Class 10 Maths?

Use Selfstudys.com to find Case Study on Pair of Equations in Two Variables Class 10 Maths. For ease, here is a step-wise procedure to download the Pair of Equations in Two Variables Case Study for Class 10 Maths in PDF for free of cost.

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Important Case Study Questions CBSE Class 10 Maths

Cbse class 10 maths important case study questions: cbse class 10 maths exam 2024 is just around the corner. case study questions can be a hard nut to track if not prepared well. check here important case study questions from class 10th maths curriculum  for cbse class 10 maths board exam 2024..

CBSE Class 10 Maths Question Paper Structure

• Section A : 20 Multiple Choice Questions (MCQs) carrying 1 mark each. 
• Section B : 5 Short Answer-I (SA-I) type questions carrying 2 marks each. 
• Section C : 6 Short Answer-II (SA-II) type questions carrying 3 marks each. 
• Section D : 4 Long Answer (LA) type questions carrying 5 marks each.
• Section E : 3 Case Based integrated units of assessment (4 marks each) with sub-parts of the values of 1, 1 and 2 marks each respectively.

CBSE Class 10 Maths Important Case Study Questions

Related:  CBSE Class 10 Maths Important Formulas for Last Minute Revision for Board Exam 2024

1 Two Friends Geeta and Sita were playing near the river. So, they decide to play a game in which they have to throw the stone in the river, and whoever will throw the stone at maximum distance, win the game. Geeta Starts first and throws the stone in the river. During her throw, her hand was making an angle of 60° with the Horizontal plane. Sita throws at 45°.

• Straight Line
• Semi circle
• Bi-Quadratic
• Parabola Open Upward
• Parabola Open Downward
• Hyperbola Open Upward
• Hyperbola Open downward
• Two Real Points
• One Real Point
• Three Real Points
• Putting y=0 in given Polynomial
• Putting y=1 in the given Polynomial
• Putting x=0 in the given Polynomial.
• Putting x=1 in the given Polynomial.

2 The department of Computer Science and Technology is conducting an International Seminar. In the seminar, the number of participants in Mathematics, Science and Computer Science are 60, 84 and 108 respectively. The coordinator has made the arrangement such that in each room, the same number of participants are to be seated and all of them being in the same subject. Also, they allotted the separate room for all the official other than participants.

(i) Find the total number of participants.

(a) 60 

(b) 84 

(c) 108 

(d) none of these

(ii) Find the LCM of 60, 84 and 108.

(a) 12 

(b) 504 

(c) 544320 

(iii) Find the HCF of 60, 84 and 108.

(iv) Find the minimum number of rooms required, if in each room, the same number of participants are to be seated and all of them being in the same subject.

(b) 20 

(c) 21 

(v) Based on the above (iv) conditions, find the minimum number of rooms required for all the participants and officials.

• In the standard form of quadratic polynomial, ax2 + bx + c, what are a, b and c ?
• If the roots of the quadratic polynomial are equal, what is the discriminant D ?
• If α and 1/α are the zeroes of the quadratic polynomial 2x2 – x + 8k, then find the value of k ?
• Represent the first situation algebraically.

a) 12x+10y=11200

b) 10x+12y=11200

c) 12x-10y=11200

d) 10x-12y=1120

2 Represent the second situation algebraically

a) 46x+55y=51400

b) 55x+46y=51400

c) 55x-46y=51400

d) 46x-55y=51400

3 The system of linear equations representing both the situations will have.

a) Infinite number of solutions

b) Unique solution

c) No Solutions

d) Exactly two solutions

4 The graph of the system of linear equations representing both the situations will be

a) Parallel lines

b) Coincident lines

c) Intersecting lines

d) None of these

• Represent algebraically the situation in hall “Rose”.

a) 50x + y = 10000

b) 50x − y = 10000

c) x + 50y = 10000

d) x − 50y = 10000

2 Represent algebraically the situation in hall “Jasmine”

a) x + 25y = 7500

b) x − 25y = 7500

c) 25x + y = 7500

d) 25x − y = 7500

3 What is the fixed rent of the halls?

4 Find the amount the hotel charged per person.

6 Riya has a field with a flowerbed and grassland. The grassland is in the shape of a rectangle while the flowerbed is in the shape of a square. The length of the grassland is found to be 3 m more than twice the length of the flowerbed. Total area of the whole land is 1260m2

(a)If the length of the square is x m then find the total length of the field i

(b) What will be the perimeter of the whole figure in terms of x?

(c )Find the value of x if the area of total field is 1260 m2

(d) Find the area of grassland and the flowerbed separately.

7 Your friend Veer wants to participate in a 200m race. He can currently run that distance in 51 seconds and with each day of practice it takes him 2 seconds less. He wants to do it in 31 seconds.

1 Write first four terms are in AP for the given situation.

2 What is the minimum number of days he needs to practice till his goal is achieved.

3 How many seconds it takes after the 5th day .

8 Helicopter Patrolling: A helicopter is hovering over a crowd of people watching a police standoff in a parking garage across the street. Stewart notices the shadow of the helicopter is lagging approximately 57 m behind a point directly below the helicopter. If he is 160 cm tall and casts a shadow of 38 cm at this time,

(i) what is the altitude of the helicopter?

(ii) What will be length of shadow of Stewart at 12:00 pm

(iii) Write the name of triangles formed for this situation.

9 Seema has a 10 m × 10 m kitchen garden attached to her kitchen. She divides it into a 10 ×10 grid and wants to grow some vegetables and herbs used in the kitchen. She puts some soil and manure in that and sow a green chilly plant at A, a coriander plant at B and a tomato plant at C. Her friend Kusum visited the garden and praised the plants grown there. She pointed out that they seem to be in a straight line. See the below diagram carefully and answer the following questions:

(i) Find the distance between A and B is

(ii) Find the mid- point of the distance AB

(iii) Find the distance between B and C

10 A heavy-duty ramp is used to winch heavy appliances from street level up to a warehouse loading dock. If the ramp is 2 meter high and the incline is 4 meter long.

(Use √3 = 1.73)

a What angle does the dock make with the street?

b How long is the base of the ramp? ( In round figure)

• If the length of the base is 12 cm and the height is 5 cm then the length of the hypotenuse of that sandwich is:
• If he increases the base length to 15 cm and the hypotenuse to 17 cm, then the height of the sandwich is :
• The value of tan 45° + cot 45°

(a) 1 (b) 2 (c) 3 (d) 4

12 A flag pole ‘h’ metres is on the top of the hemispherical dome of radius ‘r’ metres. A man is standing 7 m away from the dome. Seeing the top of the pole at an angle 45° and moving 5 m away from the dome and seeing the bottom of the pole at an angle 30°.

(i) the height of the pole

(ii) radius( height) of the dome

(iii) Is it possible to see the pole at the angle of 60 0

(iv) If the height of pole is increased, the angle elevation will .....

14 John had a farm with many animals like cows, dogs, horses etc. He had sufficient grass land for the cows and horses to graze, One day Three of his horses were tied with 7 metre long ropes at the three corners of a triangular lawn having sides 20m, 34m and 42m.

(a) Find the area of the triangular lawn .

(b) Find the area of the field that can be grazed by the horses.

(c) The area that cannot be grazed by the horses.

15 Arun, a 10th standard student, makes a project on coronavirus in science for an exhibition in his school. In this project, he picks a sphere which has volume 38808 cm3 and 11 cylindrical shapes, each of volume 1540 cm3 with length 10 cm.

Based on the above information, answer the following questions.

(i) Diameter of the base of the cylinder is

(ii) Diameter of the sphere is

(iii) Total volume of the shape formed is

(iv) Curved surface area of the one cylindrical shape is

(v) Total area covered by cylindrical shapes on the surface of sphere is

• In which age group, will the maximum number of children belong?
• Find the mode of the ages of children playing in the park?

17 Piggy bank or Money box( a coin container) is normally used by children. Piggy bank serves as a pedagogical device to teach about saving money to children. Generally, piggy banks have openings besides the slot for inserting coins but some do not have openings. We have to smash the piggy bank with a hammer or by other means, to get the money inside it. A child Shreya has a Piggybank. She saves her money in her Piggybank. One day she found that her Piggybank contains hundred 50 paisa coins, fifty 1 rupees coin, twenty 2 rupees coin, and ten 5 rupees coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down.

(a) The probability that the fallen coin will be 50 paisa coin, is -------

(b) The probability that the fallen coin will be 5 rupees coin, is---------

(c) The probability that the fallen coin will be 2 rupees coin, is---------

(d) The probability that the fallen coin will be 2 rupees coin or 5 rupees coin, is--------

• Find the probability of getting no heads
• Find the probability of getting one tail

Download answers to CBSE Class 10 Maths Important Case Study Questions

Chapter-wise important case study questions cbse class 10 maths.

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Class 10 Maths Chapter 3 Case Based Questions - Pair of Linear Equations in Two Variables

Case study - 1

Refer situation 1 : Q1: If the fixed charges of auto rickshaw be Rs x and the running charges be Rs y km/hr, the pair of linear equations representing the situation is (a) x + 10y =110, x + 15y = 75 (b) x + 10y = 75, x + 15y = 110 (c) 10x + y = 110, 15x + y = 75 (d) 10x + y = 75, 15x + y = 110 Ans: (b) Explanation:  The cost of the auto rickshaw ride consists of two components - a fixed charge (x) and a variable charge based on the distance of the journey (y per km). The total cost of the ride would therefore be the sum of these two components. In the first situation, the auto rickshaw ride in city A costs Rs 75 for a 10 km journey and Rs 110 for a 15 km journey. We can express these two situations as two linear equations: For the 10 km journey: x + 10y = 75 (equation 1) For the 15 km journey: x + 15y = 110 (equation 2) Where x is the fixed charge and y is the cost per km. Therefore, the pair of linear equations representing the situation is x + 10y = 75, x + 15y = 110 (option b).   Q2: A person travels a distance of 50km. The amount he has to pay is (a) Rs.155 (b) Rs.255 (c) Rs.355 (d) Rs.455 Ans:  (c) Explanation:  To solve this case-based problem, we first need to understand the relationship between distance traveled and the charge paid. This can be represented as a linear equation in the form of y = mx + c, where y is the charge paid, m is the rate charged per km, x is the distance traveled, and c is the fixed charge. From Situation 1, we have two equations based on the given data: 1) 75 = 10m + c 2) 110 = 15m + c Subtracting equation 1 from equation 2, we get: 35 = 5m So, m = 35/5 = 7. This means the rate charged per km in city A is Rs. 7. Substituting m = 7 in equation 1, we get: 75 = 10*7 + c So, c = 75 - 70 = 5. This means the fixed charge in city A is Rs. 5. So, the total charge for traveling a distance of x km in city A is given by the equation: Charge = 7x + 5 Now, let's calculate the charge for traveling a distance of 50km in city A: Charge = 7*50 + 5 = Rs. 355 Therefore, the answer is (c) Rs. 355.   Refer situation 2: Q3: What will a person have to pay for travelling a distance of 30km? (a) Rs.185 (b) Rs.289 (c) Rs.275 (d) Rs.305 Ans:  (b) Explanation:  To solve this problem, we first need to find the rate of the fare per kilometer and the fixed charge in city B. We can solve this problem by making two equations from the given situations and then solve them simultaneously. Let's take the fixed charge as F (in Rs.) and the rate per kilometer as R (in Rs. per km). From the given situation 2, we know that: For a journey of 8km, the charge paid is Rs 91. For a journey of 14km, the charge paid is Rs 145. So, we can formulate two equations: 1) F + 8R = 91 2) F + 14R = 145 Subtract equation 1 from equation 2 and we get 6R = 54, therefore R = 54/6 = 9 Rs per km. Substitute the value of R in equation 1, we get F = 91 - 8*9 = 91 - 72 = 19 Rs. So, the fixed charge is Rs 19 and the fare per kilometer is Rs 9 in city B. Now, to find out what a person will have to pay for traveling a distance of 30km, we add the fixed charge to the product of the fare per kilometer and the distance. Therefore, the fare for 30km = F + 30R = 19 + 30*9 = 19 + 270 = Rs 289. So, the correct answer is (b) Rs.289.  

Case Study - 2

Based on the above information, answer the following questions: Q1: Form the pair of linear equations in two variables from this situation. Ans: Area of two bedrooms= 10x sq.m Area of kitchen = 5y sq.m 10x + 5y = 95 2x + y =19 Also, x + 2+ y = 15 x + y = 13 Q2: Find the length of the outer boundary of the layout. Ans:  Length of outer boundary = 12 + 15 + 12 + 15 = 54m   Q3: Find the area of each bedroom and kitchen in the layout. Ans:  On solving two equation part(i) x = 6m and y = 7m area of bedroom = 5 x 6 = 30m area of kitchen = 5 x 7 = 35m Q4: Find the area of living room in the layout. Ans:  Area of living room = (15 x 7) – 30 = 105 – 30 = 75 sq.m   Q5: Find the cost of laying tiles in kitchen at the rate of Rs. 50 per sq.m. Ans:  Total cost of laying tiles in the kitchen = Rs.50 x 35 = Rs1750.

Case Study - 3

A test consists of ‘True’ or ‘False’ questions. One mark is awarded for every correct answer while ¼ mark is deducted for every wrong answer. A student knew answers to some of the questions. Rest of the questions he attempted by guessing. He answered 120 questions and got 90 marks.

Q2: How many questions did he guess? Ans:  Let the no of questions whose answer is known to the student x and questions attempted by cheating be y x + y =120 x – 1/4y =90 solving these two x = 96 and y = 24 He attempted 24 questions by guessing.   Q3: If answer to all questions he attempted by guessing were wrong and answered 80 correctly, then how many marks he got? Ans:  Let the no of questions whose answer is known to the student x and questions attempted by cheating be y x + y =120 x – 1/4y =90 solving these two x = 96 and y = 24 Marks = 80- ¼ 0f 40 =70

Q4: If answer to all questions he attempted by guessing were wrong, then how many questions answered correctly to score 95 marks? Ans:  Let the no of questions whose answer is known to the student x and questions attempted by cheating be y x + y =120 x – 1/4y =90 solving these two x = 96 and y = 24 x – 1/4 of (120 – x) = 95 5x = 500, x = 100

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CBSE 10th Standard Maths Subject Case Study Questions With Solution 2021 Part - II

By QB365 on 21 May, 2021

QB365 Provides the updated CASE Study Questions for Class 10 Maths, and also provide the detail solution for each and every case study questions . Case study questions are latest updated question pattern from NCERT, QB365 will helps to get  more marks in Exams

QB365 - Question Bank Software

10th Standard CBSE

Final Semester - June 2015

Case Study Questions

(ii) Proportional expense for each person is

(iii) The fixed (or constant) expense for the party is

(iv) If there would be 15 guests at the lunch party, then what amount Mr Jindal has to pay?

(v) The system of linear equations representing both the situations will have

(ii) Represent the situation faced by Suman, algebraically

(iii) The price of one Physics book is

(iv) The price of one Mathematics book is

(v) The system of linear equations represented by above situation, has

(ii) Represent algebraically the situation of day- II.

(iii) The linear equation represented by day-I, intersect the x axis at

(iv) The linear equation represented by day-II, intersect the y-axis at

(v) Linear equations represented by day-I and day -II situations, are

Amit is preparing for his upcoming semester exam. For this, he has to practice the chapter of Quadratic Equations. So he started with factorization method. Let two linear factors of  \(a x^{2}+b x+c \text { be }(p x+q) \text { and }(r x+s)\) \(\therefore a x^{2}+b x+c=(p x+q)(r x+s)=p r x^{2}+(p s+q r) x+q s .\) Now, factorize each of the following quadratic equations and find the roots. (i) 6x 2 + x - 2 = 0

(ii) 2x 2 -+ x - 300 = 0

(iii) x 2 -  8x + 16 = 0

(iv) 6x 2 -  13x + 5 = 0

(v) 100x 2 - 20x + 1 = 0

(ii) Difference of pairs of shoes in 17 th  row and 10 th row is

(iii) On next day, she arranges x pairs of shoes in 15 rows, then x =

(iv) Find the pairs of shoes in 30 th row.

(v) The total number of pairs of shoes in 5 th and 8 th row is

(ii) The number on first card is

(iii) What is the number on the 19 th card?

(iv) What is the number on the 23 rd card?

(v) The sum of numbers on the first 15 cards is 

A sequence is an ordered list of numbers. A sequence of numbers such that the difference between the consecutive terms is constant is said to be an arithmetic progression (A.P.). On the basis of above information, answer the following questions. (i) Which of the following sequence is an A.P.?

(ii) If x, y and z are in A.P., then

(iii) If a 1  a 2 , a 3  ..... , a n are in A.P., then which of the following is true?

(iv) If the n th term (n > 1) of an A.P. is smaller than the first term, then nature of its common difference (d) is

(v) Which of the following is incorrect about A.P.?

(ii) Find the radius of the core.

(iii) S 2 =

(iv) What is the diameter of roll when one tissue sheet is rolled over it?

(v) Find the thickness of each tissue sheet

(ii) Distance travelled by aeroplane towards west after   \(1 \frac{1}{2}\)   hr is

(iii) In the given figure, \(\angle\) POQ is 

(iv) Distance between aeroplanes after  \(1 \frac{1}{2}\)   hr is

(v) Area of \(\Delta\) POQ is

(ii) The value of x is

(iii) The value of PR is 

(iv) The value of RQ is 

(v) How much distance will be saved in reaching city Q after the construction of highway? 

(ii) Length of BC =

(iii) Length of AD =

(iv) Length of ED = 

(v) Length of AE = 

(ii) The value of x + y is 

(iii) Which of the following is true?

(iv) The ratio in which B divides AC is

(v) Which of the following equations is satisfied by the given points?

(ii) The value of x is equal to

(iii) If M is any point exactly in between city A and city B, then coordinates of M are

(iv) The ratio in which A divides the line segment joining the points O and M is

(v) If the person analyse the petrol at the point M(the mid point of AB), then what should be his decision?

(ii) The centre of circle is the

(iii) The radius of the circle is

(iv) The area of the circle is

(v) If  \(\left(1, \frac{\sqrt{7}}{3}\right)\)   is one of the ends of a diameter, then its other end is

(ii) The distance between A and Cis

(iii) If it is assumed that both buses have same speed, then by which bus do you want to travel from A to B?

(iv) If the fare for first bus is Rs10/km, then what will be the fare for total journey by that bus?

(v) If the fare for second bus is Rs 15/km, then what will be the fare to reach to the destination by this bus?

*****************************************

Cbse 10th standard maths subject case study questions with solution 2021 part - ii answer keys.

(i) (a): 1 st situation can be represented as x + 7y = 650 ...(i) and 2 nd situation can be represented as x + 11y = 970 ...(ii) (ii) (b): Subtracting equations (i) from (ii), we get  \(4 y=320 \Rightarrow y=80\) \(\therefore\)  Proportional expense for each person is Rs 80. (iii) (c): Puttingy = 80 in equation (i), we get x + 7 x 80 = 650 \(\Rightarrow\) x = 650 - 560 = 90 \(\therefore\)  Fixed expense for the party is Rs 90 (iv) (d): If there will be 15 guests, then amount that Mr Jindal has to pay = Rs (90 + 15 x 80) = Rs 1290 (v) (a): We have a 1  = 1, b 1  = 7, c 1  = -650 and  \(a_{2}=1, b_{2}=11, c_{2}=-970 \) \(\therefore \frac{a_{1}}{a_{2}}=1, \frac{b_{1}}{b_{2}}=\frac{7}{11}, \frac{c_{1}}{c_{2}}=\frac{-650}{-970}=\frac{65}{97}\) \(\text { Here, } \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\) Thus, system of linear equations has unique solution.

(i) (a): Situation faced by Sudhir can be represented algebraically as 2x + 3y = 850 (ii) (b): Situation faced by Suman can be represented algebraically as 3x + 2y = 900 (iii) (c) : We have 2x + 3y = 850 .........(i) and 3x + 2y = 900 .........(ii) Multiplying (i) by 3 and (ii) by 2 and subtracting, we get 5y = 750 \(\Rightarrow\)   Y = 150 Thus, price of one Physics book is Rs 150. (iv) (d): From equation (i) we have, 2x + 3 x 150 = 850 \(\Rightarrow\) 2x = 850 - 450 = 400 \(\Rightarrow\) x = 200 Hence, cost of one Mathematics book = Rs 200 (v) (a): From above, we have \(a_{1} =2, b_{1}=3, c_{1}=-850 \) \(\text { and } a_{2} =3, b_{2}=2, c_{2}=-900\) \(\therefore \quad \frac{a_{1}}{a_{2}}=\frac{2}{3}, \frac{b_{1}}{b_{2}}=\frac{3}{2}, \frac{c_{1}}{c_{2}}=\frac{-850}{-900}=\frac{17}{18} \Rightarrow \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\) Thus system of linear equations has unique solution.

(i) (b): Algebraic representation of situation of day-I is 2x + y = 1600. (ii) (a): Algebraic representation of situation of day- II is 4x + 2y = 3000 \(\Rightarrow\) 2x + y = 1500. (iii) (c) : At x-axis, y = 0 \(\therefore\)   At y = 0, 2x + y = 1600 becomes 2x = 1600 \(\Rightarrow\) x = 800 \(\therefore\) Linear equation represented by day- I intersect the x-axis at (800, 0). (iv) (d) : At y-axis, x = 0 \(\therefore\) 2x + Y = 1500 \(\Rightarrow\)  y = 1500 \(\therefore\) Linear equation represented by day-II intersect the y-axis at (0, 1500). (v) (b): We have, 2x + y = 1600 and 2x + y = 1500 Since  \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \text { i.e., } \frac{1}{1}=\frac{1}{1} \neq \frac{16}{15}\) \(\therefore\) System of equations have no solution. \(\therefore\) Lines are parallel.

(i) (b): We have  \(6 x^{2}+x-2=0\) \(\Rightarrow \quad 6 x^{2}-3 x+4 x-2=0 \) \(\Rightarrow \quad(3 x+2)(2 x-1)=0 \) \(\Rightarrow \quad x=\frac{1}{2}, \frac{-2}{3}\) (ii) (c):  \(2 x^{2}+x-300=0\) \(\Rightarrow \quad 2 x^{2}-24 x+25 x-300=0 \) \(\Rightarrow \quad(x-12)(2 x+25)=0 \) \(\Rightarrow \quad x=12, \frac{-25}{2}\) (iii) (d):   \(x^{2}-8 x+16=0\) \(\Rightarrow(x-4)^{2}=0 \Rightarrow(x-4)(x-4)=0 \Rightarrow x=4,4\) (iv) (d):   \(6 x^{2}-13 x+5=0\) \(\Rightarrow \quad 6 x^{2}-3 x-10 x+5=0 \) \(\Rightarrow \quad(2 x-1)(3 x-5)=0 \) \(\Rightarrow \quad x=\frac{1}{2}, \frac{5}{3}\) (v) (a):  \(100 x^{2}-20 x+1=0\) \(\Rightarrow(10 x-1)^{2}=0 \Rightarrow x=\frac{1}{10}, \frac{1}{10}\)  

Number of pairs of shoes in 1 st , 2 nd , 3 rd row, ... are 3,5,7, ... So, it forms an A.P. with first term a = 3, d = 5 - 3 = 2 (i) (d): Let n be the number of rows required. \(\therefore S_{n}=120 \) \(\Rightarrow \quad \frac{n}{2}[2(3)+(n-1) 2]=120 \) \(\Rightarrow \quad n^{2}+2 n-120=0 \Rightarrow n^{2}+12 n-10 n-120=0\) \(\Rightarrow \quad(n+12)(n-10)=0 \Rightarrow n=10\) So, 10 rows required to put 120 pairs. (ii) (b): No. of pairs in 1ih row = t 17 = 3 + 16(2) = 35 No. of pairs in 10th row = t 10  = 3 + 9(2) = 21 \(\therefore\) Required difference = 35 - 21 = 14 (iii) (c) : Here n = 15 \(\therefore\) t 15  = 3 + 14(2) = 3 + 28 = 31 (iv) (a): No. of pairs in 30 th row = t 30 = 3 +29(2) = 61 (v) (c): No. of pairs in 5 th row = t 5  = 3 + 4(2) = 11 No. of pairs in 8 th row = t 8  = 3 + 7(2) = 17 \(\therefore\) Required sum = 11 + 17 = 28

Let the numbers on the cards be a, a + d, a + Zd, ... According to question, We have (a + 5d) + (a + 13d) = -76 \(\Rightarrow\) 2a+18d = -76 \(\Rightarrow\) a + 9d= -38 ... (1) And (a + 7d) + (a + 15d) = -96 \(\Rightarrow\) 2a + 22d = -96 \(\Rightarrow\) a + 11d = -48 ...(2) From (1) and (2), we get 2d= -10 \(\Rightarrow\) d= -5 From (1), a + 9(-5) = -38 \(\Rightarrow\) a = 7 (i) (b): The difference between the numbers on any two consecutive cards = common difference of the A.P.=-5 (ii) (d): Number on first card = a = 7 (iii) (b): Number on 19th card = a + 18d = 7 + 18(-5) = -83 (iv) (a): Number on 23rd card = a + 22d = 7 + 22( -5) = -103 (v) (d):  \(S_{15}=\frac{15}{2}[2(7)+14(-5)]=-420\)

(i) (c) (ii) (c) (iii) (d) (iv) (b) (v) (c)

Here S n = 0.1n 2 + 7.9n (i) (c): S n -1 = 0.1(n - 1) 2 + 7.9(n - 1) = 0.1n 2 + 7.7n - 7.8 (ii) (b): S 1 = t 1  = a = 0.1(1) 2 + 7.9(1) = 8 cm = Diameter of core So, radius of the core = 4 cm (iii) (a): S 2 = 0.1(2) 2 + 7.9(2) = 16.2 (iv) (d): Required diameter = t 2 = S 2 - S 1 = 16.2 - 8 = 8.2 cm (v) (c): As d = t 2 - t 1  = 8.2 - 8 = 0.2 cm So, thickness of tissue = 0.2 \(\div\)   2 = 0.1 cm = 1 mm

(i) (a): Speed = 1200 km/hr \(\text { Time }=1 \frac{1}{2} \mathrm{hr}=\frac{3}{2} \mathrm{hr}\) \(\therefore\)  Required distance = Speed x Time \(=1200 \times \frac{3}{2}=1800 \mathrm{~km}\) (ii) (c): Speed = 1500 km/hr Time =  \(\frac{3}{2}\)  hr. \(\therefore\)  Required distance = Speed x Time \(=1500 \times \frac{3}{2}=2250 \mathrm{~km}\) (iii) (b): Clearly, directions are always perpendicular to each other. \(\therefore \quad \angle P O Q=90^{\circ}\) (iv) (a): Distance between aeroplanes after  \(1\frac{1}{2}\)   hour  \(\begin{array}{l} =\sqrt{(1800)^{2}+(2250)^{2}}=\sqrt{3240000+5062500} \\ =\sqrt{8302500}=450 \sqrt{41} \mathrm{~km} \end{array}\) (v) (d): Area of  \(\Delta\) POQ= \(\frac{1}{2}\) x base x height \(=\frac{1}{2} \times 2250 \times 1800=2250 \times 900=2025000 \mathrm{~km}^{2}\)

(i) (b) (ii) (c): Using Pythagoras theorem, we have PQ 2 = PR 2 + RQ 2 \(\Rightarrow(26)^{2}=(2 x)^{2}+(2(x+7))^{2} \Rightarrow 676=4 x^{2}+4(x+7)^{2} \) \(\Rightarrow 169=x^{2}+x^{2}+49+14 x \Rightarrow x^{2}+7 x-60=0\) \(\Rightarrow x^{2}+12 x-5 x-60=0 \) \(\Rightarrow x(x+12)-5(x+12)=0 \Rightarrow(x-5)(x+12)=0 \) \(\Rightarrow x=5, x=-12\) \(\therefore \quad x=5\)   [Since length can't be negative] (iii) (a) : PR = 2x = 2 x 5 = 10 km (iv) (b): RQ= 2(x + 7) = 2(5 + 7) = 24 km (v) (d): Since, PR + RQ = 10 + 24 = 34 km Saved distance = 34 - 26 = 8 km

(i) (b): If \(\Delta\) AED and \(\Delta\) BEC, are similar by SAS similarity rule, then their corresponding proportional sides are  \(\frac{B E}{A E}=\frac{C E}{D E}\) (ii) (c): By Pythagoras theorem, we have \(\begin{array}{l} B C=\sqrt{C E^{2}+E B^{2}}=\sqrt{4^{2}+3^{2}}=\sqrt{16+9} \\ =\sqrt{25}=5 \mathrm{~cm} \end{array}\) (iii) (a): Since \(\Delta\) ADE and \(\Delta\) BCE are similar. \(\therefore \quad \frac{\text { Perimeter of } \triangle A D E}{\text { Perimeter of } \Delta B C E}=\frac{A D}{B C} \) \(\Rightarrow \frac{2}{3}=\frac{A D}{5} \Rightarrow A D=\frac{5 \times 2}{3}=\frac{10}{3} \mathrm{~cm}\) (iv) (b): \(\frac{\text { Perimeter of } \triangle A D E}{\text { Perimeter of } \Delta B C E}=\frac{E D}{C E} \) \(\Rightarrow \frac{2}{3}=\frac{E D}{4} \Rightarrow E D=\frac{4 \times 2}{3}=\frac{8}{3} \mathrm{~cm}\) (v) (d) :   \(\frac{\text { Perimeter of } \Delta A D E}{\text { Perimeter of } \Delta B C E}=\frac{A E}{B E} \Rightarrow \frac{2}{3} B E=A E\) \(\Rightarrow A E=\frac{2}{3} \sqrt{B C^{2}-C E^{2}} \) \(\text { Also, in } \triangle A E D, A E=\sqrt{A D^{2}-D E^{2}}\)

(i) (a): We have, OA = 2 \(\sqrt{2}\) km \(\Rightarrow \sqrt{2^{2}+y^{2}}=2 \sqrt{2} \) \(\Rightarrow 4+y^{2}=8 \Rightarrow y^{2}=4 \) \(\Rightarrow y=2 \quad(\because y=-2 \text { is not possible })\) (ii) (c): We have OB = 8 \(\sqrt{2}\) \(\Rightarrow \sqrt{x^{2}+8^{2}}=8 \sqrt{2} \) \(\Rightarrow x^{2}+64=128 \Rightarrow x^{2}=64 \) \(\Rightarrow x=8 \quad(\because x=-8 \text { is not possible })\) (iii) (c) : Coordinates of A and Bare (2, 2) and (8, 8) respectively, therefore coordinates of point M are \(\left(\frac{2+8}{2}, \frac{2+8}{2}\right)\) i.e .,(5.5) (iv) (d): Let A divides OM in the ratio k: 1.Then \(2=\frac{5 k+0}{k+1} \Rightarrow 2 \mathrm{k}+2=5 k \Rightarrow 3 k=2 \Rightarrow k=\frac{2}{3}\) \(\therefore\) Required ratio = 2 : 3 (v) (b): Since M is the mid-point of A and B therefore AM = MB. Hence, he should try his luck moving towards B.

(i) (c): Required coordinates are  \(\left(0, \frac{4}{3}\right)\) (ii) (c) (iii) (a): Radius = Distance between (0,0) and  \(\left(\frac{4}{3}, 0\right)\) \(=\sqrt{\left(\frac{4}{3}\right)^{2}+0^{2}}=\frac{4}{3} \text { units }\) (iv) (b): Area of circle = \(\pi\) (radius) 2 \(=\pi\left(\frac{4}{3}\right)^{2}=\frac{16}{9} \pi \text { sq. units }\) (v) (d): Let the coordinates of the other end be (x,y). Then (0,0) will bethe mid-point of  \(\left(1, \frac{\sqrt{7}}{3}\right)\)  and (x, y). \(\therefore\left(\frac{1+x}{2}, \frac{\frac{\sqrt{7}}{3}+y}{2}\right)=(0,0) \) \(\Rightarrow \frac{1+x}{2}=0 \text { and } \frac{\frac{\sqrt{7}}{3}+y}{2}=0 \) \(\Rightarrow x=-1 \text { and } y=-\frac{\sqrt{7}}{3}\) Thus, the coordinates of other end be  \(\left(-1, \frac{-\sqrt{7}}{3}\right)\)

Coordinates of A, Band Care (-2, -3), (2, 3) and (3,2). (i) (d): Required distance  \(=\sqrt{(2+2)^{2}+(3+3)^{2}}\) \(=\sqrt{4^{2}+6^{2}}=\sqrt{16+36}=2 \sqrt{13} \mathrm{~km} \approx 7.2 \mathrm{~km}\) (ii) (d): Required distance  \(=\sqrt{(3+2)^{2}+(2+3)^{2}}\) \(=\sqrt{5^{2}+5^{2}}=5 \sqrt{2} \mathrm{~km}\) (iii) (b): Distance between Band C \(=\sqrt{(3-2)^{2}+(2-3)^{2}}=\sqrt{1+1}=\sqrt{2} \mathrm{~km}\) Thus, distance travelled by first bus to reach to B \(=A C+C B=5 \sqrt{2}+\sqrt{2}=6 \sqrt{2} \mathrm{~km} \approx 8.48 \mathrm{~km}\) and distance travelled by second bus to reach to B \(=A B=2 \sqrt{13} \mathrm{~km} \approx 7.2 \mathrm{~km}\) \(\therefore\)  Distance of first bus is greater than distance of the second bus, therefore second bus should be chosen. (iv) (d): Distance travelled by first bus = 8.48 km \(\therefore\) Total fare = 8.48 x 10 = Rs 84.80 (v) (b): Distance travelled by second bus = 7. 2 km \(\therefore\) Total fare = 7.2 x 15 = Rs 108  

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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

Download PDF for NCERT Solutions Class 10 Maths 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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Case Study Class 10 Maths Questions

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Download the app to get CBSE Sample Papers 2023-24, NCERT Solutions (Revised), Most Important Questions, Previous Year Question Bank, Mock Tests, and Detailed Notes.

Now, CBSE will ask only subjective questions in class 10 Maths case studies. But if you search over the internet or even check many books, you will get only MCQs in the class 10 Maths case study in the session 2022-23. It is not the correct pattern. Just beware of such misleading websites and books.

We advise you to visit CBSE official website ( cbseacademic.nic.in ) and go through class 10 model question papers . You will find that CBSE is asking only subjective questions under case study in class 10 Maths. We at myCBSEguide helping CBSE students for the past 15 years and are committed to providing the most authentic study material to our students.

Here, myCBSEguide is the only application that has the most relevant and updated study material for CBSE students as per the official curriculum document 2022 – 2023. You can download updated sample papers for class 10 maths .

First of all, we would like to clarify that class 10 maths case study questions are subjective and CBSE will not ask multiple-choice questions in case studies. So, you must download the myCBSEguide app to get updated model question papers having new pattern subjective case study questions for class 10 the mathematics year 2022-23.

Class 10 Maths has the following chapters.

• Real Numbers Case Study Question
• Polynomials Case Study Question
• Pair of Linear Equations in Two Variables Case Study Question
• Quadratic Equations Case Study Question
• Arithmetic Progressions Case Study Question
• Triangles Case Study Question
• Coordinate Geometry Case Study Question
• Introduction to Trigonometry Case Study Question
• Some Applications of Trigonometry Case Study Question
• Circles Case Study Question
• Area Related to Circles Case Study Question
• Surface Areas and Volumes Case Study Question
• Statistics Case Study Question
• Probability Case Study Question

Format of Maths Case-Based Questions

CBSE Class 10 Maths Case Study Questions will have one passage and four questions. As you know, CBSE has introduced Case Study Questions in class 10 and class 12 this year, the annual examination will have case-based questions in almost all major subjects. This article will help you to find sample questions based on case studies and model question papers for CBSE class 10 Board Exams.

Maths Case Study Question Paper 2023

Here is the marks distribution of the CBSE class 10 maths board exam question paper. CBSE may ask case study questions from any of the following chapters. However, Mensuration, statistics, probability and Algebra are some important chapters in this regard.

Case Study Question in Mathematics

Here are some examples of case study-based questions for class 10 Mathematics. To get more questions and model question papers for the 2021 examination, download myCBSEguide Mobile App .

Case Study Question – 1

In the month of April to June 2022, the exports of passenger cars from India increased by 26% in the corresponding quarter of 2021–22, as per a report. A car manufacturing company planned to produce 1800 cars in 4th year and 2600 cars in 8th year. Assuming that the production increases uniformly by a fixed number every year.

• Find the production in the 1 st year.
• Find the production in the 12 th year.
• Find the total production in first 10 years. OR In which year the total production will reach to 15000 cars?

Case Study Question – 2

In a GPS, The lines that run east-west are known as lines of latitude, and the lines running north-south are known as lines of longitude. The latitude and the longitude of a place are its coordinates and the distance formula is used to find the distance between two places. The distance between two parallel lines is approximately 150 km. A family from Uttar Pradesh planned a round trip from Lucknow (L) to Puri (P) via Bhuj (B) and Nashik (N) as shown in the given figure below.

• Find the distance between Lucknow (L) to Bhuj(B).
• If Kota (K), internally divide the line segment joining Lucknow (L) to Bhuj (B) into 3 : 2 then find the coordinate of Kota (K).
• Name the type of triangle formed by the places Lucknow (L), Nashik (N) and Puri (P) OR Find a place (point) on the longitude (y-axis) which is equidistant from the points Lucknow (L) and Puri (P).

Case Study Question – 3

• Find the distance PA.
• Find the distance PB
• Find the width AB of the river. OR Find the height BQ if the angle of the elevation from P to Q be 30 o .

Case Study Question – 4

• What is the length of the line segment joining points B and F?
• The centre ‘Z’ of the figure will be the point of intersection of the diagonals of quadrilateral WXOP. Then what are the coordinates of Z?
• What are the coordinates of the point on y axis equidistant from A and G? OR What is the area of area of Trapezium AFGH?

Case Study Question – 5

The school auditorium was to be constructed to accommodate at least 1500 people. The chairs are to be placed in concentric circular arrangement in such a way that each succeeding circular row has 10 seats more than the previous one.

• If the first circular row has 30 seats, how many seats will be there in the 10th row?
• For 1500 seats in the auditorium, how many rows need to be there? OR If 1500 seats are to be arranged in the auditorium, how many seats are still left to be put after 10 th row?
• If there were 17 rows in the auditorium, how many seats will be there in the middle row?

Case Study Question – 6

• Draw a neat labelled figure to show the above situation diagrammatically.

• What is the speed of the plane in km/hr.

More Case Study Questions

We have class 10 maths case study questions in every chapter. You can download them as PDFs from the myCBSEguide App or from our free student dashboard .

As you know CBSE has reduced the syllabus this year, you should be careful while downloading these case study questions from the internet. You may get outdated or irrelevant questions there. It will not only be a waste of time but also lead to confusion.

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How to Solve Case-Based Questions?

Questions based on a given case study are normally taken from real-life situations. These are certainly related to the concepts provided in the textbook but the plot of the question is always based on a day-to-day life problem. There will be all subjective-type questions in the case study. You should answer the case-based questions to the point.

What are Class 10 competency-based questions?

Competency-based questions are questions that are based on real-life situations. Case study questions are a type of competency-based questions. There may be multiple ways to assess the competencies. The case study is assumed to be one of the best methods to evaluate competencies. In class 10 maths, you will find 1-2 case study questions. We advise you to read the passage carefully before answering the questions.

Case Study Questions in Maths Question Paper

CBSE has released new model question papers for annual examinations. myCBSEguide App has also created many model papers based on the new format (reduced syllabus) for the current session and uploaded them to myCBSEguide App. We advise all the students to download the myCBSEguide app and practice case study questions for class 10 maths as much as possible.

Case Studies on CBSE’s Official Website

CBSE has uploaded many case study questions on class 10 maths. You can download them from CBSE Official Website for free. Here you will find around 40-50 case study questions in PDF format for CBSE 10th class.

10 Maths Case Studies in myCBSEguide App

You can also download chapter-wise case study questions for class 10 maths from the myCBSEguide app. These class 10 case-based questions are prepared by our team of expert teachers. We have kept the new reduced syllabus in mind while creating these case-based questions. So, you will get the updated questions only.

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

• Exercise 3.1
• Exercise 3.2
• Exercise 3.3
• Exercise 3.4
• Exercise 3.5
• Exercise 3.6
• Exercise 3.7

NCERT Solutions for Class 10 Maths Chapters:

How many exercises in Chapter 3 Pair of Linear Equations in Two Variables

At a certain time in a deer park, the number of heads and the number of legs of deer and human visitors were counted and it was found there were 39 heads & 132 legs. find the number of deer and human visitors in the park., what is graphical method of solution of a pair of linear equations, when the son will be as old as the father today their ages will add up to 126 years. when the father was old as the son is today, their ages add upto 38 years. find their present ages., contact form.

NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12

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

Get Free NCERT Solutions for Class 10 Maths Chapter 3 Ex 3.1 PDF. Pair of Linear Equations in Two Variables Class 10 Maths NCERT Solutions are extremely helpful while doing your homework. Exercise 3.1 Class 10 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of प्रश्नावली 3.1 का हल हिंदी में Class 10 Pair of Linear Equations in Two Variables Exercise 3.1 provided in NCERT TextBook.

Topics and Sub Topics in Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables:

• Pair Of Linear Equations In Two Variables Class 10 Ex 3.1
• प्रश्नावली 3.1 का हल हिंदी में
• Pair Of Linear Equations In Two Variables Class 10 Ex 3.2
• प्रश्नावली 3.2 का हल हिंदी में
• Pair Of Linear Equations In Two Variables Class 10 Ex 3.3
• प्रश्नावली 3.3 का हल हिंदी में
• Pair Of Linear Equations In Two Variables Class 10 Ex 3.4
• प्रश्नावली 3.4 का हल हिंदी में
• Pair Of Linear Equations In Two Variables Class 10 Ex 3.5
• प्रश्नावली 3.5 का हल हिंदी में
• Pair Of Linear Equations In Two Variables Class 10 Ex 3.6
• प्रश्नावली 3.6 का हल हिंदी में
• Pair Of Linear Equations In Two Variables Class 10 Ex 3.7
• प्रश्नावली 3.7 का हल हिंदी में
• Extra Questions for Class 10 Maths Linear Equations in Two Variables

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

You can also download the free PDF of  Ex 3.1 Class 10 Pair of Linear Equations in Two Variables NCERT Solutions or save the solution images and take the print out to keep it handy for your exam preparation.

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

Class 10 Maths Pair of Linear Equations in Two Variables Mind Map

System of a pair of linear equations in two variables.

Representation of Linear Equation In Two Variables

Every linear equation in two variables graphically represents a line and each solution (x, y) of a linear equation in two variables, ax + by + c = 0, corresponds to a point on the line representing the equation, and vice versa.

Ploting Linear Equation in Two Variables on the Graph

Method of Solution of a Pair of Linear Equations in Two Variables

Coordinate of the point (x, y) which satisfy the system of pair of linear equations in two variables is the required solution. This is the point where the two lines representing the two equations intersect each other. There are two methods of finding solution of a pair of Linear equations in two variables. (1) Graphical Method : This method is less convenient when point representing the solution has non-integral co-ordinates. (2) Algebraic Method : This method is more convenient when point representing the solution has non-integral co-ordinates. This method is further divided into three methods: (i) Substitution Method, (ii) Elimination Method and (iii) Cross Multiplication Method.

Consistency and Nature of the Graphs

Consider the standard form of linear equations in two variables. a 1 x + b 1 y + C 1 = 0; a 2 x + b 2 y + c 2 = 0 While solving the above system of equation following three cases arise. (i) If \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\); system is called consistent, having unique solution and pair of straight lines representing the above equations intersect at one point only (ii) If \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\) ; system is called dependent and have infinetly many solution. Pair of lines representing the equations coincide. (iii) If \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}\); system is called inconsistent and has no solution. Pair of lines representing the equations are parallel or do not intersect at any point.

Algebraic Method of Solution

Consider the following system of equation a 1 x + b 1 y + c 1 =0; a 2 x + b 2 y + c 1 =0 There are following three methods under Algebraic method to solve the above system.

(i) Substitution method

(a) Find the value of one variable, say y in terms of x or x in terms of y from one equation. (b) Substitute this value in second equation to get equation in one variable and find solution. (c) Now substitute the value/solution so obtained in step (b) in the equation got in step (a).

(ii) Elimination Method

(a) If coefficient of any one variable are not same in both the equation multiply both the equation with suitable non-zero constants to make coefficient of any one variable numerically equal. (b) Add or subtract the equations so obtained to get equation in one variable and solve it. (c) Now substitute the value of the variable got in the above step in either of the original equation to get value of the other variable.

(iii) Cross multiplication method

Equations Reducible to a Pair of Linear Equations in Two Variables

NCERT Solutions for Class 10 Maths Chapter 3 Pairs of Linear Equations in Two Variables (Hindi Medium) Ex 3.1

NCERT Solutions for Class 10 Maths

• 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

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

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