# NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables (Ex 3.6) Exercise 3.6

NCERT, or the National Council of Educational Research, is self-governing. In 1961 it was established by the Indian government itself. Its main function is to help the Central government and the state governments reach the goal of improving the quality of education in the country. The NCERT is supposed to conduct, coordinate, and promote research in fields that are related to school education. They are supposed to create and publish all the written text required for school education, like textbooks, newsletters, journals, educational kits, multimedia digital material, etc.

The NCERT Solutions like the NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6 are available for all classes on the Extramarks website. The students can easily access all of the below –

NCERT Solutions Class 12

NCERT Solutions Class 11

NCERT Solutions Class 10

NCERT Solutions Class 9

NCERT Solutions Class 8

NCERT Solutions Class 7

NCERT Solutions Class 6

NCERT Solutions Class 5

NCERT Solutions Class 4

NCERT Solutions Class 3

NCERT Solutions Class 2

NCERT Solutions Class 1

High school students frequently exhibit a strong dislike for and dread of Mathematics.

It can be concluded that the majority of students experience this fear due to an unclear understanding of the foundational concepts in Mathematics. This concern can be easily overcome if adequate time is spent explaining the foundations to the students and encouraging regular practice. The NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6 can assist in understanding the foundational concepts.

Experts in Mathematics create the NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6 for the Extramarks website. The NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6 are formulated by specialists to assist students in understanding the basics.

Students can advance with a more solid foundation by using resources like the NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6. The value of Mathematics is so great that it permeates all facets of daily life. Mathematics is crucial to human efforts to comprehend the outside world and themselves, it also forms the basis of human cognition and logic. Students can have a better understanding of Chapter 3 by using the NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6.

Mathematics is a valuable tool for cultivating mental discipline since it encourages logical reasoning and mental rigour. For assistance, students can consult the NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6. Furthermore, understanding Mathematics is necessary for studying other academic subjects like Physics, Social Studies, and to some extent, even Art.

Many students have the misperception that much of what is taught in Mathematics is useless in real-world situations. To be able to adequately study for their exams, students need to realise the value of Mathematics and reject such myths.NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6 is one of many resources available to students to aid in their preparation, and can be found on the Extramarks website. The specialists who create the NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6 are knowledgeable and have a thorough comprehension of the study material.

The most crucial thing for students to do in order to master courses like Mathematics is to practise as much as they can. For practice and revision, students can use the NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6. Students can benefit greatly from the solutions that are provided for various chapters in exercise-by-exercise order, such as the NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6. The NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6 can help the students with Class tests, Class projects, etc.

The main objective of CBSE is to enhance skill development through the integration of inputs that are job-oriented and job-linked, as well as assessment and evaluation procedures. In order for students from CBSE-affiliated schools to acquire Class 10 examination certification, the CBSE is responsible for setting test requirements and conducting exams at the end of Class 10.

The board exams for Class 10 are thought to be one of the most significant tests a student will ever take. Based on their performance in the Class 10 board exams, students get to choose the subject in which they want to continue their education.Therefore, it is crucial for students in Class 10 to take their board exams seriously.

The student’s preparation for their board exams can be aided by the NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6. The school administers pre-board exams prior to the boards in order to get Class 10 students ready for their board exams. The purpose of the pre-board exam is to familiarise students with the format of the board examination and aid in their preparation. Students can use resources like the NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6 and others to get ready for their pre-boards and boards.

To help students develop the habit of revising the chapter right away after it has been taught in class, solutions like the NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6 can be a very useful tool. This practise can help significantly in lowering the anxiety that exams cause for many students. If the students have already prepared the study materials and are confident in their preparations, the stress reduces.

Additionally, students are advised not to strive to learn everything at once before exams. This can make the situation worse for those who are afraid of examinations. Using NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6 and other similar tools can help develop the habit of revising and not leaving everything until the last minute.The NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6 can assist students in avoiding treating exams as something to be scared of or worried about. The degree to which students effectively prepare for their examinations has a major impact on how they are able to see them. Therefore, it is important that the students have easy access to the tools they need to cope with the pressure and stress of examinations. The Mathematics experts at Extramarks are aware of this and try to provide all that is needed by the students. For example tools such as the NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6. The NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6 can help students prepare well for the Class 10 Maths Chapter 3 Exercise 3.6.

**NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables (Ex 3.6) Exercise 3.6 **

Extramarks’ mathematics professionals strive to provide students with tools that can be used in a variety of ways.The NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6 is one of these resources. Students can quickly and easily access various resources, including the NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6, by simply visiting the website.They do not need to search elsewhere on the internet for information to assist them in their board examination preparation. Reliable information can be found on the Extramarks website in the form of its resources, such as the NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6. Due to the fact that these were created by Extramarks’ subject-matter experts, the students can completely rely on them for their preparations without any doubt. The NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6 can be downloaded in PDF format for offline access.

Exercise 3.6 Class 10 Maths Is a part of Chapter 3 for Class 10 Mathematics. Chapter 3 in Class 10 Mathematics is based on – Pair of Linear Equations. The 3.6 Maths Class 10 is just one part of Chapter 3. Two-variable linear equations are defined for a line that can be drawn on a graph. Points are the answers to linear equations. A straight line will result from the two-variable linear equation described in this chapter. With the help of the NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6, students can better understand the concept of graph plotting and drawing straight lines with linear equations in two variables.

Students can access the PDF for Class 10 Chapter 3 Mathematics Solutions on Extramarks to aid in their learning of the questions given in this particular chapter. Every question given in these exercises has solutions on the Extramarks website by experts in Mathematics.The NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6 have been updated to reflect the most recent CBSE Class 10 Maths Chapter 3 syllabus.

**Linear Equation –**

A linear equation is a straight-line equation. The formula is written as axe + by + c = 0, where a, b, and c are real numbers (a0 and b0), and x and y are the two variables.

In this case, the coefficients are a and b, and the equation’s constant is c.

Pair of Two Linear Equations –

The term “pair of linear equations in two variables” refers to two linear equations with the same two variables.

- a1x + b1y + c1 = 0
- a2x + b2y + c2 = 0

**Importance of a Pair of Linear Equations in Two Variables**

Linear equations have numerous uses in both Mathematics and everyday life. Equality with variables and the equal symbol (=) is known as an algebraic equation. An equation of degree one is a linear equation. Word problems are a common way for mathematicians to apply their knowledge, and solving them typically involves the use of Linear Equations.

The uses of Linear Equations in daily life are countless. To use Algebra to solve real-world issues, students have to translate the given scenario into mathematical statements that make it crystal clear how the variables relate to the information presented. The procedures for converting a situation into a mathematical statement are as follows:

- Students have to first an Algebraic Expression that accurately explains the situation by turning the problem statement into a mathematical statement.
- Then they have to determine the unknowns in the question and give them variables (a quantity whose value might vary depending on the mathematical situation).
- Then the students need to cite the information, phrases, and keywords after carefully reading the problem several times. And then sequentially arrange the facts they have collected.
- Now the students need to use the Algebraic Expression and the information in the problem statement to frame an equation and then solve it using methodical equation solving approaches.

To understand and practise this better, the students are advised to take help from the NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6.

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

The students are instructed to develop a preparation strategy that works for them and to stick with it throughout all of their lessons. If there is a strategy that can be used to deal with it, any stressful circumstance can be managed better. The use of the NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6 and other similar tools can help the students with this. The more the students use the NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6 and tools like the revision notes for each chapter, the better the students can get at the subject.

**NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.6 **

The NCERT books are organised such that the subject content is divided in a less challenging way. The NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6 and similar tools are created by the experts while keeping in mind the CBSE’s published syllabus. As the students continue to advance to the next standard, the chapters get harder. This is based on the increasing calibre of students as they move towards the next standard.

For the solutions to the questions in the NCERT books, the Extramarks Mathematics experts adhere to the same format as the NCERT books. Students can ensure that they have a firm grasp of the fundamentals of everything being taught in Chapter 3 by using the preparatory materials created by the subject specialists at Extramarks, like the NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6 and more. Students can go through the NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6 to clarify their doubts related to Chapter 3 Exercise 3.6.

All themes and topics covered in classes before Class 10 must be understood efficiently by the students. If they aim at performing excellently in Class 10 board examinations this is a must. The CBSE creates each class’s syllabus in such a way that many of the topics are the same or very similar for two different classes next to each other. The level of difficulty in which the topic is presented may vary for different standards as more and more details are added to the syllabus. Students must comprehend the significance of this method ofof dividing the syllabus.There is a reason why it is done the way it is done. This is CBSE’s way of making sure that students have a firm grasp of the material covered in earlier classes before moving on to the higher classes.

The students can make sure they have understood everything taught in Chapter 3 before going on to the next chapter by using the NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6 and the solutions for the other exercises in Chapter 3. Another method to make sure that the students have an in-depth knowledge of all that has been taught previously before moving on to use the revision notes made for Chapter 3 along with the NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6.

Students are recommended to utilise the helpful resources on the Extramarks website as much as possible for courses like Mathematics. Considering that practise and revision are essential to mastering this subject. The NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6 can help with both.

Along with the NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6 there are many other tools on the Extramarks website that students can quite simply access by visiting the website. The students are encouraged to boost their confidence during their preparation so they can master Mathematics examinations. The likelihood of a student getting good results in their final board exams is higher for those who are confident and sure about their preparation. Students that are confident may have fewer chances of making mistakes that are viewed as foolish or silly. As it is something that frequently happens when students are under pressure, and not well prepared during exams. The NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6 and other similar tools can be of much help to the students in gaining confidence in their preparations. Students can also find resources like revision notes, supplementary questions, significant questions from other chapters, etc., in addition to the NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6.

One of the subjects that many students might not enjoy learning or prepare for is Mathematics. The students are advised to take help from the NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6 and other helpful tools.

**NCERT Solutions for Class 10 Maths Chapter 3 Exercises**

If students choose wisely when preparing for Mathematics from a young age, it can be a fun subject for them to study. It is suggested that students use resources like the NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6 from a younger age in order to fully understand the subject. The likelihood that the students will be afraid of the subject decreases when they are able to understand each theme of the subject conceptually. Therefore, experts strongly suggest using tools like NCERT Solutions etc.

Students must fully understand that mathematics is nothing more than a game of practise and revision.The more the students can view it as a fun and engaging subject that is all about theorems and formulas, the more fun they can have while preparing for it. Students may easily accomplish this with the use of resources like the NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6.

It is advisable for the students to utilise the internet to the fullest extent possible given how easily accessible it has become in today’s time. When getting ready for tests in class or exams, etc., the internet can be a saviour for the students. Students can benefit from tools such as the NCERT Solutions for Class 10 Maths, Chapter 3, Exercise 3.6.The students can easily access the NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6 through the internet. All they need to do is look for them on the Extramarks website.

Tools like the NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6 are highly accessible to students and readily available. These resources, along with many others, are all offered at one destination and in the language in which the students are expected to write their exams. English or Hindi, either.

Class 10 is a very important class in the academic life of any student. The students should keep this in mind. A good score in Class 10 examinations can make it easier for the students to choose the stream of their choice for Class 11 and Class 12. The stream they choose can lay the foundation of their college education and hence plays a very vital role. The students should try not to be too stressed when preparing for the Class 10 board examinations and should try to be confident in their preparations. NCERT Solutions for Class 10 Maths, Chapter 3, Exercise 3.6, available on the Extramarks website, can assist them in achieving this goal.There are many branches of Mathematics in science that are related to numbers, including Geometric forms, Algebra, and others. Analytical thinking is aided by the study of Mathematics. Data are gathered, broken down, and then connected to answer mathematical problems.

- Thinking skills can be developed through Mathematics.
- It aids in describing how things operate.
- It fosters wisdom development.
- It accelerates intuitive processing speed.
- It contributes to the students being smarter.
- When practised as a profession, Mathematics can bring great career opportunities.
- In a world that is continuously changing, it is crucial to understand the world.
- It gives the students a chance to interact with the outside world.

**Q.1 ** ** **

$\begin{array}{l}\text{Solve the following pair of equations by reducing them to a pair of linear equations:}\\ \text{(i)\hspace{0.33em}}\frac{\text{1}}{\text{2x}}\text{\hspace{0.33em}+\hspace{0.33em}}\frac{\text{1}}{\text{3y}}\text{=2\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}(ii)\hspace{0.33em}}\frac{\text{2}}{\sqrt{\text{x}}}\text{\hspace{0.33em}+\hspace{0.33em}}\frac{\text{3}}{\sqrt{\text{y}}}\text{\hspace{0.33em}=\hspace{0.33em}2}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\frac{\text{1}}{\text{3x}}\text{\hspace{0.33em}+\hspace{0.33em}}\frac{\text{1}}{\text{2y}}\text{\hspace{0.33em}=\hspace{0.33em}}\frac{\text{13}}{\text{6}}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\frac{\text{4}}{\sqrt{\text{x}}}\text{\hspace{0.33em}}\u2013\text{\hspace{0.33em}}\frac{\text{9}}{\sqrt{\text{y}}}\text{\hspace{0.33em}=\u20131}\\ \text{(iii)\hspace{0.33em}}\frac{\text{4}}{\text{x}}\text{\hspace{0.33em}+\hspace{0.33em}3y\hspace{0.33em}=\hspace{0.33em}14\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}(iv)\hspace{0.33em}}\frac{\text{5}}{\text{x\hspace{0.33em}\u2013\hspace{0.33em}1}}\text{\hspace{0.33em}+\hspace{0.33em}}\frac{\text{1}}{\text{y\hspace{0.33em}\u2013\hspace{0.33em}2}}\text{\hspace{0.33em}=\hspace{0.33em}2}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\frac{\text{3}}{\text{x}}\text{\hspace{0.33em}}\u2013\text{\hspace{0.33em}4y\hspace{0.33em}=\hspace{0.33em}23\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\frac{\text{6}}{\text{x\hspace{0.33em}\u2013\hspace{0.33em}1}}\text{\hspace{0.33em}}\u2013\text{\hspace{0.33em}}\frac{\text{3}}{\text{y\hspace{0.33em}\u2013\hspace{0.33em}2}}\text{\hspace{0.33em}=\hspace{0.33em}1}\\ \text{(v)\hspace{0.33em}}\frac{\text{7x\hspace{0.33em}\u2013\hspace{0.33em}2y}}{\text{xy}}\text{\hspace{0.33em}=\hspace{0.33em}5\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}(vi)\hspace{0.33em}6x\hspace{0.33em}+\hspace{0.33em}3y\hspace{0.33em}=\hspace{0.33em}6xy}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\frac{\text{8x\hspace{0.33em}+\hspace{0.33em}7y}}{\text{xy}}\text{\hspace{0.33em}=\hspace{0.33em}15\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}2x\hspace{0.33em}+\hspace{0.33em}4y\hspace{0.33em}=\hspace{0.33em}5xy}\\ \text{(vii)\hspace{0.33em}}\frac{\text{10}}{\text{x\hspace{0.33em}+\hspace{0.33em}y}}\text{\hspace{0.33em}+\hspace{0.33em}}\frac{\text{2}}{\text{x\hspace{0.33em}\u2013\hspace{0.33em}y}}\text{\hspace{0.33em}=\hspace{0.33em}4\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}(viii)\hspace{0.33em}\hspace{0.33em}}\frac{\text{1}}{\text{3x\hspace{0.33em}+\hspace{0.33em}y}}\text{\hspace{0.33em}+\hspace{0.33em}}\frac{\text{1}}{\text{3x\hspace{0.33em}}\u2013\text{\hspace{0.33em}y}}\text{\hspace{0.33em}=\hspace{0.33em}}\frac{\text{3}}{\text{4}}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\frac{\text{15}}{\text{x\hspace{0.33em}+\hspace{0.33em}y}}\text{\hspace{0.33em}}\u2013\text{\hspace{0.33em}}\frac{\text{5}}{\text{x\hspace{0.33em}\u2013\hspace{0.33em}y}}\text{=\hspace{0.33em}\u20132\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\frac{\text{1}}{\text{2(3x\hspace{0.33em}+\hspace{0.33em}y)}}\text{\hspace{0.33em}}\u2013\text{\hspace{0.33em}}\frac{\text{1}}{\text{2(3x\hspace{0.33em}\u2013\hspace{0.33em}y)}}\text{\hspace{0.33em}=\hspace{0.33em}}\frac{\text{\u20131}}{\text{8}}\\ \end{array}$

**Ans.**

$\begin{array}{l}\left(\mathrm{i}\right)\\ \text{Let}\frac{1}{\mathrm{x}}=\mathrm{p}\text{and}\frac{1}{\mathrm{y}}=\mathrm{q}\\ \mathrm{then}\text{given equations can be written as:}\\ \frac{\mathrm{p}}{2}+\frac{\mathrm{q}}{3}=2\text{}\Rightarrow 3\mathrm{p}+2\mathrm{q}-12=0\text{}...\text{(1)}\\ \frac{\mathrm{p}}{3}+\frac{\mathrm{q}}{2}=\frac{13}{6}\text{}\Rightarrow 2\mathrm{p}+3\mathrm{q}-13=0\text{}...\text{(2)}\\ \text{Using cross-multiplication method, we get}\\ \frac{\mathrm{p}}{-26-(-36)}=\frac{\mathrm{q}}{-24-(-39)}=\frac{1}{9-4}\\ \text{}\Rightarrow \frac{\mathrm{p}}{10}=\frac{\mathrm{q}}{15}=\frac{1}{5}\\ \text{}\Rightarrow \frac{\mathrm{p}}{10}=\frac{1}{5}\text{and}\frac{\mathrm{q}}{15}=\frac{1}{5}\\ \text{}\Rightarrow \mathrm{p}=2\text{and q=3}\\ \text{}\Rightarrow \frac{1}{\mathrm{x}}=2\text{and}\frac{1}{\mathrm{y}}=3\\ \text{}\Rightarrow \mathrm{x}=\frac{1}{2}\text{and y=}\frac{1}{3}\\ \left(\mathrm{ii}\right)\text{}\\ \mathrm{Given}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{that}\\ \frac{2}{\sqrt{\mathrm{x}}}+\frac{3}{\sqrt{\mathrm{y}}}=2\\ \mathrm{and}\\ \frac{4}{\sqrt{\mathrm{x}}}-\frac{9}{\sqrt{\mathrm{y}}}=-1\\ \mathrm{Let}\text{}\frac{1}{\sqrt{\mathrm{x}}}=\mathrm{p}\text{and}\frac{1}{\sqrt{\mathrm{y}}}=\mathrm{q}\text{, then we get}\\ 2\mathrm{p}+3\mathrm{q}=2\text{}...\left(1\right)\\ 4\mathrm{p}-9\mathrm{q}=-1\text{}...\text{(2)}\\ \text{Multiplying equation (1) by 3, we get}\\ \text{6p + 9q = 6}\\ \text{Adding this to equation (2), we get}\\ \text{10p = 5}\\ \Rightarrow \text{p=}\frac{1}{2}\\ \text{Putting value of p in equation (1), we get}\\ \text{2}\times \frac{1}{2}+3\mathrm{q}=2\\ \Rightarrow \text{}3\mathrm{q}=1\\ \Rightarrow \text{}\mathrm{q}=\frac{1}{3}\end{array}$

$\begin{array}{l}\mathrm{Now},\\ \mathrm{p}=\frac{1}{\sqrt{\mathrm{x}}}=\frac{1}{2}\Rightarrow \sqrt{\mathrm{x}}=2\Rightarrow \mathrm{x}=4\\ \mathrm{q}=\frac{1}{\sqrt{\mathrm{y}}}=\frac{1}{3}\Rightarrow \sqrt{\mathrm{y}}=3\Rightarrow \mathrm{y}=9\\ \mathrm{Hence},\text{}\mathrm{x}=4\text{and}\mathrm{y}=9.\\ \\ \text{(iii)}\\ \text{Given that}\\ \frac{4}{\mathrm{x}}+3\mathrm{y}=14\\ \frac{3}{\mathrm{x}}-4\mathrm{y}=23\\ \text{Substituting}\frac{1}{\mathrm{x}}=\mathrm{p}\text{}\mathrm{in}\text{}\mathrm{above}\text{}\mathrm{equations},\text{we get}\\ \text{4p}+3\mathrm{y}-14=0\text{}...\text{(1)}\\ 3\mathrm{p}-4\mathrm{y}-23=0\text{}...\text{(2)}\\ \text{By cross-multiplication method, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{p}}{-69-56}=\frac{\mathrm{y}}{-42-(-92)}=\frac{1}{-16-9}\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{p}}{-125}=\frac{\mathrm{y}}{50}=\frac{1}{-25}\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{p}=5\text{and y}=-2\\ \Rightarrow \mathrm{p}=\frac{1}{\mathrm{x}}=5\Rightarrow \mathrm{x}=\frac{1}{5}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{y}=-2\end{array}$

$\begin{array}{l}\left(\mathrm{iv}\right)\\ \mathrm{Given}\text{\hspace{0.33em}}\mathrm{that}\\ \frac{5}{\mathrm{x}-1}+\frac{1}{\mathrm{y}-2}=2\\ \mathrm{and}\\ \frac{6}{\mathrm{x}-1}-\frac{3}{\mathrm{y}-2}=1\\ \text{Putting}\frac{1}{\mathrm{x}-1}=\mathrm{p}\text{and}\frac{1}{\mathrm{y}-2}=\mathrm{q}\text{, we get}\\ \text{5p}+\mathrm{q}=2\text{}...\text{(1)}\\ \text{6p}-3\mathrm{q}=1\text{}...\text{(2)}\\ \text{multiplying equation (1) by 3, we get}\\ \text{15p}+3\mathrm{q}=6\text{}...\text{(3)}\\ \text{Adding (2) and (3), we get}\\ \text{21}\mathrm{p}=7\Rightarrow \mathrm{p}=\frac{1}{3}\\ \text{Putting value of p in equation (1), we get}\\ \text{5}\times \frac{1}{3}+\mathrm{q}=2\Rightarrow \mathrm{q}=2-\frac{5}{3}=\frac{1}{3}\\ \mathrm{Now},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{p}=\frac{1}{\mathrm{x}-1}=\frac{1}{3}\Rightarrow \mathrm{x}-1=3\Rightarrow \mathrm{x}=4\\ \text{and}\mathrm{q}=\frac{1}{\mathrm{y}-2}=\frac{1}{3}\Rightarrow \mathrm{y}-2=3\Rightarrow \mathrm{y}=5\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\therefore \mathrm{x}=4\text{and y}=5.\end{array}$ \begin{array}{l}\end{array}

$\begin{array}{l}\left(\mathrm{v}\right)\\ \mathrm{We}\text{have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{7\mathrm{x}-2\mathrm{y}}{\mathrm{xy}}=5\text{}\Rightarrow \text{}\frac{7}{\mathrm{y}}-\frac{2}{\mathrm{x}}=5\text{}...\text{(1)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{8\mathrm{x}+7\mathrm{y}}{\mathrm{xy}}=15\text{}\Rightarrow \text{}\frac{8}{\mathrm{y}}+\frac{7}{\mathrm{x}}=15\text{}...\text{(2)}\\ \text{Putting}\frac{1}{\mathrm{x}}=\mathrm{p}\text{and}\frac{1}{\mathrm{y}}=\mathrm{q}\text{in equation (1) and (2), we get}\\ \text{}-2\mathrm{p}+7\mathrm{q}-5=0\text{}...\text{(3)}\\ \text{7p}+8\mathrm{q}-15=0\text{}...\text{(4)}\\ \text{By cross-multiplication method, we get}\\ \frac{\mathrm{p}}{-105-(-40)}=\frac{\mathrm{q}}{-35-30}=\frac{1}{-16-49}\\ \Rightarrow \frac{\mathrm{p}}{-65}=\frac{\mathrm{q}}{-65}=\frac{1}{-65}\\ \Rightarrow \frac{\mathrm{p}}{-65}=\frac{1}{-65}\text{and}\frac{\mathrm{q}}{-65}=\frac{1}{-65}\\ \Rightarrow \mathrm{p}=1\text{and}\mathrm{q}=1\\ \Rightarrow \mathrm{p}=\frac{1}{\mathrm{x}}=1\text{and q}=\frac{1}{\mathrm{y}}=1\\ \Rightarrow \mathrm{x}=1\text{and}\mathrm{y}=1\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}\left(\mathrm{vi}\right)\\ \text{We have}\\ 6\mathrm{x}+3\mathrm{y}=6\mathrm{xy}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{6}{\mathrm{y}}+\frac{3}{\mathrm{x}}=6\text{}...\text{(1)}\\ \text{and}\\ 2\mathrm{x}+4\mathrm{y}=5\mathrm{xy}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{2}{\mathrm{y}}+\frac{4}{\mathrm{x}}=5\text{}...\text{(2)}\\ \text{Putting}\frac{1}{\mathrm{x}}=\mathrm{p}\text{and}\frac{1}{\mathrm{y}}=\mathrm{q}\text{in above equations, we get}\\ \text{3p}+6\mathrm{q}-6=0\text{}...\text{(3)}\\ \text{4p}+2\mathrm{q}-5=0\text{}...\text{(4)}\\ \text{By cross-multiplication method, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{p}}{-30-(-12)}=\frac{\mathrm{q}}{-24-(-15)}=\frac{1}{6-24}\\ \Rightarrow \frac{\mathrm{p}}{-18}=\frac{\mathrm{q}}{-9}=\frac{1}{-18}\\ \Rightarrow \frac{\mathrm{p}}{-18}=\frac{1}{-18}\text{and}\frac{\mathrm{q}}{-9}=\frac{1}{-18}\\ \Rightarrow \mathrm{p}=1\text{and q}=\frac{1}{2}\\ \Rightarrow \mathrm{p}=\frac{1}{\mathrm{x}}=1\text{and q}=\frac{1}{\mathrm{y}}=\frac{1}{2}\\ \Rightarrow \mathrm{x}=1\text{and}\mathrm{y}=2\end{array}$ $\begin{array}{l}\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}\left(\text{vii}\right)\text{We have}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\frac{10}{\mathrm{x}+\mathrm{y}}+\frac{2}{\mathrm{x}-\mathrm{y}}=4\\ \mathrm{and}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\frac{15}{\mathrm{x}+\mathrm{y}}-\frac{5}{\mathrm{x}-\mathrm{y}}=-2\\ \text{Putting}\frac{1}{\mathrm{x}+\mathrm{y}}=\mathrm{p}\text{and}\frac{1}{\mathrm{x}-\mathrm{y}}=\mathrm{q}\text{in above equations, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} 10p}+2\mathrm{q}-4=0\text{}...\text{(3)}\\ \text{and}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} 15p}-5\mathrm{q}+2=0\text{}...\text{(4)}\\ \text{By cross-multiplication method, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{p}}{4-20}=\frac{\mathrm{q}}{-60-20}=\frac{1}{-50-30}\\ \Rightarrow \frac{\mathrm{p}}{-16}=\frac{\mathrm{q}}{-80}=\frac{1}{-80}\\ \Rightarrow \frac{\mathrm{p}}{-16}=\frac{1}{-80}\text{and}\frac{\mathrm{q}}{-80}=\frac{1}{-80}\\ \Rightarrow \mathrm{p}=\frac{1}{5}\text{and q}=1\\ \Rightarrow \mathrm{p}=\frac{1}{\mathrm{x}+\mathrm{y}}=\frac{1}{5}\text{and q}=\frac{1}{\mathrm{x}-\mathrm{y}}=1\\ \text{Thus we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}+\mathrm{y}=5\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\text{(5)}\\ \text{and x}-\mathrm{y}=1\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\text{(6)}\\ \text{Adding equations (5) and (6), we get}\\ \text{2x}=6\Rightarrow \mathrm{x}=3\end{array}$

$\begin{array}{l}\mathrm{Subtracting}\text{equation (6) from equation (5), we get}\\ \text{}\mathrm{y}=2\\ \text{Hence,}\mathrm{x}=3\text{and}\mathrm{y}=2.\\ \\ \left(\text{viii}\right)\text{}\\ \text{We have}\\ \text{}\frac{1}{3\mathrm{x}+\mathrm{y}}+\frac{1}{3\mathrm{x}-\mathrm{y}}=\frac{3}{4}\\ \text{and}\frac{1}{2(3\mathrm{x}+\mathrm{y})}-\frac{1}{2(3\mathrm{x}+\mathrm{y})}=\frac{-1}{8}\\ \text{Putting}\frac{1}{3\mathrm{x}+\mathrm{y}}=\mathrm{p}\text{and}\frac{1}{3\mathrm{x}-\mathrm{y}}=\mathrm{q}\text{in above equations, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}p}+\mathrm{q}=\frac{3}{4}\text{\hspace{0.17em}}...\text{(3)}\\ \mathrm{and}\text{}\frac{\text{p}}{2}-\frac{\mathrm{q}}{2}=\frac{-1}{8}\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{p}-\mathrm{q}=\frac{-1}{4}\text{}...\text{(4)}\\ \text{Adding equations (3) and (4), we get}\\ \text{2p}=\frac{3}{4}-\frac{1}{4}\Rightarrow \mathrm{p}=\frac{1}{4}\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}\mathrm{Now},\\ \text{}\mathrm{p}=\frac{1}{3\mathrm{x}+\mathrm{y}}=\frac{1}{4}\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}3\mathrm{x}+\mathrm{y}=4\text{}...\text{(5)}\\ \mathrm{Also},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{q}=\frac{1}{3\mathrm{x}-\mathrm{y}}=\frac{1}{2}\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}3\mathrm{x}-\mathrm{y}=2\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\text{(6)}\\ \text{Adding equations (5) and (6), we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}6x}=\text{6}\Rightarrow \mathrm{x}=1\\ \mathrm{Substitute}\text{}\mathrm{x}=1\text{in equation (5), we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}3+\mathrm{y}=4\Rightarrow \mathrm{y}=1\\ \mathrm{Hence},\text{}\mathrm{x}=1\text{and}\mathrm{y}=1.\end{array}$

**Q.2 ** Formulate the following problem as a pair of linear equations and hence find their solution:

(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 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.

**Ans.**

(i)

Let speed of Ritu in still water be x km/h and the speed of stream be y km/h.

Speed of Ritu while rowing upstream = (x– y) km/h

Speed of Ritu while rowing downstream = (x + y) km/h

According to question,

2(x + y) = 20

or x + y = 10 …(1)

Also,

2(x – y) = 4

or x – y = 2 …(2)

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

2x = 12

or x = 6

Putting this in equation (1), we get

y = 4

Hence, Ritu’s speed in still water is 6 km/h and the speed of current is 4 km/h.

(ii)

$\begin{array}{l}\text{Let the number of days taken by a woman to finish the work}\\ \mathrm{is}\text{}\mathrm{x}\text{and the number of days taken by a man to finish the}\\ \text{work}\mathrm{is}\text{}\mathrm{y}.\\ \text{Therefore,}\\ \text{Part of the work finished by a woman in 1 day =}\frac{1}{\mathrm{x}}\\ \mathrm{and}\\ \text{part of the work finished by a man in 1 day =}\frac{1}{\mathrm{y}}\\ \mathrm{Given}\text{}\mathrm{that}\text{}\\ \mathrm{the}\text{}.\\ \text{Therefore,}\\ \text{4}(\frac{2}{\mathrm{x}}+\frac{5}{\mathrm{y}}\text{) = 1}\\ \Rightarrow \text{}\frac{2}{\mathrm{x}}+\frac{5}{\mathrm{y}}=\frac{1}{4}\end{array}$

$\begin{array}{l}\mathrm{It}\text{is also given that}\\ \mathrm{the}\text{}\mathrm{work}\text{}.\\ \text{Therefore,}\\ \text{3\hspace{0.17em}\hspace{0.17em}}(\frac{3}{\mathrm{x}}+\frac{6}{\mathrm{y}})\text{= 1}\\ \Rightarrow \text{}\frac{3}{\mathrm{x}}+\frac{6}{\mathrm{y}}=\frac{1}{3}\\ \mathrm{P}\text{utting}\frac{1}{\mathrm{x}}=\mathrm{p}\text{and}\frac{1}{\mathrm{y}}=\mathrm{q}\text{in the above equations, we get}\\ \text{2p + 5q =}\frac{1}{4}\Rightarrow 8\mathrm{p}+20\mathrm{q}=1\\ \text{3p + 6q =}\frac{1}{3}\Rightarrow 9\mathrm{p}+18\mathrm{q}=1\\ \text{By cross-multiplication method, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{p}}{-20-(-18)}=\frac{\mathrm{q}}{-9-(-8)}=\frac{1}{144-180}\\ \Rightarrow \frac{\mathrm{p}}{-2}=\frac{\mathrm{q}}{-1}=\frac{1}{-36}\\ \Rightarrow \frac{\mathrm{p}}{-2}=\frac{1}{-36}\text{and}\frac{\mathrm{q}}{-1}=\frac{1}{-36}\\ \Rightarrow \mathrm{p}=\frac{1}{18}\text{and}\mathrm{q}=\frac{1}{36}\\ \Rightarrow \mathrm{p}=\frac{1}{\mathrm{x}}=\frac{1}{18}\text{and}\mathrm{q}=\frac{1}{\mathrm{y}}=\frac{1}{36}\\ \Rightarrow \mathrm{x}=18\text{and}\mathrm{y}=36\\ \text{Hence, number of days taken by a woman = 18 days}\\ \mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}number of days taken by a man = 36 days.}\end{array}$

$\begin{array}{l}\left(\text{iii}\right)\\ \text{Let the speeds of train and bus be x km/h and y km/h respectively.}\\ \text{According to question,}\\ \text{}\frac{60}{\mathrm{x}}+\frac{240}{\mathrm{y}}=4\text{}...\text{(1)}\\ \text{and}\\ \text{}\frac{100}{\mathrm{x}}+\frac{200}{\mathrm{y}}=\frac{25}{6}\text{}...\text{(2)}\\ \text{Putting}\frac{1}{\mathrm{x}}=\mathrm{p}\text{and}\frac{1}{\mathrm{y}}=\mathrm{q}\text{in these equations, we get}\\ \text{}60\mathrm{p}+240\mathrm{q}=4\text{}...\text{(3)}\\ \text{and}100\mathrm{p}+200\mathrm{q}=\frac{25}{6}\text{}\\ \Rightarrow \text{}600\mathrm{p}+1200\mathrm{q}=25\text{}...\text{(4)}\\ \text{Multiplying equation (3) by 10, we get}\\ \text{}600\mathrm{p}+2400\mathrm{q}=40\text{}...\text{(5)}\\ \text{Subtracting equation (4) from (5), we get}\\ \text{1200q = 15}\Rightarrow \mathrm{q}=\frac{1}{80}\\ \text{Substituting in equation (3), we get}\\ \text{60p = 1}\Rightarrow \mathrm{p}=\frac{1}{60}\\ \mathrm{Now},\text{}\mathrm{p}=\frac{1}{\mathrm{x}}=\frac{1}{60}\text{and q =}\frac{1}{\mathrm{y}}=\frac{1}{80}\end{array}$

$\begin{array}{l}\Rightarrow \text{x = 60 km/h and y = 80 km/h}\\ \text{Hence, speed of train = 60 km/h}\\ \text{and\hspace{0.17em}\hspace{0.17em}speed of bus = 80 km/h}\end{array}$

## FAQs (Frequently Asked Questions)

### 1. Is it a good practice to learn the step-by-step answers given in the NCERT Solutions on the Extramarks website?

The questions framed in the examinations can have different numeric values in them which can confuse the students and they may end up losing marks. It is rather suggested to understand the question and the solution to it. Then try and memorise the methods and formulas. Students can access a variety of solutions like the NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6 from the Extramarks website and mobile application.

### 2. How can the NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6 be helpful during the Class 10 board exam preparation?

The NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6 can be extremely helpful during the board exam preparation as these solutions are prepared by experts with years of experience. These solutions follow a step-by-step approach and students can rely on them as they can clarify their doubts without any help from external resources. The NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6 can be downloaded from the Extramarks website and mobile application for offline access