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Chapter 3 Class 10 Pair of Linear Equations in Two Variables

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Updated for NCERT 2023-24 Books

Get NCERT solutions of Chapter 3 Class 10 - Pair of Linear Equations in Two Variables at Teachoo. Answers to all exercise questions, examples and optional questions have been provided with video of each and every question

We studied  Linear Equations in Two Variables in Class 9, we will study pair of linear equations in this chapter.

In this chapter, we will learn

• What are Linear Equations in Two Variables
• Converting statements into Equations , and drawing graph of those linear equations
• Possible Type of Graphs for Pair of Linear Equations in Two Variables - Two Lines Intersecting, Two lines Parallel, Coincident Lines
• Finding s olution of equations from graphs
• Consistency of equations  by finding ratio of a 1 /a 2 , b 1 /b 2 , c 1 /c 2
• Intersecting Lines (Exactly one solution - unique)
• Coincident Lines (Infinitely many solutions)
• Parallel Lines ( No solutions)
• Substitution Method
• Elimination Method
• Cross Multiplication Method
• Solving complicated equations like 2/x + 3/y = 4 by substituting variables (like putting p = 1/x, q = 1/y and solving)
• Solving Statement Questions by first forming equations, and then solving

Click on exercise or topic link below to start doing the chapter

Note: When you click on a link, the first question will open. To open any other question of the exercise, go to the bottom of the page. There is a list with arrows having all the questions (with important questions also marked)

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

Class 10 Maths Exercise 3.1 Solutions Class 10 Maths Exercise 3.2 Solutions Class 10 Maths Exercise 3.3 Solutions

Revised NCERT Solutions for Class 10 Maths chapter 3 Pair of linear equations in two variables all exercises answer in Hindi and English medium updated for 2024-25. As per the new textbooks issued for academic year 2024-25, there are only three exercises in chapter 3 of class 10th mathematics.

Class 10 Maths Chapter 3 Solutions for Board Exams

• Class 10 Maths Chapter 3 Exercise 3.1
• Class 10 Maths Chapter 3 Exercise 3.2
• Class 10 Maths Chapter 3 Exercise 3.3

Class 10 Maths Chapter 3 Solutions for State Boards

• Class 10 Maths Chapter 3 Exercise 3.4
• Class 10 Maths Chapter 3 Exercise 3.5
• Class 10 Maths Chapter 3 Exercise 3.6
• Class 10 Maths Chapter 3 Exercise 3.7

Class 10th Maths Chapter 3 Solutions

• Class 10 Maths Chapter 3 NCERT Book
• Class 10 Maths NCERT Solutions
• Class 10 all Subjects NCERT Solutions

According to new syllabus the class 10th math has only three exercise for board exams. Exercise 3.1 is deleted now. The exercise 3.2 become ex. 3.1, Exercise 3.3 become ex. 3.2 and exercise 3.4 become ex. 3.3 now. The split up syllabus for 10th mathematics chapter 3 is as follows: Pair of linear equations in two variables and graphical method of their solution, consistency/inconsistency. Algebraic conditions for number of solutions. Solution of a pair of linear equations in two variables algebraically – by substitution, by elimination. Simple situational problems.

• Now, Number of exercise: 3
• Number of Period needed: 15
• Weightage of the Chapter: 7 -8

UP Board students (High School and Intermediate) are now using NCERT Textbooks for most of the subjects. Class 10 Mathematics Books for UP Board is same as the NCERT Books for Class 10 Math in CBSE Board. So, the students of Uttar Pradesh Board can download, UP Board Solutions for class 10 Math Chapter 3 from this page of Tiwari Academy. Solutions are available in Hindi and English Medium. Graphs are given for all the questions if required. Visit to Discussion Forum to ask your questions. You can reply the questions already asked by other users.

Class 10 Maths Chapter 3 Practice Tests with Answers

• Class 10 Maths Chapter 3 Practice Test 1
• Class 10 Maths Chapter 3 Practice Test 2
• Class 10 Maths Chapter 3 Practice Test 3
• Class 10 Maths Chapter 3 Practice Test 4
• Class 10 Maths Chapter 3 Practice Test 5
• Class 10 Maths Chapter 3 Practice Test 6

In NCERT Solutions Class 10, all exercises are solved in both English as well as in Hindi medium in order to help all type of students for academic session 2024-25. In Maths 10, Exercise 3.1, 3.2, and 3.3 solutions, if there is any inconvenient to understand, please inform us, we will try to solve it. All NCERT Solutions 2024-25 are made for the CBSE exam for March, 2022 based on latest CBSE Syllabus 2024-25.

Changes in CBSE Syllabus for Class 10 Maths Chapter 3

CBSE has reduced the syllabus of all subjects in all the classes. The CBSE Syllabus for Class 10 Maths is reduced to 65 percent now. The changes in 10th Maths chapter 3: Linear equations in two variables are given below.

The new CBSE Syllabus for 2024-25 Class 10 Maths Chapter 3

Pair of linear equations in two variables and graphical method of their solution, consistency /inconsistency. Algebraic conditions for number of solutions. Solution of a pair of linear equations in two variables algebraically – by substitution, by elimination. Simple situational problems. Simple problems on equations reducible to linear equations.

Deleted Section from previous Syllabus

Solution of a pair of linear equations in two variables by Cross-multiplication.

Previous Years Questions – 1 Mark or 2 Marks

1. Find whether the lines representing the following pair of linear equations intersect at a point, are parallel or coincident: 3x + y = 7 & 6x + 2y = 8. [CBSE 2016] 2. Find the value of k for which the system of equations 3𝑥 − 4𝑦 = 7; 𝑘𝑥 + 3𝑦 − 5 = 0 has no solution. [CBSE 2014] 3. A father is three times as old as his son. After five years, his age will be two and a half times as old as his son. Represent this situation algebraically only. [CBSE 2013] 4. For which value of p does the pair of equations given below have a unique solutions? 4x + py + 8 = 0; 2x + 2y + 2 = 0. [CBSE 2010, 2011, 2013] 5. For what value of k, the following system of linear equations has no solutions? 3x + y = 1; (2k – 1)x + (k – 1)y =2k + 1. [CBSE 2010, 2011, 2012]

Previous Years Questions – 3 Marks

1. Solve for x and y: 11/x – 1/y = 10 & 9/x – 4/y = 5. [CBSE 2016] 2. Solve using cross multiplication method: 5x + 4y – 4 = 0 & x – 12y – 20 = 0. [CBSE 2016] 3. A man has certain notes of denomination ₹ 20 and ₹ 5 which amount to ₹380. If the number of notes of each kind are interchanged, they amount to ₹60 less than before. Find the number of notes of each denomination. [CBSE 2015] 4. Find the value of ‘k’ for which the following system of equations represents a pair of coincident lines: 𝑥 + 2𝑦 = 3; (𝑘 − 1)𝑥 + (𝑘 + 1)𝑦 = 𝑘 + 3. [CBSE 2014] 5. Check graphically, whether the pair of equations x + 3y = 6 & 2x – 3y = 12 is consistent. If so, then solve them graphically. [CBSE 2013] 6. The path of a train A is given by the equation x + 2y – 4 = 0 and path of another train B is given by the equation 2x + 4y – 12 = 0. Represent this situation graphically and find whether the two trains meet each other at some place. [CBSE 2013] 7. Form a pair of linear equations in two variables from the data given and solve it graphically: Tina went to a book shop to get some story books and textbooks. When her friends asked her how many of each she had bought, she answered – ‘The number of textbooks is two more than twice the number of story books bought. Also, the number of textbooks is four less than four times the number of story books bought. Help her friends to find the number of textbooks and story books she had bought. [CBSE 2013] 8. Determine graphically, the coordinates of the vertices of a triangle whose sides are graphs of the equations 2x – 3y + 6 = 0, 2x + 3y – 18 = 0 and y – 2 = 0. Also, find the area of this triangle . [CBSE 2010, 2011]

Previous Years Questions – 4 Marks

1. For Uttarakhand flood victims’ two sections A and B of class X contributed ₹ 1500. If the contribution of X A was ₹ 100 less than that of X B, find graphically the amounts contributed by both the sections. [CBSE 2016] 2. Three lines 3x + 5y = 15, 6x – 5y = 30 and x = 0 are enclosing a beautiful triangular park. Find the points of intersection of the lines graphically and the area of the park if all measurements are in km. [CBSE 2016] 3. Some people collected money to be donated in two Old Age Homes. A part of the donation is fixed and remaining depends on the number of old people in the home. If they donated ₹ 14500 in the home having 60 people and ₹ 19500 with 85 people, find the fixed part of donation and the amount donated for each people. What is the inspiration behind this? [CBSE 2016] 4. While teaching about the Indian National flag, teacher asked the students that how many lines are there in Blue colour wheel? One student replies that it is 8 times the number of colours in the flag. While other says that the sum of the number colours in the flag and number of lines in the wheel of the flag is 27. Convert the statements given by the students into linear equation of two variables. Find the number of lines in the wheel. [CBSE 2015] 5. Determine the value of k for which the following system of linear equations has infinite number of solutions: (k – 3)x + 3y = k & kx + ky = 12. [CBSE 2015] 6. Draw the graph of the following pair of linear equation: x + 3y = 6 & 2x – 3y = 12. Find the ratio of the areas of the two triangles formed by first line, x = 0, y = 0 and second line, x = 0, y = 0. [CBSE 2015] 7. Places A and B are 200km apart on a high way. One car starts from A and another from B at the same time. If the cars travel in the same directions at different speeds, they meet in 10 hours. Find the speeds of the two cars. [CBSE 2014] 8. Show graphically that the system of equations𝑥 + 2𝑦 = 4 and 7𝑥 + 4𝑦 = 18 is consistent with a unique solution (2, 1). [CBSE 2014]

10th Maths Chapter 3 Questions for Practice

1. Solve for x and y: 99x + 101y = 1499; 101x + 99y = 1501. [CBSE 2010, 2011, 2012, 2013, 2014] 2. The age of father is equal to sum of ages of his 4 children. After 30 years, sum of the ages of the children will be twice the age of the father. Find the age of the father. [CBSE 2013] 3. A person can row a boat 8 km upstream and 24 km downstream in 4 hours. He can row 12 km downstream and 12 km upstream in 4 hours. Find the speed of rowing in still water and the speed of the current. [CBSE 2013] 4. Solve for x and y: 37x + 43y = 123; 43x + 37y = 117. [CBSE 2010, 2011, 2012] 5. Draw the graph of the equations: x – y + 1 = 0 & 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. Also calculate the area of the triangle so formed. [CBSE 2011] 6. The sum of a 2 digit number and number obtained by reversing the order of digits is 99. If the digits of the number differ by 3, find the number. [CBSE 2011] 7. Check graphically whether the pair of linear equations 4x – y – 8 = 0 and 2x – 3y + 6 = 0 is consistent. Also determine the vertices of the triangle form by these lines and x – axis. [CBSE 2006, 2011] 8. The sum of the digits of a two digit number is 9. Nine times this number is twice the number obtained by reversing the digits. Find the number. [CBSE 2010] 9. A leading library has a fixed charge for the first three days and an additional charge for each day thereafter. Sarita paid ₹ 27 for a book kept for seven days. While Susy paid ₹ 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day. [CBSE 2010] 10. Solve the following system of linear equations by elimination method: 6(ax + by) = 2a + 2b and 6(bx – ay) = 3b – 2a. [CBSE 2006, 2004]

Historical Facts !

History about Linear Equations with two variables. Around 4000 years ago, Babylon knew how to solve a simple linear equation with two variables. Around 200 BC, the Chinese published that “Nine Chapters of the Mathematical Art,” they displayed the ability to solve a system of equations in three variables (Perotti). Evidence from about 300 BC indicates that the Egyptians also knew how to solve problems involving a system of two equations in two unknown quantities, including quadratic equations. Euler brought to light the idea that a system of equations doesn’t necessarily have to have a solution (Perotti). He recognized the need for conditions to be placed upon unknown variables in order to find a solution. With the turn into the 19th century Gauss introduced a procedure to be used for solving a system of linear equations. Cayley, Euler, Sylvester, and others changed linear systems into the use of matrices to represent them. Gauss brought his theory to solve systems of equations proving to be the most effective basis for solving unknowns.

How many exercises are there in Class 10 Maths chapter 3?

There are in all 7 exercises in class 10 mathematics chapter 3 (Pair of linear equations in two variables).

• In second exercise (Ex 3.1), there are in all 7 questions.
• In third exercise (Ex 3.2), there are in all 3 questions.
• In fourth exercise (Ex 3.3), there are only 2 questions.
• So, there are total 12 questions in class 10 mathematics chapter 3 (Pair of linear equations in two variables).
• In chapter 3 of 10th Maths there are in all good examples. Examples 4, 5, 6 are based on Ex 3.1, Examples 7, 8, 9, 10 are based on Ex 3.2, Examples 11, 12, 13 are based on Ex 3.3.

What are the most important questions of 10th Maths Chapter 3?

• In second exercise (Ex 3.1), all questions are important.
• In third exercise (Ex 3.2), Q1 (iv, v, vi), 2, 3 are important.
• In fourth exercise (Ex 3.3), Q2 is important.
• Important examples of chapter 3 (Pair of linear equations in two variables) class 10th mathematics are examples 1, 2, 4, 5, 6, 9, 10, 11, 12, 13, 15, 16, 18, 19.

Which chapter should recall before starting class 10th Math chapter 3?

Yes, before starting class 10th mathematics chapter 3 (Pair of linear equations in two variables), students should recall chapter 4 (Linear Equations in Two Variables) of class 9th mathematics.

What are real life applications of class 10th Maths chapter 3?

Some real life applications of class 10th mathematics chapter 3 (Pair of linear equations in two variables) are: To solve age related problems. Example: 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? To solve speed, distance and time related problems. Example: 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? To solve cost related problems. Examples: The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs1750. Find the cost of each bat and each ball. To solve geometry problems. Example: In a ∆ABC, ∠ C = 3 ∠ B = 2 (∠A + ∠ B). Find the three angles.

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Chapter 4: quadratic equations ».

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

NCERT solutions for class 10 maths chapter 3 Pair of Linear Equations in Two Variables enables the students to understand and explore simple ways to solve linear equations. This is a chapter that is widely used in industrial applications. It is used to solve questions based on speed-distance, money percentage, age, force-pressure, and so on. The kids will cover algebraic methods of solving a pair of linear equations in two variables, like the elimination method, the cross-multiplication method, and equations reducible to a pair of linear equations in two variables.

The NCERT solutions class 10 maths chapter 3 Pair of Linear Equations in Two Variables teaches that the general form of a linear equation in two variables is ax + by + c = 0, where x and y are variables and a, b and c are real numbers. Now, in this equation, the constants with variables cannot be equal to zero simultaneously. Such equations have two values, each for ‘x’ and ‘y’, which make both sides of the equation equal. The class 10 maths NCERT solutions chapter 3 Pair of linear equations in two variables can be accessed through the links below and also you can find some of these in the exercises given below.

• NCERT Solutions Class 10 Maths Chapter 3 Ex 3.1
• NCERT Solutions Class 10 Maths Chapter 3 Ex 3.2
• NCERT Solutions Class 10 Maths Chapter 3 Ex 3.3
• NCERT Solutions Class 10 Maths Chapter 3 Ex 3.4
• NCERT Solutions Class 10 Maths Chapter 3 Ex 3.5
• NCERT Solutions Class 10 Maths Chapter 3 Ex 3.6
• NCERT Solutions Class 10 Maths Chapter 3 Ex 3.7

NCERT Solutions for Class 10 Maths Chapter 3 PDF

Linear equations are implemented to overcome real-world problems in a variety of fields. To evaluate this type of equation, we must either apply the substitution technique or mathematical elimination, which is explained with appropriate examples in the NCERT solutions class 10 maths chapter 3 Pair of linear equations in two variables. The pdf of the different exercises in the chapter can be found below :

☛ Download Class 10 Maths NCERT Solutions Chapter 3

NCERT Class 10 Maths Chapter 3   Download PDF

A linear equation in two variables can also be easily represented on a graph. Since the equation forms a straight line , and a line can have infinitely many solutions so each solution can be uniquely plotted on the graph . Equations aren't simply a theoretical concept; they're also incredibly useful in everyday situations; hence, it is essential that the students get a practical problem-solving approach with the help of the examples and questions presented in the chapter. A section-wise quick analysis of the exercise questions in the NCERT Solutions Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables can be seen below :

• Class 10 Maths Chapter 3 Ex 3.1 - 3 Questions
• Class 10 Maths Chapter 3 Ex 3.2 - 7 Questions
• Class 10 Maths Chapter 3 Ex 3.3 - 3 Questions
• Class 10 Maths Chapter 3 Ex 3.4 - 2 Questions
• Class 10 Maths Chapter 3 Ex 3.5 - 4 Questions
• Class 10 Maths Chapter 3 Ex 3.6 - 2 Questions
• Class 10 Maths Chapter 3 Ex 3.7 - 8 Questions

Topics Covered: The topics covered in class 10 maths NCERT Solutions chapter 3 include the definition of linear equations, graphical method of a solution of a pair of linear equations, algebraic methods of solving a pair of linear equations by elimination, substitution, and cross multiplication , as well as equations reducible to a pair of linear equations.

Total Questions: Class 10 maths chapter 3 Pair Of Linear Equations In Two Variables consists of 29 questions, of which 12 are long answers, 8 are moderate-level, and 9 are easy problems. These sums are solved in a step-wise manner so that students can get a conceptual understanding of the topics covered in this chapter.

List of Formulas in NCERT Solutions Class 10 Maths Chapter 3

Formulas give a way to simplify a tough sum and solve it more efficiently. A pair of linear equations can be solved using two methods, one is the algebraic method, and the other is the graphical method. In order to use these techniques, we need to cover topics such as the standard form of linear equations, equations for simultaneous pairs, etc. It is advised that kids make a formula chart that outlines the steps needed to be used when attempting such problems. This also helps them to revise the chapter at a steady pace and efficiently. Some of the important equations from this chapter of NCERT Solutions have been mentioned below:

• Pair of linear equations in two variables :

x = a 1 x + b 1 y + c 1 = 0 : y = a 2 x + b 2 y + c 2 = 0

The above two equations signify the general form of a system of linear equations that can be solved using elimination and substitution methods.

• Cross Multiplication Method : If a pair of linear equations given by a 1 x + b 1 y + c 1 = 0 and a 2 x + b 2 y + c 2 = 0, then the following conditions are possible :
• a 1 /a 2 ≠ b 1 /b 1 : depicting consistent pair of linear equations
• a 1 /a 2 = b 1/ b 2 ≠ c 1/ c 2 : depicting inconsistent pair of linear equations
• a 1 /a 2 = b 1 /b 2 = c 1 /c 2 : depicting dependent as well as consistent pair of linear equations

Important Questions for Class 10 Maths NCERT Solutions Chapter 3

Video Solutions for Class 10 Maths NCERT Chapter 3

FAQs on NCERT Solutions Class 10 Maths Chapter 3

Why are ncert solutions class 10 maths chapter 3 vital for scoring well.

NCERT solutions class 10 maths chapter 3 have been prepared by experts and provide practical and graphical methods of solving a pair of linear equations with examples. The important topics such as consistent, inconsistent, and dependent pair of equations in two variables are explained in a step-wise manner with appropriate examples. Also, the CBSE board recommends studying NCERT solutions making them a vital resource.

Do I Need to Practice all Questions Given in Class 10 Maths NCERT Solutions Linear Equations in Two Variables?

The NCERT solutions class 10 maths linear equations in two variables have examples and exercise questions that cover the important topics in detail and will help to build a strong foundation for students. The set of questions cover both algebraic and graphical problems. The more kids will practice all the problems, the better clarity they will get on the concepts. Hence, it's essential that the students aim to solve all the questions.

What are the Important Topics Covered in NCERT Solutions Class 10 Maths Chapter 3?

The important topics covered in the NCERT solutions class 10 maths chapter 3 are the definition of a linear equation, degrees of an equation, as well as the elimination and substitution method. NCERT solutions help the students learn how to multiply the equations. They will also come across new terms like consistent, inconsistent pair of linear equations, point of intersection, and many more.

How Many Questions are there in Class 10 Maths NCERT Solutions Chapter 3?

The NCERT Solutions contain a total of 29 well-researched questions that include both graph-based as well as algebraic sums. These 29 questions are distributed across 7 exercises. Students must go through all the examples before solving these exercises. Each question has been solved in a detailed manner with appropriate explanations in the NCERT solutions class 10 maths chapter 3.

What are the Important Formulas in NCERT Solutions Class 10 Maths Chapter 3?

The important formulas mentioned in the NCERT Solutions class 10 maths chapter 3 are formulas for the cross-multiplication and division methods. Students must also remember the standard form of linear equations in two variables, as well as their general equations. These formulas are very important from an exam perspective and can help students solve questions in minutes. In order to understand the derivation of these formulas, kids must go through the theory of the chapter.

Why Should I Practice NCERT Solutions Class 10 Maths Chapter 3?

NCERT solutions focus on equations reducible to a pair of linear equations in two variables. Practicing the elimination and substitution methods will help students master NCERT solutions class 10 maths chapter 3. Questions related to these topics are frequently asked in board exams hence it is important to go through these solutions on a regular basis in order to score excellent marks.

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.

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NCERT Solutions for Class 10 Maths Chapter 3 - Pair Of Linear Equations In Two Variables

• NCERT Solutions
• Chapter 3 Pair Of Linear Equations In Two Variables

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

The NCERT Solutions for Class 10 Maths Chapter 3 helps students understand the concept of graph plotting and forming straight lines using linear equations in two variables. The Pair Of Linear equations in two variables are used to define a line that can be plotted on a graph. The solutions to these equations are represented as points on the graph. In this chapter, we discuss linear equations in two variables from straight lines.

The free PDF of Class 10 Maths Chapter 3 Solutions pdf download is available on Vedantu, providing students with a better understanding of the problems. It covers solutions to every exercise in this chapter and is updated according to the latest CBSE syllabus . 

Glance on Maths Chapter 3 Class 10 - Pair of Linear Equations in Two Variables

Chapter 3 of Class 10 Maths deals with solving systems of two linear equations in two variables. 

There are two main methods to find the solution (values of x and y that satisfy both equations) one is Graphical Method and the other is Algebraic Method.

Types of Solutions in Pair of Linear Equations in Two Variables are:

Consistent Pair (one set of values for x and y).

Inconsistent Pair (no values of x and y satisfy both equations).

Dependent Pair (any value of x will give a corresponding y value that satisfies both equations).

This article helps you with techniques to solve these different scenarios and understand the relationship between the equations and their graphical representations.

This article contains chapter notes, important questions, exemplar solutions and exercises links for Chapter 3 - Matrices, which you can download as PDFs.

There are three exercises (12 fully solved questions) in class 10th maths chapter 3 Pair of Linear Equations in Two Variables.

Access Exercise Wise NCERT Solutions for Chapter 3 Maths Class 10

Related Chapters

Exercises under NCERT Solutions for Class 10 Maths Chapter 3

Exercise 3.1: This exercise involves solving word problems by setting up a pair of linear equations. Students will need to translate word problems into mathematical equations and solve them to find the solution.

Exercise 3.2: This exercise covers the concept of cross-multiplication and its application in solving linear equations. Students will learn to solve equations using the cross-multiplication method and will also solve problems related to this topic.

Exercise 3.3: This exercise involves solving equations that are reducible to a pair of linear equations in two variables. Students will learn to reduce equations to linear form and solve them.

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

Exercise 3.1.

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

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

Ans (i): Assuming that the number of girls and boys be \[x\] and \[y\] respectively.

Writing the algebraic representation using the information given in the question-

Solution table for \[x+y=10\]-

Solution table for \[x-y=4\]-

Graphical representation-

As we can see from the graph above, the point of intersection for the lines is \[\left( 7,3 \right)\]. Therefore, we can say that there are \[7\] girls and \[3\] boys in the class.

(ii) \[\mathbf{5}\] pencils and \[\mathbf{7}\] pens together cost \[\mathbf{Rs}\text{ }\mathbf{50}\], whereas \[\mathbf{7}\] pencils and \[\mathbf{5}\] pens together cost \[\mathbf{Rs}\text{ }\mathbf{46}\]. Find the cost of one pencil and that of one pen.

Ans (ii): Assuming that the cost of \[1\] pencil and \[1\] pen be \[x\] and \[y\] respectively.

\[5x+7y=50\]

\[7x+5y=46\]

Solution table for \[5x+7y=50\]-

\[x=\frac{50-7y}{5}\]

Solution table for \[7x+5y=46\]-

\[x=\frac{46-5y}{7}\]

As we can see from the graph above, the point of intersection for the lines is \[\left( 3,5 \right)\]. Therefore, we can say that the cost of a pencil is \[Rs\text{ }3\] and the cost of a pen is \[Rs\text{ 5}\].

2. On comparing the ratios \[\frac{{{a}_{1}}}{{{a}_{2}}},\frac{{{b}_{1}}}{{{b}_{2}}}\] and \[\frac{{{c}_{1}}}{{{c}_{2}}}\], find out whether the lines representing the following pairs of linear equations at a point, are parallel or coincident-

(i) \[\mathbf{5x}-\mathbf{4y}+\mathbf{8}=\mathbf{0}\]

\[\mathbf{7x}+\mathbf{6y}-\mathbf{9}=\mathbf{0}\]

Ans: \[5x-4y+8=0\]

\[7x+6y-9=0\]

Calculating the values of \[{{a}_{1}},{{b}_{1}},{{c}_{1}},{{a}_{2}},{{b}_{2}}\] and \[{{c}_{2}}\] by comparing the above equations with \[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0\] and \[{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0\],

\[{{a}_{1}}=5\], \[{{b}_{1}}=-4\], \[{{c}_{1}}=8\]

\[{{a}_{2}}=7\], \[{{b}_{2}}=6\], \[{{c}_{2}}=-9\]

\[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{5}{7}\]

\[\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{-4}{6}=\frac{-2}{3}\]

Since \[\frac{{{a}_{1}}}{{{a}_{2}}}\ne \frac{{{b}_{1}}}{{{b}_{2}}}\].

Therefore, the given pair of equations have a unique solution, that is, the lines intersect at exactly one point.

(ii) \[\mathbf{9x}+\mathbf{3y}+\mathbf{12}=\mathbf{0}\]

\[\mathbf{18x}+\mathbf{6y}+\mathbf{24}=\mathbf{0}\]

Ans (ii): Calculating the values of \[{{a}_{1}},{{b}_{1}},{{c}_{1}},{{a}_{2}},{{b}_{2}}\] and \[{{c}_{2}}\] by comparing the above equations with \[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0\] and \[{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0\],

\[{{a}_{1}}=9\], \[{{b}_{1}}=3\], \[{{c}_{1}}=12\]

\[{{a}_{2}}=18\], \[{{b}_{2}}=6\], \[{{c}_{2}}=24\]

\[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{1}{2}\]

\[\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{1}{2}\]

\[\frac{{{c}_{1}}}{{{c}_{2}}}=\frac{1}{2}\]

Since \[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{{{c}_{1}}}{{{c}_{2}}}\].

Therefore, the lines representing the given pair of equations have infinite solutions as they are coincident.

(iii) \[\mathbf{6x}-\mathbf{3y}+\mathbf{10}=\mathbf{0}\]

\[\mathbf{2x}-\mathbf{y}+\mathbf{9}=\mathbf{0}\]

Ans (iii): Calculating the values of \[{{a}_{1}},{{b}_{1}},{{c}_{1}},{{a}_{2}},{{b}_{2}}\] and \[{{c}_{2}}\] by comparing the above equations with \[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0\] and \[{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0\],

\[{{a}_{1}}=6\], \[{{b}_{1}}=-3\], \[{{c}_{1}}=10\]

\[{{a}_{2}}=2\], \[{{b}_{2}}=-1\], \[{{c}_{2}}=9\]

\[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{3}{1}\]

\[\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{3}{1}\]

\[\frac{{{c}_{1}}}{{{c}_{2}}}=\frac{10}{9}\]

Since \[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}\ne \frac{{{c}_{1}}}{{{c}_{2}}}\].

Therefore, the lines representing the given pair of equations have no solutions as they are parallel to each other.

3. On comparing the ratios \[\frac{{{a}_{1}}}{{{a}_{2}}},\frac{{{b}_{1}}}{{{b}_{2}}}\] and \[\frac{{{c}_{1}}}{{{c}_{2}}}\], find out whether the following pairs of linear equations are consistent, or inconsistent.

(i) \[3\mathbf{x}+2\mathbf{y}=5;2x-3y=7\]

Ans (i): For the given equations -

 \[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{3}{2}\]

\[\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{-2}{3}\]

\[\frac{{{c}_{1}}}{{{c}_{2}}}=\frac{5}{7}\]

So, the given pair of equations have a unique solution, that is, the lines intersect at exactly one point. 

Therefore, the given pair of lines is consistent.

(ii) \[2x-3y=8;4\mathbf{x}-6\mathbf{y}=9\]

Ans (ii): For the given equations -

\[\frac{{{c}_{1}}}{{{c}_{2}}}=\frac{8}{9}\]

So, the lines representing the given pair of equations have no solutions as they are parallel to each other.

Therefore, the given pair of lines is inconsistent.

(iii) \[\frac{3}{2}x+\frac{5}{3}y=7;9\mathbf{x}-10\mathbf{y}=14\]

Ans (iii): For the given equations -

\[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{1}{6}\]

\[\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{-1}{6}\]

\[\frac{{{c}_{1}}}{{{c}_{2}}}=\frac{1}{4}\]

(iv) \[5x-3y=11;-10\mathbf{x}+6\mathbf{y}=-22\]

Ans (iv): For the given equations -

\[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{-1}{2}\]

\[\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{-1}{2}\]

\[\frac{{{c}_{1}}}{{{c}_{2}}}=\frac{-1}{2}\]

So, the lines representing the given pair of equations have infinite number of solutions as they are coincident.

(v) \[\frac{4}{3}x+2y=8;2\mathbf{x}+3\mathbf{y}=12\]

Ans (v): For the given equations -

\[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{2}{3}\]

\[\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{2}{3}\]

\[\frac{{{c}_{1}}}{{{c}_{2}}}=\frac{2}{3}\]

So, the lines representing the given pair of equations have an infinite number of solutions as they are coincident.

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

(i) \[x+y=5;2\mathbf{x}+2\mathbf{y}=10\]

Solution table for \[x+y=5\]-

Solution table for \[2x+2y=10\]-

\[x=\frac{10-2y}{2}\]

As shown in the graph above, the two lines are overlapping each other. Hence, they have infinite solutions.

(ii) \[x-y=8;3\mathbf{x}-3\mathbf{y}=16\]

\[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{1}{3}\]

\[\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{1}{3}\]

(iii) \[2x+y-6=0;4\mathbf{x}-2\mathbf{y}-4=0\]

\[\frac{{{c}_{1}}}{{{c}_{2}}}=\frac{3}{2}\]

Solution table for \[2x+y-6=0\]-

Solution table for \[4x-2y-4=0\]-

\[y=\frac{4x-4}{2}\]

As shown in the graph above, the two lines intersect each other at only one point \[\left( 2,2 \right)\].

(iv) \[2x-2y-2=0;4\mathbf{x}-4\mathbf{y}-5=0\]

Ans (iv):  For the given equations -

\[\frac{{{c}_{1}}}{{{c}_{2}}}=\frac{2}{5}\]

5. Half the perimeter of a rectangular garden, whose length is \[\mathbf{4}\text{ }\mathbf{m}\] more than its width, is \[\mathbf{36}\text{ }\mathbf{m}\]. Find the dimensions of the garden.

Ans: Assuming that the width and length of the garden be \[x\] and \[y\] respectively.

Solution table for \[y-x=4\]-

Solution table for \[x+y=36\]-

As shown in the graph above, the two lines intersect each other at only one point \[\left( 16,20 \right)\]. Therefore the length of the garden is \[20\text{ }m\] and its breadth is \[\text{16 }m\].

6. Given the linear equation \[\mathbf{2x}+\mathbf{3y}-\mathbf{8}=\mathbf{0}\], write another linear equations in two variables such that the geometrical representation of the pair so formed is-

(i) Intersecting lines

Ans (i): If two lines are intersecting then-

\[\frac{{{a}_{1}}}{{{a}_{2}}}\ne \frac{{{b}_{1}}}{{{b}_{2}}}\]

So, taking the second line as \[2x+4y-6=0\],

\[\frac{{{a}_{1}}}{{{a}_{2}}}=1\]

\[\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{3}{4}\]

As \[\frac{{{a}_{1}}}{{{a}_{2}}}\ne \frac{{{b}_{1}}}{{{b}_{2}}}\], the two lines are interesting each other.

(ii) Parallel lines

Ans: If two lines are intersecting then-

\[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}\ne \frac{{{c}_{1}}}{{{c}_{2}}}\]

So, taking the second line as \[4x+6y-8=0\],

\[\frac{{{c}_{1}}}{{{c}_{2}}}=1\]

As \[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}\ne \frac{{{c}_{1}}}{{{c}_{2}}}\], the two lines are parallel to each other.

(iii) Coincident lines

\[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{{{c}_{1}}}{{{c}_{2}}}\]

So, taking the second line as \[6x+9y-24=0\],

\[\frac{{{c}_{1}}}{{{c}_{2}}}=\frac{1}{3}\]

As \[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{{{c}_{1}}}{{{c}_{2}}}\], the two lines are coincident.

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

Ans: Solution table for \[x-y+1=0\]-

Solution table for \[3x+2y-12=0\]-

\[x=\frac{12-2y}{3}\]

As shown in the graph above, the lines are intersecting each other at point \[\left( 2,3 \right)\] and \[x-axis\] at \[\left( -1,0 \right)\] and \[\left( 4,0 \right)\]. So the obtained triangle has vertices \[\left( 2,3 \right)\], \[\left( -1,0 \right)\] and \[\left( 4,0 \right)\].

Exercise 3.2

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

(i) \[x+y=14;x-y=4\]

Ans (i): The given equations are-

\[x+y=14\]           …… (i)

\[x-y=4\]             …… (ii)

From equation (i)-

\[x=14-y\]          …… (iii)

Substituting (iii) in equation (ii), we get

\[\left( 14-y \right)-y=4\]

\[14-2y=4\]

\[y=5\]                  …… (iv)

Substituting (iv) in (iii), we get

Therefore, \[x=9\] and \[y=5\].

(ii) \[s-t=3;\frac{s}{3}+\frac{t}{2}=6\]

Ans (ii): The given equations are-

\[s-t=3\]           …… (i)

\[\frac{s}{3}+\frac{t}{2}=6\]       …… (ii)

\[s=t+3\]          …… (iii)

\[\frac{t+3}{3}+\frac{t}{2}=6\]

\[2t+6+3t=36\]

\[t=6\]                  …… (iv)

Therefore, \[s=9\] and \[t=6\].

(iii) \[3x-y=3;9x-3y=9\]

Ans (iii): The given equations are-

\[3x-y=3\]           …… (i)

\[9x-3y=9\]             …… (ii)

\[y=3x-3\]          …… (iii)

\[9x-3\left( 3x-3 \right)=9\]

\[9x-9x+9=9\]

For all \[x\] and \[y\].

Therefore, the given equations have infinite solutions. One of the solution is \[x=1,y=0\].

(iv) \[0.2x-0.3y=1.3;0.4x+0.5y=2.3\]

Ans (iv): The given equations are-

\[0.2x-0.3y=1.3\]           …… (i)

\[0.4x+0.5y=2.3\]             …… (ii)

\[x=\frac{1.3-0.3y}{0.2}\]          …… (iii)

\[0.4\left( \frac{1.3-0.3y}{0.2} \right)-0.5y=2.3\]

\[2.6-0.6y+0.5y=2.3\]

\[2.6-2.3=0.1y\]

\[y=3\]                  …… (iv)

\[x=\frac{1.3-0.3\left( 3 \right)}{0.2}\]

Therefore, \[x=2\] and \[y=3\].

(v) \[\sqrt{2}x-\sqrt{3}y=0;\sqrt{3}x-\sqrt{8}y=0\]

Ans (v): The given equations are-

\[\sqrt{2}x-\sqrt{3}y=0\]           …… (i)

\[\sqrt{3}x-\sqrt{8}y=0\]             …… (ii)

\[x=\frac{-\sqrt{3}y}{\sqrt{2}}\]          …… (iii)

\[\sqrt{3}\left( \frac{-\sqrt{3}y}{\sqrt{2}} \right)-\sqrt{8}y=0\]

\[\frac{-\sqrt{3}y}{\sqrt{2}}-2\sqrt{2}y=0\]

\[y\left( \frac{-\sqrt{3}}{\sqrt{2}}-2\sqrt{2} \right)=0\]

\[y=0\]                  …… (iv)

Therefore, \[x=0\] and \[y=0\].

(vi) \[\frac{3x}{2}-\frac{5y}{3}=-2;\frac{x}{3}+\frac{y}{2}=\frac{13}{6}\]

Ans (vi): The given equations are-

\[\frac{3x}{2}-\frac{5y}{3}=-2\]           …… (i)

\[\frac{x}{3}+\frac{y}{2}=\frac{13}{6}\]             …… (ii)

\[x=\frac{-12+10y}{9}\]          …… (iii)

\[\frac{\left( \frac{-12+10y}{9} \right)}{3}+\frac{y}{2}=\frac{13}{6}\]

\[\frac{-12+10y}{27}+\frac{y}{2}=\frac{13}{6}\]

\[\frac{-24+20y+27y}{54}=\frac{13}{6}\]

\[47y=141\]

Therefore, \[x=0\] and \[y=3\].

2. Solve \[\mathbf{2x}+\mathbf{3y}=\mathbf{11}\] and \[\mathbf{2x}-\mathbf{4y}=-\mathbf{24}\] and hence find the value of ‘\[m\]’ for which \[\mathbf{y}=\mathbf{mx}+\mathbf{3}\].

Ans: The given equations are-

\[2x+3y=11\]           …… (i)

\[2x-4y=-24\]             …… (ii)

\[x=\frac{11-3y}{2}\]          …… (iii)

\[2\left( \frac{11-3y}{2} \right)-4y=-24\]

\[11-3y-4y=-24\]

\[-7y=-35\]

Therefore, \[x=-2\] and \[y=5\].

Calculating the value of \[m\]-

\[5=-2m+3\]

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 \[\mathbf{26}\] and one number is three times the other. Find them.

Ans (i): Assuming one number be \[x\] and another number be \[y\] such that \[y>x\],

\[y=3x\]             …… (i)

\[y-x=26\]    …… (ii)

Substituting the value of \[y\] from equation (i) in equation (ii), we get

\[3x-x=26\]

\[x=13\]                  …… (iii)

Substituting (iii) in (i), we get

Therefore, \[x=13\] and \[y=39\].

(ii) The larger of two supplementary angles exceeds the smaller by \[\mathbf{18}\] degrees. Find them.

Ans (ii): Assuming the larger angle be \[x\] and smaller angle be \[y\]. 

The sum of a pair of supplementary angles is always \[{{180}^{\circ }}\].

\[x+y=180\]             …… (i)

\[x-y=18\]    …… (ii)

Substituting the value of \[x\] from equation (i) in equation (ii), we get

\[180-y-y=18\]

\[y=81\]                  …… (iii)

Therefore, the two angles are  \[x={{99}^{\circ }}\] and \[y={{81}^{\circ }}\].

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

Ans (iii): Assuming the cost of a bat is \[x\] and the cost of a ball is \[y\].

\[7x+6y=3800\]             …… (i)

\[3x+5y=1750\]    …… (ii)

\[y=\frac{3800-7x}{6}\]       …… (iii)

Substituting (iii) in equation (ii)- 

\[3x+5\left( \frac{3800-7x}{6} \right)=1750\]

\[3x+\frac{9500}{3}-\frac{35x}{6}=1750\]

\[3x-\frac{35x}{6}=1750-\frac{9500}{3}\]

\[\frac{18x-35x}{6}=\frac{5250-9500}{3}\]

\[\frac{17x}{6}=\frac{-4250}{3}\]

\[x=500\]                  …… (iv)

\[y=\frac{3800-7\left( 500 \right)}{6}\]

Therefore, the bat costs \[Rs\text{ }500\] and the ball costs \[Rs\text{ }50\].

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

Ans (iv): Assuming the fixed charge be \[Rs\text{ }x\] and the per km charge be \[Rs\text{ y}\].

\[x+10y=105\]             …… (i)

\[x+15y=155\]    …… (ii)

\[x=105-10y\]       …… (iii)

\[105-10y+15y=155\]

\[y=10\]                  …… (iv)

\[x=105-10\left( 10 \right)\]

Therefore, the fixed charge is \[Rs\text{ }5\] and the per km charge is \[Rs\text{ 10}\].

So, charge for \[25\text{ }km\] will be-

\[=Rs\text{ }\left( x+25y \right)\]

\[=Rs\text{ 255}\]

(v) A fraction becomes \[\frac{9}{11}\], if \[\mathbf{2}\] is added to both the numerator and the denominator. If, \[\mathbf{3}\] is added to both the numerator and the denominator it becomes \[\frac{5}{6}\]. Find the fraction.

Ans (v): Assuming the fraction be \[\frac{x}{y}\].

\[\frac{x+2}{y+2}=\frac{9}{11}\]

\[11x+22=9y+18\]

\[11x-9y=-4\]             …… (i)

\[\frac{x+3}{y+3}=\frac{5}{6}\]

\[6x+18=5y+15\]

\[6x-5y=-3\]    …… (ii)

\[x=\frac{-4+9y}{11}\]       …… (iii)

\[6\left( \frac{-4+9y}{11} \right)-5y=-3\]

\[-24+54y-55y=-33\]

\[y=9\]                  …… (iv)

\[x=\frac{-4+9\left( 9 \right)}{11}\]

Therefore, the fraction is \[\frac{7}{9}\].

(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?

Ans (vi): Assuming the age of Jacob be \[x\] and the age of his son be \[y\].

Writing the algebraic representation using the information given in the .question-

\[\left( x+5 \right)=3\left( y+5 \right)\]

\[x-3y=10\]             …… (i)

\[\left( x-5 \right)=7\left( y-5 \right)\]

\[x-7y=-30\]    …… (ii)

\[x=3y+10\]       …… (iii)

\[3y+10-7y=-30\]

\[-4y=-40\]

\[x=3\left( 10 \right)+10\]

Therefore, Jacob’s present age is \[40\] years and his son’s present age is \[10\] years.

Exercise 3.3

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

(i) \[\mathbf{x}+\mathbf{y}=\mathbf{5}\] and \[\mathbf{2x}\mathbf{3y}=\mathbf{4}\] 

Ans (i): 

Elimination method

The given equations are-

\[x+y=5\]          …… (i)

\[2x-3y=4\]   …… (ii)

Multiplying equation (ii) by \[2\], we get

\[2x+2y=10\]   …… (iii)

Subtracting equation (ii) from equation (iii), we obtain

\[y=\frac{6}{5}\]              …… (iv)

Substituting the value of (iv) in equation (i), we get

\[x=5-\frac{6}{5}\]

\[x=\frac{19}{5}\]

Therefore, \[x=\frac{19}{5}\] and \[y=\frac{6}{5}\].

Substitution method-

From equation (i) we get

\[x=5-y\]      …… (v)

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

\[2\left( 5-y \right)-3y=4\]

\[y=\frac{6}{5}\]            …… (vi)

Substituting (vi) in equation (v), we obtain

(ii) \[\mathbf{3x}+\mathbf{4y}=\mathbf{10}\] and \[\mathbf{2x}\mathbf{2y}=\mathbf{2}\]

Ans (ii): Elimination method

\[3x+4y=10\]          …… (i)

\[2x-2y=2\]   …… (ii)

\[4x-4y=4\]   …… (iii)

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

\[x=2\]              …… (iv)

\[6+4y=10\]

Therefore, \[x=2\] and \[y=1\].

From equation (ii) we get

\[x=1+y\]      …… (v)

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

\[3\left( 1+y \right)+4y=10\]

\[y=1\]            …… (vi)

(iii) \[\mathbf{3x}\mathbf{5y}\mathbf{4}=\mathbf{0}\] and \[\mathbf{9x}=\mathbf{2y}+\mathbf{7}\]

Ans (iii): Elimination method

\[3x-5y-4=0\]          …… (i)

\[9x=2y+7\]

\[9x-2y=7\]   …… (ii)

Multiplying equation (i) by \[3\], we get

\[9x-15y-12=0\]   …… (iii)

Subtracting equation (iii) from equation (ii), we obtain

\[y=-\frac{5}{13}\]              …… (iv)

\[3x+\frac{25}{13}-4=0\]

\[3x=\frac{27}{13}\]

\[x=\frac{9}{13}\]

Therefore, \[x=\frac{9}{13}\] and \[y=-\frac{5}{13}\].

\[x=\frac{5y+4}{3}\]      …… (v)

\[9\left( \frac{5y+4}{3} \right)-2y-7=0\]

\[y=\frac{-5}{13}\]            …… (vi)

\[x=\frac{5\left( \frac{-5}{13} \right)+4}{3}\]

Therefore, \[x=\frac{9}{13}\] and \[y=\frac{-5}{13}\].

(iv) \[\frac{x}{2}+\frac{2y}{3}=-1\] and \[x-\frac{y}{3}=3\]

Ans (iv): Elimination method

\[\frac{x}{2}+\frac{2y}{3}=-1\]

\[3x+4y=-6\]          …… (i)

\[x-\frac{y}{3}=3\]

\[3x-y=9\]   …… (ii)

Subtracting equation (ii) from equation (i), we obtain

\[y=-3\]              …… (iv)

\[3x+4\left( -3 \right)=-6\]

Therefore, \[x=2\] and \[y=-3\].

\[x=\frac{y+9}{3}\]      …… (v)

\[3\left( \frac{y+9}{3} \right)+4y=-6\]

\[y=-3\]            …… (vi)

\[x=\frac{-3+9}{3}\]

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 \[\mathbf{1}\] to the numerator and subtract \[\mathbf{1}\] from the denominator, a fraction reduces to \[\mathbf{1}\]. It becomes \[\frac{\mathbf{1}}{2}\] if we only add \[\mathbf{1}\] to the denominator. What is the fraction?

Ans (i): Assuming the fraction be \[\frac{x}{y}\].

\[\frac{x+1}{y-1}=1\]

\[x-y=-2\]             …… (i)

\[\frac{x}{y+1}=1\]

\[2x-y=1\]    …… (ii)

Subtracting equation (i) from equation (ii), we obtain

\[x=3\]              …… (iii)

Substituting the value of (iii) in equation (i), we get

Hence the fraction is \[\frac{3}{5}\].

(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?

Ans (ii): Assuming the present age of Nuri be \[x\] and present age of Sonu be \[y\].

\[\left( x-5 \right)=3\left( y-5 \right)\]

\[x-3y=-10\]             …… (i)

\[\left( x+10 \right)=2\left( y+10 \right)\]

\[x-2y=10\]    …… (ii)

\[y=20\]              …… (iii)

\[x-60=-10\]

Therefore, \[x=50\] and \[y=20\].

Hence Nuri’s present age is \[50\] years and Sonu’s present age is \[20\] years.

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

Ans (iii): Assuming the unit digit of the number be \[x\] and the tens digit be \[y\].

Therefore, the number is \[10y+x\]

The number after reversing the digits is \[10x+y\].

\[x+y=9\]             …… (i)

\[9\left( 10y+x \right)=2\left( 10x+y \right)\]

\[-x+8y=0\]    …… (ii)

Adding equation (i) and (ii), we obtain

\[y=1\]              …… (iii)

Therefore, \[x=8\] and \[y=1\].

Hence the number is \[10y+x=18\].

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

Ans (iv): Assuming the number of \[Rs\text{ }50\] notes be \[x\] and the number of \[Rs\text{ }100\] be \[y\].

\[x+y=25\]             …… (i)

\[50x+100y=2000\]    …… (ii)

Multiplying equation (i) by \[50\], we obtain

\[50x+50y=1250\]     …… (iii)

\[50y=750\]

\[y=15\]              …… (iv)

Therefore, \[x=10\] and \[y=15\].

Hence Meena has \[10\] notes of \[Rs\text{ }50\] and \[15\] notes of \[Rs\text{ }100\].

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

Ans (v): Assuming that the charge for first three days is \[Rs\text{ }x\] and the charge for each day thereafter is \[Rs\text{ y}\].

\[x+4y=27\]             …… (i)

\[x+2y=21\]    …… (ii)

\[y=3\]     …… (iii)

Subtracting equation (iii) from equation (i), we obtain

\[x+12=27\]

\[x=15\]              …… (iv)

Therefore, \[x=15\] and \[y=3\].

Hence, fixed charges are \[Rs\text{ 15}\] and charges per day are \[Rs\text{ 3}\].

Also You Can Find the Solutions of All the Maths Chapters Below.

3. introduction.

A Linear Equation is an equation of straight line. It is in the form of ax + by + c = 0 where a, b and c are the real numbers (a≠0 and b≠0) and x and y are the two variables,

Here a and b are the coefficients and c is the constant of the equation.

3.1 Importance

The key points that state why there is a need for NCERT solutions of chapter 3 maths class 10 are:

Our subject experts have prepared these solutions in a detailed step-wise manner.

All the important concepts of this chapter are covered in these solutions.

Every sum given in the exercises is solved in a self-explanatory manner so that students can prepare effectively for the exams.

3.2 Pair of Linear Equations

Two Linear Equations having two same variables are known as the pair of Linear Equations in two variables. A linear equation in two variables involves variables x and y, and is expressed as ax + by + c = 0, where a, b, and c are real numbers (a and b ≠ 0).

a 1 x + b 1 y + c 1 = 0

a 2 x + b 2 y + c 2 = 0

3.3 Graphical Method of Solution of a Pair of Linear Equations

When two lines intersect each other at only one point, then we conclude that there is one and only one solution. It means that a unique solution exists for this pair of linear equations in two variables. This type of pair of linear equations is called a consistent pair of Linear equations. 

If the two lines are coincident, we can say that the pair of linear equations will have infinitely many solutions. This type of pair of linear equations can be called an inconsistent pair of Linear equations. 

If the two lines are parallel to each other, which means they do not meet at all, then we can say that the two linear equations will not have any common solution. This type of pair of linear equations will be called the dependent pair of Linear equations. 

3.4 Algebraic Methods of Solving a Pair of Linear Equations

The solution of a pair of linear equations is of the form (x,y), and it satisfies both equations at the same time. Using following methods, we can find the solution to a consistent pair of linear equations:

3.4.1 Substitution Method:

(a) Use one equation to find the value of a single variable, such as y in terms of x or x in terms of y.

(b) To get the equation in one variable and determine the solution, substitute this value in the second equation.

(c) Now, change the value or solution that was found in step (b) to the one found in step (a) of the equation.

3.4.2 Elimination Method:

(a) To make the coefficient of any one variable numerically equal, multiply both equations by appropriate non-zero constants if the coefficient of any one variable differs in either one.

(b) Solve the equations in one variable by adding or subtracting the ones that were so obtained.

(c) To obtain the value of the second variable, now substitute the value of the variable obtained in the previous step into either of the original equations.

3.4.3 Cross-multiplication Method:

The two linear equations in two variables are as follows: 

$a_1x + b_1y + c_1 = 0$

$a_2x + b_2y + c_2 = 0$

Take a look at the diagram below.

Determine the solution by solving it, given that $a_1b_2 - a_2b_1 \neq 0$.

3.5 Solving Word Problems:

Carefully read the problem and identify the unknown quantities. Give these unknown quantities: x, y, r, s, t, and so on.

Find the variables that must be determined.

Carefully read the problem and transform the equations into terms of the variables to be determined.

Solve the equations using any of the three methods listed above.

3.6 Equations Reducible to a Pair of Linear Equations in Two Variables

Sometimes pairs of equations are not linear (or not in standard form), then they are altered so that they reduce to a pair of linear equations in standard form. For example

Here we substitute $\dfrac{1}{x} = p$ & $\dfrac{1}{y} = q$, the above equations reduces to: $a_1p + b_1q = c_1$ ; $a_2p - b_2q = c_2$ Now we can use any method to solve them.

Summary of Pair of Linear Equations in Two Variables

The main ideas that should be committed to memory in order to answer the practice questions in the chapter Pair of Linear Equations in Two Variables are included in the Summary section. You can review every idea covered in this chapter by referring to the points in this section.

A pair of linear equations in two variables is defined as two linear equations in the same two variables. Both a graphic and an algebraic representation are possible for the pair of linear equations in two variables. Two lines can be used to depict the graph:

A pair of equations is considered consistent if and only if the lines cross at that moment.

The two equations are dependent if the lines coincide.

The two equations are inconsistent if the lines are parallel to one another.

The two linear equations in two variables can be solved algebraically using the following techniques:

Substitution Method

Elimination Method

Cross-multiplication Method

Overview of Deleted Syllabus for CBSE Class 10 Maths Chapter - 3

Class 10 Maths Chapter 3: Exercise Breakdown

The NCERT Solutions for Class 10 Maths Chapter 3 - Pair Of Linear Equations In Two Variables provided by Vedantu offer comprehensive explanations to tackle problems effectively. It's crucial to grasp the concept of linear equations and their graphical representation. Focus on understanding the methods like substitution and elimination for solving pairs of linear equations. Previous year question papers usually include around 6-8 questions from this chapter. Practice these questions to gain proficiency. Overall, these solutions serve as a valuable resource to enhance your understanding and problem-solving skills in this topic.

Other Related Links for CBSE Class 10 Maths Chapter 3

Chapter-Specific NCERT Solutions for Class 10 Maths

Given below are the chapter-wise NCERT Solutions for Class 10 Maths . Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

FAQs on NCERT Solutions for Class 10 Maths Chapter 3 - Pair Of Linear Equations In Two Variables

1. Where can I get the Best NCERT Maths Class 10 Chapter 3 Solutions Online?

You can get the best NCERT Solutions for class 10 chapter 3 on Vedantu for free download in the form of a PDF. Subject experts prepare the solutions to let students get maximum benefits of it. Also, every answer is explained with detailed step-wise explanations so students can refer to them to clear their doubts. All the material is prepared by our subject-matter experts keeping in mind the new syllabus.

2. How Many Solutions can you Find for a Linear Equation in Two Variables?

A linear equation in two variables, it can be represented in the form of a line on the graph. A line is endless, and hence it can have infinitely many solutions. Each solution of the linear equation in two variables will be uniquely identified and plotted on the graph. Each point on the line will be the solution to the linear equation defining that line. Thus there is no end to the solutions of a linear equation in two variables. Suppose if we have two different lines of linear equation, then those either intersect, coincide, or are parallel to each other.

3. How many exercises are there in Class 10 Maths Chapter 3?

NCERT Class 10 Maths Chapter 3 is Pair of Linear Equations in Two Variables. This chapter has a total of seven exercises. Vedantu offers solutions for all the exercises. The NCERT Solutions for Class 10 Maths Chapter 3 have been created by subject matter experts. You need to download them in PDF format so that you can study them anytime and anywhere. 

4.  What is a Linear Equation in Two variables Class 10?

A Linear equation in Two variables is written in the form of ax + by + c=0. In this, a, b, and c are real numbers where c is a constant and a and b are the coefficients of x and y, respectively. It is important that a and b should not be zero.

Examples of Linear equations in Two variables are x+ 2y = 14, 10x - 8y = 7, etc. Linear equations in Two variables have two values as a solution, one for x and the other for y. Vedantu has explained these concepts in detail and provided the solutions to exercises from NCERT textbooks.

5. What are the most important questions of 10th Maths Chapter 3?

Here are some important questions of Class 10th Maths Chapter 3 - Pair of Linear Equations in Two Variables:

Linear Equations

Forms of Linear Equations

The Standard Form of Linear Equations

Solving Linear Equations with One Variable

Solving Linear Equations with Two Variables

You can find the solutions to all the exercises of 10th Maths Chapter 3 in Vedantu’s  NCERT Solutions for Class 10 Maths . All the questions present there are accurately answered, and the concepts are properly explained.

6. How can I download the Class 10 Maths Chapter 3 solutions?

Class 10 Maths Chapter 3, A Linear Equation in Two variables, is one of the scoring chapters. If you want to get the solutions for this chapter, you can visit the official website of Vedantu. We offer the NCERT Solutions for Class 10 Maths in PDF format so that you can download them for offline use. It will allow you to study them anywhere and at any time. You can also download the solutions from the Vedantu App. All you have to do is download the app from Google play store, follow the instructions to sign in, and download the study material. The best part is that all these solutions are available for free. 

7. How do you solve Linear Equations in Class 10?

Following are the different methods of solving Linear Equations:

Cross multiplication method

Substitution method

Graphical method

8. What are the topics covered in class 10 chapter 3 maths?

The topics covered in class 10 chapter 3 maths from Pair Of Linear Equations In Two Variables are:

Basics of linear equations: This lays the foundation, defining linear equations and introducing the concept of solving them in pairs.

Graphical method: You'll learn how to visualize the solutions of a pair of equations by graphing them.

Algebraic methods: Here, you'll explore different techniques to solve linear equations algebraically, including:

Cross-multiplication method

Applications: You'll learn how to use these methods to solve real-world problems that can be modeled by a pair of linear equations.

9. What is the formula of the substitution method?

The substitution method is a methodical approach to solving a system of two linear equations. Nevertheless, it lacks a single formula. This is how the process is broken down:

Express one variable in terms of the other:  In one of the equations, rearrange the equation to isolate (solve for) one variable (x or y) in terms of the other variable.

Take the expression you just got for that variable and plug it into the other equation. This replaces that variable entirely with the expression you obtained.

Solve for the remaining variable:  Now you have a single equation with just one variable remaining. Solve this equation like you would normally solve any linear equation.

Solve for the first variable:  Once you find the value of the remaining variable, plug it back into the equation you isolated it from in step 1. Now you can solve for the first variable's value.

Substitute both your solutions (the values of x and y) back into the original equations. If both sides evaluate to the same value, then your solution is correct.

10. What is the name of Chapter 3 Maths Class 10?

Pair of Linear Equations in Two Variables is the title of Maths Class 10 Chapter 3. For this chapter, most curricula use this standard name. It appropriately conveys the material, which is on comprehending and resolving equation systems with two unknowns denoted by variables.

11. What is another name for the substitution method?

In the context of solving linear equations, the substitution method is the most commonly used word, but it lacks a commonly accepted replacement name.

It may be referred to as the replacement method in some sources. This highlights the fact that an expression comprising the other variable is being used in place of one variable.

But in mathematics teaching, "substitution method" is the accepted and commonly used phrase.

12. What is the main goal of the substitution method?

Converting a system of two linear equations with two variables into a single equation with one variable is the primary objective of the substitution method.

Finding an Isolated Variable: To begin, examine both equations and attempt to find an isolated variable (x or y) in relation to the other variable in one of the equations.This could require simple algebraic operations.

Replacing the Expression: After a variable has been identified, you take that expression in its entirety and replace it in the other equation.This just substitutes the expression you got in step 1 for one variable.

Finding the Single Variable: At this point, there is just one variable left in your equation.To determine the value of that variable, solve this equation just like you would any other single-variable problem.

Identifying the Additional Variable: You re-insert the value of one variable into the original equations (where the substitution was not done).You may now solve for the other variable as a result.

13. Does Maths Class 10 Chapter 3 NCERT Solutions require knowledge of all three approaches in order to solve pairs of linear equations in two variables?

Yes, mastering all three techniques is required in order to solve two linear equations in two variables found in NCERT Solutions for Class 10 Maths Chapter 3. These subjects will be covered in further education and might even come up in their Class 10 final exams. BYJU'S does a fantastic job of explaining these ideas. Therefore, the primary goal of these solutions developed by BYJU'S experts is to impart information on the foundational ideas of mathematics, which helps students grasp each idea with clarity.

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

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables:  The Class 10 Maths NCERT Solutions Chapter 3 deals with the Pair of Linear Equations in Two Variables topic. The topic is easy, and students can obtain full marks if they practice answering the NCERT questions from this chapter. It covers all the subtopics of Pair of Linear Equations in Two Variables.

Embibe provides over 2400 questions for practice. Moreover, it provides 10 books for students’ reference. They involve questions that test and improve your logical reasoning and analytical abilities for speed and accuracy. Scroll down for more details related to the Class 10 Maths Chapter 3 Solutions.

NCERT Solutions for Class 10 Maths Chapter 3: Important Topics

NCERT Solutions for Class 10 Maths Chapter 3 discusses Pair of Linear Equations in Two Variables and provides a good understanding of the problems in the chapter. The solutions for NCERT Chapter 3 Maths Class 10 Linear Equations Two Variables are according to the latest CBSE syllabus and exam pattern guidelines. The NCERT Solutions Class 10 Maths Chapter 3 is very interesting, and students can easily score full marks in this chapter. Get the exercise-wise NCERT Solutions for Class 10 Maths.

Scroll down for the list of subtopics covered in chapter 3:

NCERT Solutions for Class 10 Maths Chapter 3: Points to Remember

Below we have provided some of the important points to remember for NCERT Class 10 Maths Chapter 3 to ace your exams:

• We must first draw the lines they represent to solve a pair of linear equations in two variables using the Graphical method.
• If the two lines intersect at a point, we say the pair is consistent, and the intersection point coordinates provide a unique solution.
• When the two lines are parallel, the pair has no solution and is referred to as an inconsistent pair of equations.
• If the two lines coincide, the problem has an infinite number of solutions. Every point on the line represents a solution. In this case, we say that the pair of linear equations have an infinite number of solutions.
• The following are the algebraic solutions to a pair of linear equations in two variables: Substitution, Elimination, and Cross-multiplication.

NCERT Solutions for Class 10 Maths: All Chapters

The detailed NCERT Class 10 Maths Solutions for all chapters are provided below for your reference:

• 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

FAQs on NCERT Solutions for Class 10 Maths Chapter 3

Ans. Students can get NCERT Solutions for Class 10 Maths Chapter 3 at Embibe.

Ans: There are over 2400 questions available for practice at Embibe.

Ans: The name of Class 10 Maths Chapter 3 is Pair of Linear Equations in Two Variables.

Ans: No, class 10 Maths is not tough to crack. Students just need to prepare for exams strategically.

Ans: There are six topics which are covered in Class 10 chapter 3 Maths.

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Mathematics class 10 lab manual | 21 lab activities, math assignment class x ch-3 | linear equations in two variables.

  CHAPTER 3    CLASS   X

PAIR OF LINEAR EQUATIONS IN TWO VARIABLE

Extra questions of chapter 3 class 10 : Pair of Linear Equations in Two Variable with answer and  hints . Useful math assignment for the students of class 10

For better results

• Students should learn all the  basic points of  Chapter 3 Pair of Linear Equations in two variable .
• Student should revise N C E R T book thoroughly with examples.
• Now revise this assignment. This assignment integrate the knowledge of the students.

ASSIGNMENT BASED ON CH-3 CLASS 10

a)   2x - 3y = 1 ;  x + 2y = 2

Ans: Unique solution, intersecting lines, consistent

b)   3x - y = 6 ;  6x - 2y = 4

Ans: No solution, parallel lines, inconsistent

c)   x + 3y = 8 ;  2x - 3y = 12

d)   9x + 3y + 12 = 0 ;  18x + 6y = - 24

Ans: Many solution, coincident lines, consistent

e)   6x - 3y +10 = 0 ;  2x – y + 9 = 0

a)    2x + y - 5 = 0;  6x + 3y + p = 0       Ans [p = -15]

b)   x + 2y + 7 = 0;   2x + py + 14 = 0     Ans [p = 4]

c)  3x + 4y + 2 = 0;  9x + 12y + p = 0     Ans[p = 6]

d)    x + 5y - 7 = 0 ;  4x + 20y + p = 0     Ans [p = -28]

a)   8x + 5y = 9;   kx + 10y = 15                                          Ans [k = 16 ]

b) (3k + 1)x + 3y - 2 = 0;    (k 2  + 1)x + (k - 2)y  -  5 = 0     Ans[k = -1]

c)   3x + y = 1;   (2k - 1)x + (k - 1)y = 2k + 1                      Ans [k = 2]

d) kx + 3y = k  –  3;  12x + ky = k                                      Ans [k = -6]

a)  (a + b)x  - 2by = 5a + 2b + 1;    3x - y = 14             Ans [a = 5, b =  1]

b)     2x - 3y = 7;  (a + b)x  +  (a + b)y = 4a + b           Ans[-5, -1]

c)  2x + 3y = 7;    (a - b)x - (a + b - 3)y = 3a + b - 2    Ans[a = 5, b = 1]

d) (2a - 1)x + 3y – 5 = 0;   3x + (b - 1)y – 2 = 0          Ans[a = 17/4,   b= 11/5]

e)   kx + 3y = 2k + 1;  2(k + 1)x + 9y = 7k + 1            Ans [k = 2]

f)   2x + 3y = 7;   (k – 1)x + (k + 2)y = 3k                  Ans [k = 7]

a) x + y = 3;   2x + 5y = 12                        Ans [1, 2]

b)  2x + 3y + 5 = 0;   3x - 2y - 12 = 0        Ans [2,-3]

a) x + y = 5;      2x - y = - 2                     

[Sol. (1, 4) & Area= 12]

b) x - y + 1 = 0;    3x + 2y - 12 = 0         

  [Area= 7.5]

c)  2x + y = 6;      2x - y + 2 = 0               

a)     x+ y = 7    ;   3x - 2y = 1                  Ans [3, 4]

b)     3x + 2y = 10  ; 4x - y = 6                Ans [2, 2]

d) 2x + y – 35 = 0   ;  3x + 4y - 65 = 0      Ans [15, 5]

e)  x/a - y/b = 0 ;  bx + ay = 4ab             Ans[2a, 2b]

f ) - 6x + 5y = 2 ;    - 5x + 6y = 9              Ans[3, 4]

g)  23x - 29y = 98  ;   29x - 23y = 110      Ans [3, - 1]

h)   99x + 101y =  499 ; 101x +  99y = 501      Ans [x = 3, y = 2]

i)    a(x + y) + b(x - y) = a 2  – ab + b 2  ;   a(x + y) - b(x - y) = a 2  + ab + b 2           

j) ax + by = a - b ;   bx - ay = a + b            Ans [1, - 1]

Ans [45 and 30]

Ans [Rs1800 & Rs 1400]

Ans  [Rs 6.50 and Rs 1.50]

Ans  [93]

Ans [47 or 74]

Ans   [64]  

Ans  [12/25]

Ans  [7/18]

Ans  [42, 10]  

Ans  [10 km/h and  2 km/h]

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Classroom Assignment: 3 | Class 10 BBC Compacta Solutions PDF Download

On the basis of your understanding of the passage, answer the following questions. Q1: The initial use of computers was made in which of the following fields?  1. Business activities  2. Defense forces  3. Education and research  4. International networking  (a) Options 1 and 2  (b) Options 2 and 3  (c) Options 3 and 4  (d) Options 1 and 4 Ans: (b) Options 2 and 3

Q2: On the basis of paragraph 2, which of the following can be considered to be MOST PROBABLY TRUE ?   (a) Berners-Lee was not initially interested in studying science. (b) Berners-Lee had little interest in Electronics and Maths. (c) Berners-Lee's decision to study Science proved to be a disaster for him. (d)  Berners-Lee inherited his interest in computers from his parents. Ans: (d)  Berners-Lee inherited his interest in computers from his parents.

Q3: Which of the following statements do you think Berners-Lee in all likelihood could have made as a school boy? (a) "When will I ever get to see a train?" (b)  "There goes my favorite engine!" (c)  "Watching trains is such a boring activity!" (d)  "One day I will make a computer that will run trains!" Ans: (b) "There goes my favorite engine!"

Q4: Which of the following sentences illustrates the incorrect use of the word 'pragmatic' as used in paragraph 2?   (a) In business, the pragmatic approach to problems is often more successful than an idealistic one. (b) Coach, do you have a more pragmatic plan to beat the opposition? (c)  The prime-minister's decision to not resign was very pragmatic . (d) She was a dreamer-not a pragmatic . Ans: (d) She was a dreamer-not a pragmatic. 

Q5: What was the twin purpose of the world's first website http://info.cern.ch ? Ans: The twin purpose of the world's first website http://info.cern.ch was to explain the World Wide Web concept and to give users an introduction to getting started with their own websites.

Q6: According to Berners-Lee's website, the Web was intended to ___________________________ . Ans:  According to Berners-Lee's website, the Web was intended to be used as a communication tool and to help people understand each other .

Q7: Choose the option that correctly identifies the following statements as TRUE and FALSE.   1. Before Berners-Lee the internet used to be just a collection of academic and military computer system. 2. The internet came into being 30 years after Berners-Lee had first thought of it. 3. Berners-Lee's decision to study Science opened up a world of knowledge for him. 4. Berners-Lee was already a teenager when he took to trainspotting. 5. Berners-Lee transformed an old computer into a television set. 6. The world's first website was launched on 6th August 1991. 7. Berners-Lee completed his project to decentralise the Internet in 2020. (a)  TRUE: 1,2 and 6; FALSE: 3,4,5 and 7 (b)  TRUE: 1,5 and 6; FALSE: 2,3,4, and 7 (c)  TRUE: 1,3 and 6; FALSE: 2,4,5 and 7   (d) TRUE: 1,3 and 7; FALSE: 2,4,5 and 6 Ans: (b)  TRUE: 1,5 and 6; FALSE: 2,3,4, and 7

Q8: 'To infringe upon' (paragraph 5) means the same as which of the following? (a)  follow (b) protect (c) violate (d) destroy Ans: (c) violate

Q9: Complete the given sentence with an appropriate inference with respect to the following: Tim Berners-Lee decided to make a comeback in 2020 so that __________________________________ Ans:  Tim Berners-Lee decided to make a comeback in 2020 so that he could address the issues of privacy infringement and centralized control of the Internet by tech giants, with the aim of decentralizing the Internet and ensuring users' privacy, ultimately restoring the Internet to his original vision of a democratic and equal network of information.

Q10: The chief complaint that Sir Berners-Lee has against the tech-giants ________________________ Ans:  The chief complaint that Sir Tim Berners-Lee has against the tech giants is their infringement upon the privacy of Internet users and their concentration of control over the Internet, which has led to inequality and division .

Q11: Based on your reading of the passage, say which of the following statements Tim Berners-Lee is LEAST LIKELY to make? (a)  "My dream is to see the internet turn once again into a democratic and equal network of information." (b)  "I'm glad that the internet is now controlled by a few tech giants." (c)  "How dare anyone use your personal information without your consent!" (d)  "Did I create the World Wide Web to allow organization to spy on people?" Ans: (b) "I'm glad that the internet is now controlled by a few tech giants."

Q12: Which of the following is the MOST SUITABLE TITLE for the passage? (a)  Berners-Lee and Tech Giants (b)  The Train Spotter! (c)  The Internet and Your Privacy (d) The WWW Man! Ans: (a)  Berners-Lee and Tech Giants

Q13: Tick (✓) the option that correctly matches the personality traits of Tim Berners-Lee's character with the pieces of textual evidence that these traits exemplify.

(a)  A-(iv); B-(iii); C-(i), D-(ii) (b)  A-(iii); B-(i); C-(ii), D-(iv) (c) A-(iv); B-(i); C-(iii),D-(ii) (d) A-(ii); B-(iv); C-(i), D-(iii) Ans: (a)  A-(iv); B-(iii); C-(i), D-(ii)

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NCERT Solutions for Class 10 Science Chapter 3 Metals and Non-metals

September 27, 2019 by Veerendra

Metals and nonmetals class 10 : The NCERT Solutions for Class 10 Science is designed to provide a strong foundation for further studies in science.  Candidates who are searching for NCERT Solutions for Class 10 Science Chapter 3 Metals And Non Metals questions and answers can refer to the class 10 science chapter 3 notes below.  We have included all the important questions from Metals and Non Metals similar to that of your chapter 3 science Class 10 notes  to help you secure a good score in Class 10 science exam

Chapter 3 Metals and Non Metals ,  is derived from the NCERT textbook of Class 10 Science as prescribed in CBSE Schools of India. These CBSE NCERT Solutions will not only help in your Class 10 exam preparation but also in clearing other competitive exams. Read further to find out everything about NCERT Solutions for Class 10 Science Chapter 3 Metals And Non Metals.

Metals and nonmetals Class 10

Before going through NCERT Solutions for class 10 science chapter 3 answers, let’s have a look at the list of topics and subtopics under NCERT solutions for class 10 science chapter 3 Notes for Metals and Non-Metals :

• Physical Properties of Metals And Non-Metals
• Chemical Properties Of Metals
• What happens when metals are burnt in air?
• What happens when metals react with water?
• What happens when metals react with acids?
• How do metals react with solutions of other Metal Salts
• The Reactivity Series
• How do Metals and Non-Metals react?
• Properties of Ionic Compounds
• Extraction of metals
• Enrichment of Ores

Extracting Metals Low in the Activity Series

• Extracting Metals in the Middle of the activities Series
• Extracting Metals towards the top of the Activity Series
• Refining of Metals
• Prevention of Corrosion

Free download NCERT Solutions for Class 10 Science Chapter 3 Metals and Non-Metals PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

• Metals and Non-metals
• धातु और अधातु कक्षा 10 विज्ञान हिंदी में
• Class 10 Metals and Non-metals Important Questions
• Metals and Non-metals Class 10 Notes
• Metals and Non-metals NCERT Exemplar Solutions
• Metals and Non-metals Class 10 Extra Questions
• Class 10 Metals and Non Metals Mind Map

NCERT Solutions for Class 10 Science Chapter 3 Intext Questions

Class 10 Metals and Non Metals NCERT Book Page Number: 40

Question 1 Give an example of a metal which : (i) is a liquid at room temperature. (ii) can be easily cut with a knife. (iii) is the best conductor of heat. (iv) is a poor conductor of heat. Answer: (i) Mercury (ii) Sodium (iii) Silver (iv) Lead

Question 2 Explain the meanings of malleable and ductile. Answer: Malleable : A metal that can be beaten into thin sheets on hammering is called malleable. Ductile : A metal which can be drawn into thin wires is called ductile.

Malleable Meaning in Hindi

कुछ धातुओ को पीटकर पतली चादर बनाया जा सकता है | इस गुणधर्म को आघातवर्ध्य कहते है | कुछ धातुओ के पतले तार के रूप में खीचने कि क्षमता को तन्यता कहते है |

Class 10 Metals and Non Metals NCERT Book Page Number: 46

Question 1 Why is sodium kept immersed in kerosene oil ? Answer: Sodium is highly reactive. So it is kept immersed in kerosene oil to prevent its reaction with oxygen, moisture and carbon dioxide of air to prevent accidental fires.

Question 3 Samples of four metals A, B, C and D were taken and added to the following solution one by one. The results obtained have been tabulated as follows :

Use the Table above to answer the following questions about metals A, B, C and D. (i) Which is the most reactive metal ? (ii) What would you observe if B is added to a solution of copper (II) sulphate? (iii) Arrange the metals A, B, C and D in the order of decreasing reactivity. Answer: (i) B is the most reactive metal because it gives displacement reaction with iron (II) sulphate. (ii) When metal B is added to copper (II) sulphate solution, a displacement reaction will take place due to which the blue colour of copper (II) sulphate solution will fade and a red-brown deposit of copper will be formed on metal B. (iii) Metal B is the most reactive because it displaces iron from its salt solution. Metal A is less reactive because it displaces copper from its salt solution. Metal C is still less reactive because it can displace only silver from its salt solution and metal D is the least reactive because it cannot displace any metal from its salt solution. Hence, the decreasing order of reactivity of the metals is B > A > C > D.

Question 4 Which gas is produced when dilute hydrochloric acid is added to a reactive metal ? Write the chemical reaction when iron reacts with dilute H 2 SO 4 . Answer: Hydrogen gas is produced when dilute hydrochloric acid is added to a reactive metal. Chemical reaction when iron reacts with dilute H2SO4 : Fe(s) + H 2 SO 4 (aq) → FeSO 4 (aq) + H 2 (g)

Class 10 Metals and Non Metals NCERT Book Page Number: 49

(iii) In Na 2 O, ions present are Na + and O 2- . In MgO, ions present are Mg 2+ and O 2- .

Question 2 Why do ionic compounds have high melting points ? (iii) What are ions present in these compounds? Answer: The ionic compounds are made up of positive and negative ions. There is a strong force of attraction between the oppositely charged ions, so a lot of heat energy is required to break this force of attraction and melt the ionic compound. Due to this, ionic compounds have high melting points.

Class 10 Metals and Non Metals NCERT Book Page Number: 53

Question 1 Define the following terms : (i) Mineral, (ii) Ore and (iii) Gangue. Answer: (i) Mineral : The natural materials in which the metals or their compounds are found in earth are called minerals. (ii) Ore : Those minerals from which the metals can be extracted conveniently and profitably are called ores. (iii) Gangue : The unwanted impurities like sand, rocky material, earth particles, lime stone, mica, etc in an ore are called gangue.

Question 2 Name two metals which are found in nature in the free state.

Answer: Gold and platinum

Question 3 What chemical process is used for obtaining a metal from its oxide. Answer: Reduction process is used for obtaining a metal from its oxide. For example, zinc oxide is reduced to metallic zinc by heating with carbon. ZnO(s) + C(s) → Zn(s) + CO(g)

Besides carbon, highly reactive metals like sodium, calcium, aluminium etc. are used as reducing agents. These displace metals of low reactivity from their oxides. For example, Fe 2 O 3 (s) + 2Al(s) → 2Fe(l) + Al 2 O 3 (s) + Heat

Gold is Metal or Nonmetal ?

Gold is a metal found in nature in the free state

Class 10 Metals and Non Metals NCERT Book Page Number: 55

Question 1 Metallic oxides of zinc, magnesium and copper were heated with the following metals :

In which cases will you find displacement reactions taking place ? Answer: A more reactive metal can displace a less reactive metal from its oxide. But out of zinc, magnesium, and copper metals, magnesium is the most reactive, zinc is less reactive whereas copper is the least reactive metal.

The displacement will take place in the following cases :

Question 2 Which metals do not corrode easily ? Answer: Gold and Platinum.

Question 3 What are alloys ? Answer: An alloy is a homogeneous mixture of two or more metals, or a metal and a non-metal. For example, bronze is an alloy of copper and tin.

NCERT Solutions for Class 10 Science Chapter 3 Textbook Chapter End Questions

Metals and Nonmetals Class 10 Question 1. Which of the following pairs will give displacement reactions ? (a) NaCl solution and copper metal. (b) MgCl 2 solution and aluminium metal. (c) FeSO 4 solution and silver metal. (d) AgNO 3 solution and copper metal. Answer: (d) AgNO 3 solution and copper metal.

Question 2. Which of the following methods is suitable for preventing an iron frying pan from rusting ? (a) Applying grease (b) Applying paint. (c) Applying a coating of zinc (d) All the above. Answer: (c) Applying a coating of zinc.

Question 3. An element reacts with oxygen to give a compound with a high melting point. This compound is also soluble in water. The element is likely to be (a) calcium (b) carbon (c) silicon (d) iron Answer: (a) Calcium.

Question 4. Food cans are coated with tin and not with zinc because (a) zinc is costlier than tin (b) zinc has a higher melting point than tin (c) zinc is less reactive than tin (d) zinc is more reactive than tin. Answer: (d) Zinc is more reactive than tin.

Metals and Non metals Class 10 Question 5. You are given a hammer, a battery, a bulb, wires and a switch. (a) How could you use them to distinguish between samples of metals and non-metals? (b) Assess the usefulness of these tests in distinguishing between metals and non-metals. Answer: (a) Metals can be beaten into thin sheets with a hammer without breaking. Non-metals cannot be beaten with a hammer to form thin sheets. Non-metals break into pieces when hammered. Metals are malleable, while non-metals are non-melleable. When metals are connected into circuit using a battery, bulb, wires and switch, current passes through the circuit and the bulb glows. When non-metals (like sulphur) are connected, the bulb does not light up at all. Metals are good conductors of electricity. (b) Because of malleability, metals can be casted into sheets. Metals are good conductors of electricity so these can be used for electrical cables.

Question 7. Name two metals which will displace hydrogen from dilute acids and two metals which will not. Answer: (i) Metals above hydrogen in the activity series like sodium and magnesium displace hydrogen from dilute acids. (ii) Metals below hydrogen in the activity series like copper, silver do not displace hydrogen from dilute acids.

Question 8. In the electrolytic refining of a metal M, what would you take as the anode, the cathode and the electrolyte ? Answer: Cathode           –      Pure metal Anode               –     Impure metal Electrolyte       –     Metal salt solution

Question 10. State two ways to prevent the rusting of iron. Answer: Ways to prevent rusting of iron are : (a) By painting (b) By galvanizing

Question 12. Give reasons : (a) Platinum, gold and silver are used to make jewellery. (b) Sodium, potassium and lithium are stored under oil. (c) Aluminium is a highly reactive metal, yet it is used to make utensils for cooking. (d) Carbonate and sulphide ores are usually converted into oxides during the process of extraction. Answer: (a) Platinum, gold and silver are used to make jewellery because these are malleable and ductile. These are highly resistant to corrosion. (b) Sodium, potassium and lithium are very reactive and catch fire when exposed to air. This is due to their low ignition temperature and high reactivity. (c) Aluminium forms a non-reactive layer of aluminium oxide on its surface. This layer prevents aluminium to react with other substances. That’s why aluminium is used to make cooking utensils. (d) It is easier to reduce a metal oxide into free metal. Since it is easier to obtain metals from their oxides than from their carbonates or sulphides directly, therefore, the carbonate and sulphide ores are first converted to oxides for extracting the metals.

Question 13. You must have seen tarnished copper vessels being cleaned with lemon or tamarind juice. Explain why these sour substances are effective in cleaning the vessels. Answer: The sour substances such as lemon or tamarind juice contain acids. These acids dissolve the coating of copper oxide or basic copper carbonate present on the surface of tarnished copper vessels and makes them shining red-brown again.

Question 14. Differentiate between metal and non-metal on the basis of their chemical properties. [CBSE 2017 (Delhi)] Answer: Difference between metals and non-metals

Question 15. A man went door-to door posing as a goldsmith. He promised to bring back the glitter of old and dull gold ornaments. An unsuspecting lady gave a set of gold bangles to him which he dipped in a particular solution. The bangles sparkled like new but their weight was reduced drastically. The lady was upset but after a futile argument the man beat a hasty repeat. Can you play the detective to find out the nature of the solution he has used ? Answer: The dishonest goldsmith dipped the gold bangles in aqua-regia (which contains 1 part of concentrated nitric acid and 3 parts of concentrated hydrochloric acid, by volume). Aqua-regia dissolved a considerable amount of gold from gold bangles and hence reduced their weight drastically. The dishonest goldsmith can recover the dissolved gold from aqua-regia by a suitable treatment.

Question 16. Give reasons why copper is used to make hot water tanks and not steel (analloy of iron). Answer: (i) Copper is a better conductor of heat than steel. (ii) Copper does not corrode easily. But steel corrodes easily. (iii) Copper does not react with water at any temperature, whereas iron reacts with water on heating.

Metals and non metals: Properties of metals and non-metals, reactivity series, Formation and properties of ionic compounds, Basic metallurgical processes, corrosion and its prevention.

Question 1 What are amphoteric oxides? Give two examples of amphoteric oxides. Solution: Amphoteric oxides are the oxides, which react with both acids and bases to form salt and water. E.g. ZnO and Al 2 O 3 .

Question 2 Name two metals, which will displace hydrogen from dilute acids, and two metals which will not. Solution: Very reactive metals like Zn and Mg displace hydrogen from dilute acids. On the other hand less reactive metals like Cu, Ag, etc. do not displace hydrogen from dilute acids.

Question 3 In the electrolytic refining of a metal M, what would you take as the anode, the cathode and the electrolyte? Solution: Anode is impure, thick block of metal M. Cathode is a thin strip/wire of pure metal M. Electrolyte is a suitable salt solution of metal M.

Metals and nonmetals Class 10 PDF

Question 4 State two ways to prevent the rusting of iron. Solution: By coating the surface of iron by rust proof paints. By applying oil or grease to the surface of iron objects so that supply of air consisting of moisture is cut off form the surface.

Question 5 What types of oxides are formed when non-metals combine with oxygen? Solution: When non-metals combine with oxygen it forms either neutral or acidic oxides. CO is a neutral oxide; N 2 O 5 or N 2 O 3 is an acidic oxide.

extraction of metals from ores class 10 Question 6 Give reason i. Metals replace hydrogen from dilute acids, where as non-metals do not. ii. Carbonate and sulphide ores are usually converted into oxides during the process of extraction. Solution: i. Metals are electropositive in nature. They readily lose electrons. These electrons reduce the protons liberated from the acid to liberate hydrogen gas, where as non-metals possess a tendency to gain electrons and hence they do not furnish electrons to protons liberated from acids. Hence H 2 gas is not liberated. ii. As it is easier to reduce metal oxides to metal, prior to reduction, metal sulphides and carbonates must be converted to oxides.

Question 8 Explain why the surface of some metals acquires a dull appearance when exposed to air for a long time. Solution: This is due to the surface oxidation of metals when exposed to moist air. For e.g. copper turns green on its surface due to the formation of basic copper carbonate Cu(OH) 2 . CuCO 3 . Similarly silver becomes black due to the formation of black Ag 2 S and Aluminium forms a white coating of Al2O3 on its surface.

Question 10 Name a non-metallic element, which conducts electricity. Solution: Carbon in the form of graphite conducts electricity, as there is a free electron in each carbon atom, which moves freely in between the hexagonal layers.

Question 11 Which metals do not corrode easily? Solution: Gold and platinum and other noble metals do not corrode in air.

Question 12 What are alloys? Solution: Alloys are homogeneous mixtures of two or more metals, or a metal and a non-metal.E.g. steel, brass, bronze, etc.

Question 13 Define the following terms. (i) Minerals (ii) Ores (iii) Gangue Solution: (i) Minerals All compounds or elements, which occur naturally in the earth’s crust, are called minerals. Example: Alums, K 2 SO 4 .Al 2 (SO 4 ) 3 . 24 H 2 O, Bauxite Al 2 O 3 .2H 2 O (ii) Ores Those minerals from which a metal can be profitably extracted are called ores. Bauxite (Al 2 O 3 .2H 2 O) is the ore of Al, copper pyrite CuFeS 2 . All minerals are not ores but all ores are minerals. (iii) Gangue When an ore is mined from the earth, it is always found to be contaminated with sand rocky materials. The impurity of sand and rock materials present in the ore is known as gangue.

Question 14 Name two metals that are found in nature in the free state. Solution: Gold and platinum are found in the free state in nature.

Question 16 Name two metals, which can form hydrides with metals. Solution: Sodium and calcium form stable hydrides on reacting with hydrogen.

Question 17 Does every mineral have a definite and a fixed composition? Explain. Solution: Yes, every mineral has a definite and a fixed composition. Minerals are widely distributed in the earth’s crust in the form of oxides, carbonates, sulphides, sulphates, nitrates, etc. These minerals are formed as a result of chemical changes taking place during the formation of earth.

Class 10 metals and nonmetals Question 18 Explain the meaning of malleable and ductile. Solution: Malleable is being able to be beaten/hammered into thin sheets. Ductile is being able to be drawn into thin wires.

iii. The ions present in sodium oxide compound (Na 2 0) aie sodium ions (2Na + and oxide ions (O 2- ). The ions present in Magnesium oxide compound (MgO) are magnesiumions Mg 2+ and oxide ions (O 2- ).

Question 20 You must have seen tarnished copper vessels being cleaned with lemon or tamarind juice. Explain why these sour substances are effective in cleaning the vessels. Solution: The sour substances such as lemon (or tamarind juice) contain acids. These acids dissolve the coating of copper oxide or basic copper carbonate present on the surface of tarnished copper vessels and make them shining red-brown again.

Question 21 Give an example of a metal which i. is a liquid at room temperature. ii. can be easily cut with a knife. iii. is the best conductor of heat. iv. is a poor conductor of heat. Solution: i. Mercury is in liquid state at room temperature. ii. Sodium and potassium are soft metals which can be easily cut with a knife. iii. Silver is the best conductor of electricity. iv. Mercury is a poor conductor of heat.

Question 22 Why is sodium kept immersed in kerosene? Solution: Sodium metal is kept immersed in kerosene to prevent their reaction with oxygen, moisture and carbon dioxide of air.

Question 23 Why do ionic compounds have high melting points? Solution: These compounds are made up of positive and negative ions. There is a strong force of attraction between the oppositively charged ions, so a lot of heat energy is required to break this force of attraction and melt the ionic compounds. This is why ionic compounds have high melting points.

Question 24 A man went door to door posing as a goldsmith. He promised to bring back the glitter of old and dull gold ornaments. An unsuspecting lady gave a set of gold bangles to him which he dipped in a particular solution. The bangles sparkled like new but their weight was reduced drastically. The lady was upset but after a futile argument the man beat a hasty retreat. Can you play the detective to find out the nature of the solution he had used? Solution: Aqua regia (By volume, this contains 3 parts of concentrated hydrochloric acid and 1 part of concentrated nitric acid) is the solution, which is used to sparkle the bangles like new, but their weight will be reduced drastically.

Metals and nonmetals class 10  Question 27 Pratyush took sulphur powder on a spatula and heated it. He collected the gas evolved by inverting a test-tube over the burning sulphur. What will be the action of this gas on: Dry litmus paper? Moist litmus paper? Write a balanced chemical equation for the reaction taking place. Solution: a) When sulphur is brunt in air then sulphur dioxide gas is formed. (i) Sulphur dioxide gas has no action on dry litmus paper. (ii) Sulphur dioxide gas turns moist blue litmus paper to red. (b) S (s) + O 2(g) —> SO 2(g)

Multiple Choice Questions (MCQs) [1 Mark each]

Metals and nonmetals class 10 Question 1. What is the colour of aqueous solution of CuSO 4 and FeSO 4 as observed in the laboratory? (a) CuSO 4 – blue; FeSO 4 – light green (b) CuSO 4 – blue; FeSO 4 – dark green (c) CuSO 4 – green; FeSO 4 – blue (d) CuSO 4 – green; FeSO 4 – colourless Answer: (a) Colour of CuSO 4 solution is blue and FeSO 4 solution is light green.

Metals and nonmetals class 10 Question 3. Aluminium sulphate and copper sulphate solutions were taken in two test tubes I and II respectively. A few pieces of iron filings were then added to both the solutions. The four students A, B, C and D recorded their observations in the form of a table as given below:

Metals and nonmetals class 10 Question 4. Aqueous solutions of zinc sulphate and iron sulphate were taken in test tubes I and II by four students A, B, C and D. Metal pieces of iron and zinc were dropped in the two solutions and observations made after several hours were recorded in the form of table as given below:

Metals and nonmetals class 10 Question 5. 2 mL each of cone. HCl, cone. HNO 3 and a mixture of cone. HCl and cone. HNO 3 in the ratio of 3 : 1 were taken in test tubes labelled as A, B and C. A small piece of metal was put in each test tube. No change occurred in test tubes A’and Bbut the metal got dissolved in test tube C. The metal could be [NCERT Exemplar] (a) Al (b) Au (c) Cu (d) Pt Answer: (b, d) A mixture of cone. HCl and cone. HNO 3 in the ratio of 3 : 1 is known as aqua-regia. Gold (Au) and platinum (Pt) dissolve only in aqua-regia as these metals are very less reactive.

Metals and nonmetals class 10 Question 7. Aluminium is used for making cooking utensils. Which of the following properties of aluminium are responsible for the same? (i) Good thermal conductivity (ii) Good electrical conductivity (iii) Ductility (iv) Fligh melting point [NCERT Exemplar] (a) (i) and (ii) (b) (i) and (iii) (c) (ii) and (iii) (d) (i) and (iv) Answer: (d) Good thermal conductivity, malleability, light weight and high melting point are the properties/of aluminium due to which it-is used for making cooking utensils.

NCERT Solutions for Class 10 Science Chapter 3 Metals and Non-metals (Hindi Medium)

Class 10 Science Metals and Non-metals Mind Maps

Metals and Non-Metals There 92 well known naturally occurring minerals of which 70 are metals and rest 20 are the non-metals.

Physical Properties of Metals & Non-Metals

Reaction between Metals and Non Metals

• Reactivity of an element can be explained as tendency to attain a completely filled outermost shell.
• Metals have 1, 2 or 3 e- in outermost shell and thus it is easier for them to loss e- rather than to gain. They loss e- & gains positive charge & are tenned as cation.
• In contrast, non-metals have 4-8 e- in outermost shell & thus they gain e- to achieve their octet. They gain e- as well as negative charge & tenned as anion.
• Cations & anions attract each other & are held by strong electrostatic force of attraction.

Occurrence of Metals The elements or compounds, which occur naturally in the earth’s crust, are known as minerals. At some places, minerals contain a very high percentage of a particular metal and the metal can be profitably extracted from it. These minerals arc called ores.

Corrosion Corrosion is the deterioration of materials by chemical interaction with their environment for e.g. darkening of silver articles when exposed to air, gaining of green coat on copper, rusting of iron. Prevention: The rusting of iron can be prevented by painting, oiling, greasing, galvanising, chrome plating, anodising or making alloys. • Galvanisation is a method of protecting steel and iron from rusting by coating them with a thin layer of zinc. • Alloy is a homogeneous mixture of two or more metals, or a metal & nonmetal. For e.g. stainless steel (alloy of Fe. Ni, & Cr), amalgam (alloy of Hg), brass (alloy of Cu & Zn) etc. The electrical conductivity & melting point of an alloy is less than that of pure metals.

Enrichment of Ores Ores mined from the earth are usually contaminated with large amounts of impurities such as soil, sand, etc., called gangue. The impurities must be removed from the ore prior to the extraction of the metal. The processes used for removing the gangue from the ore are based on the differences between physical or chemical properties of the gangue and the ore.

Extracting Metals towards the Top of the Activity Series These metals are highly reactive & are obtained by electrolytic reduction. For e.g. Na, Mg, & Ca are obtained by the electrolysis of their molten chlorides. The metals are deposited at the cathode whereas chlorine is liberated at anode. At cathode  Na +  + e –  →Na At anode  2Cl –  → Cl 2  + 2e – Similarly, aluminium is obtained by the electrolytic reduction of aluminium oxide.

• These metals are the least reactive & are often found in a free state for e.g. An, Ag. Pt & Cu are found in the free state.
• However, Cu & Ag are also found in the combined state as their sulphide or oxide ores.
• The oxides of these metals can be reduced to metals by heating alone. For e.g. cinnabar (HgS), ore of mercury it is heated in air to converted it in mercuric oxide (HgO) which is then reduced to mercury by further heating.
• 2HgS(s) + 3O 2 (g) Heat 2HgO(s) + 2SO 2 (g)
• 2HgO(s) Heat 2Hg(l) + O(g)
• Another instance is reduction of Cu 2 S (ore of copper) to copper by heating.
• 2Cu 2 S + 3O 2 (g)Heat 2Cu 2 O(s) + 2SO 2 (g)
• 2CU 2 O + Cu 2 S Heat 6Cu(s) + SO 2

Extracting Metals Middle in the Activity Series

• These metals such as Fe, Zn, Pb, Cu, etc are moderately reactive & are usually present as sulphides or carbonates in nature.
• The sulphide ores are converted into oxides by heating strongly in the presence of excess air which is known as roasting.
• The carbonate ores are changed into oxides by heating strongly in limited air which is known as calcination.
• The metal oxides are then reduced to the corresponding metals by using suitable reducing agents such as carbon.
• For e.g. extraction of Zn
• Roasting: 2ZnS(s) + 3O 2 (g) Heat . 2ZnO(s) + 2SO 2 (g)
• Calcination: ZnCO 3 (s) Heat ZnO(s) + CO 2 (g)
• Reduction: ZnO(s) + C(s) → Zn(s) + CO(g)
• Sometimes displacement reactions can also be used in place of reduction & highly reactive metals such as Na, Ca, Al, etc., are used as reducing agents.
• For e.g. 3MnO 2 (s) + 4Al(s) → 3Mn(l) + 2Al 2 O 3 (s) + Heat
• Fe 2 O 3 (s) + 2Al(s) → 2Fe(l)+Al 2 O 3 (s) + Heat
• This reaction is used to join railway tracks or cracked machine parts and is known as the thermit reaction.

Now that you are provided all the necessary information regarding NCERT solutions for class 10 science chapter 3 Metals and Non Metals and we hope this detailed article on metal and nonmetal class 10 notes is helpful. If you find any doubts regarding this article or NCERT solutions for class 10 science chapter 3 Metals and Non Metals, leave your comments in the comment section below and we will get back to you as soon as possible.

NCERT Solutions for Class 10 Science All Chapters

• Chapter 1 Chemical Reactions and Equations
• Chapter 2 Acids, Bases and Salts
• Chapter 3 Metals and Non-metals
• Chapter 4 Carbon and Its Compounds
• Chapter 5 Periodic Classification of Elements
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Class 10 Science Chapter 3 All Activities Solutions | Metals and Non-metals | Activities in Text Book with Solution

Metal and non- metals, activities in textbook with solution.

In this post, you will find class 10 science chapter 3 Metals and non-metals all activities solutions. Class 10 Metals and non-metals all activities contain a complete explanation, and conclusion with suitable diagrams.

 After reading class 10 activities, you will understand the Properties of Metals and Non- metals.

    1.   Activity 3.1 Class 10 Science

    2.   Activity 3.2 Class 10 Science

    3.   Activity 3.3 Class 10 Science

    4.   Activity 3.4 Class 10 Science

    5.   Activity 3.5 Class 10 Science

    6.   Activity 3.6 Class 10 Science

    7.   Activity 3.7 Class 10 Science

    8.   Activity 3.8 Class 10 Science

    9.   Activity 3.9 Class 10 Science

   10.   Activity3.10 Class 10 Science

   11.   Activity3.11 Class 10 Science

   12. Activity3.12 Class 10 Science

   13.   Activity3.13 Class 10 Science

   14.   Activity3.14 Class 10 Science

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Cal Football: Safety Tre' Harrison Commits to Bears' Class of '25

Jeff faraudo | jun 29, 2024.

• California Golden Bears

Cal football received the 13th commitment to its 2025 recruiting class when Tre’ Harrison, a three-star prospect from Junipero Serra High School in Gardena, announced his college choice on social media Friday.

Harrison, at 5-foot-11, 165 pounds, is listed on recruiting sites as an athlete, but the coaches on the Cal staff who recruited him were Terrence Brown and Tre Watson, who work with the Bears’ defensive backs.

He is expected to play safety when he arrives at Berkeley in the fall of 2025.

“The different pathways after football strike my eye the most,” Harrison wrote on social media on why he chose Cal. “They are in the heart of the Bay and have very close ties to Silicon Valley, which is on the rise of bringing a lot more income in today’s society.

“The culture is hard to come by as well. Everyone is connected and in the loop.”

Harrison visited the Cal campus two weeks ago, and also took official visits to San Diego State, Harvard and Columbia. He also had scholarship offers from Notre Dame, Washington State and Boston College, among other schools. 

Rivals rates Harrison as the No. 95 prospect in California while 247Sports lists him at No. 52 in the state. Among safeties nationwide, 247Sports slots him as No. 42.

Besides football, Harrison plays basketball and runs track and field, where is specializes in the sprints and hurdles.

JEFF FARAUDO

Jeff Faraudo was a sports writer for Bay Area daily newspapers since he was 17 years old, and was the Oakland Tribune's Cal beat writer for 24 years. He covered eight Final Fours, four NBA Finals and four Summer Olympics. 

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NCERT Solutions Class 10 Science

Ncert solutions for class 10 science updated for 2023-24 free pdf.

NCERT Solutions Science Class 10 contains very important information that helps the students understand the complex topics and helps them in preparation for the Class 10 board examination. Studying the answers to the questions in the textbook will ensure your understanding of a particular topic and help you determine your strengths and weaknesses. These NCERT Solutions for Class 10 Science are prepared by our subject experts in such a way that the students understand all the topics covered in the syllabus of CBSE 10 Science quite effectively.

Access NCERT Solutions of Class 10 Science all chapters in PDF

The following chapters have been removed from the NCERT Class 10 Science textbook 2023-24. Periodic Classification Of Elements Sources of Energy Sustainable Management of Natural Resources

The Solutions of NCERT Books also helps students to understand the topic thoroughly, which is very important not just from the point of view of the Class 10 examination. Better understanding lays a great foundation for their future studies. Quite often,  questions from textbooks are also asked in competitive examinations. We have covered both in-text and exercise questions in detail.

NCERT Solutions Class 10 Science Book All Chapters Brief:

Chapter 1 – chemical reactions and equations.

The first chapter of Class 10 NCERT Science will teach the students about chemical reactions and how to write equations, how to conduct combination and decomposition reactions and more. In the previous classes, we have learned about physical and chemical changes in matter. Whenever a chemical change occurs, we can say that a chemical reaction has taken place. A complete chemical reaction represents the reactants, products and their physical states symbolically. Students will also study how to write a chemical reaction , which is a symbolic representation of a chemical reaction. The chapter also explains how various chemical equations can be balanced in different states.

The next subtopic teaches various chemical reactions such as Combination Reaction, Decomposition reaction, Displacement Reaction, and Double Displacement Reaction along with various examples and chemical reactions. On the basis of energy, exothermic and endothermic reactions are explained. Exothermic reactions are those reactions in which heat is given out along with the products, and endothermic reactions are those reactions in which energy is absorbed. Then redox reaction is explained, which is a combination of reduction reaction and oxidation reaction. The chapter explains all types of reactions with suitable examples with their respective chemical equations.

Topics Covered in Class 10 Science Chapter 1 Chemical Reactions and Equations :

Chemical reactions: Chemical equation, Balanced chemical equation, implications of a balanced chemical equation, types of chemical reactions: combination, decomposition, displacement, double displacement, precipitation, endothermic exothermic reactions, oxidation and reduction.

Also, access the following resources for Class 10 Chapter 1 Chemical Reactions and Equations, at BYJU’S:

• CBSE Science Notes For Class 10 Chapter 1
• Important Questions for Class 10 Science Chapter 1 – Chemical Reactions And Equations
• NCERT Exemplar Class 10 Science Solutions for Chapter 1 – Chemical Reactions And Equations

Chapter 2 – Acids, Bases and Salts

NCERT Class 10 Science Chapter 2 is all about acids, bases and salts . In previous classes, students have learned that the sour and bitter tastes of food are due to acids and bases, respectively, present in them. We all know that acids are sour in taste and change the colour of blue litmus to red, whereas, bases are bitter and change the colour of the red litmus to blue. In this chapter, we will study the reactions of acids and bases, how acids and bases cancel out each other’s effects and many more interesting things that we use in our day-to-day life. Students will get to understand the chemical properties of acids and bases, how acids and bases react with metals, how  metal carbonates and metal hydrogen carbonates react with acids, how acids and bases react with each other, reaction of metallic oxides with acids and reaction of a non-metallic oxide with base explained with suitable examples and various chemical reactions.

The chapter then explains what all acids and bases have in common with a suitable example which gives the conclusion that acid solution in water conducts electricity. Students get to learn various experiments on what happens to acid or a base in a water solution and how strong are acid or base solutions by making use of universal indication. Along with it, students will get to learn about the importance of pH in everyday life. The chapter ends with a detailed explanation of salt preparation, properties and its uses.

Topics Covered in Class 10 Science Chapter 2 Acids, Bases and Salts :

Acids, bases and salts: Their definitions in terms of furnishing of H+ and OH– ions, General properties, examples and uses, neutralization, concept of pH scale (Definition relating to logarithm not required), importance of pH in everyday life; preparation and uses of Sodium Hydroxide, Bleaching powder, Baking soda, Washing soda and Plaster of Paris.

Also, access the following resources for Class 10 Chapter 2 Acids, Bases and Salts at BYJU’S:

• Acids, Bases and Salts Class 10 Chapter 2 Notes
• Chapter 2 -Acids, Bases and Salts
• Revision Notes For Class 10 Science Chapter 2 – Acids, Bases and Salts
• NCERT Exemplar Class 10 Science Solutions for Chapter 2 – Acids Bases And Salts

Chapter 3 – Metals and Non-metals

In previous classes, students must have learned about various elements that can be classified as metals or non-metals on the basis of their properties. Here in Chapter 3 of Class 10 Science, students will learn about the physical properties of metals and non-metals . Metals are lustrous, malleable, ductile and are good conductors of heat and electricity. They are solid at room temperature, except mercury which is a liquid. The physical properties of metals are explained on various parameters such as ductility, malleability, tensile nature, strength, etc. On the basis of physical properties, metals and non-metals are differentiated. Some of the examples of non-metals are carbon, sulphur, iodine, oxygen, hydrogen, etc. The non-metals are either solids or gases except bromine which is a liquid. Under the subtopic chemical properties of metals, chemical reactions are discussed with oxygen gas, water, acids and other metal salts. The reactions and conditions depend on the reactivity series. The reactivity series tops potassium as the most reactive and gold as the least reactive.

The compounds formed in this manner by the transfer of electrons from a metal to a non-metal are known as ionic compounds or electrovalent compounds. Some of the general properties for ionic compounds are physical nature, melting and boiling points, solubility and conduction of electricity. Metal is extracted from its ore and then refining them for use is known as metallurgy. Metals are refined by using the method of electrolytic refining. The end topic explains corrosion and how it can be prevented.

Topics Covered in Class 10 Science Chapter 3 Metals and Non-metals :

Metals and nonmetals: Properties of metals and non-metals; Reactivity series; Formation and properties of ionic compounds; Basic metallurgical processes; Corrosion and its prevention.

Also, access the following resources for Class 10 Chapter 3 Metals and Non-metals, at BYJU’S:

• CBSE Class 10 Chapter 3 Metals and Non-metals Notes
• Revision Notes For Class 10 Science Chapter 3 – Metals and Non-metals
• NCERT Exemplar Class 10 Science Solutions for Chapter 3 – Metals And Non Metals

Chapter 4 – Carbon and its Compounds

In the previous chapter, we discussed compounds of importance to us. In this chapter, we will study some more interesting compounds and their properties. Also, we shall be learning about carbon, an element which is of immense significance to us in both its elemental form and in the combined form. Carbon is a versatile element that forms the basis for all living organisms and many of the things we use. Covalent bonds are formed by the sharing of electrons between two atoms so that both can achieve a completely filled outermost shell. Carbon forms covalent bonds with itself and other elements such as hydrogen, oxygen, sulphur, nitrogen and chlorine. Organic compounds are categorized into saturated and unsaturated carbon compounds. Saturated compounds are compounds with only a single bond. Unsaturated carbon compounds are compounds with a double or triple bond. The saturated compounds of carbon and hydrogen are methane, ethane, propane, butane, pentane and hexane.

The chapter also explains some of the chemical properties of carbon and its compounds such as combustion, oxidation, addition reaction, substitution reaction. Ethanol and ethanoic acid are carbon compounds of importance in our daily lives. Soap and detergents are studied with their chemical structures and properties and their difference is also discussed. Detergents are usually used to make shampoos and products for cleaning clothes.

Topics Covered in Class 10 Science Chapter 4 Carbon and its Compounds :

Carbon compounds: Covalent bonding in carbon compounds. Versatile nature of carbon. Homologous series. Nomenclature of carbon compounds containing functional groups (halogens, alcohol, ketones, aldehydes, alkanes and alkynes), difference between saturated hydro carbons and unsaturated hydrocarbons. Chemical properties of carbon compounds (combustion, oxidation, addition and substitution reaction). Ethanol and Ethanoic acid (only properties and uses), soaps and detergents.

Also, access the following resources for Class 10 Chapter 4 Carbon and its Compounds at BYJU’S:

• CBSE Class 10 Science Chapter 4 Carbon and its Compounds Notes
• NCERT Exemplar Class 10 Science Solutions for Chapter 4 – Carbon And Its Compounds

Chapter 5 – Periodic Classification of Elements

In Standard 9 we have learned that matter around us is present in the form of elements, compounds and mixtures, and the elements contain atoms of only one type. The early attempts at the classification of elements resulted in grouping the then-known elements as metals and non-metals. Dobereiner grouped the elements into triads and Newlands gave the Law of Octaves. Mendeleev arranged the elements in increasing order of their atomic masses and according to their chemical properties. He even predicted the existence of some yet-to-be-discovered elements on the basis of gaps in his Periodic table. The modern periodic table came into existence. Mendeleev’s Periodic Law was modified, and the atomic number was adopted as the basis of the Modern Periodic Table and the Modern Periodic Law can be stated as follows: ‘Properties of elements are a periodic function of their atomic number’.

In this chapter, Periodic Classification of Elements, elements in the Modern Periodic Table are arranged in 18 vertical columns called groups and 7 horizontal rows called periods. Elements thus arranged show periodicity of properties including atomic size, valency or combining capacity and metallic and non-metallic character. The valency of an element is determined by the number of valence electrons present in the outermost shell of its atom. The term atomic size refers to the radius of an atom.

Also, access the following resources for Class 10 Chapter 5 Periodic Classification of Elements at BYJU’S:

• CBSE Class 10 Chapter 5 Periodic Classification of Elements Notes
• Revision Notes For Class 10 Science Chapter 5 – Periodic Classification of Elements
• Chapter 5 – Periodic Classification of Elements
• NCERT Exemplar Class 10 Science Solutions for Chapter 5 – Periodic Classification Of Elements

Chapter 6 – Life Processes

NCERT Class 10 Science Chapter 6 explains Life Processes . There are six life processes that all living organisms perform. They are movement, respiration, growth, reproduction, excretion and nutrition. The chapter also teaches about nutrition which means the process of taking in food and using it for growth, metabolism and repair. Nutritional stages are ingestion, digestion, absorption, transport, assimilation, and excretion. Nutrition is further divided into Autotrophic Nutrition and Heterotrophic Nutrition. Autotrophic nutrition involves the intake of simple inorganic materials from the environment and using an external energy source like the Sun to synthesize complex high-energy organic material. Heterotrophic nutrition involves the intake of complex material prepared by other organisms. Different types of heterotrophic nutrition are parasitic nutrition, saprophytic nutrition and holozoic nutrition. The next topic is nutrition in human beings. The various steps of nutrition are ingestion, digestion, Oesophagus, stomach, small intestine, bile, absorption, assimilation and egestion. The next subtopic is respiration in which the human respiratory system is explained beautifully. The different elements of the human respiratory system are lungs, bronchi, larynx, pharynx, etc. During the process of respiration, organic compounds such as glucose are broken down to provide energy in the form of ATP. ATP is used to provide energy for other reactions in the cell.

Respiration may be aerobic or anaerobic. Aerobic respiration makes more energy available to the organism. For plants, the soil is the nearest and richest source of raw materials like nitrogen, phosphorus and other minerals. In human beings, the transport of materials such as oxygen, carbon dioxide, food and excretory products is a function of the circulatory system. The circulatory system consists of the heart, blood and blood vessels. In highly differentiated plants, transport of water, minerals, food and other materials is a function of the vascular tissue which consists of xylem and phloem. In human beings, excretory products in the form of soluble nitrogen compounds are removed by the nephrons in the kidneys. Plants use a variety of techniques to get rid of waste material.

Topics Covered in Class 10 Science Chapter 6 Life Processes :

Life processes: ‘Living Being’. Basic concept of nutrition, respiration, transport and excretion in plants and animals.

Also access the following resources for Class 10 Chapter 6 Life Processes at BYJU’S:

• CBSE Class 10 Science Notes Chapter 6 Life Processes
• NCERT Exemplar Class 10 Science Solutions for Chapter 6 – Life Processes

Chapter 7 – Control and Coordination

Chapter 7 of Class 10 teaches about control and coordination , which are the functions of the nervous system and hormones in our bodies. The responses of the nervous system can be classified as a reflex action, voluntary action or involuntary action. The nervous system uses electrical impulses to transmit messages. It gets information from our sense organs and acts through our muscles. Chemical coordination is seen in both plants and animals. Hormones produced in one part of an organism move to another part to achieve the desired effect. A feedback mechanism regulates the action of the hormones.

Topics Covered in Class 10 Science Chapter 7 Control and Coordination :

Control and co-ordination in animals and plants: Tropic movements in plants; Introduction of plant hormones; Control and co-ordination in animals: Nervous system; Voluntary, involuntary and reflex action; Chemical co-ordination: animal hormones.

Also, access the following resources for Class 10 Chapter 7 Control and Coordination at BYJU’S:

• CBSE Class 10 Science Chapter 7 Control and Coordination Notes
• NCERT Exemplar Class 10 Science Solutions for Chapter 7 – Control And Coordination

Chapter 8 – How do Organisms Reproduce

Unlike other life processes, reproduction is not essential to maintain the life of an individual organism. How do Organisms Reproduce chapter involves the creation of a DNA copy and additional cellular apparatus by the cell involved in the process. Depending on their body design, various organisms use different modes of reproduction. In fission, many bacteria and protozoa simply divide into two or more daughter cells. Organisms such as hydra can regenerate if they are broken into pieces. They can also give out buds which mature into new individuals. Roots, stems and leaves of some plants develop into new plants through vegetative propagation. These are examples of asexual reproduction where new generations are created from a single individual. Sexual reproduction involves two individuals for the creation of a new individual. DNA copying mechanisms create variations which are useful for ensuring the survival of the species. Modes of sexual reproduction allow for greater variation to be generated.

Reproduction in flowering plants involves the transfer of pollen grains from the anther to the stigma which is referred to as pollination. This is followed by fertilisation. Changes in the body during puberty, such as an increase in breast size in girls and new facial hair growth in boys, are signs of sexual maturation. The male reproductive system in human beings consists of testes which produce sperms, vas deferens, seminal vesicles, prostate gland, urethra and penis. The female reproductive system in human beings consists of ovaries, fallopian tubes, uterus and vagina. Sexual reproduction in human beings involves the introduction of sperm in the vagina of the female. Fertilisation occurs in the fallopian tube. Contraception to avoid pregnancy can be achieved by the use of condoms, oral pills, copper -T and other methods.

Topics Covered in Class 10 Science Chapter 8 How Do Organisms Reproduce :

Reproduction: Reproduction in animals and plants (asexual and sexual) reproductive health – need and methods of family planning. Safe sex vs HIV/AIDS. Child bearing and women’s health.

Also, access the following resources for Class 10 Chapter 8 How do Organisms Reproduce at BYJU’S:

• CBSE Class 10 Science Chapter 8 How Do Organisms Reproduce Notes
• Chapter 8 -How do Organisms Reproduce?
• NCERT Exemplar Class 10 Science Solutions for Chapter 8 – How Do Organisms Reproduce

Chapter 9 – Heredity And Evolution

In this chapter, we will learn about Heredity and Evolution . We have seen that reproductive processes give rise to new individuals that are similar, but subtly different. We have discussed how some amount of variation is produced even during asexual reproduction. The Rules for the Inheritance of Traits in human beings relate to the fact that both the father and the mother contribute practically equal amounts of genetic material to the child. This means that each trait can be influenced by both paternal and maternal DNA. Sex can be determined by different factors in various species. Changes in the non-reproductive tissues caused by environmental factors are not inheritable. Speciation may take place when the variation is combined with geographical isolation. Evolutionary relationships are traced in the classification of organisms. Tracing common ancestors back in time leads us to the idea that at some point in time, non-living material must have given rise to life.

Evolution can be worked out by the study of not just living species, but also fossils. Complex organs may have evolved because of the survival advantage of even the intermediate stages. Organs or features may be adapted to new functions during the course of evolution. Evolution cannot be said to progress from lower forms to higher forms. Rather, evolution seems to have given rise to more complex body designs even while the simpler body designs continue to flourish. Study of the evolution of human beings indicates that all of us belong to a single species that evolved in Africa and spread across the world in stages.

Topics Covered in Class 10 Science Chapter 9 Heredity and Evolution :

Heredity and Evolution: Heredity; Mendel’s contribution – Laws for inheritance of traits: Sex determination: brief introduction: (topics excluded – evolution; evolution and classification and evolution should not be equated with progress).

Also, access the following resources for Class 10 Chapter 9 Heredity and Evolution, at BYJU’S:

• CBSE Class 10 Science Chapter 9 Heredity And Evolution Notes
• NCERT Exemplar Class 10 Science Solutions for Chapter 9 – Heredity And Evolution

Chapter 10 – Light Reflection and Refraction

In  NCERT Class 10 Science Chapter 10, we will study the phenomena of reflection and refraction of light using the straight-line propagation of light. These basic concepts will help us in the study of some of the optical phenomena in nature. The chapter also discusses the reflection of light by spherical mirrors and refraction of light and their application in real life. Light is a source of energy which generates a sensation of vision in human beings. Light seems to travel in straight lines. The different types of a spherical mirror, convex and concave are taught. The various terms related to spherical mirrors like the centre of curvature, the radius of curvature, etc., focus, pole, etc. are discussed with ray diagrams. Uses of a spherical mirror are also discussed in this chapter. Mirror formula gives the relationship between the object-distance, image-distance, and focal length of a spherical mirror. The focal length of a spherical mirror is equal to half its radius of curvature.

Refraction is the bending of a wave when it enters a medium where its speed is different. The refraction of light when it passes from a fast medium to a slow medium bends the light rays toward the normal to the boundary between the two media. The phenomena of refraction can be understood easily by the concepts of the refractive index and optical density. The refractive index of a transparent medium is the ratio of the speed of light in a vacuum to that in the medium. In case of a rectangular glass slab, the refraction takes place at both the air-glass interface and glass-air interface. The emergent ray is parallel to the direction of the incident ray. Lens formula gives the relationship between the object-distance, image-distance, and the focal length of a spherical lens. Power of a lens is the reciprocal of its focal length. The SI unit of power of a lens is dioptre.

Topics Covered in Class 10 Science Chapter 10 Light Reflection and Refraction :

Reflection of light by curved surfaces; Images formed by spherical mirrors, centre of curvature, principal axis, principal focus, focal length, mirror formula (Derivation not required),magnification. Refraction; Laws of refraction, refractive index. Refraction of light by spherical lens; Image formed by spherical lenses; Lens formula(Derivation not required); Magnification. Power of a lens.

Also, access the following resources for Class 10 Chapter 10 Light Reflection and Refraction at BYJU’S:

• CBSE Class 10 Science Chapter 10 Light – Reflection and Refraction Notes
• Chapter 10 -Light: Reflection and Refraction
• NCERT Exemplar Class 10 Science Solutions for Chapter 10 – Light Reflection And Refraction

Chapter 11 – The Human Eye and Colorful World

In the previous chapter, we learnt about light and some of its properties. In this chapter, we will study some of the optical phenomena in nature. The chapter also discusses the rainbow formation, splitting of white light and blue colour of the sky. The human eye is one of the most valuable and sensitive sense organs. It enables us to see the wonderful world and the colours around us. The ability of the eye to focus on both near and distant objects, by adjusting its focal length, is called the accommodation of the eye. The smallest distance, at which the eye can see objects clearly without strain, is called the near point of the eye or the least distance of distinct vision. For a young adult with normal vision, it is about 25cm. The common refractive defects of vision include myopia, hypermetropia and presbyopia. Myopia, short-sightedness-the image of distant objects is focused before the retina is corrected by using a concave lens of suitable power. Hypermetropia (far-sightedness-the image of nearby objects is focussed beyond the retina) is corrected by using a convex lens of suitable power. The eye loses its power of accommodation at old age. The splitting of white light into its component colours is called dispersion. Scattering of light causes the blue colour of the sky and the reddening of the Sun at sunrise and sunset.

Topics Covered in Class 10 Science Chapter 11 The Human Eye and Colorful World :

Functioning of a lens in human eye, defects of vision and their corrections, applications of spherical mirrors and lenses. Refraction of light through a prism, dispersion of light, scattering of light, applications in daily life (excluding colour of the sun at sunrise and sunset).

Also, access the following resources for Class 10 Chapter 11 The Human Eye and Colorful World, at BYJU’S:

• CBSE Class 10 Science Chapter 11 The Human Eye and the Colourful World Notes
• NCERT Exemplar Class 10 Science Solutions for Chapter 11 – Human Eye And Colourful World

Chapter 12 – Electricity

Electricity has an important place in modern society. It is a controllable and convenient form of energy for a variety of uses in homes, schools, hospitals, industries and so on. It is a phenomenon related to the flow of charge. A stream of electrons moving through a conductor constitutes an electric current. Conventionally, the direction of current is taken opposite to the direction of flow of electrons. The SI unit of electric current is ampere. To set the electrons in motion in an electric circuit, we use a cell or a battery. A cell generates a potential difference across its terminals. It is measured in volts (V). Resistance is a property that resists the flow of electrons in a conductor. It controls the magnitude of the current. The SI unit of resistance is Ohm. Ohm’s law: the potential difference across the ends of a resistor is directly proportional to the current through it, provided its temperature remains the same. The resistance of a conductor depends directly on its length, inversely on its areas of cross-section, and also on the material of the conductor. The equivalent resistance of several resistors in series is equal to the sum of their individual resistances. The electrical energy dissipated in a resistor is given by W=V x I x t. The unit of power is watt (W). One watt of power is consumed when 1 A of current flows at a potential difference of 1 V. The commercial unit of electrical energy is kilowatt-hour (kWh). 1kW h = 3,6000,000 J = 3.6 x 10 6 J.

Topics Covered in Class 10 Science Chapter 12 Electricity :

Electric current, potential difference and electric current. Ohm’s law; Resistance, Resistivity, Factors on which the resistance of a conductor depends. Series combination of resistors, parallel combination of resistors and its applications in daily life. Heating effect of electric current and its applications in daily life. Electric power, Interrelation between P, V, I and R.

Also, access the following resources for Class 10 Chapter 12 Electricity at BYJU’S:

• Electricity Class 10 Notes
• NCERT Exemplar Class 10 Science Solutions for Chapter 12 – Electricity

Chapter 13 – Magnetic Effects of Electric Current

In this chapter, students will study magnetic fields and such electromagnetic effects, along with electromagnets and electric motors, which involve the magnetic effect of electric current, and electric generators, which involve the electric effect of moving magnets. A compass needle is a small magnet. Its one end, which points towards the north, is called a north pole, and the other hand, which points towards the south, is called a south pole. A magnetic field exists in the region surrounding a magnet in which the force of the magnet can be detected. Field lines are used to represent a magnetic field. A field line is a path along which a hypothetical free north pole would tend to move. The direction of the magnetic field at a point is given by the direction that a north pole placed at that point would take. Field lines are shown closer together where the magnetic field is greater. A metallic wire carrying an electric current has associated with it a magnetic field. The field lines about the wire consist of a series of concentric circles whose direction is given by the right-hand rule. The pattern of the magnetic field around a conductor due to an electric current flowing through it depends on the shape of the conductor. The magnetic field of a solenoid carrying a current is similar to that of a bar magnet. An electromagnet consists of a core of soft iron wrapped around a coil of insulated copper wire. A current-carrying conductor, when placed in a magnetic field, experiences a force. If the direction of the field and that of the current are mutually perpendicular to each other, then the force acting on the conductor will be perpendicular to both and will be given by Fleming’s left-hand rule. This is the basis of an electric motor. An electric motor is a device that converts electric energy into mechanical energy.

The phenomenon of electromagnetic induction is the production of induced current in a coil placed in a region where the magnetic field changes with time. The magnetic field may change due to relative motion between the coil and a magnet placed near to the coil. If the coil is placed near a current-carrying conductor, the magnetic field may change either due to a change in the current through the conductor or due to the relative motion between the coil and conductor, the magnetic field may change either due to a change in the current through the conductor or due to the relative motion between the coil and the conductor. The direction of the induced current is given by Fleming’s right-hand rule. A generator converts mechanical energy into electrical energy. It works on the basis of electromagnetic induction. There are 2 types of generator AC and DC generator. Fuse is the most important safety device, used for protecting the circuits due to short-circuiting or overloading of the circuits.

Topics Covered in Class 10 Science Chapter 13 Magnetic Effects of Electric Current:

Magnetic effects of current: Magnetic field, field lines, field due to a current carrying conductor, field due to current carrying coil or solenoid; Force on current carrying conductor, Fleming’s Left Hand Rule, Direct current. Alternating current: frequency of AC. Advantage of AC over DC. Domestic electric circuits.

Also, access the following resources for Class 10 Chapter 13 Magnetic Effects of Electric Current at BYJU’S:

• CBSE Class 10 Science Chapter 13 Magnetic Effects of Electric Current Notes
• Revision Notes For Class 10 Science Chapter 13 – Magnetic Effects of Electric Current
• NCERT Exemplar Class 10 Science Solutions for Chapter 13 – Magnetic Effects Of Electric Current

Chapter 14 – Sources of Energy

Our energy requirements increase with our standard of living. In order to fulfil our energy requirements, we try to improve the efficiency of energy usage and also try and exploit new sources of energy. The chapter discusses different sources of energy, and they are conventional sources of energy, which we keep on using for many years. It includes fossil fuels, thermal power plants and hydropower plants. The advantages and disadvantages are also discussed. After those improvements in the technology for using conventional sources of energy is also discussed such as Biomass and wind energy. Next, students will study the topic of alternative or non-conventional energy resources. It includes solar energy, in which energy is generated through the solar cell and solar panel. Energy can be generated from the sea, such as Tidal energy, wave energy and ocean thermal energy. Energy can also be generated from the earth’s crust, known as geothermal energy . Nuclear energy is energy in the nucleus (core) of an atom. The energy source we select would depend on factors like the ease and cost of extracting energy from the source, the efficiency of the technology available for using that source of energy and the environmental impact of using that source. Many of the sources ultimately derive their energy from the Sun. All the topics are explained with the advantages and disadvantages of it.

Also, access the following resources for Class 10 Chapter 14 Sources of Energy at BYJU’S:

• CBSE Class 10 Science Chapter 14 Sources of Energy Notes
• NCERT Exemplar Class 10 Science Solutions for Chapter 14 – Sources Of Energy

Chapter 15 – Our Environment

This chapter discusses how various components in the environment interact with each other and how we impact the environment. The various components of an ecosystem are interdependent. The producers make the energy from sunlight available to the rest of the ecosystem. There is a loss of energy as we go from one trophic level to the next, this limits the number of trophic levels in a food chain. The food-chain is explained in detail with examples according to nature, such as in the forest, in grassland and in the pond. Human activities have an impact on the environment. The use of chemicals like CFCs has endangered the ozone layer. Since the ozone layer protects against the ultraviolet radiation from the Sun, this could damage the environment. The waste we generate may be biodegradable or non-biodegradable . The disposal of the waste we generate is causing serious environmental problems.

Topics Covered in Class 10 Science Chapter 15 Our Environment :

Our environment: Eco-system, Environmental problems, Ozone depletion, waste production and their solutions. Biodegradable and non-biodegradable substances.

Also, access the following resources for Class 10 Chapter 15 Our Environment at BYJU’S:

• CBSE Class 10 Chapter 15 Our Environment Notes
• Revision Notes For Class 10 Science Chapter 15 – Our Environment
• Chapter 15 -Our Environment
• NCERT Exemplar Class 10 Science Solutions for Chapter 15 – Our Environment

Chapter 16 – Sustainable Management of Natural Resources

In the previous class, we have learned about some natural resources like soil, air and water and how various components are cycled over and over again in nature. In this chapter, we will look at some of our resources and how we are using them. Our resources, like forests, wildlife, water, coal and petroleum, need to be used in a sustainable manner. We can reduce pressure on the environment by sincerely applying the maxim of ‘Refuse, Reduce, Reuse, Repurpose and Recycle’ in our lives. Management of forest resources has to take into account the interests of various stakeholders. The harnessing of water resources by building dams has social, economic and environmental implications. Alternatives to large dams exist. These are locale-specific and may be developed so as to give local people control over their local resources. The fossil fuels, coal and petroleum, will ultimately be exhausted. Owing to this and because their combustion pollutes our environment, we need to use these resources judiciously.

Also, access the following resources for Class 10 Chapter 16 Sustainable Management of Natural Resources at BYJU’S:

• CBSE Class 10 Science Chapter 16 Sustainable Management of Natural Resources Notes
• Revision Notes For Class 10 Science Chapter 16 – Sustainable Management of Natural Resources
• NCERT Exemplar Class 10 Science Solutions for Chapter 16 – Management Of Natural Resources

CBSE Class 10 Science Evaluation Scheme (Theory) –

How NCERT Solutions Class 10 Science is helpful while preparing for the exam?

Class 10 Science is an important subject for those students who want to pursue their future studies in this field. To score good marks in this subject, students need to follow the NCERT textbook of Class 10 Science and should be thorough with it. For each chapter, there will be exercise questions for practice which students need to write after they finish completing each topic. It will help them to revise the topics and get to know how much they have understood the concepts. After completing the exercise question, they can refer to the NCERT Solutions of Class 10 Science to cross-check whether they answered all the questions correctly. These solutions work as a guide for the students so that they don’t repeat the mistakes and make sure they answer all the questions correctly.

Preparing for the Class 10 exam needs a lot of attention and commitment as it is considered the turning point of the educational journey. Students need to have an overall understanding of individual chapters, and the process of it demands hard work towards studies and an effective approach to getting through the solutions. A significant role is played by Class 10 NCERT Science Solutions, which helps them to prepare effectively for their Class 10 board exams.

These solutions are based on the CBSE syllabus of Class 10 Science, which provides solutions to all the exercise questions of each chapter mentioned in the NCERT textbook of Class 10 Science. By referring to the solutions, students get to know which topic to focus more upon which will help students to learn faster. Students must know the right technique to answer all the questions given in the NCERT textbook of Class 10 Science . So, in order to help them prepare for their exams, we have provided all the chapter-wise NCERT Solutions. Students can also refer to the NCERT Workbook Solutions Class 10 Science for their further preparation.

Features of NCERT Class 10 Science Solutions

NCERT Science Solutions Class 10 is the best resource study material for students as it delivers a wide range of solutions to all the NCERT questions in the syllabus. It gives detailed solutions to the three dimensions of science, which are Physics, Chemistry and Biology. With the help of the solutions, students can instantly solve their doubts. These solutions are designed as per the NCERT curriculum to help students prepare for their Class 10 board exam. Some of the features of NCERT Solutions of Class 10 Science  are mentioned below:

• The solutions are solved in easy-to-understand language so that students don’t get confused while referring to them.
• The answers are explained elaborately for all the exercise questions mentioned in each chapter.
• Solving these solutions will help students to solve the sample papers as well as the previous year question papers.
• It boosts the student’s confidence level and also helps them work on their weak points.
• All the answers are explained with proper diagrams related to the question.

Other NCERT Resources for Class 10 Science

Students of Class 10 should refer to other study materials like NCERT Class 10 Science Exemplar, NCERT Class 10 Science textbook, and NCERT Class 10 Science Syllabus besides studying from NCERT Class 10 Science Solutions. These study materials are also prepared as per the CBSE Class 10 science syllabus and help them while preparing for the board exams. Students can refer to these study materials after they complete the entire syllabus, which will help in quick revision before the exam.

Below we have provided the links of the other NCERT Resources for Class 10 Science to help you prepare effectively for the Class 10 board exam.

Download NCERT Solutions Class 10 Science Book APP

Students of Class 10 can download the NCERT Class 10 chapter-wise Science solutions from the BYJU’S app, which will help them while preparing for their exam. Apart from the NCERT Solutions for Class 10 , students can also utilize the BYJU’S App for other study materials such as previous year question papers, syllabus, important questions, etc. The BYJU’S App will make your learning easier as you can access it from anywhere you want, by downloading it on your smart device. The solutions of NCERT Class 10 Science are prepared by our highly experienced subject experts, as per the latest CBSE Class 10 Science syllabus.

Frequently Asked Questions on NCERT Solutions for Class 10 Science

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