case study questions for class 8 maths chapter 2

• Linear Equations in One Variable Class 8 Case Study Questions Maths Chapter 2

Last Updated on April 29, 2024 by XAM CONTENT

Hello students, we are providing case study questions for class 8 maths. Case study questions are the new question format that is introduced in CBSE board. The resources for case study questions are very less. So, to help students we have created chapterwise case study questions for class 8 maths. In this article, you will find case study questions for CBSE Class 8 Maths Chapter 2 Linear Equations in One Variable. It is a part of Case Study Questions for CBSE Class 8 Maths Series.

Table of Contents

Case Study Questions on Linear Equations in One Variable

Passage 1: It is common that government revises fares from time to time based on various factors such as taxes, economy and inflation, for various vehicles like auto-rickshaw, taxis and radio cab etc. The auto and Taxi charge in a city comprise of fixed charge and the charge for the distance covered. Few situations are given below in the form questions. Find the correct option.

Difficulty Level: Medium

Q. 1. If the fixed charge in a city is ₹ x and charge per km is ₹5 and total fare is ₹60 then find the linear equation for the journey of 10 km. (a) x + 50 = 60 (b) x – 50 = 60 (c) x + 50 = 50 (d) None of these

Q. 2. In the above question, what is the value of fixed charge? (a) ₹20 (b) ₹5 (c) ₹10 (d) ₹15

Q. 3. If in a city a person has to pay ₹110 for a journey of 15 km and fixed charge is ₹20 then what is the charge per km is? (a) ₹12 (b) ₹6 (c) ₹8 (d) no fixed charge

Q. 4. If in a city fixed charge is double of the charge per km and a person paid ₹75 for a journey of 1 km, then the linear equation for the following situation is?

Q. 5. According to the given equation: 2x + 17 = 85, if ₹17 is the fixed charge and the total fare is ₹85 for a journey of 2km then what is the charge per kilometre?

Understanding Quadrilaterals Class 8 Case Study Questions Maths Chapter 3

Rational numbers class 8 case study questions maths chapter 1, topics from which case study questions may be asked.

• Linear Equation in One Variable
• Solving an Equation
• Solving Equation having variable on both sides
• Reducing Equation to Simpler Form

Case study questions from the above given topic may be asked.

Frequently Asked Questions (FAQs) on Linear Equations in One Variable Case Study

Q1: what is an algebraic expression.

A1: A combination of constants and variables connected by some or all of the four fundamental operators such as +, –, × and ÷ is called algebraic expression.

Q2: What is an equation?

A2: An equation is like a scale, while manipulating equations we have to keep the scale balanced by doing the same thing to both sides.

Q3: What is linear equation in one variable?

A3: An equation is called linear equation if it has only one degree i.e., the highest power of the variable appearing in equation is 1. e.g., ax + b = 0

Q4: What is solution of an equation?

A4: A number which satisfies an equation is called the solution of the equation.

Q5: How do you solve a linear equation in one variable?

A5: To solve a linear equation in one variable, isolate the variable on one side of the equation by performing inverse operations (such as addition, subtraction, multiplication, or division) on both sides until the variable is alone.

Q6: What is the solution of a linear equation?

A6: The solution of a linear equation is the value of the variable that satisfies the equation. It is the value that, when substituted into the equation, makes the equation true.

Q7: What does the term ‘one variable’ mean in linear equations?

A7: ‘One variable’ means that the equation contains only a single unknown quantity, typically represented by the letter ‘x’. Other terms or constants may be present, but there is only one variable.

Q8: Can linear equations be represented graphically?

A8: Yes, linear equations can be represented graphically as straight lines on the coordinate plane. The slope-intercept form (y = mx + b) is particularly useful for graphing linear equations.

Q9: Are there any online resources or tools available for practicing linear equations in one variable case study questions?

A9: We provide case study questions for CBSE Class 8 Maths on our  website . Students can visit the website and practice sufficient case study questions and prepare for their exams. If you need more case study questions, then you can visit  Physics Gurukul  website. they are having a large collection of case study questions for all classes.

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Chapter 2 Class 8 Linear Equations in One Variable

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We studied Equations and Expressions in Algebra ( Chapter 9 Class 8 ). Equation has an = sign, expression does not have it

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• What is a Linear Equation
• What is a Linear Equation in one variable
• Solving Linear Equations - by taking variables on one side, numbers on the side
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• Questions involving Age

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Home » 8th Class » NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable (PDF)

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable (PDF)

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable has been published by Aglasem. You can now download the Class 8 Maths Ch 2 Questions and Answers PDF here. This NCERT Solutions for Class 8 Maths contains answers of all questions asked in Chapter 2 in textbook, Mathematics . Therefore you can refer it to solve Linear Equations in One Variable exercise questions and learn more about the topic.

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable

Class – Class 8 Subject – Maths Chapter – Ch 2 Chapter Name – Linear Equations in One Variable Book – Mathematics Study Material – NCERT Solutions

NCERT Solutions for Class 8 Maths Chapter 2 PDF

While you can read NCERT Solutions for Class 8 Maths Ch 2 for all exercises here on aglasem. You can also download this NCERT Solutions PDF to refer at anytime when you study Linear Equations in One Variable. Here it is.

NCERT Solutions for Class 8 Maths Chapter 2 PDF Download Link – Click Here to Download Solutions PDF

How to download NCERT Solutions for Class 8 Maths Chapter 2 PDF?

You can download the complete NCERT solutions for chapter 2 of this NCERT Book i.e. Mathematics with following steps.

• First search NCERT Solutions for Class 8 Maths Ch 2 PDF aglasem and come to this page.
• Now you will see the exercise questions answers of Linear Equations in One Variable and download pdf link on it.
• Click the Download PDF link to obtain the Linear Equations in One Variable questions with answers document.

NCERT Solutions for Class 8 Maths

There are more chapters to study besides Linear Equations in One Variable in this subject. So here are NCERT solutions for all topics of Maths taught in 8th class here at aglasem.

• Chapter 1 Rational Numbers
• Chapter 2 Linear Equations in One Variable
• Chapter 3 Understanding Quadrilaterals
• Chapter 4 Data Handling
• Chapter 5 Squares and Square Roots
• Chapter 6 Cubes and Cube Roots
• Chapter 7 Comparing Quantities
• Chapter 8 Algebraic Expressions and Identities
• Chapter 9 Mensuration
• Chapter 10 Exponents and Powers
• Chapter 11 Direct and Inverse Proportions
• Chapter 12 Factorisation
• Chapter 13 Introduction to Graphs

NCERT Solutions for Class 8

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NCERT Solutions for Class 8 Maths Chapter 2 – An Overview

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NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable

February 24, 2021 by Sastry CBSE

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.1

• Class 8 Maths Linear Equations in One Variable Exercise 2.1
• Class 8 Maths Linear Equations in One Variable Exercise 2.2
• Class 8 Maths Linear Equations in One Variable Exercise 2.3
• Class 8 Maths Linear Equations in One Variable Exercise 2.4
• Class 8 Maths Linear Equations in One Variable Exercise 2.5
• Class 8 Maths Linear Equations in One Variable Exercise 2.6
• Linear Equations in One Variable Class 8 Extra Questions

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.1

Ex 2.1 Class 8 Maths Question 1. Solve the equation: x – 2 = 7. Solution: Given: x – 2 = 7 ⇒ x – 2 + 2 = 7 + 2 (adding 2 on both sides) ⇒ x = 9 (Required solution)

Ex 2.1 Class 8 Maths Question 2. Solve the equation: y + 3 = 10. Given: y + 3 = 10 ⇒ y + 3 – 3 = 10 – 3 (subtracting 3 from each side) ⇒ y = 7 (Required solution)

Ex 2.1 Class 8 Maths Question 3. Solve the equation: 6 = z + 2 Solution: We have 6 = z + 2 ⇒ 6 – 2 = z + 2 – 2 (subtracting 2 from each side) ⇒ 4 = z Thus, z = 4 is the required solution.

Ex 2.1 Class 8 Maths Question 5. Solve the equation 6x = 12. Solution: We have 6x = 12 ⇒ 6x ÷ 6 = 12 ÷ 6 (dividing each side by 6) ⇒ x = 2 Thus, x = 2 is the required solution.

Ex 2.1 Class 8 Maths Question 6. Solve the equation \(\frac { t }{ 5 }\) = 10. Solution: Given \(\frac { t }{ 5 }\) = 10 ⇒ \(\frac { t }{ 5 }\) × 5 = 10 × 5 (multiplying both sides by 5) ⇒ t = 50 Thus, t = 50 is the required solution.

Ex 2.1 Class 8 Maths Question 7. Solve the equation \(\frac { 2x }{ 3 }\) = 18. Solution: We have \(\frac { 2x }{ 3 }\) = 18 ⇒ \(\frac { 2x }{ 3 }\) × 3 = 18 × 3 (multiplying both sides by 3) ⇒ 2x = 54 ⇒ 2x ÷ 2 = 54 ÷ 2 (dividing both sides by 2) ⇒ x = 27 Thus, x = 27 is the required solution.

Ex 2.1 Class 8 Maths Question 8. Solve the equation 1.6 = \(\frac { y }{ 1.5 }\) Solution: Given: 1.6 = \(\frac { y }{ 1.5 }\) ⇒ 1.6 × 1.5 = \(\frac { y }{ 1.5 }\) × 1.5 (multiplying both sides by 1.5) ⇒ 2.40 = y Thus, y = 2.40 is the required solution.

Ex 2.1 Class 8 Maths Question 9. Solve the equation 7x – 9 = 16. Solution: We have 7x – 9 = 16 ⇒ 7x – 9 + 9 = 16 + 9 (adding 9 to both sides) ⇒ 7x = 25 ⇒ 7x ÷ 7 = 25 ÷ 7 (dividing both sides by 7) ⇒ x = \(\frac { 25 }{ 7 }\) Thus, x = \(\frac { 25 }{ 7 }\) is the required solution.

Ex 2.1 Class 8 Maths Question 10. Solve the equation 14y – 8 = 13. Solution: We have 14y – 8 = 13 ⇒ 14y – 8 + 8 = 13 + 8 (adding 8 to both sides) ⇒ 14y = 21 ⇒ 14y ÷ 14 = 21 ÷ 14 (dividing both sides by 14) ⇒ y = \(\frac { 21 }{ 14 }\) ⇒ y = \(\frac { 3 }{ 2 }\) Thus, y = \(\frac { 3 }{ 2 }\) is the required solution.

Ex 2.1 Class 8 Maths Question 11. Solve the equation 17 + 6p = 9. Solution: We have, 17 + 6p = 9 ⇒ 17 – 17 + 6p = 9 – 17 (subtracting 17 from both sides) ⇒ 6p = -8 ⇒ 6p ÷ 6 = -8 ÷ 6 (dividing both sides by 6) ⇒ p = \(\frac { -8 }{ 6 }\) ⇒ p = \(\frac { -4 }{ 3 }\) Thus, p = \(\frac { -4 }{ 3 }\) is the required solution.

More CBSE Class 8 Study Material

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NCERT Solutions for Class 8 Maths Chapter 2 – Linear Equations in One Variable

First of all, NCERT solutions for class 8 Maths Chapter 2 is an excellent helping tool. Experts of high-quality prepare these solutions. Most noteworthy, these experts have superior skills and rich experience in Maths. The experts make these solutions student-friendly. Difficult concepts are explained in an easy manner. There is the breaking of difficult parts into smaller parts.

One efficient way to learn NCERT solutions is by Toppr. It’s certainly rare to find such a well-equipped learning platform. Toppr helps students gain maximum advantage from NCERT solutions.

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Download NCERT Solutions for  Class 8 maths chapterwise here .

CBSE Class 8 Maths Chapter 2 Linear Equations in One Variable NCERT Solutions

NCERT solutions for class 8 maths chapter 2 deal with linear expression in one variable. A Linear Equation is an equation of a straight line and written in one variable. Furthermore, Linear Equations in one variable may assume the form ax + b = 0. Their solution is with the help of basic algebraic expressions. The most noteworthy, Linear equation is of three types.

Sub-topics covered under NCERT Solutions for Class 8 Maths Chapter 2

• Ex. 2.1 Introduction
• Ex. 2.2 Solving Equations which have Linear Expressions on one Side and Numbers on the other side.
• Ex. 2.3 Some Applications
• Ex. 2.4 Solving Equations having the Variable on both sides
• Ex. 2.5 Some more applications
• Ex. 2.6 Reducing Equation to Simpler to a simpler form
• Ex. 2.7 Equations reducible to the Linear form.

NCERT Solutions for Class 8 Maths Chapter 2

Students probably know that an algebraic equation is an equality involving variables. It certainly has an equality sign. The expression on the left of this equality sign is the Left-hand side. Similarly, the expression on the right of equality sign is the Right-hand side. In an equation the values of expressions on Right and Left-hand side are equal. However, this is true only for certain values. The solution of the equation are these values. Linear equations form a large of Algebra. Hence, this chapter will help students with a lot of other Algebra topics. Let us see what will we learn from each chapter.

Ex. 2.1 Introduction– First of all, a general idea of Linear equations. The students get basic information of Linear Equations in One Variable.

Ex. 2.2 Solving Equations which have Linear Expressions on one Side and Numbers on the other side– This is an important part.

Following is an example from this topic –

Solve 2x−3=52x−3=5

Transporting 3 to the other side 2x=5+32x=5+3 2x=82x=8 Dividing both the sides by 2 X=4 Hence, there is a linear expression on one side and numbers on the other side.

Ex. 2.3. Some Applications– Here some Applications of the above are there.

Ex. 2.4. Solving Equations having the Variable on both sides – Here students must transpose both the variable and number to the side. This is so that one side has the number and the other side has the variable. Hence, an equation comes into existence.

Ex. 2.5. Some more applications– Once again some more applications of the above.

Ex. 2.6. Reducing Equation to Simpler form– Sometimes Linear Equation may have a number in Denominator. Therefore, a student should take LCM of the denominator of both the left and right-hand side. Furthermore, a student must multiply on both sides. This will solve the problem. Hence, Equation is reduced to a simpler form.

Ex. 2.7. Equations reducible to the Linear form– Finally, this is the last sub-unit. Here an equation which is not Linear can be reduced to Linear Equation.

You can download NCERT Solutions for Class 8 Maths Chapter 2 by clicking on the button below

Solved Questions for You

Question 1: Solve the equation:

Answer: Transporting 2 to RHS, we obtain x =7+2 x =9

Question 2: Sum of two numbers is 95. If one exceeds the other number 15, find the numbers

Answer: Let one number = x ,

∴ The other will be:  x +15 As per the question:

x + x +15=95,2 x =80 x =4

One number ⇒40

Other number will be:  x +15⇒40+15 =55

Question 3: Fifteen years from now Ravi’s age will be four times his present age. What is Ravi’s present age?

Answer : Let Ravi’s present age :  x Fifteen years later, Ravi’s age =4× x x +15=4 x 15=4 x − x 3 x =15 ∴ x =5 Ravi’s present age 5 years

Question 4: Solve the equation:

Answer: Transporting 3 to RHS, we obtain y =10−3=7 ∴ y =7

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NCERT Solutions for Class 8 Maths

• NCERT Solutions for Class 8 Maths Chapter 6 Free PDF Download
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Important Questions for CBSE Class 8 Maths 2023-24

Home » CBSE » Important Questions for CBSE Class 8 Maths 2023-24

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Maths Important Questions for Class 8

Maths is a vast subject. The concept students learn in the lower classes helps in developing skills for higher classes. Thus, solving important questions for class 8 Maths helps students to score good marks.

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Class 8 Maths gives the student a chapter-wise plan to prepare the student for the CBSE class 8 examination. Important questions for class 8 Maths will help the students with practice for exam preparation. The questions are prepared by expert teachers of the subject so that all the topics and concepts are covered.

The important questions for class 8 Mathes will help students to gauge their understanding of the major concepts that are taught in class 8 Maths. By practising them, students will get an idea of what type of questions will be asked in the exam. Solving the important questions of Maths for class 8 CBSE will help students understand the concept well.

To access the important questions for class 8 Maths, students may register at Extramarks.

Class 8 Maths Important Questions – Chapters Covered

There are 16 Chapters and sub-topics that have important questions for class 8 Maths. The list here is created as per the NCERT books and the latest CBSE syllabus and is mentioned below.

 Chapter 1: Rational Numbers

Rational numbers are the numbers that can be represented in the form of p/q, where p and q are integers, and q is not equal to 0.

• Ex 1.1 Introduction to Rational Numbers
• Ex 1.2 Properties of Rational Numbers
• Ex 1.3 Representation of Rational Numbers on the Number Line
• Ex 1.4 Rational Numbers between Two Rational Numbers

Chapter 2: Linear Equations in One Variable

Linear equations in one variable are algebraic expressions that have only one variable.

• Ex 2.1 Introduction to Linear Equations in One Variable
• Ex 2.2 Solving Equations which have Linear Expressions on one Side and Numbers on the Other Side
• Ex 2.3 Some Applications
• Ex 2.4 Solving Equations having the Variable on Both Sides
• Ex 2.5 Some More Applications
• Ex 2.6 Reducing Equations to Simpler Form
• Ex 2.7 Equations Reducible to the Linear Form

Chapter 3: Understanding Quadrilaterals

Understanding quadrilaterals teaches different types of quadrilaterals and their properties.

• Ex 3.1 Introduction to Quadrilaterals
• Ex 3.2 Polygons
• Ex 3.3 Sum of the Measures of the Exterior Angles of a Polygon
• Ex 3.4 Kinds of Quadrilaterals
• Ex 3.5 Some Special Parallelograms

Chapter 4: Practical Geometry

In this chapter, students will learn to construct the quadrilaterals based on certain given conditions, such as when the dimensions are given for its sides and diagonals or for angles.

• Ex 4.1 Introduction to Practical Geometry
• Ex 4.2 Constructing a Quadrilateral
• Ex 4.3 Some Special Cases

Chapter 5: Data Handling

Students will understand the significance of data and its management in small and large industries.

• Ex 5.1 Looking for Information – Data Handling
• Ex 5.2 Organising Data
• Ex 5.3 Grouping Data
• Ex 5.4 Circle Graph or Pie Chart
• Ex 5.5 Chance and Probability

Chapter 6: Squares and Square Roots.

Students will learn how to find squares and square roots for different numbers.

• Ex 6.1 Introduction to Square and Square Roots
• Ex 6.2 Properties of Square Numbers
• Ex 6.3 Some More Interesting Patterns
• Ex 6.4 Finding the Square of a Number
• Ex 6.5 Square Roots
• Ex 6.6 Square Roots of Decimals
• Ex 6.7 Estimating Square Root

Chapter 7: Cubes and Cube Roots

In this chapter, students will learn to find cube and cube roots for different numbers.

• Ex 7.1 Introduction to Cubes and Cube Roots
• Ex 7.2 Cubes
• Ex 7.3 Cube Roots

Chapter 8: Comparing Quantities

This chapter is based on finding percentage, cost price, selling price, profit, loss, and compound interest.

• Ex 8.1 Recalling Ratios and Percentages
• Ex 8.2 Finding the Increase or Decrease in percent
• Ex 8.3 Finding Discounts
• Ex 8.4 Prices Related to Buying and Selling (Profit and Loss)
• Ex 8.5 Sales Tax/Value Added Tax/Goods and Services Tax
• Ex 8.6 Compound Interest
• Ex 8.7 Deducing a Formula for Compound Interest
• Ex 8.8 Rate Compounded Annually or Half Yearly (Semi-Annually)
• Ex 8.9 Applications of Compound Interest Formula

Chapter 9: Algebraic Expressions and Identities

The chapter teaches students to find the degree of the polynomial, dividing a monomial by monomial, dividing a polynomial by a monomial, and dividing polynomial by polynomial. It also teaches the students to divide the algebraic expression employing factorisation.

• Ex 9.1 What are Expressions?
• Ex 9.2 Terms, Factors and Coefficients
• Ex 9.3 Monomials , Binomials and Polynomials
• Ex 9.4 Like and Unlike Terms
• Ex 9.5 Addition and Subtraction of Algebraic Expressions
• Ex 9.6 Multiplication of Algebraic Expressions: Introduction
• Ex 9.7 Multiplying a Monomial by a Monomial
• Ex 9.8 Multiplying a Monomial by a Polynomial
• Ex 9.9 Multiplying a Polynomial by a Polynomial
• Ex 9.10 What is an Identity?
• Ex 9.11 Standard Identities
• Ex 9.12 Applying Identities

Chapter 10: Visualising Solid Shapes

The chapter introduces some very interesting concepts, such as visualising solid shapes, and views of 3d-shapes like cubes, cylinders, cones, spheres, etc.

• Ex 10.1 Introduction to Visualising Solid Shapes
• Ex 10.2 Views of 3D-Shapes
• Ex 10.3 Mapping Space Around Us
• Ex 10.4 Faces, Edges and Vertices

Chapter 11: Mensuration

This chapter deals with the measurement of area, perimeter, surface area and volume of different types of shapes.

• Ex 11.2 Introduction to Mensuration
• Ex 11.2 Let us Recall
• Ex 11.3 Area of Trapezium
• Ex 11.4 Area of a General Quadrilateral
• Ex 11.5 Area of a Polygon
• Ex 11.6 Solid Shapes
• Ex 11.7 Surface Area of Cube, Cuboid and Cylinder
• Ex 11.8 Volume of Cube, Cuboid and Cylinder
• Ex 11.9 Volume and Capacity

Chapter 12: Exponents and Powers

Here students will learn how to write large numbers easily using exponents and powers.

• Ex 12.1 Introduction to Exponents and Powers
• Ex 12.2 Powers with Negative Exponents
• Ex 12.3 Laws of Exponents
• EX 12.4 Use of Exponents to Express Small Numbers in Standard Form

Chapter 13: Direct and Inverse Proportions.

A direct and inverse proportion are applied to determine how the quantities and amounts are correlated with each other.

• Ex 13.1 Introduction to Direct and Inverse Proportions
• Ex 13.2 Direct Proportion
• Ex 13.3 Inverse Proportion

Chapter 14: Factorisation

This chapter deals with writing a given expression as the product of two or more factors is called factorisation.

• Ex 14.1 Introduction to Factorisation
• Ex 14.2 What is Factorisation?
• Ex 14.3 Division of Algebraic Expressions
• Ex 14.4 Division of Algebraic Expressions Continued (Polynomial ÷ Polynomial)
• Ex 14.5 Can you Find the Error?

Chapter 15: Introduction to Graphs 

This chapter deals primarily with the representation of data using different graph diagrams.

• Ex 15.1 Introduction
• Ex 15.2 Linear Graphs
• Ex 15.3 Some Applications

Chapter 16: Playing with Numbers

Playing with Numbers, introducing Numbers in General Form, Games with Numbers and Divisibility test by some numbers.

• Ex 16.1 Introduction
• Ex 16.2 Numbers in General Form
• Ex 16.3 Games with Numbers
• Ex 16.4 Letters for Digits
• Ex 16.5 Tests of Divisibility

Important Questions For Class 8 Maths Weightage

Practicing important questions for class 8 Maths will build the student’s confidence which will help them to score better during examinations. It will help in building a strong foundation for students. While class 8 Maths’ important questions are crucial, it is also vital to revise all topics under the chapter before trying to solve the important questions. 

All important questions for class 8 Maths are prepared according to the latest syllabus and follow the guidelines set by CBSE, India. Solving these Maths questions for class 8 will assist the students in judging their understanding of the chapter.

Important Questions for Class 8 Maths Chapter Wise

Extramarks important questions for class 8 Maths are equipped with chapter-wise plans for the students to prepare well for the CBSE class 8 examination. The solutions provided for the Maths important questions for class 8 are prepared by subject experts who have got very good teaching experience. These questions are compiled with reference to NCERT Books, several years of examination papers and based on how crucial the topic is for higher studies.

Students can click on the link for chapter-wise important questions for class 8 Maths with solutions.

• Chapter 1. Rational Numbers
• Chapter 2. Linear Equation in One Variable
• Chapter 3. Understanding Quadrilaterals
• Chapter 4. Practical Geometry
• Chapter 5. Data Handling
• Chapter 6. Squares and Square Roots
• Chapter 7. Cubes and Cube Roots 
• Chapter 8. Comparing Quantities
• Chapter 9. Algebraic Expressions and Identities
• Chapter 10. Visualising Solid Shapes
• Chapter 11. Mensuration
• Chapter 12. Exponents and Powers
• Chapter 13. Direct and Inverse Proportions
• Chapter 14. Factorisation
• Chapter 15. Introduction to Graphs
• Chapter 16. Playing with Numbers

Maths Questions for Class 8 with Answers – Types of Questions

The important questions for class 8 Maths include MCQs, short answer type questions, long answer type questions, and case study-based questions. These questions are a vital part of the study schedule. The chapter-wise important questions are good for students for self-study, which will help them in preparing for examinations. In addition to it, solving sample papers will be a great help for students.

Along with important questions for class 8 Maths, students can use various study materials during revision by clicking the link below.

Important Questions For Class 8 Maths With Solutions

• CBSE revision notes
• CBSE Sample papers
• CBSE previous year question papers
• CBSE extra questions

Benefits Of Solving Maths Questions for Class 8

Some of the benefits of Extramarks important questions for class 8 Maths include the following.

• Solving these questions increases the confidence in the student, which in turn helps them to score good marks.
• Practicing important questions for class 8 Maths helps students to improve in their weak topics.
•  Solving them helps students to learn the formulae and understand the concepts well.
• It covers all the important topics that will be asked in the exam.
• Solving important questions for class 8 Maths increases students’ problem-solving skills and critical thinking abilities.
• Students will understand the pattern of the question paper and the marking scheme of the exam.
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CBSE Class 8 Maths Important Questions

Chapter 1 - rational numbers.

Chapter 2 - Linear Equations in One Variable

Chapter 3 - understanding quadrilaterals, chapter 4 - practical geometry, chapter 5 - data handling, chapter 6 - squares and square roots, chapter 7 - cubes and cube roots, chapter 8 - comparing quantities, chapter 9 - algebraic expressions and identities, chapter 10 - visualising solid shapes, chapter 11 - mensuration, chapter 12 - exponents and powers, chapter 13 - direct and inverse proportions, chapter 14 - factorisation, chapter 15 - introduction to graphs, chapter 16 - playing with numbers, faqs (frequently asked questions), 1. how are chapter-wise important questions for class 8 maths helpful for students.

The chapter-wise important questions for class 8 Maths can help the students to clear their doubts, improve in their weak topics and prepare themselves for the exam in a better way. They also help with board exam preparation.

2. Which chapter is interesting in class 8 Maths?

Playing with numbers is an interesting chapter in class 8 Mathcs. Students will enjoy puzzles of missing numbers.

3. How to score the highest marks in class 8 Maths?

One can score 90% and above in class 8 Maths by practising important questions for class 8 Maths, solving past  years’ question papers, revision notes, and sample papers.

CBSE Related Links

• NCERT Solutions
• NCERT Class 8
• NCERT 8 Maths
• Chapter 2: Linear Equation
• Exercise 2.2

NCERT Solutions for Class 8 Maths Exercise 2.2 Chapter 2 - Linear Equations in One Variable

NCERT Solutions for Class 8 Maths Exercise 2.2 of Chapter 2 Linear Equations in One Variable are a set of all the answers to the questions listed under the exercise. NCERT Solutions cover each concept thoroughly using simple language. Class 8 students are encouraged to practise the concepts learned as much as possible. Without practising the CBSE Class 8 NCERT solutions, students may not be exam ready, which is essential. Therefore, NCERT Solutions act as a great resource for scoring well on the exam.

NCERT Solutions for Class 8 Maths Chapter 2 – Linear Equations in One Variable Exercise 2.2

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Access other exercise solutions of Class 8 Maths Chapter 2 – Linear Equations in One Variable

Exercise 2.1 Solutions 12 Questions (12 Short Answer Questions) Exercise 2.3 Solutions 10 Questions (3 Long Answer Questions, 7 Short Answer Questions) Exercise 2.4 Solutions 10 Questions (4 Long Answer Questions, 6 Short Answer Questions) Exercise 2.5 Solutions 10 Questions (1 Long Answer Question, 9 Short Answer Questions) Exercise 2.6 Solutions 7 Questions (1 Long Answer Question, 6 Short Answer Questions)

Access answers of Maths NCERT Class 8 Chapter 2 – Linear Equations in One Variable 2.2 Page Number 28

1. If you subtract ½ from a number and multiply the result by ½, you get 1/8, what is the number?

Let the number be x.

According to the question,

(x – 1/2) × ½ = 1/8

x/2 – ¼ = 1/8

x/2 = 1/8 + ¼

x/2 = 1/8 + 2/8

x/2 = (1+ 2)/8

x = (3/8) × 2

2. The perimeter of a rectangular swimming pool is 154 m. Its length is 2 m, more than twice its breadth. What are the length and breadth of the pool?

Given that,

The perimeter of the rectangular swimming pool = 154 m. Let the breadth of the rectangle be = x

Length of the rectangle = 2x + 2 We know that,

Perimeter = 2(length + breadth)

⇒ 2(2x + 2 + x) = 154 m

⇒ 2(3x + 2) = 154

⇒ 3x +2 = 154/2

⇒ 3x = 77 – 2

Therefore, Breadth = x = 25 cm

Length = 2x + 2

= (2 × 25) + 2

Base of isosceles triangle = 4/3 cm

Let the length of the equal sides of the triangle be x.

4/3 + x + x = 62/15 cm

⇒ 2x = (62/15 – 4/3) cm

⇒ 2x = (62 – 20)/15 cm

⇒ 2x = 42/15 cm

⇒ x = (42/30) × (½)

⇒ x = 42/30 cm

⇒ x = 7/5 cm

The length of either of the remaining equal sides is 7/5 cm.

4. Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.

Let one of the numbers be= x.

Then, the other number becomes x + 15. According to the question,

x + x + 15 = 95

⇒ 2x + 15 = 95

⇒ 2x = 95 – 15

First number = x = 40

And, other number = x + 15 = 40 + 15 = 55

5. Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers?

Let the two numbers be 5x and 3x. According to the question,

5x – 3x = 18

The numbers are 5x = 5 × 9 = 45

And 3x = 3 × 9 = 27

6. Three consecutive integers add up to 51. What are these integers?

Let the three consecutive integers be x, x+1 and x+2. According to the question,

x + (x+1) + (x+2) = 51

⇒ 3x + 3 = 51

⇒ 3x = 51 – 3

Thus, the integers are

7. The sum of three consecutive multiples of 8 is 888. Find the multiples.

Let the three consecutive multiples of 8 by 8x, 8(x+1) and 8(x+2). According to the question,

8x + 8(x+1) + 8(x+2) = 888

⇒ 8 (x + x+1 + x+2) = 888 (Taking 8 as common)

⇒ 8 (3x + 3) = 888

⇒ 3x + 3 = 888/8

⇒ 3x + 3 = 111

⇒ 3x = 111 – 3

⇒ x = 108/3

Thus, the three consecutive multiples of 8 are

8x = 8 × 36 = 288

8(x + 1) = 8 × (36 + 1) = 8 × 37 = 296

8(x + 2) = 8 × (36 + 2) = 8 × 38 = 304

8. Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3 and 4, respectively, they add up to 74. Find these numbers.

2x + 3(x+1) + 4(x+2) = 74

⇒ 2x + 3x +3 + 4x + 8 = 74

⇒ 9x + 11 = 74

⇒ 9x = 74 – 11

Thus, the numbers are

9. The ages of Rahul and Haroon are in the ratio 5:7. Four years later the sum of their ages will be 56 years. What are their present ages?

Let the ages of Rahul and Haroon be 5x and 7x. Four years later,

The ages of Rahul and Haroon will be (5x + 4) and (7x + 4), respectively. According to the question,

(5x + 4) + (7x + 4) = 56

⇒ 5x + 4 + 7x + 4 = 56

⇒ 12x + 8 = 56

⇒ 12x = 56 – 8

⇒ x = 48/12

Therefore, the present age of Rahul = 5x = 5×4 = 20

And, the present age of Haroon = 7x = 7×4 = 28

10. The number of boys and girls in a class is in the ratio of 7:5. The number of boys is 8 more than the number of girls. What is the total class strength?

Let the number of boys be 7x and girls be 5x.

7x = 5x + 8

⇒ 7x – 5x = 8

Therefore, the number of boys = 7×4 = 28

And, the number of girls = 5×4 = 20

Total number of students = 20+28 = 48

11. Baichung’s father is 26 years younger than Baichung’s grandfather and 29 years older than Baichung. The sum of the ages of all three is 135 years. What is the age of each one of them?

Let the age of Baichung’s father be x.

Then, the age of Baichung’s grandfather = (x+26)

and, age of Baichung = (x-29). According to the question,

x + (x+26) + (x-29) = 135

⇒ 3x + 26 – 29 = 135

⇒ 3x – 3 = 135

⇒ 3x = 135 + 3

⇒ x = 138/3

Age of Baichung’s father = x = 46

Age of Baichung’s grandfather = (x+26) = 46 + 26 = 72

Age of Baichung = (x-29) = 46 – 29 = 17

12. Fifteen years from now, Ravi’s age will be four times his present age. What is Ravi’s present age?

Let the present age of Ravi be x.

Fifteen years later, Ravi’s age will be x+15 years. According to the question,

x + 15 = 4x

⇒ 4x – x = 15

Therefore, present age of Ravi = 5 years.

13. A rational number is such that when you multiply it by 5/2 and add 2/3 to the product, you get -7/12. What is the number?

Let the rational be x.

x × (5/2) + 2/3 = -7/12

⇒ 5x/2 + 2/3 = -7/12

⇒ 5x/2 = -7/12 – 2/3

⇒ 5x/2 = (-7- 8)/12

⇒ 5x/2 = -15/12

⇒ 5x/2 = -5/4

⇒ x = (-5/4) × (2/5)

⇒ x = – 10/20

Therefore, the rational number is -½.

14. Lakshmi is a cashier in a bank. She has currency notes of denominations ₹100, ₹50 and ₹10, respectively. The ratio of the number of these notes is 2:3:5. The total cash with Lakshmi is ₹4,00,000. How many notes of each denomination does she have?

Let the numbers of notes of ₹100, ₹50 and ₹10 be 2x, 3x and 5x, respectively.

Value of ₹100 = 2x × 100 = 200x

Value of ₹50 = 3x × 50 = 150x

Value of ₹10 = 5x × 10 = 50x

200x + 150x + 50x = 4,00,000

⇒ 400x = 4,00,000

⇒ x = 400000/400

Numbers of ₹100 notes = 2x = 2000

Numbers of ₹50 notes = 3x = 3000

Numbers of ₹10 notes = 5x = 5000

15. I have a total of ₹300 in coins of denomination ₹1, ₹2 and ₹5. The number of ₹2 coins is 3 times the number of ₹5 coins. The total number of coins is 160. How many coins of each denomination are with me?

Let the number of ₹5 coins be x.

Number ₹2 coins = 3x

and, number of ₹1 coins = (160 – 4x)

Value of ₹5 coins = x × 5 = 5x

Value of ₹2 coins = 3x × 2 = 6x

Value of ₹1 coins = (160 – 4x) × 1 = (160 – 4x)

5x + 6x + (160 – 4x) = 300

⇒ 11x + 160 – 4x = 300

⇒ x = 140/7

Number of ₹5 coins = x = 20

Number of ₹2 coins = 3x = 60

Number of ₹1 coins = (160 – 4x) = 160 – 80 = 80

16. The organisers of an essay competition decide that a winner in the competition gets a prize of ₹100 and a participant who does not win gets a prize of ₹25. The total prize money distributed is ₹3,000. Find the number of winners, if the total number of participants is 63.

Let the number of winners be x.

Then, the number of participants who didn’t win = 63 – x

Total money given to the winner = x × 100 = 100x

Total money given to the participant who didn’t win = 25×(63-x)

100x + 25×(63-x) = 3,000

⇒ 100x + 1575 – 25x = 3,000

⇒ 75x = 3,000 – 1575

⇒ 75x = 1425

⇒ x = 1425/75

Therefore, the numbers of winners are 19.

Exercise 2.2 of NCERT Solutions for Class 8 Maths Chapter 2 – Linear Equations in One Variable is based on some Applications of Linear Equations in One Variable. This will help the students understand the context and situations in which the concepts they learned in the chapter, Linear Equations in One Variable, can be applied.

Also, explore – 

NCERT Solutions for Class 8 Maths

NCERT Solutions for Class 8 

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CBSE Class 8th Maths Value Based Questions PDF Download

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CBSE Class 8th Maths Value Based Questions are the easiest questions which you see in your question paper and the scoring one all student who attempt it surely get they are just little bit difficult and examine your basic knowledge regarding the particular chapter. Maths Value Based Questions for Class 8th are available here at Free of cost. These questions are expected to be asked in the Class 8th board examination. These Maths Value Based Questions are from complete CBSE Syllabus.

CBSE Class 8th Maths Value Based Questions

Most of these Maths Value Based Questions are quite easy and students need only a basic knowledge of the chapter to answer these questions. Download CBSE Maths Value Based Questions for board examinations. These Maths Value Based Questions are prepared by Directorate of Education, Delhi.

CBSE Maths Value Based Questions Class 8th PDF

The purpose of the Maths Value Based Questions is to make students aware of how basic values are needed in the analysis of different situations and how students require to recognize those values in their daily lives. Some questions are subject related. But even if they are not, that one-minute awareness of what we write about value without any specific preparation is a good step indeed.

CBSE Maths Value Based Questions for Class 8th download here in PDF format. The most CBSE Maths Value Based Questions for annual examination are given here for free of cost. The additional questions for practice the Class 8th exam are collected from various sources. It covers questions asked in previous year examinations.

CBSE Maths Value Based Questions for Class 8th Free PDF

Class 8th books have many questions. These questions are regularly asked in exams in one or other way. Practising such most CBSE Maths Value Based Questions certainly help students to obtain good marks in the examinations. 

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NCERT Solutions for Class 8 Maths Chapter 2: Linear Equations in One Variable - Exercise 2.2

• NCERT Solutions
• Chapter 2 Linear Equations In One Variable Exercise 2.2

NCERT Solutions for Class 8 Maths Chapter 2 (EX 2.2)

Chapter 2 of the NCERT Class 8 Maths textbook focuses on " Linear Equations in One Variable ." Exercise 2.2 is deleted from the current CBSE syllabus for the academic year 2024-25. This exercise specifically deals with solving linear equations that may include variables on both sides of the equation. This exercise helps students practice the fundamental techniques needed to isolate the variable and solve the equation. The key points to focus on in Exercise 2.2 includes Transposing Terms, Combining Like Terms, Balancing Equations and Checking Solutions. Through solving various problems in this exercise, students gain confidence in handling linear equations, which is a foundational skill for more advanced algebraic concepts. This practice will also enhance their problem-solving abilities and prepare them for future mathematical challenges.

Access Exercise Wise NCERT Solutions for Chapter 2 Maths Class 8

Glance on Class 8 Maths Exercise 2.2 Linear Equations in One Variable Chapter 2

Class 8 Maths Exercise 2.2 from Chapter 2 Linear Equations in One Variable solves word problems that linear equations can model in one variable.

There are general approach to solving these equations. This approach typically involves:

Assigning a variable to represent the unknown quantity.

Translating the word problem into an equation using mathematical terms and operations.

Manipulating the equation algebraically to isolate the variable.

Solving for the variable and interpreting the solution in the context of the word problem.

Exercise 2.2 Class 8 maths NCERT Solutions has overall 16 Questions 

Access PDF for Maths NCERT Chapter 2 Exercise 2.2 Class 8 (Not in the Current Syllabus)

Exercise 2.2.

1. If you subtract \[\frac{\text{1}}{\text{2}}\] from a number and multiply the result by \[\frac{\text{1}}{\text{2}}\], you get \[\frac{\text{1}}{\text{8}}\], What is the number?

Ans: To solve the question, we will assume the number

Let the number be \[\text{x}\], now according to the question

\[\left( \text{x-}\frac{\text{1}}{\text{2}} \right)\text{ }\!\!\times\!\!\text{ }\frac{\text{1}}{\text{2}}\text{=}\frac{\text{1}}{\text{8}}\]

Multiplying the equation by \[\text{2}\], we get

\[\text{x-}\frac{\text{1}}{\text{2}}\text{=}\frac{\text{2}}{\text{8}}\]

\[\text{x-}\frac{\text{1}}{\text{2}}\text{=}\frac{\text{1}}{\text{4}}\]

Now shifting \[\frac{\text{1}}{\text{2}}\] to right hand side, we get

\[\text{x=}\frac{\text{1}}{\text{4}}\text{+}\frac{\text{1}}{\text{2}}\]

\[\Rightarrow \text{x=}\frac{\text{1+2}}{\text{4}}\]

\[\Rightarrow \text{x=}\frac{\text{3}}{\text{4}}\]

Therefore, the required number is \[\frac{\text{3}}{\text{4}}\].

2. The perimeter of a rectangular swimming pool is \[\text{154}\] m. Its length is \[\text{2}\] m more than twice its breadth. What are the length and the breadth of the pool?

Ans: To solve the question, we will assume the breadth of the pool

Let the breadth \[\left( \text{b} \right)\] of the pool be \[\text{x}\], now according to the question the length \[\left( \text{l} \right)\] of the pool is \[\text{2x+2}\] m.

According to the question,

Perimeter of the rectangular pool \[\text{=154}\]

\[\Rightarrow \text{2}\left( \text{l+b} \right)\text{=154}\]

\[\Rightarrow \text{2}\left( \text{2x+2+x} \right)\text{=154}\]

\[\Rightarrow \text{2}\left( \text{3x+2} \right)\text{=154}\]

Dividing the equation by \[\text{2}\], we get

\[\text{3x+2=77}\]

Shifting \[\text{2}\] to right hand side, we get

\[\Rightarrow \text{3x=77-2}\]

\[\Rightarrow \text{3x=75}\]

Dividing the equation by \[\text{3}\], we get

\[\text{x=25}\], now

\[\text{l=2x+2}\]

\[\text{l=52}\]

Therefore, the length and breadth of the pool is \[\text{52}\] m and \[\text{25}\] m respectively.

3. The base of an isosceles triangle is \[\frac{\text{4}}{\text{3}}\] cm. The perimeter of the triangle is \[\text{4}\frac{\text{2}}{\text{15}}\] cm. What is the length of either of the remaining equal sides?

Ans: Let the length \[\left( \text{l} \right)\] of equal sides be \[\text{x}\] cm,

According to the question

Perimeter of triangle \[\text{=4}\frac{\text{2}}{\text{15}}\]

\[\Rightarrow \text{x+x+}\frac{\text{4}}{\text{3}}\text{=}\frac{\text{62}}{\text{15}}\]

\[\Rightarrow \text{2x+}\frac{\text{4}}{\text{3}}\text{=}\frac{\text{62}}{\text{15}}\]

Shifting \[\frac{\text{4}}{\text{3}}\] to left hand side, we get

\[\Rightarrow \text{2x=}\frac{\text{62}}{\text{15}}\text{-}\frac{\text{4}}{\text{3}}\]

\[\Rightarrow \text{2x=}\frac{\text{62-20}}{\text{15}}\]

\[\Rightarrow \text{2x=}\frac{\text{42}}{\text{15}}\]

\[\Rightarrow \text{x=}\frac{\text{21}}{\text{15}}\]

\[\Rightarrow \text{x=}\frac{\text{7}}{\text{5}}\]

Therefore, the length of remaining equal sides is \[\frac{\text{7}}{\text{5}}\] cm.

4. Sum of the two numbers is \[\text{95}\]. If one exceeds the other by \[\text{15}\], find the numbers.

Ans: Let us assume a number \[\text{x}\], then the other number would be \[\text{x+15}\]

Now, according to question

\[\text{x+x+15=95}\]

\[\Rightarrow \text{2x+15=95}\]

Shifting \[\text{15}\] to right hand side, we get

\[\text{2x=95-15}\]

\[\text{2x=80}\]

\[\Rightarrow \text{x=40}\], therefore, other number would be

\[\Rightarrow \text{x+15=40+15}\]

\[\Rightarrow \text{55}\]

Therefore, the required numbers are \[40\] and \[55\].

5. Two numbers are in the ratio \[\text{5:3}\]. If they differ by \[\text{18}\], what are the numbers?

Ans: Let us assume the number, let the number be \[\text{x}\]

Then the two numbers would be \[\text{5x}\] and \[\text{3x}\]

According to questions the difference between the two numbers is \[18\], therefore, 

\[\text{5x-3x=18}\]

\[\text{2x=18}\]

\[\text{x=9}\]

Therefore, the first number is \[\text{5x=45}\] and the second number is \[\text{3x=27}\].

6. Three consecutive integer add up to \[\text{51}\]. What are these integers?

Ans: Let us assume the three consecutive integers be \[\text{x}\], \[\text{x+1}\] and \[\text{x+2}\]

\[\text{x+}\left( \text{x+1} \right)\text{+}\left( \text{x+2} \right)\text{=51}\]

\[\Rightarrow \text{3x+3=51}\]

Shifting \[\text{3}\] to right hand side, we get

\[\text{3x=51-3}\]

\[\Rightarrow \text{3x=48}\]

\[\Rightarrow \text{x=16}\]

First integer is \[\text{x=16}\],

Second integer is \[\text{x+1=17}\]

Third integer is \[\text{x+2=18}\]

Therefore, the three consecutive integers are \[\text{16}\], \[\text{17}\] and \[\text{18}\].

7. The sum of three consecutive multiples of \[\text{8}\] is \[\text{888}\]. Find the multiples.

Ans: Let us assume the three consecutive multiples of \[\text{8}\] be \[\text{8x}\], \[\text{8}\left( \text{x+1} \right)\] and \[\text{8}\left( \text{x+2} \right)\]

According to the question the sum of multiples is \[\text{888}\]

\[\Rightarrow \text{8x+8}\left( \text{x+1} \right)\text{+8}\left( \text{x+2} \right)\text{=888}\]

\[\Rightarrow \text{8}\left( \text{x+x+1+x+2} \right)\text{=888}\]

\[\Rightarrow \text{8}\left( \text{3x+3} \right)\text{=888}\]

Dividing the equation by \[\text{8}\], we get

\[\text{3x+3=111}\]

\[\Rightarrow \text{3x=111-3}\]

\[\Rightarrow \text{3x=108}\]

\[\text{x=36}\]

First multiple is \[\text{8x=288}\]

Second multiple is \[\text{8}\left( \text{x+1} \right)\text{=296}\]

Third multiple is \[\text{8}\left( \text{x+2} \right)\text{=304}\]

Therefore, the required three consecutive multiples of 8 are \[288\], \[296\] and \[304\].

8. Three consecutive integers are such that when they are taken in increasing order and multiplied by \[\text{2}\], \[\text{3}\] and \[\text{4}\] respectively, they add up to \[\text{74}\]. Find these numbers.

Ans: Let us assume the three consecutive integers be \[x\], \[\text{x+1}\] and \[\text{x+2}\]

\[\Rightarrow \text{2x+3}\left( \text{x+1} \right)\text{+4}\left( \text{x+2} \right)\text{=74}\]

\[\Rightarrow \text{2x+3x+3+4x+8=74}\]

\[\Rightarrow \text{9x+11=74}\]

Shifting \[\text{11}\] to right hand side, we get

\[\text{9x=74-11}\]

\[\Rightarrow \text{9x=63}\]

Dividing the equation by \[\text{9}\], we get

\[\Rightarrow \text{x=7}\]

\[\Rightarrow \text{x+1=8}\]

\[\Rightarrow \text{x+2=9}\]

Therefore, the required consecutive integers are \[\text{7}\], \[\text{8}\] and \[\text{9}\].

9. The ages of Rahul and Haroon are in the ratio \[\text{5:7}\]. Four years later the sum of their ages will be \[\text{56}\] years. What are their present ages?

Ans: Let us assume the age be \[\text{x}\], therefore, the age of Rahul is \[\text{5x}\] and the age of Haroon is \[\text{7x}\].

Age of Rahul and Haroon after four years will be \[\text{5x+4}\] and\[\text{7x+4}\]

\[\left( \text{5x+4} \right)\text{+}\left( \text{7x+4} \right)\text{=56}\]

\[\text{12x+8=56}\]

Shifting \[\text{8}\] to right hand side, we get

\[\text{12x=56-8}\]

\[\text{12x=48}\]

Dividing the equation by \[\text{12}\], we get

\[\text{x=4}\]

Therefore, the current age of Rahul and Haroon would be \[\text{20}\] years and \[\text{28}\] years respectively.

10. The number of boys and girls in a class are in the ratio \[\text{7:5}\]. The number of boys is \[\text{8}\] more than the number of girls. What is the total class strength?

Ans: Let the number of boys and girls be \[\text{x}\], therefore, the number of boys and girls will be \[\text{7x}\] and \[\text{5x}\] respectively

\[\text{7x=5x+8}\]

Shifting \[\text{5x}\] on left hand side, we get

\[\text{2x=8}\]

Number of boys is \[\text{7x=28}\] and number of girls is \[\text{5x=20}\] 

Therefore, the total strength of class is \[\text{48}\] students.

11. Baichung’s father is \[\text{26}\] years younger than Baichung’s grandfather and \[\text{29}\] years older than Baichung. The sum of the ages of all the three is \[\text{135}\] years. What is the age of each one of them?

Ans: Let us assume the age of Baichung’s father be \[\text{x}\] years, therefore, the age of Baichung and Baichung’s grandfather would be \[\left( \text{x-29} \right)\] years and \[\left( \text{x+26} \right)\] years respectively.

\[\text{x+}\left( \text{x-29} \right)\text{+}\left( \text{x+26} \right)\text{=135}\]

\[\text{3x-3=135}\]

\[\text{3x=138}\]

\[\text{x=46}\] 

Therefore, the age of Baichung’s father is \[\text{46}\] years, age of Baichung’s grandfather is \[\text{46+26=72}\] years and the age of Baichung is \[\text{46-29=17}\] years.

12. Fifteen years from now Ravi’s age will be four times his present age. What is Ravi’s present age?

Ans: Let us assume the present age of Ravi be \[\text{x}\] years

\[\text{x+15=4x}\]

Shifting \[\text{x}\] to right hand side, we get

\[\text{15=4x-x}\]

\[\text{15=3x}\]

\[\text{x=5}\]

Therefore, the present age of Ravi is \[\text{5}\] years.

13. A rational number is such that when you multiply it by \[\frac{\text{5}}{\text{2}}\] and add \[\frac{\text{2}}{\text{3}}\] to the product, you get \[\text{-}\frac{\text{7}}{\text{12}}\]. What is the number?

Ans: Let us assume the rational number \[\text{x}\]

\[\left( \text{x }\!\!\times\!\!\text{ }\frac{\text{5}}{\text{2}} \right)\text{+}\frac{\text{2}}{\text{3}}\text{=-}\frac{\text{7}}{\text{12}}\]

\[\Rightarrow \frac{\text{5}}{\text{2}}\text{x+}\frac{\text{2}}{\text{3}}\text{=-}\frac{\text{7}}{\text{12}}\]

Shifting \[\frac{\text{2}}{\text{3}}\] to right hand side, we get

\[\Rightarrow \frac{\text{5}}{\text{2}}\text{x=-}\frac{\text{7}}{\text{12}}\text{-}\frac{\text{2}}{\text{3}}\]

\[\Rightarrow \frac{\text{5}}{\text{2}}\text{x=}\frac{\text{-7-8}}{\text{12}}\]

\[\Rightarrow \frac{\text{5}}{\text{2}}\text{x=}\frac{\text{-15}}{\text{12}}\]

Multiplying \[\frac{\text{2}}{5}\] to the equation, we get

\[\text{x=}\frac{\text{-3}}{\text{6}}\]

\[\Rightarrow \text{x=}\frac{\text{-1}}{2}\]

 Therefore, the required rational number is \[\frac{\text{-1}}{2}\].

14. Lakshmi is a cashier in a bank. She has currency notes of denominations of Rs. \[\text{100}\], Rs. \[\text{50}\] and Rs. \[\text{10}\]. The ratios of these notes is \[\text{2:3:5}\]. The total cash with Lakshmi is Rs. \[\text{4,00,000}\]. How many notes of each denomination does she have?

Ans: Let us assume the number of notes be \[\text{x}\], therefore, the number of Rs. \[\text{100}\] notes, Rs. \[\text{50}\] notes and Rs. \[\text{10}\] notes with Lakshmi is \[\text{2x}\], \[\text{3x}\]and \[\text{5x}\] respectively.

Amount of Rs. \[100\] notes are Rs. \[\text{200x}\]

Amount of Rs. \[50\] notes are Rs. \[\text{150x}\]

Amount of Rs. \[10\] notes are Rs. \[\text{50x}\]

\[\Rightarrow \text{200x+150x+50x=400000}\]

\[\Rightarrow \text{400x=400000}\]

Dividing the equation by \[\text{400}\], we get

\[\text{x=1000}\]

Therefore, the number of notes she has,

Rs. \[100\] notes \[\text{=2000}\]

Rs. \[50\] notes \[\text{=3000}\]

Rs. \[10\] notes \[\text{=5000}\]

15. I have a total of Rs. \[\text{300}\] in coins of denomination Rs. \[\text{1}\], Rs. \[\text{2}\] and Rs. \[\text{5}\]. The number of Rs. \[\text{2}\]coins is \[\text{3}\]times the number of Rs. \[\text{5}\] coins. The total number of coins is \[\text{160}\]. How many coins of each denomination are with me?

Ans: Let us assume the number of Rs. \[\text{5}\] coins be \[\text{x}\],

The number of Rs. \[\text{2}\] coins be\[\text{3x}\] and number of Rs. \[\text{1}\] coins be \[\text{160-4x}\left( \text{160-x-3x} \right)\].

Amount of Rs. \[1\] coins are Rs. \[\text{1 }\!\!\times\!\!\text{ }\left( \text{160-4x} \right)\text{=160-4x}\]

Amount of Rs. \[2\] coins are Rs. \[\left( \text{2 }\!\!\times\!\!\text{ 3x} \right)\text{=6x}\]

Amount of Rs. \[5\] coins are Rs. \[\left( \text{5 }\!\!\times\!\!\text{ x} \right)\text{=5x}\]

According to the question, total number of coins is Rs. \[\text{300}\], therefore,

\[\Rightarrow \left( \text{160-4x} \right)\text{+6x+5x=300}\]

\[\Rightarrow \text{160+7x=300}\]

Shifting \[\text{160}\] to right hand side, we get

\[\Rightarrow \text{7x=300-160}\]

\[\Rightarrow \text{7x=140}\]

\[\Rightarrow \text{x=20}\] 

Therefore, Number of Rs. \[1\] coins is \[\text{80}\]

Number of Rs. \[2\] coins is \[\text{60}\]

Number of Rs. \[5\] coins is \[\text{20}\]

16. The organizers of an essay competition decide that a winner in the competition gets a prize of Rs. \[\text{100}\] and a participant who does not win gets a prize of Rs. \[\text{25}\]. The total prize money distributed is Rs. \[\text{3000}\]. Find the number of winners, if the total number of participants is \[\text{63}\].

Ans: Let us assume the number of winners be \[\text{x}\] 

Total number of participants \[\text{=63}\]

Number of people who didn’t win the competition will be \[\text{63-x}\]

Amount given to winner would be Rs. \[\text{100x}\]

Amount given to the participants who didn’t won would be Rs. \[\text{25}\left( \text{63-x} \right)\text{=1575-25x}\]

\[\Rightarrow \text{100x+1575-25x=3000}\]

Shifting 1575 to right hand side, we get

\[\Rightarrow \text{100x-25x=3000-1575}\]

\[\Rightarrow \text{75x=1425}\]

Dividing the equation by\[\text{75}\], we get

\[\Rightarrow \text{x=19}\]

Therefore, the number of the winners is \[\text{19}\].

To sum up, Class 8 Maths Chapter 2 Exercise 2.2 Solutions , "Linear Equations in One Variable Exercises," covers the process of solving equations using a single variable. Understanding ideas like creating equations from actual situations, utilising equality properties to solve equations, and validating answers are essential. Pay close attention to comprehending the transposition and balancing techniques used in the solution of linear equations. Exams from prior years usually have 10-12 questions that cover these subjects. To gain confidence and familiarity with the principles, practise solving a range of equations. These are essential abilities for any future mathematical studies.

Other Related Links for Chapter 2  Linear Equations in One Variable of Class 8

Chapter-Specific NCERT Solutions for Class 8 Maths

Given below are the chapter-wise NCERT Solutions for Class 8 Maths . You can use it as your 8th maths guide. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

FAQs on NCERT Solutions for Class 8 Maths Chapter 2: Linear Equations in One Variable - Exercise 2.2

1. What do NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.2 include?

NCERT Solutions for Class 8 Maths Exercise 2.2 of Chapter 2 Linear Equations in One Variable includes all the answers to the questions given in the exercise. NCERT Solutions cover each concept thoroughly and provide step-by-step explanations to the exercise problems. Class 8 students need to practice the concepts as much as possible to have in-depth knowledge and score well in exams. Practising the CBSE Class 8 NCERT exercise-wise textbook problems for Maths Chapter 2 as well as other chapters prepares students for the exam. In case of any doubts, they can refer to the solutions which include answers prepared by experts.

2. What are the advantages of studying from exercise-wise NCERT Solutions for Class 8 Maths Chapter 2?

NCERT Solutions for Class 8 Maths is a great study material for scoring well in exams. It provides the clarity of concepts which is required for students in order to be able to solve exercise problems. Linear Equations in One Variable Chapter 2 is an important chapter of Mathematics as it allows students to have the foundation of solving problems based on linear equations. This knowledge will be really helpful while solving higher classes maths problems. Hence, students should try to solve all the exercise problems on their own and then refer to NCERT Solutions which will clear all their doubts.

3. How is Exercise 2.2 of Linear Equations in One Variable helpful?

Exercise 2.2 of Class 8 Maths Chapter 2 Linear Equations in One Variable deals with the application-based numerical problems. Linear equations concepts find its application in solving those problems. This exercise is extremely helpful in teaching students how to write equations based on a problem statement and then put the values to find the answer. This is a really interesting exercise as it allows students to use their understanding of concepts in solving application-based questions. Students facing any difficulty in solving the exercise can refer to NCERT Solutions for the same designed by Vedantu.

4. How will students be able to solve the questions given in Exercise 2.2 correctly?

In order to be able to solve application-based problems on Linear Equations in One Variable of Class 8 Maths, it is important to first understand how linear equations are formed and have a thorough practice of solving simple problems on linear equations. If students are familiar with the concepts and practise the previous exercise properly, solving Exercise 2.2 will be a piece of cake. In case of any doubts, students can register on Vedantu’s site and avail the master classes on the site for the guidance by experts. Students can also download NCERT Solutions for this exercise.

5. How to solve a  linear equation in one variable question?

To solve a question which consists of linear equations in one variable you have to study the questions carefully. The question itself provides you with all the information you require to form an equation with only one assumption for an unknown digit to be denoted by any alphabetic variable. Then, you solve the equation formed and get the value for x which is usually the answer. In some cases the answer may be n times of x or n added or subtracted from x. It will be mentioned in the question.

6. Is Chapter 2 Linear Equations in One Variable of Class 8 Maths difficult?

Linear Equations in One Variable is a chapter that students find troublesome, but it becomes easier once you understand the applications and solve all the questions. Checking your progress as you go using NCERT Solutions helps you prepare better. It is crucial that you comprehend the theory and the application of the methods used to solve the problem questions. NCERT Solutions are your reference book for this purpose. You can download them for later use as well.

7. How many exercises are there in Chapter 2 Linear Equations in One Variable of Class 8 Maths?

There are six exercises in Chapter 2 Linear Equations in One Variable of Class 8 Maths. These exercises have a total of 65 questions divided into 12, 16, 10, 10, 10, and seven questions in each exercise respectively. Students should try to solve all these questions by themselves and also the examples so they get a fair idea about the type of questions from this particular chapter. It is an easy sail for revisions once you are well prepared.

8. Do I need to practice all the questions provided in Exercise 2.2 of Chapter 2 Linear Equations in One Variable of Class 8 Maths?

Practising all the questions given in Exercise 2.2 will guide you to have clear concepts. It also helps you understand the variety of questions that can come in exams from the chapter. Mathematics, regardless of the class, requires regular practice. It will be less burdensome to prepare the chapter with a disciplined study schedule. You can set a time you wish to dedicate to each question or chapter and work accordingly. Achieving your goal is smooth with a set routine and determination.

9. Where can I get NCERT Solutions for Chapter 2 Linear Equations in One Variable of Class 8 Maths?

You can get NCERT Solutions for Chapter 2 Linear Equations in One Variable of Class 8 Maths from Vedantu. You can download the PDF free of cost. You can also install the Vedantu mobile application for more study material for all the subjects. Class 8 is an important growth step in school life. It is essential that you study well and have clear concepts for each subject. Hence, NCERT Solutions play the role of helping hand for clarification of doubts and quick revision before the examinations.

NCERT Solutions for Class 8

Cbse study materials for class 8, cbse study materials.