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

Points to learn before Exercise 2.2:

• Algebraic expressions are any expressions that include a constant, a variable, and some operations such as addition, multiplication, and so on.

• A variable is an unknown number that is typically represented by a letter like x, y, n, and so on.

• Constant refers to any number that has no variables.

• The Coefficient of that variable is a number followed by a variable.

• Any number or variable separated by operators is referred to as a term.

• Equation is a statement that says the two expressions are equal.

• A linear expression is one in which the variable's greatest power is one.

• The equation of a straight line will be the linear equation. It can be in one variable or two variables. In case there is only one variable in the equation, then it can be called a linear equation in one variable. 

• In the linear equation, there is an equality sign. The LHS (left-hand side) is the expression to the left of the equal sign, and the RHS (right-hand side) is the expression to the right of the equal sign (right-hand side).

• The LHS is equal to the RHS in a linear equation, but only for some values, and these values are the solutions to these linear equations.

Solving Equations which have Linear Expressions on one Side and Numbers on the other Side. There are two methods to solve such type of problems:

1. Balancing Method: We can use this method to find the solution by adding or subtracting the same number on both sides without changing the balance.

2. Transposing Method: In this method, we must transpose constants or variables from one side to the other until the solution is obtained. The sign will change if we transpose the terms.

Solving Equations having the Variable on both Sides

We should be able to solve such problems because the equation can have variables on both sides. In this type of problem, we must arrange all of the constants on one side and all of the terms with variables on the other. Then they can be easily solved.

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

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

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

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

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

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

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

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

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

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

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

\[\Rightarrow \text{x=40}\], 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, 

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

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

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

\[\text{x=9}\]

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 \[\text{x+2}\]

According to the question,

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

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

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

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

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

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

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

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

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

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

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

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

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

\[\text{x=36}\]

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 \[\text{x+2}\]

According to the question,

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

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

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

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\[\text{7x+4}\]

According to the question,

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

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

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

According to the question,

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

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

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

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

\[\text{x=4}\]

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.

According to the question,

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

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

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

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

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

\[\text{x=46}\] 

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

According to the question,

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

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

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

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

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

\[\text{x=5}\]

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 \[\text{x}\]

According to the question,

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

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

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. \[\text{50x}\]

According to the question,

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

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

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 \[\text{160-4x}\left( \text{160-x-3x} \right)\].

According to the question,

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. \[\text{25}\left( \text{63-x} \right)\text{=1575-25x}\]

According to the question,

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

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 \[\text{19}\].

What you’ll learn in NCERT Class 8 Maths Chapter 2 Exercise 2.2

This chapter will comprise of seven exercises that will cover the major topics including- Ex 2.1 will cover the introduction to linear equations, Ex 2.2 is all about solving equations that have linear expressions on one side and numbers on another side, Ex 2.3 will cover the applications, Ex 2.4 includes solving equations that have variables on both sides. Ex 2.5 cover some more applications, Ex 2.6 is about reducing equations to a simpler form, and Ex 2.7 will cover equations reducible to the linear form. Mathematics might be the favourite subject for some of the students while others simply fear this subject. Well, if you are one amongst the second category, Vedantu notes can help you come out of subject fear and help in enhancing your Mathematics skills. Now, if you are clear with the NCERT questions, you can easily score a good grade in your exams because of all the school level exams - whether unit tests, half-yearly or annual exams, all of them cover the majority of the questions from NCERT textbook itself. 

In your NCERT textbook, after you learned about Rational Numbers in Chapter 1, you would now look at the next chapter which is Linear Equations in One Variable. And the students are looking for solved NCERT problems of 8th Class Maths 2nd Chapter 2.2 Exercise must refer to Vedantu notes. 

With Vedantu, you will get a complete detailed explanation and solved solutions of Class 8th Maths Chapter 2 Exercise 2.2. All the solutions provided in the notes will be solved and verified by our expert subject teachers. You can make use of these solved problems and brief notes of Chapter 2- Linear Equations in One Variable to even complete your homework and assignments on time and further score good marks in the coming exams. Every single question of this chapter is going to be explained in a step by step manner with diagrams and relevant figures to make the concept easy to understand for the students.

NCERT Maths Class 8 Chapter 2 Exercise 2.2: Get Detailed Solutions

Many students are not at all fond of the subject Mathematics and study it just for the sake of getting passing marks. In many cases, parents even complained that their kids show no excitement and find it difficult to understand the formulae and other concepts. But if the students get the right study material and notes that explain the difficult topics in a simple way, mathematics can become your favourite subject. Vedantu offers detailed NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.2 Questions that are prepared by expert teachers who are highly qualified and hold relevant certifications in Mathematics. 

During the 8th grade, if you have the Maths concepts clear, it’s a green flag for you if you are planning to choose the Non-Medical stream in 11th standard. Opting for Vedantu notes makes the Mathematics concept clear for you while explaining the difficult topics in the easiest way. Our subject-specific experts ensure that the notes are prepared in the simplest way to make learning fun and easy for the students. We follow a step-by-step approach while explaining the complex formulae in the simplest order and solve the problems without any complexities. Vedantu notes not only specifically emphasize on Maths Class 8 Chapter 2 Exercise 2.2, but even explain formulae, algorithms, identities, examples, and other problems in an easy way. 

About Chapter 2 of NCERT Textbook: Linear Equations in One Variable 

Starting with the basics of the chapter, Vedantu NCERT Solutions will explain the concept of Class 8 Maths Chapter 2 Exercise 2.2 in a very simple way that is easy to memorize for the students. Basically, a linear equation can be defined as an equation of a straight line that is written in one variable. Linear equations can be classified into three types- identity, conditional, and inconsistent. Linear equations in one variable might take the form ax+b=0 and solved with the help of basic algebraic operations. The solutions will even cover the basic topics so that it becomes easy to understand the topic in depth instead of cramming the exercise. 

All the solutions of NCERT Solution of Maths Class 8 Chapter 2 are prepared as per the CBSE guidelines and we update out study material from time to time so that you can get the latest notes and gain good grades in the exams.

Download NCERT Solutions for Class 8 Maths to help you to revise complete syllabus ans score more marks in your examinations.

Free PDF Downloads for Offline Access

To access the solutions in the offline mode, Vedantu provides free PDF downloads of the respective chapters and subjects. So, students can prepare the study material and solutions in advance before the exam preparation. This will allow you to prepare for the exams without the stress of gathering the NCERT Solutions from here and there. So, you can download NCERT Solution for Class 8 Maths Chapter 2 Exercise 2.2 from Vedantu and study the solutions in offline mode as well. Apart from Chapter 2 of your Maths NCERT textbook, Vedantu comes up with study material of other chapters as well including- quadrilaterals, data handling, square and square roots, cube and cube roots, comparing quantities, factorization, exponents, and powers, etc. 

You can download PDF files from Vedantu for other chapters as well and other subjects too, including- English, Hindi, Science, SST, and so on. At Vedantu, we ensure that all the NCERT Solutions delivered to you are easy to understand while covering every important topic of the exercise. Download the Maths Class 8 Chapter 2 Exercise 2.2 NCERT solutions to prepare for your exams.

NCERT Solutions Class 8 Maths Chapter 2 All Exercises