# Important Questions for CBSE Class 8 Maths Chapter 2 – Linear Equations in One Variable

## CBSE Class 8 Maths Important Questions for Linear Equations in One Variable

Maths is used worldwide as a necessary element in many different fields. There are several uses of Maths, and having a solid foundation in the subject will benefit a variety of careers. Since Maths is entirely based on numbers and is not a language-based subject, you can either get the answer right or wrong, which increases your chance of receiving full marks for this subject.

Chapter 2 of Class 8 Maths is called ‘Linear Equations in One Variable’. These are the basic equations used to represent and resolve an unknown quantity. It is represented with the help of graphs by using straight lines. A linear equation is a simple way of representing maths equations. The unknown quantities can be represented using the variable ‘X’. A linear equation can be solved in various simple methods. All the variables are placed on one side, and the rest is isolated on the other to obtain the value of the unknown quantity. In a linear equation, the degree of the equation is exactly equal to 1. Similarly, there is only one variable, and we can only obtain one solution. It is drawn in a graph, and it happens to appear either vertically or horizontally.

Extramarks is one of the best learning platforms for students to achieve excellent scores. It helps students by providing them with important questions from the CBSE syllabus. The Extramarks team refers to NCERT books, CBSE sample papers, CBSE revision notes, CBSE past year’s questions, etc., and provides you with the solutions to important questions. Students can get a good hold on this chapter, by solving our class 8 linear equations extra questions.

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## Get Access to CBSE Class 8 Maths Important Questions 2022-23 with Chapter-Wise Solutions

You can also find CBSE Class 8 Maths Chapter-by-Chapter Important Questions here:

 CBSE Class 8 Maths Important Questions Sr No. Chapters Chapters Name 1 Chapter 1 Rational Numbers 2 Chapter 2 Linear Equations in One Variable 3 Chapter 3 Understanding Quadrilaterals 4 Chapter 4 Practical Geometry 5 Chapter 5 Data Handling 6 Chapter 6 Squares and Square Roots 7 Chapter 7 Cubes and Cube Roots 8 Chapter 8 Comparing Quantities 9 Chapter 9 Algebraic Expressions and Identities 10 Chapter 10 Visualising Solid Shapes 11 Chapter 11 Mensuration 12 Chapter 12 Exponents and Powers 13 Chapter 13 Direct and Inverse Proportions 14 Chapter 14 Factorisation 15 Chapter 15 Introduction to Graphs 16 Chapter 16 Playing with Numbers

## Linear Equations Class 8 Extra Questions with Solution

Mentioned below are a few of the questions and their answers from our Chapter 2 Class 8 Maths important questions.

Question 1: The perimeter of a rectangular swimming pool is 154m. Its length is

2m, more than twice its breadth. What is the length and the breadth of the pool?

Answer 1: Let the breadth of the swimming pool be x m.

The length of the swimming pool will be = (2x + 2) m.

Perimeter of swimming pool:- 2 (l + b) =154

2 (2x + 2 + x)=154

2 (3x + 2)=154

∴Dividing both sides by 2, we obtain

(3x + 2)=77

On transporting two on the R.H.S., we get

3x = 77 – 2

3x = 75

x= 75/3

x= 25 m

Hence, the breadth of the swimming pool is x= 25m

The length of the swimming pool will be= (2x + 2) m.

=(2 х 25 + 2) m

=(50 + 2) m

=52 m

Thus, the length of the swimming pool is 52m, and the breadth of the swimming pool is 25m.

Question 2: What is the share of A when Rs 25 are divided between A and B so that A gets Rs 8 more than B is 16.5?

Answer 2: Let the share of B be x.

Let the share of A be (x + 8).

From this, we get,

x + x + 8 = 25

2x = 25 – 8

2x= 17

x = 17/2

x = 8.5

Therefore, A’s share will be 8.5.

Question 3: Find three consecutive odd numbers whose sum is 147.

Answer 3: Let the first, second, and third consecutive odd numbers be (2x +1),(2x + 3) and (2x +  5), respectively.

Hence the sum of the consecutive odd numbers is

(2x + 1) + (2x + 3) + (2x + 5)= 147.

On further simplifying, we get

2x + 2x + 2x + 1 + 3 + 5=147

6x + 9= 147.

On rearranging, we obtain

6x= 147 – 9

6x= 138

X=  138/6=23,

So the three consecutive odd numbers are (2x + 1)= 47

(2x + 3)= 49

(2x + 5)= 51.

Question 4: Ram’s father is 26 years younger than Ram’s grandfather and 29 years older than Ram. The sum of the ages of all three is 135 years. What is the age of each one of them?

Answer 4: Let Ram’s present age be x years

Rams father’s present age is = (x + 29) years

Rams grandfather’s present age =(x + 29 + 26) years

The sum of all three ages adds up to 135 years

Hence,

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

x + x + x + 29 + 29 + 26 =135

3x + 84= 135

3x = 135-84

3x = 51

x= 51/3

x= 17

Hence, Ram’s present age is x=17 years

Ram’s father’s present age =(x + 29)

=(17 + 29)

=46 years

Ram’s grandfather’s age =(x + 29 + 26)

=(17 + 29 + 26)= 72 years

Question 5: If 8x – 3 +17x, then x ________.

• is a fraction
• is an integer
• is a rational number
• cannot be solved

Answer 5: (C) A rational number

Given:- 8x-3=25+17x

Moving -3 to R.H.S. and becomes 3 and 17x to L.H.S.

We obtain,

8x – 17x = 25 +3

-9x=28

x=-28/9

Thus, x is a rational number.

Question 6: 3x+2/3=2x+1

Answer 6: 3x+ 2/3 = 2x +1

By transposing the above equation, we get

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

3x+2=6x+3

By moving all the variables on the L.H.S., we get,

3x-6x=3-2

-3x=1

x=-1/3

Question 7: The angles of a triangle are in the ratio 2 : 3: 4. Find the angles of the triangle.

Answer 7: Let the angles of the triangle be 2x°, 3x° and 4x°.

From the given question, we get,

2x + 3x + 4x = 180

∵ The sum of all the angles of a triangle is 180°)

⇒ 9x = 180

⇒ x = 20……….. (Transposing 9 to R.H.S.)

Hence, The angles of the given triangle are

2× 20 = 40°,

3 × 20 = 60°,

4 × 20 = 80°.

Question 8: The sum of the two numbers is 95. If one exceeds the other by 15, find the numbers.

Answer 8: Let the smaller number be x.

Then, the larger number =x +15.

According to the question,

the sum of the two numbers is 95

x + (x + 15) =95

2x + 15 =95 ………..(transposing 15 to the R.H.S.)

2x= 80

x=80/2

x= 40

Hence, the smaller number is 40

The larger number is (x + 15)= 40 +15=55

Hence, the required numbers are 40 and 55

Question 9: If (5x/3) – 4 = (2x/5), then the numerical value of 2x – 7 is

(A) 19/13

(B) -13/19

(C) 0

(D) 13/19

Given :- (5x/3) – 4 = (2x/5)

(5x/3) – (2x/5) = 4

L.C.M. of 3 and 5 is 15

(25x  –  6x)/15 = 4

19x = 4 × 15

19x = 60

X = 60/19

Substituting x=60/19 in the given equation,

= (2 × (60/19)) – 7

= (120/19) – 7

= (120  –  133)/19

= – 13/19

Question 10:  9x + 5 = 4(x – 2)+ 8

Answer 10: 9x + 5= 4(x – 2) + 8,

By transposing the above equation, we get,

9x + 5= 4x – 8 + 8

9x – 4x =5

Again by transposing

5x=5

X=5/5

X=1

Question 11: The sum of three consecutive multiples of 8 is 888. Find the multiple.

Answer 11: Let the three consecutive multiples be x, x + 8, x + 16

According to the given question,

The sum of three consecutive multiples of 8 is 888

x + x + 8 + x + 16 = 888

3x + 24 = 888

3x= 888 – 24

3x =864

x =864/3

x =288

Therefore the three consecutive multiples are:

x =288

x + 8=296

x + 16=304, respectively.

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

Answer 12: Let the rational number be x

According to the question,

X x (5/2) + 2/3 =-7/12

5x/2 +2/3 =-7/12

5x/2=  -7/12 – 2/3

Taking L.C.M. on the R.H.S.

5x/2 = (-7-8)/12

5x/2 = -15/12

5x/2 = -5/4

x= (-5/4) x (2/5)

x=-10/20

x= -1/2

Therefore, the rational number is -1/2

Question 13: Find the number whose fifth part increased by 5 is equal to its fourth part diminished by 5.

Answer 13: Let the number be x.

According to the question, we get

(1/5)x + 5 = (1/4) x – 5

On rearranging the given equation,

(1/5) x – (1/4) x = -5-5

(1/5) x – (1/4)x =-10

By taking L.C.M., we will get,

(4x-5x)/20=-10

Again by transposing

-x= -200

x= 200

Question 14: The sum of two numbers is 11, and their difference is 5. Find the numbers.

Answer 14: Let one of the numbers from the two numbers be x.

Let the other number = 11 – x.

As per the given conditions, we have

x – (11 – x) = 5

⇒ x – 11 + x =5

⇒ 2x – 11 = 5

⇒ 2x = 5 + 11……………… (Transposing 11 to R.H.S.)

⇒ 2x = 16

⇒ x = 8

Hence, the required numbers for the given question are 8 and 11 – 8 = 3, respectively.

Question 15: 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?

Answer 15: Let the number of boys in a class be 7x

Let the number of girls in a class be 5x

According to the question,

7x = 5x + 8

7x -5x = 8

2x = 8

x=8/2

x = 4

Therefore, Number of boys = 7 x 4 =28

Number of girls = 5 x 4 =20

Total number of students = 20+ 28= 48

Question 16:- Linear equation in one variable has ________.

• only one variable with any power
• only one term with a variable
• only one variable with power 1
• only constant term

Answer 16:- (C) Only one variable with power 1

Question 17:- Arpita’s present age is thrice of Shilpa. If Shilpa’s age three years ago was x. Then Arpita’s present age is

• 3(x – 3)
• 3x + 3
• 3x – 9
• 3(x + 3)

Answer 17:- (D) 3(X + 3)

Given:- Shilpa’s age three years ago was x

Then, Shilpa’s present age is= x + 3

Arpita’s present age is thrice of Shilpa’s age =3(x + 3)

Question 18: Verify that x = 2 is the solution of the equation 4.4x – 3.8 = 5.

Answer 18: 4.4x – 3.8 = 5

Putting x = 2, we have

4.4 × 2 – 3.8 = 5

⇒ 8.8 – 3.8 = 5

⇒ 5 = 5

L.H.S. = R.H.S.

Hence, it is verified.

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The subject of Maths requires a lot of practice. The Maths of Classes 8, 9, and 10 are the milestones of fundamental knowledge. We suggest students access our platform Extramarks to access linear equations class 8 extra questions. By systematically solving questions and going through the required solutions, students will gain confidence to solve complex questions from linear equations with one variable.

Given below are a few benefits of frequently solving important questions Class 8 Maths Chapter 2:

• Our experienced Maths subject teachers have carefully assembled the most important questions Class 8 Maths Chapter 2 by analysing many past exam questions.
• The questions and solutions provided are based on the latest CBSE syllabus and as per CBSE guidelines. So students can completely count on it.
• The questions covered in our set of linear equations class 8 extra questions are based on various topics covered in linear equations in one variable. It is recommended for students to revise and clear all their doubts before solving these important questions.
• By solving our important questions Class 8 Maths Chapter 2, students will get a gist of how the paper will be prepared. Practising questions similar to the exam questions would help the students gain confidence, perform better in their exams, and eventually ace them.

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Q.1 Five years ago, the age of Neeraj and Neera was in the ratio 4:5. The ratio of their present ages is 5:6. Find their present ages.

Marks:4
Ans

$\begin{array}{l}\mathrm{Let}\text{the present age of Neeraj and Neera be}\\ \text{5x and 6x years respectively.}\\ \text{F}\mathrm{ive}\text{years before,}\\ \mathrm{Neeraj}‘\mathrm{s}\text{age=5x-5 years}\\ \mathrm{Neera}‘\mathrm{s}\text{age=6x-5 years}\\ \text{According to the question,}\\ \text{}\frac{\text{5x-5}}{\text{6x-5}}=\frac{4}{5}\\ 5\left(5\mathrm{x}5\right)=4\left(6\mathrm{x}5\right)\\ 25\mathrm{x}25=24\mathrm{x}20\\ 25\mathrm{x}24\mathrm{x}=2520\\ \mathrm{x}=5\\ \mathrm{Therefore},\text{}\\ \text{the present age of Neeraj=5x=5}×\text{5=25 years and}\\ \text{the present age of Neera=6x=6}×\text{5=30 years.}\end{array}$

Q.2 The sum of two numbers is 184. One-third of one number exceeds one-seventh of the other number by 8. Find the two numbers.

Marks:3
Ans

$\begin{array}{l}\mathrm{Let}\text{}\mathrm{one}\mathrm{number}\text{}\mathrm{be}\text{}\mathrm{x},\text{}\mathrm{then}\text{}\mathrm{the}\text{}\mathrm{other}\mathrm{be}\text{}\left(184\text{}\text{}\mathrm{x}\right).\\ \mathrm{According}\text{}\mathrm{to}\text{}\mathrm{the}\text{}\mathrm{question},\\ \frac{1}{3}\mathrm{x}=\frac{1}{7}\left(184\mathrm{x}\right)+8\\ \text{}\frac{\mathrm{x}}{3}=\frac{184\mathrm{x}+56}{7}\\ \text{}7\mathrm{x}\text{}=\text{}3\left(240\mathrm{x}\right)\\ \text{}7\mathrm{x}=7203\mathrm{x}\\ 10\mathrm{x}=720\\ \text{}\mathrm{x}\text{}=\text{}\frac{720}{10}\\ \text{}\mathrm{x}=\text{}72\\ \mathrm{So},\\ \text{}\mathrm{One}\text{}\mathrm{number}\text{}=\text{}72\\ \text{Secondnumber=184}\text{}72\\ \text{}=\text{}112.\end{array}$

Q.3

$\text{Solve the equation:}\frac{\frac{3}{4}y+7}{\frac{2}{5}y4}=\frac{5}{4}$

Marks:2
Ans

$\begin{array}{l}\frac{3y+28}{4}×\frac{5}{2y20}=\frac{5}{4}\\ \text{}20\left(3y+28\right)=20\left(2y20\right)\\ \text{}3y+28=2y20\\ \text{}y=48\end{array}$

Q.4 The sum of three consecutive multiples of 7 is 777, find the numbers.

Marks:2
Ans

Let the first multiple of 7 be x.

Other two will be (x + 7) and (x + 14).

According to the question,

$\begin{array}{l}\mathrm{x}+\left(\mathrm{x}+7\right)+\left(\mathrm{x}+14\right)=777\\ \text{}3\mathrm{x}=756\\ \text{}\mathrm{x}=252\\ \text{Hence, the three consecutive multiples of 7 are 252, 259 and 266.}\end{array}$

## 1. How many solutions does a linear equation in one variable have?

A linear equation with one variable has only one solution.

## 2. What is the formula of a linear equation in one variable?

The standard equation of a linear equation in one variable is given by ax+b=0.

There is only a single variable which is x.

## 3. What can I get from the Extramarks’ website?

Extramarks is one of the best educational platforms as it has its own archive of its educational resources, which assists students in acing their exams. You can get all the NCERT-related material like NCERT solutions, solved exemplar solutions, NCERT based mock tests, CBSE revision notes, important questions Class 8 Maths Chapter 3 on the Extramarks website. Apart from this, you can get comprehensive guidance from our subject experts and doubt-clearing sessions once sign up on our official website for any study resources.