Important Questions for CBSE Class 8 Maths Chapter 2 – 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.   

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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

Answer 9: (B) -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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