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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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 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
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.
Benefits Of Solving Important Questions Class 8 Maths Chapter 2
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