Class 8 Maths Chapter 2 Linear Equation in One Variable Soln

Chapter 2 of Maths book of Class 8 is “Linear Equation in one Variable”. Here you will get the NCERT Solutions of Maths Chapter 2 Linear Equation in One Variable Class 8

Class 8 Maths Chapter 2 : Linear Equation in One Variable

Class 8 Maths Chapter 2 is “Linear Equation in One Variable” . Linear Equation in One Variable Class 8 Maths is the most important chapter for exam point of view. The ncert exercise of this chapter consist of the questions which are related to the topics – Linear Equation, Finding value of x . Solution of problems based on linear polynomials, etc

Class 8 Maths Chapter 2 NCERT Solutions

Chapter Chapter 2 : Linear Equation in One Variable
Exercise 2.1
Exercise 2.2
SubjectMathematics
Class 8 ( Class – VIII )
Study MaterialNCERT Solution ( Intext and End Exercise )
Number of QuestionsQuestions in Exercise 2.1 : 11 Questions
Question in Exercise 2.2 : 4 Questions
Total 15 Questions
Text Book NameMathematics ( NCERT Text Book for Class 8 )

End Exercise NCERT Solutions

EXERCISE 2.1

Solve the Following

Q1. x – 2 = 7
Solution :
Given equation is as follows

x – 2 = 7
Take  -2 on the right-hand side , It becomes plus 2  (Transposing  – 2 to the right-hand side)
We get,
X = 7 + 2
X = 9
Solution of the above question is  X = 9 .

Q2. Y + 3 = 10
Solution :
Given equation is as follows
Y – 3 = 10
 Transpose   – 3 to the Right-Hand Side
 We get ,
Y = 10  +  3
Y = 13
Solution of the above question is   Y= 13

Q3. 6 = z + 2
Solution :
Given equation is as follows
6 = z + 2
Transpose 2 on the Left-Hand Side
We get ,
6 – 2 = z
4 = z

Z= 4
Solution of the above question is Z = 4

Q4. 3/7 + x  = 17/7
Solution :
Given equation is as follows
3/7 + x = 17/7
Transpose 3/7 on the Right-hand Side
We get ,
X = 17/7  – 3/7
X=  (17 – 3 ) / 7  ————— By taking LCM of 7 and  7
X=  (14 ) / 7
X=  2
Solution of the above question is  X = 2

Q5. 6x = 12
Solution :
Given equation is as follows
6x = 12
Dividing  by six on both sides
We get,
X= 12/6
X = 2
Solution of the above question is X = 2

Q6. t/5 = 10
Solution :
Given equation is as follows
t/5 = 10
Multiplying by 5 on both side
We get,
t  = 10 * 5
t = 50
Solution of the  above question is  t = 50

Q7. 2x / 3 = 18
Solution : 

Given equation is as follows
2x / 3 = 18
Multiplying by  3 on Both side
We get,
2x  = 18 * 3
2x = 54
Dividing by 2 on both sides
X = 54 / 2
X = 27
Solution of the above question is x = 27

Q8. 1.6  =  y/1.5
Solution :
Given equation is as follows
1.6  = y / 1.5
Multiplying by 1.5 on both sides
We get,
1.6 * 1.5 = y
2.4 = y
y = 2.4
Solution of the above question is  y = 2.4 

Q9. 7x – 9 = 16
Solution :
Given equation is as follow
7x – 9 = 16
Transposing -9 to the Right-Hand Side
We get,
7x = 16 + 9
7x = 25
Dividing by 7 on both sides
We get,
X = 25/7
Solution of the above question is X = 25/7

Q10. 14y – 8 = 13
Solution :
Given equation is as follow
14y – 8 = 13
Transposing -8 on the Right-Hand Side
 We get,
14y = 13 + 8
14y = 21
Dividing by 14 on both sides
We get,
Y = 21 / 14
Solving further
We get,
Y= 3/2
Solution of the above question is  y = 3/2

Q11. 17 + 6p = 9
Solution :
Given equation is as follows:
17 + 6p = 19
Transpose 17 on the right-hand side
We get,
6p = 19 – 17 = 2
Dividing  by 6 on both sides
We get,
P = 2/6
P = 1/3
Solution of the above  question is p= 13

Q12. x/3 + 1  = 7/15
Solution :
Given equation is as follows
x/3 + 1 = 7/ 15
Transpose 1 on the right-hand side
We get,
x/3 = 7/15 – 1
x/3 = (7-15) /15                ————– By taking LCM of 15 and 1
x/3 = 8/15
Multiplying by 3 on both sides|
We get,
X= (8/15) * 3
X= 8/5
Solution of the above question is  x = 8/5

EXERCISE 2.2

Q1. If you subtract 1 /2 from a number and multiply the result by 1/ 2 , you get 1 /8 . What is the number?
Solution :
Let the number be x .
According to question,
(x – 1/2)*  (1/2) = 1/8
Multiplying 2 on both sides           
We get,
x – ½ = (1/8) *2
x- ½ = ¼
Transposing ½  on the Right-hand side
We get,
x = ¼ + ½
X = ¾
Solution of the above question is  x= ¾ .

Q2. 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 the breadth of the pool?
Solution:
According to the question
Perimeter of swimming pool : 154 m
Length =  2 * breadth +2
l  = 2b +2
Now,
We know that perimeter of a rectangle is given by
Perimeter (p) = 2*(length + breadth )

Let length of the swimming pool  = l
Let breadth of the swimming pool  = b
Now,
p = 154
2*( l + b ) = 154
2*[ (3b + 2) + b ] = 154  ——————– Given l =2b +2
(6b + 4 ) = 154
Transposing  4 on the other side
We get,
6b = 154 -4
6b = 150
Dividing by 6 on both sides
We get,
b = 150/6
b= 25
Solution  of the given question is
Breadth  = 25
Length = 2b +2
            = 2*25 +2
            = 52

Q3. The base of an isosceles triangle is 4/3 cm . The perimeter of the triangle is 62/15 cm . What is the length of either of the remaining equal sides?
Solution :

class 8 maths exercise 2.2 question 3

According to question
Base of isosceles triangle is 4/3 cm.
Perimeter of isosceles triangle = 62/15 cm
We have to find the other sides of the triangle   
We know that the two sides of isosceles triangle are equal
Therefore,
a = c
b= 4/3 cm   —————- [ Given ]
Perimeter p = 62/15
Now ,
Perimeter = a+b+c
62/15 = a+b+c
62/15 = c + 4/3 + c        ————– [Replacing ‘a’ by ‘c’ because a = c ]
62/15 = 2c + 4/3
Transposing 4/3 on the other side
62/15  – 4/3  = 2c
(62 – 5*4)/ 15  =  2c  —————– [Solving by taking the LCM of 15 and 3 ]
42/15 = 2c
Dividing by 2 on both sides
We get,
(42/15 ) / 2 = c
c = 7/5
Solution of the  above question is
a = c = 7/5 cm

Q4. Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.
Solution
Let the first number be x .
Let the other number be y .
According to question,
x + y = 95 —— 1. [ Given]
x = y +15  —— 2. [Given ]
Now ,
Putting the values of x from equation 2 in the equation 1
We get,
x + y = 95
y + 15 + y = 95
2y + 15  = 95
Transposing 15 on the other side
We get,
2y = 95 – 15
2y = 80
Dividing by 2 on both sides
We get,
y = 80 / 2 = 40
y  = 40
x  = y + 15
x  = 40 + 15
Solution of the above question is
Y = 40
x  = 55