*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 VariableExercise 2.1 Exercise 2.2 |

Subject | Mathematics |

Class | 8 ( Class – VIII ) |

Study Material | NCERT Solution ( Intext and End Exercise ) |

Number of Questions | Questions in Exercise 2.1 : 11 Questions Question in Exercise 2.2 : 4 Questions Total 15 Questions |

Text Book Name | Mathematics ( 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 :**

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**