NCERT Solutions for Class 8 Maths Chapter 2 - Linear Equations in One Variable Exercise 2.5

The NCERT Solutions for Class 8 Maths Chapter 2 – Linear Equations in One Variable are prepared by the subject experts at BYJU’S to help the students fall in love with the subject. These NCERT Solutions make the students familiar with various concepts. CBSE Class 8 Maths Chapter 2 consists of questions that comprise various topics carrying different marks, so students must be well aware of all concepts to score well in the annual exam. Practising the NCERT solutions repeatedly is an easy way to learn the concepts covered in Class 8 Maths Chapter 2.

NCERT Solutions for Class 8 Maths Chapter 2 – Linear Equations in One Variable Exercise 2.5

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Access Other Exercise Solutions of Class 8 Maths Chapter 2 – Linear Equations in One Variable

Exercise 2.1 Solutions 12 Questions (12 Short Answer Questions)
Exercise 2.2 Solutions 16 Questions (6 Long Answer Questions, 10 Short Answer Questions)
Exercise 2.3 Solutions 10 Questions (3 Long Answer Questions, 7 Short Answer Questions)
Exercise 2.4 Solutions 10 Questions (4 Long Answer Questions, 6 Short Answer Questions)
Exercise 2.6 Solutions 7 Questions (1 Long Answer Question, 6 Short Answer Questions)

Access Answers to NCERT Class 8 Maths Chapter 2 – Linear Equations in One Variable Exercise 2.5 Page Number 33

Solve the following linear equations.

1. x/2 – 1/5 = x/3 + ¼

Solution:

x/2 – 1/5 = x/3 + ¼

⇒ x/2 – x/3 = ¼+ 1/5

⇒ (3x – 2x)/6 = (5 + 4)/20

⇒ 3x – 2x = 9/20 × 6

⇒ x = 54/20

⇒ x = 27/10

2. n/2 – 3n/4 + 5n/6 = 21

Solution:

n/2 – 3n/4 + 5n/6 = 21

⇒ (6n – 9n + 10n)/12 = 21

⇒ 7n/12 = 21

⇒ 7n = 21 × 12

⇒ n = 252/7

⇒ n = 36

3. x + 7 – 8x/3 = 17/6 – 5x/2

Solution:

x + 7 – 8x/3 = 17/6 – 5x/2

⇒ x – 8x/3 + 5x/2 = 17/6 – 7

⇒ (6x – 16x + 15x)/6 = (17 – 42)/6

⇒ 5x/6 = – 25/6

⇒ 5x = – 25

⇒ x = – 5

4. (x – 5)/3 = (x – 3)/5

Solution:

(x – 5)/3 = (x – 3)/5

⇒ 5(x-5) = 3(x-3)

⇒ 5x-25 = 3x-9

⇒ 5x – 3x = -9+25

⇒ 2x = 16

⇒ x = 8

5. (3t – 2)/4 – (2t + 3)/3 = 2/3 – t

Solution:

(3t – 2)/4 – (2t + 3)/3 = 2/3 – t

⇒ ((3t – 2)/4) × 12 – ((2t + 3)/3) × 12

⇒ (3t – 2) × 3 – (2t + 3) × 4 = 2 × 4 – 12t

⇒ 9t – 6 – 8t – 12 = 8 – 12t

⇒ 9t – 6 – 8t – 12 = 8 – 12t

⇒ t – 18 = 8 – 12t

⇒ t + 12t = 8 + 18

⇒ 13t = 26

⇒ t = 2

6. m – (m – 1)/2 = 1 – (m – 2)/3

Solution:

m – (m – 1)/2 = 1 – (m – 2)/3

⇒ m – m/2 – 1/2 = 1 – (m/3 – 2/3)

⇒ m – m/2 + ½ = 1 – m/3 + 2/3

⇒ m – m/2 + m/3 = 1 + 2/3 – ½

⇒ m/2 + m/3 = ½ + 2/3

⇒ (3m + 2m)/6 = (3 + 4)/6

⇒ 5m/6 = 7/6

⇒ m = 7/6 × 6/5

⇒ m = 7/5

Simplify and solve the following linear equations.

7. 3(t – 3) = 5(2t + 1)

Solution:

3(t – 3) = 5(2t + 1)

⇒ 3t – 9 = 10t + 5

⇒ 3t – 10t = 5 + 9

⇒ -7t = 14

⇒ t = 14/-7

⇒ t = -2

8. 15(y – 4) –2(y – 9) + 5(y + 6) = 0

Solution:

15(y – 4) –2(y – 9) + 5(y + 6) = 0

⇒ 15y – 60 -2y + 18 + 5y + 30 = 0

⇒ 15y – 2y + 5y = 60 – 18 – 30

⇒ 18y = 12

⇒ y = 12/18

⇒ y = 2/3

9. 3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17

Solution:

3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17

⇒ 15z – 21 – 18z + 22 = 32z – 52 – 17

⇒ 15z – 18z – 32z = -52 – 17 + 21 – 22

⇒ -35z = -70

⇒ z = -70/-35

⇒ z = 2

10. 0.25(4f – 3) = 0.05(10f – 9)

Solution:

0.25(4f – 3) = 0.05(10f – 9)

⇒ f – 0.75 = 0.5f – 0.45

⇒ f – 0.5f = -0.45 + 0.75

⇒ 0.5f = 0.30

⇒ f = 0.30/0.5

⇒ f = 3/5

⇒ f = 0.6

Exercise 2.5 of NCERT Solutions for Class 8 Maths Chapter 2 – Linear Equations in One Variable is based on the method of reducing the equations to a simpler form. The reduction of equations to simpler forms helps the students in solving problems in an effective way. Students can understand the method of reducing linear equations thoroughly by solving the questions in Exercise 2.5.

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NCERT Solutions for Class 8 Maths

NCERT Solutions for Class 8