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

The NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.5 is meant to provide practice on questions related to ages in word problems. In this exercise, you will be able to read about situations regarding ages in real life and turn them into equations to solve for ages.

The exercise is according to the latest NCERT syllabus and will cover questions about a present age, future age or age difference between two persons in real life. Regular practice of the step by step solutions provided will help you understand how to form equations from word problems which will help in the school exams and other competitive exams like the Olympiads.

1.0Download NCERT Solutions Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.5: Free PDF

The  NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.5 contains age-related word problems provided in a step by step manner. Download the PDF for free from below:

2.0Key Concepts in Exercise 2.5 of Class 8 Maths Chapter 2

This exercise focuses on word problems related to age and the method of solving them using linear equations. Key concepts in this exercise include:

• Forming equations from age-based word problems
• Using one variable to represent age
• Applying basic operations to solve the equation
• Reading and understanding real-life age scenarios
• Verifying answers using the given conditions

3.0NCERT Class 8 Maths Chapter 2: Other Exercises

4.0NCERT Class 8 Maths Chapter 2 Exercise 2.5: Detailed Solutions

1. Solve the following linear equations:

1. x/2 − 1/5 = x/3 + 1/4
2. n/2 − 3n/4 + 5n/6 = 21
3. x+7 − 8x/3 = 17/6 − 5x/2
4. (x−5)/3 = (x−3)/5
5. (3t−2)/4 − (2t+3)/3 = 2/3 − t
6. m−(m−1)/2 = 1−(m−2)/3

Sol.

1. x/2 − 1/5 = x/3 + 1/4
   The denominators are 2, 5, 3, and 4. The LCM of these denominators is 60.
   Multiply both sides of the equation by 60:
   => 60(x/2 − 1/5) = 60(x/3 + 1/4)
   => (60 × x/2) − (60 × 1/5) = (60 × x/3) + (60 × 1/4)
   => 30x − 12 = 20x + 15
   Collect x terms on one side and constants on the other:
   => 30x − 20x = 15 + 12
   => 10x = 27
   => x=27/10
2. n/2 − 3n/4 + 5n/6 = 21
   The denominators are 2, 4, and 6. The LCM of these denominators is 12.
   Multiply both sides of the equation by 12:
   => 12(n/2 − 3n/4 + 5n/6) = 12(21)
   => (12 × n/2) − (12 × 3n/4) + (12 × 5n/6) = 252
   => 6n − 9n + 10n = 252
   Combine like terms:
   => (6−9+10)n = 252
   => 7n = 252
   => n = 252/7
   => n=36
3. x+7 − 8x/3 = 17/6 − 5x/2
   The denominators are 3, 6, and 2. The LCM of these denominators is 6.
   Multiply both sides of the equation by 6:
   => 6(x+7 − 8x/3) = 6(17/6 − 5x/2)
   => (6 × x) + (6 × 7) − (6 × 8x/3) = (6 × 17/6) − (6 × 5x/2)
   => 6x + 42 − 16x = 17 − 15x
   Combine like terms on each side:
   => (6x−16x) + 42 = 17 − 15x
   => −10x + 42 = 17 − 15x
   Collect x terms on one side and constants on the other:
   => −10x + 15x = 17 − 42
   => 5x = −25
   => x = −25/5
   => x=−5
4. (x−5)/3 = (x−3)/5
   Use cross multiplication:
   => 5(x−5) = 3(x−3)
   Distribute the numbers:
   => 5x − 25 = 3x − 9
   Collect x terms on one side and constants on the other:
   => 5x − 3x = −9 + 25
   => 2x = 16
   => x = 16/2
   => x=8
5. (3t−2)/4 − (2t+3)/3 = 2/3 − t
   The denominators are 4, 3, and 3. The LCM of these denominators is 12.
   Multiply both sides of the equation by 12:
   => 12((3t−2)/4) − 12((2t+3)/3) = 12(2/3) − 12(t)
   => 3(3t−2) − 4(2t+3) = 8 − 12t
   Distribute the numbers:
   => 9t − 6 − 8t − 12 = 8 − 12t
   Combine like terms on the LHS:
   => (9t−8t) + (−6−12) = 8 − 12t
   => t − 18 = 8 − 12t
   Collect t terms on one side and constants on the other:
   => t + 12t = 8 + 18
   => 13t = 26
   => t = 26/13
   => t=2
6. m−(m−1)/2 = 1−(m−2)/3
   The denominators are 2 and 3. The LCM of these denominators is 6.
   Multiply both sides of the equation by 6:
   => 6m − 6((m−1)/2) = 6(1) − 6((m−2)/3)
   => 6m − 3(m−1) = 6 − 2(m−2)
   Distribute the numbers:
   => 6m − 3m + 3 = 6 − 2m + 4
   Combine like terms on each side:
   => 3m + 3 = 10 − 2m
   Collect m terms on one side and constants on the other:
   => 3m + 2m = 10 − 3
   => 5m = 7
   => m=7/5

2. Simplify and solve the following linear equations:

1. 3(t−3)=5(2t+1)

2. 15(y−4)−2(y−9)+5(y+6)=0

3. 3(5z−7)−2(9z−11)=4(8z−13)−17

4. 0.25(4f−3)=0.05(10f−9)

Sol.

1. 3(t−3)=5(2t+1)

Distribute the numbers:

=> 3t − 9 = 10t + 5

Collect t terms on one side and constants on the other:

=> −9 − 5 = 10t − 3t

=> −14 = 7t

=> t = −14/7

=> t=−2

2. 15(y−4)−2(y−9)+5(y+6)=0
Distribute the numbers:
=> 15y − 60 − 2y + 18 + 5y + 30 = 0
Combine like terms:
=> (15y − 2y + 5y) + (−60 + 18 + 30) = 0
=> 18y + (−12) = 0
=> 18y − 12 = 0
=> 18y = 12
=> y = 12/18
=> y=2/3

3. 3(5z−7)−2(9z−11)=4(8z−13)−17
Distribute the numbers:
=> 15z − 21 − 18z + 22 = 32z − 52 − 17
Combine like terms on each side:
=> (15z − 18z) + (−21 + 22) = 32z + (−52 − 17)
=> −3z + 1 = 32z − 69
Collect z terms on one side and constants on the other:
=> 1 + 69 = 32z + 3z
=> 70 = 35z
=> z = 70/35
=> z=2

4. 0.25(4f−3)=0.05(10f−9)
Distribute the numbers:
=> 0.25 × 4f − 0.25 × 3 = 0.05 × 10f − 0.05 × 9
=> f − 0.75 = 0.5f − 0.45
Collect f terms on one side and constants on the other:
=> f − 0.5f = −0.45 + 0.75
=> 0.5f = 0.30
=> f = 0.30/0.5
To remove decimals, multiply numerator and denominator by 100:
=> f = 30/50
=> f = 3/5
=> f=0.6

5.0Key Features and Benefits of Class 8 Maths Chapter 2 Exercise 2.5

• Demonstrates how to solve age-related problems encountered in real-life situations with linear equations.
• The exercise is based on the latest NCERT syllabus for CBSE Class 8 Maths.
• Every solution is explained using step-by-step manner for easy understanding.
• They are also useful for preparing for any Maths Olympiad where age and logic questions may arise.
• Regular practice develops excellent problem solving skills and confidence in solving such problems.