Class 9 Maths Chapter 2 Exercise 2.5 Solutions (NCERT Ganita Manjari 2026)

Master the concept of Linear Relationships through these easy-to-understand solutions based on the latest NCERT Ganita Manjari (2026). In this Class 9 Maths Chapter 2 Exercise 2.5 Solutions, you will learn how to determine the values of a and b in the equation y = ax + b using real-life situations.

📚 Table of Contents

This Class 9 Maths Chapter 2 Exercise 2.5 Solutions contains 3 important questions based on Linear Relationships (y = ax + b). Click on any question below to jump directly to its complete step-by-step solution.

About Class 9 Maths Chapter 2 Exercise 2.5

These Class 9 Maths Chapter 2 Exercise 2.5 Solutions help students understand how to represent a linear relationship using the equation y = ax + b. In this exercise, you will learn how to determine the values of a and b from two given observations and write the required linear equation step by step according to the latest NCERT Ganita Manjari (2026) syllabus.

The questions are based on real-life applications of linear relationships such as online learning platform charges, gym membership fees, and the Celsius–Fahrenheit temperature relationship. By solving these problems, students learn how fixed charges and variable charges combine to form the equation y = ax + b, an important concept for higher algebra and CBSE examinations.

Important Concepts Covered

• Linear Relationship (y = ax + b)
• Finding the values of a and b
• Fixed Cost and Variable Cost
• Forming Linear Equations from Real-Life Situations
• Online Learning Platform Billing
• Gym Membership and Hourly Charges
• Celsius–Fahrenheit Conversion Formula
• CBSE Step-by-Step Algebraic Method

Class 9 Maths Chapter 2 Exercise 2.5 Solutions (Step-by-Step)

Question 1

A learning platform charges a fixed monthly fee and an additional cost per digital learning module accessed. A student observes that when she accessed 10 modules, her bill was ₹400. When she accessed 14 modules, her bill was ₹500. If the monthly bill y depends on the number of modules accessed, x, according to the relation y = ax + b, find the values of a and b.

---

Given:

• x represents the number of digital learning modules accessed.
• y represents the monthly bill (in ₹).
• When x = 10, y = 400
• When x = 14, y = 500
• Relation: y = ax + b

To Find:

The values of a and b.

---

Solution:

Step 1: Form the two linear equations

The given linear relation is

y = ax + b

Substituting x = 10 and y = 400,

400 = 10a + b

…(1)

Substituting x = 14 and y = 500,

500 = 14a + b

…(2)

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Step 2: Find the value of a

Subtract equation (1) from equation (2).

500 − 400 = (14a + b) − (10a + b)

100 = 4a

a = 25

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Step 3: Find the value of b

Substitute a = 25 in equation (1).

400 = 10(25) + b

400 = 250 + b

b = 150

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Step 4: Write the required linear relationship

Therefore,

a = 25

b = 150

Hence, the required linear relationship is

y = 25x + 150

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Verification:

For x = 10,

y = 25(10) + 150 = 400 ✓

For x = 14,

y = 25(14) + 150 = 500 ✓

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Key Concept Used: Substitute the given values into the linear equation y = ax + b, form two linear equations, eliminate one variable, and then find the values of a and b.

Common Mistake: Students often substitute the values of x and y incorrectly or forget to subtract the two equations to eliminate one variable.

Exam Tip: Whenever two observations and the relation y = ax + b are given, first form two equations, eliminate one variable, and finally substitute back to find the other variable.

Question 2

A gym charges a fixed monthly fee and an additional cost per hour for using the badminton court. A student using the gym observed that when she used the badminton court for 10 hours, her bill was ₹800. When she used it for 15 hours, her bill was ₹1100. If the monthly bill y depends on the hours of the use of the badminton court, x, according to the relation y = ax + b, find the values of a and b.

---

Given:

• x represents the number of hours the badminton court is used.
• y represents the monthly bill (in ₹).
• When x = 10, y = 800
• When x = 15, y = 1100
• Relation: y = ax + b

To Find:

The values of a and b.

---

Solution:

Step 1: Form the two linear equations

The given linear relation is

y = ax + b

Substituting x = 10 and y = 800,

800 = 10a + b

…(1)

Substituting x = 15 and y = 1100,

1100 = 15a + b

…(2)

---

Step 2: Find the value of a

Subtract equation (1) from equation (2).

1100 − 800 = (15a + b) − (10a + b)

300 = 5a

a = 60

---

Step 3: Find the value of b

Substitute a = 60 in equation (1).

800 = 10(60) + b

800 = 600 + b

b = 200

---

Step 4: Write the required linear relationship

Therefore,

a = 60

b = 200

Hence, the required linear relationship is

y = 60x + 200

---

Verification:

For x = 10,

y = 60(10) + 200 = 800 ✓

For x = 15,

y = 60(15) + 200 = 1100 ✓

---

Key Concept Used: Substitute the given values into the linear equation y = ax + b, form two linear equations, eliminate one variable, and then find the values of a and b.

Common Mistake: Students often substitute the values of x and y incorrectly or make mistakes while subtracting the two equations to eliminate one variable.

Exam Tip: Whenever two observations and the relation y = ax + b are given, first form two equations, eliminate one variable, and finally substitute the obtained value into either equation to find the other variable.

Question 3

Consider the relationship between temperature measured in degrees Celsius (°C) and degrees Fahrenheit (°F), which is given by

°C = a°F + b

Find the values of a and b, given that ice melts at 0°C and 32°F, and water boils at 100°C and 212°F.

---

Given:

• Temperature relation: °C = a°F + b
• When °F = 32, °C = 0
• When °F = 212, °C = 100

To Find:

The values of a and b.

---

Solution:

Step 1: Form the two linear equations

The given linear relation is

°C = a°F + b

Substituting °F = 32 and °C = 0,

0 = 32a + b

…(1)

Substituting °F = 212 and °C = 100,

100 = 212a + b

…(2)

---

Step 2: Find the value of a

Subtract equation (1) from equation (2).

100 − 0 = (212a + b) − (32a + b)

100 = 180a

a = 100 180 = 5 9

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Step 3: Find the value of b

Substitute 5 9 for a in equation (1).

0 = 32 × 5 9 + b

0 = 160 9 + b

b = − 160 9

---

Step 4: Write the required linear relationship

a = 5 9

b = − 160 9

Hence, the required linear relationship is

°C = 5 9 °F − 160 9

---

Verification:

When °F = 32,

°C = 5 9 × 32 − 160 9 = 0 ✓

When °F = 212,

°C = 5 9 × 212 − 160 9 = 100 ✓

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Key Concept Used: Form two linear equations by substituting the given temperature values into the relation °C = a°F + b. Solve the equations to find a and b.

Common Mistake: Students often interchange the values of Celsius and Fahrenheit while substituting into the given relation.

Exam Tip: Always substitute the temperatures exactly as given. Here, Celsius is expressed in terms of Fahrenheit, so Fahrenheit values are substituted first.

📘 What We Have Learnt in Class 9 Maths Chapter 2 Exercise 2.5 Solutions

In this Class 9 Maths Chapter 2 Exercise 2.5 Solutions, we learned how to determine the values of a and b in the linear relationship y = ax + b using two given observations. We also understood how linear equations can represent practical situations such as monthly bills, service charges, and temperature conversion.

Frequently Asked Questions (FAQs)

1. What is the main concept taught in Exercise 2.5?

Exercise 2.5 teaches how to determine the values of a and b in the linear relationship y = ax + b using two given observations.

2. Why do we substitute two sets of values in y = ax + b?

Two observations help us form two linear equations. Solving these equations gives the values of a and b.

3. Which method is used to solve Exercise 2.5 questions?

NCERT uses the elimination method. First form two equations, eliminate one variable, and then substitute back to find the other variable.

4. What does the constant a represent in y = ax + b?

The constant a represents the rate of change or variable charge, while b represents the fixed value or initial amount.

5. Why is the Celsius–Fahrenheit question included in this exercise?

It shows that temperature conversion can also be represented by a linear relationship and solved using the same algebraic method.

6. Is Exercise 2.5 important for CBSE Class 9 exams?

Yes. Questions based on forming linear equations and finding unknown constants are important for school examinations and help build concepts for higher algebra.

7. What are the common mistakes students make in Exercise 2.5?

Students often substitute the values incorrectly, subtract the equations in the wrong order, or forget to write the final linear relationship after finding a and b.

8. How can I solve Exercise 2.5 questions quickly in the exam?

Write the given relation first, form two equations carefully, eliminate one variable, substitute back to find the other, and finally verify the obtained equation whenever possible.

Official Learning Resources

For additional reference and official curriculum information, students can visit:

• NCERT Official Website
• CBSE Academic Official Website

These official educational resources provide authentic syllabus information, curriculum updates, and academic guidelines.

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