Class 9 Maths Chapter 2 Exercise 2.4 Solutions | Linear Growth and Linear Decay | NCERT Ganita Manjari 2026

NCERT GANITA MANJARI 2026 • CLASS 9 MATHEMATICS

Looking for complete Class 9 Maths Chapter 2 Exercise 2.4 Solutions? Here you will get step-by-step solutions of all questions from the latest NCERT Ganita Manjari (2026) textbook. These solutions explain Linear Growth and Linear Decay patterns through real-life examples such as plant growth, mobile depreciation, village population growth, and telecom prepaid balance reduction.

these strong>Exercise 2.4 Solutions help students understand linear expressions, constant rate of change, table patterns, and the formation of linear models in an easy and exam-oriented manner.

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In Class 9 Maths Chapter 2 Exercise 2.4, we study Linear Growth and Linear Decay through practical examples.

About Class 9 Maths Chapter 2 Exercise 2.4

Class 9 Maths Chapter 2 Exercise 2.4 Solutions introduce students to the concepts of Linear Growth and Linear Decay through practical real-life situations. In this exercise, students learn how quantities increase or decrease by a fixed amount over equal intervals and how such situations can be represented using linear expressions.

The questions in Exercise 2.4 of NCERT Ganita Manjari (2026) are based on everyday examples such as plant growth, mobile phone depreciation, village population growth, and telecom prepaid balance reduction. These examples of Class 9 Maths Chapter 2 Exercise 2.4 Solutions help students understand the practical applications of linear polynomials and constant rate of change.

Class 9 Maths Chapter 2 Exercise 2.4 Solutions

Question 1

Suppose a plant has height 1.75 feet and it grows by 0.5 feet each month.

(i) Find the height after 7 months.

(ii) Make a table of values for t varying from 0 to 10 months and show how the height, h, increases every month.

(iii) Find an expression that relates h and t, and explain why it represents linear growth.

Given:

• Initial height of the plant = 1.75 feet
• Monthly growth = 0.5 feet

To Find:

• The height after 7 months.
• The table showing the growth pattern.
• The linear expression relating h and t.

Solution:

Initially, the height of the plant is

1.75 feet

The plant grows by

0.5 feet every month.

(i) Height after 7 months

h = 1.75 + (7 × 0.5)

= 1.75 + 3.5

= 5.25 feet

Therefore, the height after 7 months is

5.25 feet

(ii) Table of Values

Therefore, after t months,

h = 1.75 + 0.5t

(iii) Why does this represent Linear Growth?

The height of the plant increases by a fixed amount of

0.5 feet every month.

Since the increase is constant, the relation

h = 1.75 + 0.5t

represents a Linear Growth Pattern.

Key Concept Used: A quantity that increases by a fixed amount over equal intervals represents linear growth.

Common Mistake: Students often multiply the initial value by the number of months instead of adding the monthly increase.

Exam Tip: In linear growth problems, first write the initial value and then add the fixed increase for each time interval.

Question 2

A mobile phone is bought for ₹10,000. Its value decreases by ₹800 every year.

(i) Find the value of the phone after 3 years.

(ii) Make a table of values for t varying from 0 to 8 years and show how the value of the phone, v, depreciates with time.

(iii) Find an expression that relates v and t, and explain why it represents linear decay.

Given:

• Initial value of the mobile phone = ₹10,000
• Decrease in value every year = ₹800

To Find:

• The value of the phone after 3 years.
• The table showing the depreciation pattern.
• The linear expression relating v and t.

Solution:

Initially, the value of the mobile phone is

₹10,000

The value decreases by

₹800 every year.

(i) Value after 3 years

v = 10000 − (3 × 800)

= 10000 − 2400

= ₹7600

Therefore, the value of the mobile phone after 3 years is

₹7600

(ii) Table of Values

Therefore, after t years,

v = 10000 − 800t

(iii) Why does this represent Linear Decay?

The value of the mobile phone decreases by a fixed amount of

₹800 every year.

Since the decrease is constant, the relation

v = 10000 − 800t

represents a Linear Decay Pattern.

Key Concept Used: A quantity that decreases by a fixed amount over equal intervals represents linear decay.

Common Mistake: Students often add ₹800 instead of subtracting it while calculating depreciation.

Exam Tip: In depreciation problems, always start with the original value and subtract the fixed decrease occurring in each time interval.

Question 3

The initial population of a village is 750. Every year, 50 people move from a nearby city to the village.

(i) Find the population of the village after 6 years.

(ii) Make a table of values for t varying from 0 to 10 years and show how the population, P, increases every year.

(iii) Find an expression that relates P and t, and explain why it represents linear growth.

Given:

• Initial population of the village = 750
• Increase in population every year = 50 people

To Find:

• The population after 6 years.
• The table showing the population growth.
• The linear expression relating P and t.

Solution:

Initially, the population of the village is

750

Every year, the population increases by

50 people.

(i) Population after 6 years

P = 750 + (6 × 50)

= 750 + 300

= 1050

Therefore, the population of the village after 6 years is

1050

(ii) Table of Values

Therefore, after t years,

P = 750 + 50t

(iii) Why does this represent Linear Growth?

The population of the village increases by a fixed number of

50 people every year.

Since the increase is constant, the relation

P = 750 + 50t

represents a Linear Growth Pattern.

Key Concept Used: A quantity that increases by a fixed amount over equal intervals represents linear growth.

Common Mistake: Students often forget to include the initial population while writing the linear expression.

Exam Tip: In growth problems, always write the initial value first and then add the fixed increase occurring in each time interval.

Question 4

A telecom company charges ₹600 for a certain recharge scheme. This prepaid balance is reduced by ₹15 each day after the recharge.

(i) Write an equation that models the remaining balance b(x) after using the scheme for x days. Explain why it represents linear decay.

(ii) After how many days will the balance run out?

(iii) Make a table of values for x varying from 1 to 10 days and show how the balance b(x) reduces with time.

Given:

• Initial prepaid balance = ₹600
• Balance decreases by ₹15 every day

To Find:

• The equation representing the remaining balance.
• The number of days after which the balance becomes zero.
• A table showing the decrease in balance.

Solution:

Initially, the prepaid balance is

₹600

The balance decreases by

₹15 every day.

(i) Equation of the Remaining Balance

After 1 day,

Balance = 600 − 15

= ₹585

After 2 days,

Balance = 600 − 2 × 15

= ₹570

Similarly, after x days,

b(x) = 600 − 15x

Since the balance decreases by a fixed amount of ₹15 every day, the relation

b(x) = 600 − 15x

represents a Linear Decay Pattern.

(ii) When will the balance run out?

The balance becomes zero when

b(x) = 0

Substituting in the equation,

600 − 15x = 0

15x = 600

x = 40

Therefore, the prepaid balance will run out after

40 days

(iii) Table of Values

Key Concept Used: A quantity that decreases by a fixed amount over equal intervals represents linear decay.

Common Mistake: Students often write 600 + 15x instead of 600 − 15x while forming the equation.

Exam Tip: For linear decay problems, always begin with the initial value and subtract the fixed amount that decreases in each time interval.

📚What We Learned in Class 9 Maths Chapter 2 Exercise 2.4 solutions

Frequently Asked Questions (FAQs)

What is the answer of Question 1 of Class 9 Maths Chapter 2 Exercise 2.4?

In Question 1, the height of the plant after 7 months is 5.25 feet and the linear growth equation is h = 1.75 + 0.5t.

What is the linear decay equation in Question 2 of Exercise 2.4?

The value of the mobile phone decreases by ₹800 every year. Therefore, the linear decay equation is v = 10000 − 800t.

What is the population equation in Question 3 of Exercise 2.4?

The population increases by 50 people every year. Hence, the linear growth equation is P = 750 + 50t.

After how many days will the prepaid balance become zero in Question 4?

The prepaid balance becomes zero after 40 days using the equation b(x)=600−15x.

What is linear growth in Class 9 Maths?

Linear growth occurs when a quantity increases by a fixed amount after equal intervals of time. Examples include plant growth and population growth.

What is linear decay in Class 9 Maths?

Linear decay occurs when a quantity decreases by a fixed amount after equal intervals of time. Examples include mobile depreciation and prepaid balance reduction.

How can we identify a linear pattern?

A pattern is linear if the increase or decrease between consecutive values remains constant.

Why is Exercise 2.4 important in Class 9 Maths Chapter 2?

Exercise 2.4 helps students understand linear growth, linear decay, linear expressions, and real-life applications of linear polynomials.

What are some real-life examples of linear growth and linear decay?

Examples include plant growth, village population increase, mobile phone depreciation, salary increments, and prepaid balance reduction.

Are questions from Exercise 2.4 important for CBSE exams?

Yes. Questions based on linear growth, linear decay, tables, and forming linear expressions are frequently used in CBSE examinations and higher mathematics.

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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