Class 9 Maths Ganita Manjari Chapter 2 Exercise 2.4 Solutions: Introduction to Linear Polynomials

Class 9 Maths Ganita Manjari Chapter 2 Exercise 2.4 Solutions cover linear growth and linear decay from Introduction to Linear Polynomials. This exercise helps students write linear expressions for situations where a quantity increases or decreases by a fixed amount over equal intervals.

In Ganita Manjari Class 9 Chapter 2 Exercise 2.4, students work with plant height, mobile phone value, village population and prepaid balance. These examples show how linear polynomial Class 9 concepts are used in real-life situations. The solutions below explain each answer step by step so students can understand linear growth Class 9, linear decay Class 9 and related algebraic expressions Class 9 clearly.

Key Takeaways

Linear Growth: A quantity increases by the same amount at equal intervals.
Linear Decay: A quantity decreases by the same amount at equal intervals.
Table of Values: Tables show how a linear expression changes over time.
Zero Value: In decay questions, setting the expression equal to 0 gives the time when the quantity runs out.

Class 9 Maths Ganita Manjari Chapter 2 Exercise 2.4 Solutions Structure 2026

Class 9 Maths Ganita Manjari Chapter 2 Exercise 2.4 Solutions

Exercise Set 2.4 is based on the idea that a linear expression can model growth or decline.

A situation shows linear growth when a quantity increases by a fixed amount over equal intervals. A situation shows linear decay when a quantity decreases by a fixed amount over equal intervals. These Class 9 Maths linear growth and decay answers help students practise how to make tables, write expressions and identify whether a pattern is increasing or decreasing.

Exercise 2.4 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.

Solution:

Initial height of the plant = 1.75 feet

Growth per month = 0.5 feet

Height after 7 months:

Height = 1.75 + 0.5 × 7

Height = 1.75 + 3.5

Height = 5.25 feet

Answer: The height of the plant after 7 months is 5.25 feet.

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

Solution:

The height increases by 0.5 feet every month.

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

Solution:

Initial height = 1.75 feet

Growth after t months = 0.5t

So,

h = 1.75 + 0.5t

This represents linear growth because the height increases by the same amount, 0.5 feet, every month.

Answer: The expression is h = 1.75 + 0.5t. It represents linear growth because the plant height increases by a fixed amount every month.

Exercise 2.4 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.

Solution:

Initial value of phone = ₹10,000

Decrease every year = ₹800

Decrease after 3 years:

800 × 3 = 2400

Value after 3 years:

10000 − 2400 = 7600

Answer: The value of the phone after 3 years is ₹7,600.

(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.

Solution:

The phone value decreases by ₹800 every year.

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

Solution:

Initial value = ₹10,000

Decrease after t years = 800t

So,

v = 10000 − 800t

This represents linear decay because the phone value decreases by the same amount, ₹800, every year.

Answer: The expression is v = 10000 − 800t. It represents linear decay because the value decreases by a fixed amount every year.

Exercise 2.4 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.

Solution:

Initial population = 750

Increase every year = 50

Increase after 6 years:

50 × 6 = 300

Population after 6 years:

750 + 300 = 1050

Answer: The population of the village after 6 years is 1050.

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

Solution:

The population increases by 50 every year.

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

Solution:

Initial population = 750

Increase after t years = 50t

So,

P = 750 + 50t

This represents linear growth because the population increases by the same number, 50 people, every year.

Answer: The expression is P = 750 + 50t. It represents linear growth because the population increases by a fixed amount every year.

Exercise 2.4 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.

Solution:

Initial prepaid balance = ₹600

Balance reduced each day = ₹15

Balance reduced after x days = 15x

So, the remaining balance is:

b(x) = 600 − 15x

This represents linear decay because the balance decreases by the same amount, ₹15, every day.

Answer: The equation is b(x) = 600 − 15x. It represents linear decay because the balance reduces by a fixed amount each day.

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

Solution:

The balance runs out when:

b(x) = 0

So,

600 − 15x = 0

15x = 600

x = 600 / 15

x = 40

Answer: The balance will run out after 40 days.

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

Solution:

Using:

b(x) = 600 − 15x

Answer: The balance decreases by ₹15 every day, so it shows linear decay.

Final Answers for Exercise 2.4

Concept Used in Introduction to Linear Polynomials Exercise 2.4

Introduction to Linear Polynomials Exercise 2.4 focuses on linear growth and linear decay. These are situations where a value changes by a constant amount over equal time intervals.

Important points:

• Linear growth means a quantity increases by a fixed amount.
• Linear decay means a quantity decreases by a fixed amount.
• A growth expression usually has a plus sign, such as h = 1.75 + 0.5t.
• A decay expression usually has a minus sign, such as v = 10000 − 800t.
• These expressions are examples of linear expression Class 9 because the variable has power 1.
• Such expressions are also connected to linear polynomial Class 9 and later graph-based linear relationships.

These ideas are important in Class 9 Ganita Manjari linear polynomials solutions because they show how algebra can describe real-life change.

About Class 9 Maths Ganita Manjari Chapter 2 Exercise 2.4

Ganita Manjari Class 9 Chapter 2 Exercise 2.4 appears in the section on Linear Growth and Linear Decay. It comes after students practise linear patterns in Exercise 2.3.

These Class 9 Maths Chapter 2 Exercise 2.4 Solutions prepare students for:

• linear growth,
• linear decay,
• linear expressions,
• real-life algebraic modelling,
• linear relationships,
• tables of values,
• and equations of the form y = ax + b.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 2