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.

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

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 2