NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.4 | 2026-27

2.4. Linear growth and linear decay

Think and Reflect 

1. What is the cost for travelling 15 km? For how many kilometres will the cost of the journey be ₹ 700?

Solution:

The cost of the journey is represented by the linear function:

C(d) = 100 + 60d

where C represents the total cost in rupees and d represents the distance travelled in kilometres.

For travelling 15 km, substitute d = 15 in the function:

C(15) = 100 + 60 × 15
= 100 + 900
= ₹1000

Therefore, the cost of travelling 15 km is ₹1000.

Now, if the total cost of the journey is ₹700:

700 = 100 + 60d

700 - 100 = 60d

600 = 60d

d = 10

Hence, the cost of the journey will be ₹700 for travelling 10 km.

Think and Reflect 

1. What will be the height of the water at the end of 5 months?

Solution:

The height of water h (in metres) after t months is represented by the linear function:

h(t) = 3 – 0.5t

To find the height of water after 5 months, substitute t = 5 in the function:

h(5) = 3 – 0.5 × 5
= 3 – 2.5
= 0.5 m

Therefore, the height of the water at the end of 5 months is 0.5 m.

Exercise Set 2.4 

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 of the plant per month = 0.5 feet

Height after 7 months = Initial height + (Growth per month × Number of months)

∴ h = 1.75 + (0.5 × 7)
= 1.75 + 3.5
= 5.25 feet

Therefore, 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 of the plant after t months can be represented as:

h = 1.75 + 0.5t

Using this formula, the values of height for t = 0 to 10 months are calculated as follows:

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

Solution:

The height h of the plant after t months can be represented by the expression:

h(t) = 1.75 + 0.5t

This expression represents linear growth because the height of the plant increases by a constant value of 0.5 feet for every increase of 1 month in time. 

Since the rate of growth remains constant, the relationship between h and t is linear.

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 the phone = ₹ 10,000
Decrease in value per year = ₹ 800

Value after 3 years = Initial value – (Decrease per year × Number of years)

∴ Value (v) = ₹ 10000 – (₹ 800 × 3)
= ₹ 10000 – ₹ 2400
= ₹ 7600

Therefore, the value of the phone after 3 years is ₹ 7600.

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

Solution:

The value of the phone after t years can be represented as:

v(t) = 10000 – 800t

Using this expression, the values of the phone for t = 0 to 8 years can be calculated as follows:

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

Solution:

The value v of the phone after t years can be represented by the expression:

v(t) = 10000 – 800t

This expression represents linear decay because the phone's value decreases by a fixed amount of ₹ 800 each year. 

Since the rate of decrease remains constant, the relationship between v and t is linear.

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 of the village = 750
Increase in population per year = 50

Population after 6 years = Initial population + (Increase per year × Number of years)

∴ Population (P) = 750 + (50 × 6)
= 750 + 300
= 1050

Therefore, the village population 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 of the village after t years can be represented by the expression:

P(t) = 750 + 50t

Using this expression, the population values for t = 0 to 10 years are calculated as follows:  

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

Solution:

The population P after t years can be represented by the expression:

P(t) = 750 + 50t

This expression represents linear growth because the population increases by a fixed amount of 50 people every year. 

Since the rate of increase remains constant, the relationship between P and t is linear.

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 balance = ₹ 600
Reduction in balance per day = ₹ 15

The remaining balance after x days can be represented as:

b(x) = 600 – 15x

This expression represents linear decay because the balance decreases by a constant amount of ₹ 15 every day. 

Since the rate of decrease remains the same, the relationship between b(x) and x is linear.

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

Solution:

To find when the balance becomes zero, substitute b(x) = 0.

⇒ 600 – 15x = 0

⇒ 15x = 600

⇒ x = 40

Therefore, 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:

Key Benefits of NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.4

• Helps students understand linear polynomials, zeroes, and their relationship with coefficients covered in Class 9 Maths Chapter 2 Exercise 2.4.

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• Supports effective exam preparation by covering important exercise concepts from Class 9 Maths NCERT Solutions Chapter 2 Exercise 2.4.

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• Provides a useful learning PDF resource for students preparing for Class 9 Maths exams with detailed explanations and systematic methods.

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