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.

Ex 2.4 Class 9 Ganita Manjari Chapter 2Solutions

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 each month = 0.5 feet
Growth in 7 months:
0.5 × 7 = 3.5 feet
Height after 7 months:
1.75 + 3.5 = 5.25 feet
Answer: The height of the plant after 7 months will be 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 initial height of the plant is 1.75 feet, and it grows by 0.5 feet every month.

The height after t months is:

h = 1.75 + 0.5t

Answer: The height increases by 0.5 feet every month. The linear expression is h = 1.75 + 0.5t.

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

Solution:

The height h after t months can be expressed as: h(t)= 1.75 + 0.5;
This expression represents linear growth because the height increases by a constant amount (0.5 feet) for each unit increase in time (months).

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:

Find the value of the phone after 3 years.
Initial value of the phone = ₹10,000
Decrease in value every year = ₹800
Decrease in 3 years:
₹800 × 3 = ₹2,400
Value after 3 years:
₹10,000 - ₹2,400 = ₹7,600
Answer: The value of the phone after 3 years is ₹7,600.

(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 initial value of the phone is ₹10,000, and it decreases by ₹800 every year.

The value decreases by ₹800 every year.

So, the linear pattern is:

v = 10,000 - 800t

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

Solution:

The initial value of the phone is ₹10,000, and its value decreases by ₹800 every year.
Therefore, the expression relating v and t is:
v = 10,000 - 800t
where:
v = value of the phone in rupees
t = time in years
This represents linear decay because the value decreases by the same amount, ₹800, every year.
Answer: v = 10,000 - 800t. It represents linear decay because the rate of decrease is constant.

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 people
Increase in 6 years:
50 × 6 = 300
Population after 6 years:
750 + 300 = 1,050
Answer: The population of the village after 6 years will be 1,050.

(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 initial population of the village is 750, and it increases by 50 people every year.

The population increases by 50 people every year.

The linear expression is:

P = 750 + 50t

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

Solution:

The initial population of the village is 750, and it increases by 50 people every year.
Therefore, the expression relating P and t is:
P = 750 + 50t
where:
P = population of the village
t = time in years
It represents linear growth because the population increases by the same amount, 50 people, every year.
Answer: P = 750 + 50t. It represents linear growth because the rate of increase is constant.

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 forx days. Explain why it represents linear decay.

Solution:

The initial prepaid balance is ₹600.
The balance decreases by ₹15 each day.
So, after x days:
b(x) = 600 - 15x
where:
b(x) = remaining balance after x days
x = number of days
It represents linear decay because the balance decreases by the same amount, ₹15, every day.
Answer: b(x) = 600 - 15x

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

Solution:

The balance runs out when:
b(x) = 0
Using:
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 the equation:

b(x) = 600 - 15x

we get the following values:

Thus, the balance decreases by ₹15 every day.

Related Resource:

• Exercise 2.1 – Degree, Coefficients and Constants
• Exercise 2.2 – Polynomial Values and Linear Equations
• Exercise 2.3 – Linear Patterns
• Exercise 2.5 – Linear Relationships
• Exercise 2.6 – Graphs of Linear Relationships
• End-of-Chapter Exercises – Introduction to Linear Polynomials