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Class 9 Ganita ManjariClass 9 Maths Solutions | 2026–27
Chapter 2: Introduction to Linear PolynomialsPage 25 – 27All 4 Questions on This Page

Class 9 Maths Chapter 2 Exercise Set 2.4 Solutions

Complete step-by-step solutions for Class 9 Mathematics Ganita Manjari Chapter 2 Exercise Set 2.4 (Questions 1–4). Covers modeling linear growth and linear decay through changing quantities, constructing value tables over time intervals, and establishing linear polynomial expressions.

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Key Concepts for Exercise Set 2.4

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1

Linear Growth: Constant Increase Over Equal Intervals

When a quantity starts at an initial value and increases by the same fixed amount over equal intervals of time or steps, it exhibits linear growth. The rate of change is positive, giving an expression of the form y = y₀ + m · t.

y = y₀ + m · t (m > 0) ⇒ Linear Growth
2

Linear Decay: Constant Decrease Over Equal Intervals

When a quantity decreases by the same fixed amount over equal intervals of time, it exhibits linear decay. The rate of change is negative, giving an expression of the form y = y₀ − m · t.

y = y₀ − m · t (m > 0) ⇒ Linear Decay
3

Modeling Real-Life Quantities with Linear Polynomials

In linear models, the starting value independent of time is the constant term, while the uniform rate of increase or decrease per unit time is the coefficient of the variable.

Quantity = Initial Value ± (Rate of Change) × (Time)
4

Constructing and Interpreting Value Tables

A value table records the state of a changing quantity at successive intervals. For a linear relationship, consecutive differences between table values remain constant.

Consecutive Difference = Constant ⇒ Linear Relationship
Question 1
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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.

Solution
Given:
Initial height = 1.75 feet
Growth each month = 0.5 feet

Let t be the number of months.

(i) Height after 7 months

In 7 months, the increase in height is:

7 × 0.5 = 3.5 feet
Therefore:
Height after 7 months = 1.75 + 3.5 = 5.25 feet
Therefore:

5.25 feet

(ii) Table of values for t varying from 0 to 10 months

Show the height month by month in a horizontal table:

Month t 0 1 2 3 4 5 6 7 8 9 10
Height h (feet) 1.75 2.25 2.75 3.25 3.75 4.25 4.75 5.25 5.75 6.25 6.75

The table clearly shows that the height increases by 0.5 feet every month:

At t = 1: h = 1.75 + 0.5 = 2.25

At t = 2: h = 1.75 + 2(0.5) = 2.75

(iii) Expression relating h and t

Initially, the height is 1.75 feet.

After t months, the total increase is:

0.5t feet

Therefore:
h = 1.75 + 0.5t

This represents linear growth because the height increases by the same fixed amount, 0.5 feet, for every one-month increase in t.

Answer:(i) 5.25 feet | (ii) Table values for t = 0 to 10: 1.75, 2.25, 2.75, 3.25, 3.75, 4.25, 4.75, 5.25, 5.75, 6.25, 6.75 feet | (iii) h = 1.75 + 0.5t (linear growth)
Question 2
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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.

Solution
Given:
Initial value = ₹10,000
Decrease each year = ₹800

Let t be the number of years.

(i) Value after 3 years

Decrease in 3 years:

3 × ₹800 = ₹2,400
Therefore:

Value after 3 years = ₹10,000 − ₹2,400 = ₹7,600

Therefore:
₹7,600
(ii) Table of values for t varying from 0 to 8 years

Show the depreciating value year by year in a horizontal table:

Year t 0 1 2 3 4 5 6 7 8
Value v (₹) 10,000 9,200 8,400 7,600 6,800 6,000 5,200 4,400 3,600

The table clearly shows that the value decreases by ₹800 each year:

At t = 1: v = 10,000 − 800 = 9,200

At t = 2: v = 10,000 − 2(800) = 8,400

(iii) Expression relating v and t

Initially, the phone is worth ₹10,000.

After t years, the total decrease is:

₹800t

Therefore:
v = 10,000 − 800t

This represents linear decay because the value decreases by the same fixed amount, ₹800, for every one-year increase in t.

Answer:(i) ₹7,600 | (ii) Table values for t = 0 to 8: ₹10,000, ₹9,200, ₹8,400, ₹7,600, ₹6,800, ₹6,000, ₹5,200, ₹4,400, ₹3,600 | (iii) v = 10,000 − 800t (linear decay)
Question 3
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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.

Solution

Initially, the population is:

750 people

Every year, 50 people move into the village.

Let t be the number of years.

(i) Population after 6 years

Increase in 6 years:

6 × 50 = 300
Therefore:

Population after 6 years = 750 + 300 = 1,050

Therefore:

1,050 people

(ii) Table of values for t varying from 0 to 10 years

Show the population growth year by year in a horizontal table:

Year t 0 1 2 3 4 5 6 7 8 9 10
Population P 750 800 850 900 950 1,000 1,050 1,100 1,150 1,200 1,250

The table clearly shows the population increasing by 50 people every year:

At t = 1: P = 750 + 50 = 800

At t = 2: P = 750 + 2(50) = 850

(iii) Expression relating P and t

Initially, the population is 750.

After t years, the increase is:

50t
Therefore:
P = 750 + 50t

This represents linear growth because the population increases by the same fixed amount, 50 people, for every one-year increase in t.

Answer:(i) 1,050 people | (ii) Table values for t = 0 to 10: 750, 800, 850, 900, 950, 1,000, 1,050, 1,100, 1,150, 1,200, 1,250 | (iii) P = 750 + 50t (linear growth)
Question 4
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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.

Solution

Initially, the prepaid balance is:

₹600

The balance decreases by ₹15 each day.

Let x be the number of days.

(i) Equation for the remaining balance

After x days, the total decrease is:

₹15x

Therefore:
b(x) = 600 − 15x

This represents linear decay because the balance decreases by the same fixed amount, ₹15, for every one-day increase in x.

Therefore:
b(x) = 600 − 15x
(ii) When will the balance run out?

The balance runs out when:

b(x) = 0
Therefore:
600 − 15x = 0
15x = 600
x = 40
Hence, the balance will run out after:

40 days

(iii) Table of values for x varying from 1 to 10 days

Show the balance day by day in a horizontal table:

Day x 1 2 3 4 5 6 7 8 9 10
Balance b(x) (₹) 585 570 555 540 525 510 495 480 465 450

The table clearly shows that the balance decreases by ₹15 each day:

b(1) = 600 − 15 = 585
b(2) = 600 − 30 = 570

The balance values from day 1 to day 10 are:

585, 570, 555, 540, 525, 510, 495, 480, 465, 450

Answer:(i) b(x) = 600 − 15x (linear decay) | (ii) 40 days | (iii) Balance from day 1 to 10: ₹585, ₹570, ₹555, ₹540, ₹525, ₹510, ₹495, ₹480, ₹465, ₹450

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