Welcome to Sid Classes. In this detailed guide, we provide accurate, step-by-step solutions for NCERT Class 9 Maths Ganita Manjari Chapter 2 Exercise 2.4. This exercise explores the practical concepts of Linear Growth and Linear Decay using real-world scenarios, tabular values, and algebraic expressions.
Our solutions utilize clean standard mathematical notation (\(\text{LaTeX}\)) and breakdown logic to ensure you score 100% marks in your school tests and board examinations.
🔑 Key Concepts to Remember:
- Linear Growth: Occurs when a quantity increases by a constant, fixed amount over equal intervals of time (e.g., plant growth, population addition).
- Linear Decay: Occurs when a quantity decreases or depreciates by a constant, fixed amount over equal intervals of time (e.g., asset depreciation, phone balance reductions).
- Tables of Values: Used to map variables systematically to visualize how patterns scale over discrete periods.
NCERT Class 9 Maths Ganita Manjari Chapter 2 Exercise 2.4 Solutions
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.
(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:
(i) Find the height after 7 months:
Given, Initial Height \(= 1.75\text{ feet}\)
Growth rate per month \(= 0.5\text{ feet}\)
$$\text{Growth in 7 months} = 7 \times 0.5 = 3.5\text{ feet}$$
$$\text{Total Height after 7 months} = 1.75 + 3.5 = \mathbf{5.25\text{ feet}}$$
(ii) Table of values for \(t\) varying from 0 to 10 months:
| Time \(t\) (months) | 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 |
(iii) Find an expression that relates \(h\) and \(t\), and explain why it represents linear growth:
The formula relating the height \(h\) to time \(t\) is:
$$\mathbf{h = 1.75 + 0.5t}$$
Explanation: This expression represents linear growth because the height increases by a constant, fixed addition of \(0.5\text{ feet}\) every single month, and the highest power of the variable \(t\) is exactly 1.
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.
(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:
(i) Find the value of the phone after 3 years:
Given, Initial Value \(= \text{₹}10,000\)
Depreciation rate per year \(= \text{₹}800\)
$$\text{Value lost in 3 years} = 3 \times 800 = \text{₹}2,400$$
$$\text{Value after 3 years} = 10000 - 2400 = \mathbf{\text{₹}7,600}$$
(ii) Table of values for \(t\) varying from 0 to 8 years:
| Time \(t\) (years) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|---|
| Value \(v\) (₹) | 10000 | 9200 | 8400 | 7600 | 6800 | 6000 | 5200 | 4400 | 3600 |
(iii) Find an expression that relates \(v\) and \(t\), and explain why it represents linear decay:
The formula mapping the changing value \(v\) over years \(t\) is:
$$\mathbf{v = 10000 - 800t}$$
Explanation: This expression represents linear decay because the valuation drops subtractionally by a fixed, unchanging quantity of \(₹800\) every year, showing an inverse constant rate of change with a maximum variable power of 1.
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.
(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:
(i) Find the population of the village after 6 years:
Given, Initial Population \(= 750\)
Growth addition rate per year \(= 50\text{ people}\)
$$\text{New migration in 6 years} = 6 \times 50 = 300\text{ people}$$
$$\text{Total Population after 6 years} = 750 + 300 = \mathbf{1050}$$
(ii) Table of values for \(t\) varying from 0 to 10 years:
| Time \(t\) (years) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| Population \(P\) | 750 | 800 | 850 | 900 | 950 | 1000 | 1050 | 1100 | 1150 | 1200 | 1250 |
(iii) Find an expression that relates \(P\) and \(t\), and explain why it represents linear growth:
The relation is expressed as:
$$\mathbf{P = 750 + 50t}$$
Explanation: This represents linear growth because the total count increases by a constant common sum addition of exactly \(50\) citizens each year without compounding fluctuations.
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.
(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:
(i) Write an equation that models the remaining balance \(b(x)\):
$$\mathbf{b(x) = 600 - 15x}$$
Explanation: This represents linear decay because the original balance value is consistently stripped away at a constant uniform rate of \(₹15\) each passing day.
(ii) After how many days will the balance run out?
The balance runs out completely when \(b(x) = 0\). Setting the equation to zero:
$$0 = 600 - 15x$$
$$15x = 600$$
$$x = \frac{600}{15} = \mathbf{40\text{ days}}$$
Hence, the prepaid balance runs out fully after exactly 40 days.
(iii) Table of values for \(x\) varying from 1 to 10 days:
| Days \(x\) | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| Balance \(b(x)\) (₹) | 585 | 570 | 555 | 540 | 525 | 510 | 495 | 480 | 465 | 450 |
Frequently Asked Questions (FAQs)
Q1: What is the difference between linear growth and linear decay?
A: Linear growth means a quantity is increasing by adding a fixed constant value over time intervals, while linear decay means a value is decreasing by subtracting a fixed value continuously.
Q2: How can we identify a linear function from a table of values?
A: If the differences between consecutive output row values remain perfectly identical while inputs step up by equal units, the relation is linear.
Q3: What does the 'constant rate of change' mean?
A: It refers to the fixed addition or subtraction value per unit of time (the slope), such as the plant growing exactly \(0.5\text{ feet}\) every month or a balance dropping by ₹15 every day.
Q4: Why are these equations written with the variable power as 1?
A: A algebraic model can only represent a straight line graph (linear) if the degree power of its independent variable stays exactly equal to 1.
Q5: How do these HTML tables render in dark mode on mobile devices?
A: The layout uses transparent gray border parameters (`rgba`), which auto-adjust visually matching dark/light screens cleanly without breaking style layouts.
Q6: Are these solutions strictly matched to the 2026-27 CBSE class blueprint requirements?
A: Yes, these solution patterns strictly adhere to the updated standard syllabus parameters specified for Grade 9 Ganita Manjari NCERT publications.
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