Looking for complete Class 9 Maths Chapter 2 Exercise 2.3 Solutions? Here you will get step-by-step solutions of all questions from the latest NCERT Ganita Manjari 2026 textbook. These Class 9 Maths Chapter 2 Solutions cover linear patterns, linear growth, linear decay, and real-life applications of Introduction to Linear Polynomials in a simple and exam-oriented manner.
The NCERT Class 9 Maths Chapter 2 Exercise 2.3 Solutions include questions based on savings patterns, decreasing quantities, area and volume relationships, and linear expressions. Each solution is explained with proper steps, formulas, patterns, and final answers according to the latest CBSE and NCERT syllabus.
- How to identify and write linear patterns
- How to form linear expressions from real-life situations
- Applications of linear growth and linear decay
- How to find the nth term of a linear pattern
- Step-by-step NCERT solutions with exam tips
📚 Table of Contents
- About Class 9 Maths Chapter 2 Exercise 2.3
- Question 1 – Savings Account Linear Pattern
- Question 2 – Rally Members Linear Decay
- Question 3 – Area of Rectangle Linear Pattern
- Question 4 – Volume of Rectangular Box Pattern
- Question 5 – Book Reading Linear Pattern
- What We Learned in Exercise 2.3
- Frequently Asked Questions
About Class 9 Maths Chapter 2 Exercise 2.3 Solutions
These Class 9 Maths Chapter 2 Exercise 2.3 Solutions are prepared according to the latest NCERT Ganita Manjari 2026 textbook. In this exercise, students learn how to identify and represent linear patterns, solve problems involving linear growth and linear decay, and express real-life situations using linear expressions.
The NCERT Class 9 Maths Chapter 2 Exercise 2.3 Solutions help students understand how algebra is applied in practical situations such as savings, decreasing quantities, area, volume, and daily-life patterns. These concepts form the foundation for higher mathematics and future chapters involving sequences, functions, and algebraic relationships.
Important Topics Covered in Exercise 2.3
- Linear Patterns and Sequences
- Linear Growth and Linear Decay
- Formation of Linear Expressions
- Real-Life Applications of Linear Polynomials
- Savings and Expenditure Patterns
- Area and Volume Relationships
- Pattern Recognition and Generalization
- Finding the nth Term of a Linear Pattern
Exercise 2.3 of Introduction to Linear Polynomials introduces students to mathematical modeling and pattern formation. These concepts are frequently used in higher algebra, coordinate geometry, sequences, and real-life problem-solving.
Quick Revision: Linear Patterns and Linear Expressions
Before solving these Class 9 Maths Chapter 2 Exercise 2.3 Solutions, let us quickly revise the important concepts of linear patterns, linear growth, and linear decay.
| Concept | General Form |
|---|---|
| Linear Growth Pattern | A = Initial Value + (Rate × n) |
| Linear Decay Pattern | A = Initial Value − (Rate × n) |
| Linear Expression | ax + b |
A linear pattern is a sequence in which the difference between consecutive terms remains constant. Most questions in Exercise 2.3 Solutions use this concept to form a linear expression or determine the nth term of the pattern.
How to Solve Class 9 Maths Chapter 2 Exercise 2.3 Questions?
The questions in Class 9 Maths Chapter 2 Exercise 2.3 Solutions are mainly based on identifying patterns and forming linear expressions. To solve these questions correctly, students should follow a step-by-step approach instead of directly applying formulas.
Step-by-Step Method
- Identify the quantity that changes regularly.
- Find the increase or decrease occurring in each step.
- Represent the changing quantity using a variable (usually n).
- Form the required linear expression or linear equation.
- Substitute the given value and verify the answer.
By practicing these Class 9 Maths Chapter 2 Exercise 2.3 NCERT Solutions, students can improve their understanding of linear polynomials, pattern recognition, and real-life applications of algebra.
Class 9 Maths Chapter 2 Exercise 2.3 Question-wise Solutions
Question 1
A student has ₹500 in her savings bank account. She gets ₹150 every month as pocket money. How much money will she have at the end of every month from the second month onwards? Find a linear expression to represent the amount she will have in the nth month.
Given:
- Initial amount in the savings account = ₹500
- Pocket money received every month = ₹150
To Find:
- The amount of money at the end of each month.
- A linear expression for the amount in the nth month.
Solution:
At the beginning, the student has
₹500
At the end of the first month, she receives ₹150.
Amount after first month
= 500 + 150
= ₹650
Similarly,
| Month | Amount (₹) |
| 1 | 500 + 1 × 150 = 650 |
| 2 | 500 + 2 × 150 = 800 |
| 3 | 500 + 3 × 150 = 950 |
| 4 | 500 + 4 × 150 = 1100 |
| … | … |
Observe the pattern:
1st month = 500 + 1 × 150
2nd month = 500 + 2 × 150
3rd month = 500 + 3 × 150
4th month = 500 + 4 × 150
Therefore, in the nth month,
Amount = 500 + n × 150
A(n) = 500 + 150n
The amount of money in the nth month is
A(n) = 500 + 150n
Key Concept Used: Observe the increasing pattern and express the quantity in terms of n.
Common Mistake: Students often write 500 + (n−1)×150. Here, the amount after the first month already includes one installment of ₹150.
Exam Tip: Write the first few terms of the pattern clearly before finding the expression for the nth month.
Question 2
A rally starts with 120 members. Each hour, 9 members drop out of the group. How many members will remain after 1, 2, 3, … hours? Find a linear expression to represent the number of members at the end of the nth hour.
Given:
- Initial number of members = 120
- Number of members dropping out every hour = 9
To Find:
- The number of members remaining after each hour.
- The linear expression for the number of members remaining after the nth hour.
Solution:
Initially, the rally has
120 members
After 1 hour, 9 members leave the rally.
Number of members after 1 hour
= 120 − 9
= 111
Similarly, we can continue the pattern:
| Hour | Number of Members |
| 1 | 120 − 1 × 9 = 111 |
| 2 | 120 − 2 × 9 = 102 |
| 3 | 120 − 3 × 9 = 93 |
| 4 | 120 − 4 × 9 = 84 |
| … | … |
Observe the Linear Pattern:
After 1 hour = 120 − 1 × 9
After 2 hours = 120 − 2 × 9
After 3 hours = 120 − 3 × 9
After 4 hours = 120 − 4 × 9
Since 9 members leave the rally every hour, the number of members decreases by 9 at each step.
Therefore, the number of members remaining after the nth hour is:
M(n) = 120 − n × 9
M(n) = 120 − 9n
M(n) = 120 − 9n
Key Concept Used: Identify the linear decay pattern and express it using a linear expression in terms of n.
Common Mistake: Students often write 120 + 9n instead of 120 − 9n, forgetting that members leave the rally every hour.
Exam Tip: When a quantity decreases by a fixed amount at every step, use subtraction while writing the expression for the nth term.
Question 3
Suppose the length of a rectangle is 13 cm. Find the area if the breadth is (i) 12 cm, (ii) 10 cm, (iii) 8 cm. Find the linear pattern representing the area.
Given:
- Length of the rectangle = 13 cm
- Breadth = 12 cm, 10 cm and 8 cm
To Find:
- The area of the rectangle for the given breadths.
- The linear pattern representing the area.
Solution:
We know that,
Area of a Rectangle = Length × Breadth
(i) When breadth = 12 cm
A = 13 × 12
= 156 cm²
(ii) When breadth = 10 cm
A = 13 × 10
= 130 cm²
(iii) When breadth = 8 cm
A = 13 × 8
= 104 cm²
Arrange the values in a table:
| Breadth (cm) | 12 | 10 | 8 | … | b |
| Area (cm²) | 156 | 130 | 104 | … | 13b |
Observe the Linear Pattern:
A = 13 × 12 = 156
A = 13 × 10 = 130
A = 13 × 8 = 104
If the breadth of the rectangle is
b cm,
then the area of the rectangle will be
A = 13 × b
A = 13b
- When breadth = 12 cm, Area = 156 cm²
- When breadth = 10 cm, Area = 130 cm²
- When breadth = 8 cm, Area = 104 cm²
Linear Pattern:
A = 13b
Key Concept Used: When one dimension of a rectangle is fixed, the area varies directly with the other dimension and forms a linear pattern.
Common Mistake: Students often try to identify a numerical pattern instead of applying the formula for the area of a rectangle.
Exam Tip: Whenever one quantity depends directly on another quantity, first write the mathematical formula and then express it using a variable.
Question 4
Suppose the length of a rectangular box is 7 cm and breadth is 11 cm. Find the volume if the height is (i) 5 cm, (ii) 9 cm, (iii) 13 cm. Find the linear pattern representing the volume of the rectangular box.
Given:
- Length of the rectangular box = 7 cm
- Breadth of the rectangular box = 11 cm
- Height = 5 cm, 9 cm and 13 cm
To Find:
- The volume of the rectangular box for the given heights.
- The linear pattern representing the volume.
Solution:
We know that,
Volume of a Rectangular Box
= Length × Breadth × Height
(i) When height = 5 cm
V = 7 × 11 × 5
= 77 × 5
= 385 cm³
(ii) When height = 9 cm
V = 7 × 11 × 9
= 77 × 9
= 693 cm³
(iii) When height = 13 cm
V = 7 × 11 × 13
= 77 × 13
= 1001 cm³
Arrange the values in a table:
| Height (cm) | 5 | 9 | 13 | … | h |
| Volume (cm³) | 385 | 693 | 1001 | … | 77h |
Observe the Linear Pattern:
V = 7 × 11 × 5 = 385
V = 7 × 11 × 9 = 693
V = 7 × 11 × 13 = 1001
If the height of the rectangular box is
h cm,
then the volume of the rectangular box will be
V = 7 × 11 × h
V = 77h
- When height = 5 cm, Volume = 385 cm³
- When height = 9 cm, Volume = 693 cm³
- When height = 13 cm, Volume = 1001 cm³
Linear Pattern:
V = 77h
Key Concept Used: When the length and breadth are fixed, the volume of a rectangular box varies directly with its height and forms a linear pattern.
Common Mistake: Students often substitute the height correctly but forget to multiply the fixed dimensions first.
Exam Tip: When some dimensions are fixed, first multiply them together and then express the formula in terms of the variable quantity.
Question 5
Sarita is reading a book of 500 pages. She reads 20 pages every day. How many pages will be left after 15 days? Express this as a linear pattern.
Given:
- Total number of pages in the book = 500
- Number of pages read every day = 20
To Find:
- The number of pages left after 15 days.
- The linear pattern representing the pages left after n days.
Solution:
Initially, the book contains
500 pages
After 1 day, Sarita reads 20 pages.
Pages left after 1 day
= 500 − 20
= 480
Similarly,
| Day | 1 | 2 | 3 | … | n |
| Pages Left | 500 − 1 × 20 = 480 | 500 − 2 × 20 = 460 | 500 − 3 × 20 = 440 | … | 500 − 20n |
Observe the pattern:
After 1 day = 500 − 1 × 20
After 2 days = 500 − 2 × 20
After 3 days = 500 − 3 × 20
Therefore, after n days, the number of pages left will be
P = 500 − 20n
Pages left after 15 days
Substituting n = 15,
P = 500 − 20(15)
= 500 − 300
= 200
Therefore, after 15 days,
Pages left = 200
- The number of pages left after 15 days is 200.
Linear pattern:
P = 500 − 20n
Key Concept Used: When a fixed quantity decreases by the same amount repeatedly, it can be represented by a linear expression.
Common Mistake: Students often write 500 + 20n instead of 500 − 20n, forgetting that the number of pages is decreasing.
Exam Tip: For decreasing patterns, always start with the initial quantity and subtract the amount reduced at each step.
What We Learned in Class 9 Maths Chapter 2 Exercise 2.3 Solutions
In these Class 9 Maths Chapter 2 Exercise 2.3 Solutions, we learned how to identify linear patterns, represent real-life situations using linear expressions, and solve problems involving linear growth and linear decay. These concepts help students understand how algebra can be applied to practical situations and form the foundation for higher mathematics.
| Question | Concept Learned |
|---|---|
| Question 1 | Savings Pattern and Linear Growth |
| Question 2 | Linear Decay Pattern |
| Question 3 | Area and Linear Relationship |
| Question 4 | Volume and Linear Pattern |
| Question 5 | Book Reading and Linear Decay |
- A linear pattern has a constant difference between consecutive terms.
- Linear growth is represented by addition, while linear decay is represented by subtraction.
- Express practical situations using variables before forming linear expressions.
- Always verify the obtained pattern by substituting values.
- Questions based on the nth term of a pattern are important for examinations.
Frequently Asked Questions (FAQs)
Q1. What is taught in Class 9 Maths Chapter 2 Exercise 2.3?
Class 9 Maths Chapter 2 Exercise 2.3 teaches students how to identify linear patterns, form linear expressions, and solve real-life problems involving linear growth and linear decay.
Q2. How do you identify a linear pattern?
A pattern is called a linear pattern when the difference between consecutive terms remains constant throughout the sequence.
Q3. Is Exercise 2.3 important for CBSE Class 9 exams?
Yes. Exercise 2.3 is important because questions based on linear patterns, algebraic expressions, and real-life applications are frequently asked in school examinations.
Q4. What is the general form of a linear expression?
The general form of a linear expression is ax + b, where a and b are constants and x is a variable.
Q5. Where can I get complete Class 9 Maths Chapter 2 Solutions?
You can access complete Class 9 Maths Chapter 2 Solutions, including Exercise 2.1, Exercise 2.2, Exercise 2.3, and End Exercise solutions, on Maths Gurukulam with step-by-step explanations.
Official Learning Resources
For additional reference and official curriculum information, students can visit:
These official educational resources provide authentic syllabus information, curriculum updates, and learning guidelines.
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📘 Chapter 2 Hub
Complete Chapter 2 Resources
✅ Exercise 2.1
Polynomial Basics
✅ Exercise 2.2
Linear Polynomials
📍 Exercise 2.3
You Are Here
✅ End Exercise
Chapter Assessment
📝 Notes PDF
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⭐ Important Questions
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⬅ Chapter 1
Coordinates Solutions
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These Class 9 Maths Chapter 2 Exercise 2.3 Solutions have been carefully prepared according to the latest NCERT Ganita Manjari (2026) textbook to help students understand linear patterns and real-life applications step-by-step.