NCERT Solutions for Class 9 Maths Chapter 2 Introduction to Linear Polynomials

NCERT Solutions for Class 9 Maths Chapter 2 - Introduction to Linear Polynomials are designed to help students understand the chapter concepts clearly and solve textbook questions with confidence. At Infinity Learn, these solutions are prepared by experienced subject experts and provide detailed, step-by-step explanations for every exercise question based on the latest NCERT textbook.

The NCERT Solutions Class 9 Maths Chapter 2 Introduction to Linear Polynomials help students strengthen their conceptual understanding, improve problem-solving skills, and prepare effectively for school examinations. Each solution follows the latest CBSE curriculum and is presented in a simple and easy-to-understand format, making learning more effective.

With Infinity Learn NCERT Solutions for Class 9 Maths, students can practice important questions, clear their doubts, and build a strong foundation in Mathematics. These NCERT solutions are an excellent resource for revision, homework assistance, and exam preparation while ensuring complete coverage of the latest CBSE syllabus and guidelines.

NCERT Solutions Class 9 Maths Chapter 2 Introduction to Linear Polynomials

NCERT Solutions for Class 9 Maths Chapter 2 – Introduction to Linear Polynomials provide comprehensive solutions for Exercise 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, and the End-of-Chapter Exercises from the latest NCERT Ganita Manjari Grade 9 Part I (Session 2026-27) textbook. These solutions are designed to help students understand every concept clearly and solve questions with confidence.

Class 9 maths chapter 2 Introduction to Linear Polynomials, introduces students to the fundamental concepts of algebra and polynomial expressions. The chapter explains important terms such as variables, coefficients, constants, and the degree of a polynomial through relatable real-life situations. Students learn how linear polynomials are formed and how they can be used to represent patterns, relationships, and changing quantities in everyday life.

Download NCERT Solutions Class 9 Maths Chapter 2 Introduction to Linear Polynomials PDF 2026-27

NCERT Class 9 Maths Ganita Manjari Chapter 2 Introduction to Linear Polynomials Solutions

NCERT solutions class 9 maths chapter 2 exercise 2.1

1. Find the degrees of the following polynomials:

(i) 2x² − 5x + 3

(ii) y³ + 2y − 1

(iii) −9

(iv) 4z − 3

Solutions: 

(i) 2x² - 5x + 3

Powers of x are x², x¹, and constant 3.

Highest power = 2

Answer: Degree = 2

(ii) y³ + 2y - 1

Powers of y are y³, y¹, and constant -1.

Highest power = 3

Answer: Degree = 3

(iii) -9

This is a constant polynomial. A non-zero constant polynomial has degree 0.

Answer: Degree = 0

(iv) 4z - 3

Powers of z are z¹ and constant -3.

Highest power = 1

Answer: Degree = 1

2. Write polynomials of degrees 1, 2 and 3

Solutions:

Degree 1 polynomial: x + 5

Degree 2 polynomial: x² + 3x + 2

Degree 3 polynomial: x³ - 2x² + x + 4

3. What are the coefficients of x² and x³ in the polynomial x⁴ − 3x³ + 6x² − 2x + 7?

Solutions: Given polynomial:

x⁴ - 3x³ + 6x² - 2x + 7

The x³ term is -3x³, so the coefficient of x³ is -3.

The x² term is 6x², so the coefficient of x² is 6.

Coefficient of x² = 6

Coefficient of x³ = -3

4. What is the coefficient of z in the polynomial 4z³ + 5z² − 11?

Solutions: Given polynomial: 4z³ + 5z² − 11

Step 1: Identify the term containing z.

In this polynomial, the terms are:

• 4z³
• 5z²
• −11

Step 2: Check whether there is any term with only z.

There is no term like az in the given polynomial.

The coefficient of z is 0.

5. What is the constant term of the polynomial 9x³ + 5x² − 8x − 10?

Given polynomial: 9x³ + 5x² − 8x − 10

Step 1: Identify the term without any variable.

In this polynomial, the terms are:

• 9x³
• 5x²
• −8x
• −10

Step 2: The term without x is the constant term.

Here, −10 has no variable.

The constant term is −10.

NCERT solutions class 9 maths chapter 2 exercise 2.2

1. Find the value of the linear polynomial 5x − 3 if:

(i) x = 0

(ii) x = −1

(iii) x = 2

Solutions: Given polynomial: 5x − 3

(i) x = 0

= 5(0) − 3 = −3

(ii) x = −1

= 5(−1) − 3 = −5 − 3 = −8

(iii) x = 2

= 5(2) − 3 = 10 − 3 = 7

2. Find the value of the quadratic polynomial 7s² − 4s + 6 if:

(i) s = 0

(ii) s = −3

(iii) s = 4

Solutions: (i) For s = 0

7s² − 4s + 6

= 7(0)² − 4(0) + 6 = 6

(ii) For s = −3

7s² − 4s + 6

= 7(−3)² − 4(−3) + 6

= 7(9) + 12 + 6

= 63 + 12 + 6 = 81

(iii) For s = 4

7s² − 4s + 6

= 7(4)² − 4(4) + 6

= 7(16) − 16 + 6

= 112 − 16 + 6 = 102

 3. The present age of Salil’s mother is three times Salil’s present age. After 5 years, their ages will add up to 70 years. Find their present ages.

Solution: Let Salil's present age = x years

Since Salil's mother is three times his age,

Mother's present age = 3x years

Step 1: Find their ages after 5 years.

• Salil's age after 5 years = x + 5
• Mother's age after 5 years = 3x + 5

Step 2: According to the question, the sum of their ages after 5 years will be 70 years.

(x + 5) + (3x + 5) = 70

Step 3: Simplify the equation.

4x + 10 = 70

4x = 70 − 10

4x = 60

x = 15

Step 4: Find their present ages.

• Salil's present age = 15 years
• Mother's present age = 3 × 15 = 45 years

Salil is 15 years old and his mother is 45 years old.

4. The difference between two positive integers is 63. The ratio of the two integers is 2:5. Find the two integers.

Solution: Let the integers be 2x and 5x

Difference: 5x − 2x = 63

3x = 63

x = 21

Integers: 2x = 42

5x = 105

Required integers are 42 and 105.

5. Ruby has 3 times as many two-rupee coins as she has five-rupee coins. If she has a total ₹88, how many coins does she have of each type?

Solution: 

Let number of five-rupee coins = x

Number of two-rupee coins = 3x

Total value of the coins = ₹88

Therefore,

5x + 2(3x) = 88

5x + 6x = 88

11x = 88

x = 8

Hence,

Number of five-rupee coins = 8

Number of two-rupee coins = 3 × 8 = 24

Answer: Five-rupee coins = 8 and Two-rupee coins = 24.

6. A farmer cuts a 300 feet fence into two pieces of different sizes. The longer piece is four times as long as the shorter piece. How long are the two pieces?

Solution: Let the length of the shorter piece be x feet.

According to the question, the longer piece is four times the shorter piece.

Therefore,

Length of the longer piece = 4x feet

Step 1: Form the equation

The total length of both pieces is 300 feet.

So,

x + 4x = 300

Step 2: Solve the equation

5x = 300

x = 300 ÷ 5

x = 60

Step 3: Find the lengths of both pieces

Shorter piece = 60 feet

Longer piece = 4 × 60 = 240 feet

The shorter piece is 60 feet long and the longer piece is 240 feet long.

7. If the length of a rectangle is three more than twice its width and its perimeter is 24 cm, what are the dimensions of the rectangle?

Solution: Let width = x cm

Length = 2x + 3 cm

Perimeter = 2(length + width)

2[(2x + 3) + x] = 24

2(3x + 3) = 24

6x + 6 = 24

6x = 18

x = 3

Therefore:

Width = 3 cm

Length = 2(3) + 3 = 9 cm

NCERT solutions class 9 maths chapter 2 exercise 2.3

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.

Solution: The amount in the bank starts at ₹500, and ₹150 is added every month as pocket money. Therefore, the total amount increases by ₹150 each month.

If Aₙ represents the amount in the bank at the end of the nᵗʰ month, then the amount can be represented by the linear expression:

Aₙ = 500 + 150n

or

Aₙ = 150n + 500

Thus, the required linear expression is Aₙ = 150n + 500.

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.

Solution: Initial members = 120

Members leaving each hour = 9

After:

1 hour = 120 − 9 = 111

2 hours = 120 − 18 = 102

3 hours = 120 − 27 = 93

and so on...

Let the number of members after n hours be Mₙ.

Then,

Mₙ = 120 − 9n

Thus, the required linear expression is:

Mₙ = 120 − 9n.

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 of the rectangle.

Solution: Length = 13 cm

Area = Length × Breadth

(i) Breadth = 12 cm

Area = 13 × 12 = 156 cm²

(ii) Breadth = 10 cm

Area = 13 × 10 = 130 cm²

(iii) Breadth = 8 cm

Area = 13 × 8 = 104 cm²

Let breadth = x cm

Area = 13x

Thus, linear pattern is A = 13x.

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.

Solution: Length = 7 cm, Breadth = 11 cm

Volume of the cuboid = Length × Breadth × Height

= 7 × 11 × h

= 77h

For different values of height:

When h = 5 cm, Volume = 77 × 5 = 385 cm³

When h = 9 cm, Volume = 77 × 9 = 693 cm³

When h = 13 cm, Volume = 77 × 13 = 1001 cm³

If the height is represented by h cm, then the volume is given by:

V = 77h

Thus, the required linear pattern is:

V = 77h.

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.

Solution: Total pages = 500

Pages read per day = 20

Pages read in 15 days = 20 × 15 = 300

Pages left = 500 − 300 = 200

Let pages left after n days = Pₙ

So, Pₙ = 500 − 20n

Thus, linear pattern is Pₙ = 500 − 20n.

NCERT solutions class 9 maths chapter 2 exercise 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.

(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: Initial height = 1.75 feet

Growth per month = 0.5 feet

(i) Height growth in each month = 0.5 feet

So, Height after 7 months:

h = 1.75 + (0.5 × 7)

= 1.75 + 3.5

= 5.25 feet

(ii) Table of values:

(iii) Expression:

height after t months = h

h = 1.75 + 0.5t

This represents linear growth because height increases by a constant amount (0.5 feet) every month.

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.

Solutions: Initial value = ₹10,000

Decrease per year = ₹800

(i) Decrease in value after 1 year = ₹800

So, Value after 3 years:

v = 10000 − (800 × 3)

= 10000 − 2400

= ₹7600

(ii) Table of values:

 (iii) Expression:

Let value after t years = v

v = 10000 − 800t

Thus it represents linear decay because the value decreases by a constant amount (₹800) every year.

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) Population after 6 years:

P = 750 + (50 × 6)

= 750 + 300

= 1050

(ii) Table of values:

 (iii) Expression:

Let population after t years = P

P = 750 + 50t

Thus it represents linear growth because the population increases by a constant number (50 people) per year.

4. A telecom company charges ₹600 for a certain recharge scheme. This prepaid balance is reduced by ₹15 each day after 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.

Solutions: (i) Expression:

Let remaining balance after x days = b(x)

b(x) = 600 − 15x

This represents a linear decay because the balance decreases by a fixed amount of ₹15 per day.

(ii) When the balance becomes zero:

b(x) = 0

600 − 15x = 0

⇒ 15x = 600

⇒ x = 40

Hence, the balance will run out after 40 days.

(iii) Table of values:

 NCERT solutions class 9 maths chapter 2 exercise 2.5

1. A learning platform charges a fixed monthly fee and an additional cost per digital learning module accessed. A student observed that when she accessed 10 modules, her bill was ₹400. When she accessed 14 modules, her bill was ₹500. If the monthly bill y depends on the number of modules accessed, x, according to the relation y = ax + b, find the values of a and b.

Solution: When x = 10, y = 400

When x = 14, y = 500

Using the linear equation:

y = ax + b

Substituting x = 10 and y = 400:

400 = 10a + b ...(1)

Substituting x = 14 and y = 500:

500 = 14a + b ...(2)

Subtracting equation (1) from equation (2):

500 − 400 = 14a − 10a

⇒ 100 = 4a

⇒ a = 25

Substituting a = 25 into equation (1):

400 = 10 × 25 + b

⇒ 400 = 250 + b

⇒ b = 150

Therefore,

a = 25 and b = 150.

2. A gym charges a fixed monthly fee and an additional cost per hour for using the badminton court. A student using the gym observed that when she used the badminton court for 10 hours, her bill was ₹800. When she used it for 15 hours, her bill was ₹1100. If the monthly bill y depends on the hours of the use of the badminton court, x, according to the relation y = ax + b, find the values of a and b.

Solution: Given:

When x = 10, y = 800

When x = 15, y = 1100

Using the linear equation:

y = ax + b

Substituting the given values:

800 = 10a + b ...(1)

1100 = 15a + b ...(2)

Subtracting equation (1) from equation (2):

1100 − 800 = 15a − 10a

⇒ 300 = 5a

⇒ a = 60

Substituting a = 60 into equation (1):

800 = 10 × 60 + b

⇒ 800 = 600 + b

⇒ b = 200

Therefore,

a = 60 and b = 200.

3. Consider the relationship between temperature measured in degrees Celsius (°C) and degrees Fahrenheit (°F), which is given by °C = a°F + b. Find a and b, given that ice melts at 0 degrees Celsius and 32 degrees Fahrenheit, and water boils at 100 degrees Celsius and 212 degrees Fahrenheit.

(Hint: When °C = 0, °F = 32 and when °C = 100, °F = 212. Use this information to find a and b, and thus, the linear relationship between °C and °F.)

Solution: Given the relation:

°C = a°F + b

Using the point (32, 0):

0 = 32a + b ...(1)

Using the point (212, 100):

100 = 212a + b ...(2)

Subtracting equation (1) from equation (2):

100 − 0 = 212a − 32a

⇒ 100 = 180a

⇒ a = 100/180

⇒ a = 5/9

Substituting a = 5/9 into equation (1):

0 = 32 × (5/9) + b

⇒ 0 = 160/9 + b

⇒ b = −160/9

Therefore,

a = 5/9 and b = −160/9

Hence, the linear relationship between Celsius and Fahrenheit is:

°C = (5/9)°F − 160/9

or equivalently,

°C = (5/9)(°F − 32).

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