NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 2 – Introduction to Linear Polynomials are available here for Session 2026–27. Exercise Sets 2.1, 2.2, 2.3, 2.4, 2.5, 2.6 and End of Chapter Exercises are explained with step-by-step solutions, clear methods and student-friendly language based on the latest NCERT syllabus under NEP 2020 and NCF 2023.
In this chapter, students learn the basic concepts of linear polynomials, variables, coefficients, constants and degree of a polynomial through simple daily-life situations. The chapter also explains linear equations, linear patterns, linear growth and decay, and relationships of the form y = ax + b. Students further understand how linear relationships are represented on graphs using straight lines, slope and y-intercept concepts.
All questions are solved in an easy and exam-oriented format to help students improve algebraic understanding, logical thinking and problem-solving skills. These solutions are useful for homework, revision, school exams and concept clarity for CBSE Class 9 Mathematics Session 2026–27.
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Exercise Set 2.1
Exercise Set 2.2
Exercise Set 2.3
Exercise Set 2.4 will be upload shortly
Exercise Set 2.5
Exercise Set 2.6 will be upload shortly
End of Chapter Exercise will be upload shortly
Exercise Set 2.1
1. Find the degrees of the following polynomials:
(i) 2x² − 5x + 3
(ii) y³ + 2y − 1
(iii) −9
(iv) 4z – 3
Answer:
(i) 2x² − 5x + 3
Degree: 2
Explanation: The highest power of the variable x in this expression is 2.
(ii) y³ + 2y − 1
Degree: 3
Explanation: The highest power of the variable y in this expression is 3.
(iii) −9
Degree: 0
Explanation: This is a constant polynomial, which can be written as -9x⁰ where the power of the variable is 0.
(iv) 4z − 3
Degree: 1
Explanation: The highest power of the variable z in this expression is 1 (4z¹).
2. Write polynomials of degrees 1, 2 and 3.
Answer:
Polynomial of degree 1 (Linear Polynomial)
Example: 3z + 7
Explanation: The highest power of the variable z is 1.
Polynomial of degree 2 (Quadratic Polynomial)
Example: x² + 5x + 1
Explanation: The highest power of the variable x is 2.
Polynomial of degree 3 (Cubic Polynomial)
Example: 5y³ + y² + 2y − 1
Explanation: The highest power of the variable y is 3.
3. What are the coefficients of x² and x³ in the polynomial x⁴ − 3x³ + 6x² − 2x + 7?
Answer:
In the polynomial x⁴ − 3x³ + 6x² − 2x + 7:
The coefficient of x² is 6.
Explanation: The number multiplied by x² is 6.
The coefficient of x³ is −3.
Explanation: The number multiplied by x³ is −3 (including its negative sign).
4. What is the coefficient of z in the polynomial 4z³ + 5z² − 11?
Answer:
In the polynomial 4z³ + 5z² − 11:
The coefficient of z is 0.
Explanation: There is no z term written in the polynomial, which means it can be thought of as 0z. Therefore, its coefficient is 0
5. What is the constant term of the polynomial 9x³ + 5x² − 8x − 10?
Recall that polynomials of degree 1 are called linear polynomials. In this chapter, we shall study linear polynomials.
Answer:
In the polynomial 9x³ + 5x² − 8x − 10:
The constant term is −10.
Explanation: The constant term is the term that does not have any variable attached to it. In this expression, the number at the end is −10 (including its negative sign).
Exercise Set 2.2
1. Find the value of the linear polynomial 5x − 3 if:
(i) x = 0
(ii) x = −1
(iii) x = 2
Answer:
Here are the values of the linear polynomial 5x – 3 for each case:
(i) For x = 0
Answer: −3
Explanation: Substitute x = 0 in the polynomial:
5(0) − 3 = 0 − 3 = −3
(ii) For x = −1
Answer: −8
Explanation:
Substitute x = −1 in the polynomial:
5(−1) − 3 = −5 − 3 = −8
(iii) For x = 2
Answer: 7
Explanation: Substitute x = 2 in the polynomial:
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
Answer:
(i) For s = 0
Answer: 6
Explanation: Substitute s = 0 in the polynomial:
7(0)² − 4(0) + 6 = 0 − 0 + 6 = 6
(ii) For s = −3
Answer: 81
Explanation: Substitute s = −3 in the polynomial:
7(−3)² − 4(−3) + 6
= 7(9) + 12 + 6 = 63 + 12 + 6 = 81
(iii) For s = 4
Answer: 102
Explanation: Substitute s = 4 in the polynomial:
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.
Answer:
Let Salil’s present age = x years
Mother’s present age = 3x years
After 5 years: Salil’s age = x + 5
Mother’s age = 3x + 5
According to question: (x + 5) + (3x + 5) = 70
⇒ 4x + 10 = 70 ⇒ 4x = 60
⇒ x = 15
Therefore, Salil’s age = 15 years
Mother’s age = 45 years
4. The difference between two positive integers is 63. The ratio of the two integers is 2:5. Find the two integers.
Answer:
Let the common ratio between the two integers be x.
Therefore, the two integers are 2x and 5x.
According to the question, the difference between them is 63:
5x − 2x = 63
⇒ 3x = 63
⇒ x = 21
Now, substituting the value of x:
First integer = 2x = 2 × 21 = 42
Second integer = 5x = 5 × 21 = 105
Therefore, the two 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?
Answer:
Let the number of five-rupee coins = x
Then, the number of two-rupee coins = 3x
Value of five-rupee coins = 5 × x = 5x
Value of two-rupee coins = 2 × 3x = 6x
According to the question, the total amount is ₹88:
5x + 6x = 88
⇒ 11x = 88
⇒ x = 8
Now, calculating the number of coins of each type:
Number of five-rupee coins = x = 8
Number of two-rupee coins = 3x = 3 × 8 = 24
Therefore, Ruby has 8 five-rupee coins and 24 two-rupee coins.
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?
Answer:
Let the length of the shorter piece = x feet
Then, the length of the longer piece = 4x feet
According to the question, the total length of the fence is 300 feet:
x + 4x = 300
⇒ 5x = 300
⇒ x = 60
Now, calculating the length of each piece:
Length of the shorter piece = x = 60 feet
Length of the longer piece = 4x = 4 × 60 = 240 feet
Therefore, the two pieces are 60 feet and 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?
Answer:
Let the width of the rectangle = x cm
Then, the length of the rectangle = 2x + 3 cm
According to the question, the perimeter of the rectangle is 24 cm:
Perimeter = 2 × (length + width)
24 = 2 × ((2x + 3) + x)
⇒ 24 = 2 × (3x + 3)
⇒ 24 = 6x + 6
⇒ 24 − 6 = 6x
⇒ 18 = 6x
⇒ x = 3
Now, calculating the dimensions:
Width = x = 3 cm
Length = 2x + 3 = 2(3) + 3 = 6 + 3 = 9 cm
Therefore, the dimensions of the rectangle are 9 cm (length) and 3 cm (width).
Exercise Set 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.
Answer:
Let n represent the month number.
Initial amount in account = ₹500 Amount added every month = ₹150
The amount of money she will have at the end of each month is as follows:
End of 1st month = 500 + 150 = ₹650
End of 2nd month = 650 + 150 = ₹800
End of 3rd month = 800 + 150 = ₹950
End of 4th month = 950 + 150 = ₹1100
From the second month onwards, the sequence of money she will have at the end of every month is ₹800, ₹950, ₹1100, …
To find the linear expression for the amount in the nth month:
Amount = Initial amount + (Amount per month × n) Linear expression = 500 + 150n
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.
Answer:
Initial number of members = 120 Members dropping out each hour = 9
The number of members remaining after the first few hours:
After 1 hour = 120 − 9 = 111 members
After 2 hours = 111 − 9 = 102 members
After 3 hours = 102 − 9 = 93 members
Therefore, the number of members remaining after 1, 2, 3, … hours is 111, 102, 93, …
To find the linear expression for the number of members at the end of the nth hour:
Members remaining = Initial members − (Members dropping out per hour × n) Linear expression = 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.
Answer:
Length of the rectangle = 13 cm
The area of the rectangle (Area = length × breadth) for each breadth is:
(i) If breadth = 12 cm Area = 13 × 12 = 156 cm²
(ii) If breadth = 10 cm Area = 13 × 10 = 130 cm²
(iii) If breadth = 8 cm Area = 13 × 8 = 104 cm²
To find the linear pattern representing the area: Let the breadth of the rectangle be x cm.
Linear pattern for the area = 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.
Answer:
Length of the rectangular box = 7 cm Breadth of the rectangular box = 11 cm
The volume of a rectangular box is given by the formula:
Volume = length × breadth × height.
Therefore,
Volume = 7 × 11 × height = 77 × height.
The volume for each given height is:
(i) If height = 5 cm Volume = 77 × 5 = 385 cm³
(ii) If height = 9 cm Volume = 77 × 9 = 693 cm³
(iii) If height = 13 cm Volume = 77 × 13 = 1001 cm³
To find the linear pattern representing the volume:
Let the height of the rectangular box be h cm.
Linear pattern for the volume = 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.
Answer:
Total pages in the book = 500 Pages read every day = 20
The number of pages left after 15 days:
Pages read in 15 days = 20 × 15 = 300 pages
Pages left = 500 − 300 = 200 pages
Therefore,
200 pages will be left after 15 days.
To express this as a linear pattern:
Let d represent the number of days.
Linear pattern for pages left = 500 − 20d
Exercise Set 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.
Answer:
Initial amount in savings account = ₹500 Pocket money received every month = ₹150
The amount she will have at the end of each month is calculated by adding ₹150 for each passing month:
Amount at the end of 1st month = 500 + 150 = ₹650
Amount at the end of 2nd month = 650 + 150 = ₹800
Amount at the end of 3rd month = 800 + 150 = ₹950
Amount at the end of 4th month = 950 + 150 = ₹1100
Therefore, from the second month onwards, the amount of money she will have at the end of every month is ₹800, ₹950, ₹1100, …
To represent the amount she will have in the nth month as a linear expression:
Linear expression = 500 + 150n
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.
Answer:
Initial number of members = 120
Members dropping out each hour = 9
The number of members remaining after the first few hours is calculated as follows:
After 1 hour = 120 − 9 = 111 members
After 2 hours = 111 − 9 = 102 members
After 3 hours = 102 − 9 = 93 members
Therefore, the number of members remaining after 1, 2, 3, … hours is 111, 102, 93, …
To find the linear expression to represent the number of members at the end of the nth hour:
Linear expression = 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.
Answer:
Length of the rectangle = 13 cm
The area of the rectangle (Area = length × breadth) for each given breadth is:
(i) If breadth = 12 cm Area = 13 × 12 = 156 cm²
(ii) If breadth = 10 cm Area = 13 × 10 = 130 cm²
(iii) If breadth = 8 cm Area = 13 × 8 = 104 cm²
To find the linear pattern representing the area of the rectangle:
Let the breadth of the rectangle be x cm.
Linear pattern for the area = 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.
Answer:
Length of the rectangular box = 7 cm
Breadth of the rectangular box = 11 cm
The volume of a rectangular box is given by the formula:
Volume = length × breadth × height.
Therefore, Volume = 7 × 11 × height = 77 × height.
The volume for each given height is:
(i) If height = 5 cm Volume = 77 × 5 = 385 cm³
(ii) If height = 9 cm Volume = 77 × 9 = 693 cm³
(iii) If height = 13 cm Volume = 77 × 13 = 1001 cm³
To find the linear pattern representing the volume:
Let the height of the rectangular box be h cm.
Linear pattern for the volume = 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.
Answer:
Total pages in the book = 500
Pages read every day = 20
Pages read in 15 days = 20 × 15 = 300
Pages left after 15 days = 500 – 300 = 200
Therefore, 200 pages will be left after 15 days.
To express this as a linear pattern:
Let d represent the number of days.
Linear pattern for pages left = 500 – 20d
Let pages left after n days = Pₙ
So, Pₙ = 500 − 20n
Thus, linear pattern is Pₙ = 500 − 20n.
(Exercise Set 2.4 will be upload shortly)
Exercise Set 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.
Answer:
According to the question, the relation between the monthly bill (y) and the number of modules (x) is given by:
y = ax + b
From the given observations:
For x = 10, y = 400:
⇒ 400 = 10a + b — (Equation 1)
For x = 14, y = 500:
⇒ 500 = 14a + b — (Equation 2)
Subtracting Equation 1 from Equation 2:
(14a + b) − (10a + b) = 500 − 400
⇒ 4a = 100
⇒ a = 100 / 4
⇒ a = 25
Substituting the value of a = 25 in Equation 1:
400 = 10(25) + b
⇒ 400 = 250 + b
⇒ b = 400 − 250
⇒ b = 150
Therefore, the values are 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.
Answer:
According to the question, the relation between the monthly bill (y) and the hours of badminton court use (x) is given by: y = ax + b
From the given observations:
1. For x = 10, y = 800: 800 = 10a + b — (Equation 1)
2. For x = 15, y = 1100: 1100 = 15a + b — (Equation 2)
Subtracting Equation 1 from Equation 2:
(15a + b) − (10a + b) = 1100 − 800
⇒ 5a = 300
⇒ a = 300 / 5
⇒ a = 60
Substituting the value of a = 60 in Equation 1:
800 = 10(60) + b
⇒ 800 = 600 + b
⇒ b = 800 − 600
⇒ b = 200
Therefore, the values are 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.
Answer:
According to the question, the relation between the temperature in Celsius (°C) and Fahrenheit (°F) is given by: °C = a(°F) + b
From the given observations:
Ice melts at 0°C and 32°F (For °F = 32, °C = 0):
0 = a(32) + b
⇒ 32a + b = 0 — (Equation 1)
Water boils at 100°C and 212°F (For °F = 212, °C = 100):
100 = a(212) + b
⇒ 212a + b = 100 — (Equation 2)
Subtracting Equation 1 from Equation 2:
(212a + b) − (32a + b) = 100 − 0
⇒ 180a = 100
⇒ a = 100 / 180
⇒ a = 5 / 9
Substituting the value of a = 5 / 9 in Equation 1:
32(5 / 9) + b = 0
⇒ 160 / 9 + b = 0
⇒ b = −160 / 9
Therefore, the values are a = 5 / 9 and b = −160 / 9.
