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

Linear polynomials are algebraic expressions in one variable whose highest power is 1. In Class 9 Maths Chapter 2, students learn how variables, coefficients, constants, degree, linear patterns, linear growth, linear decay, slope and y-intercept help represent real-life situations using expressions and graphs.

NCERT Solutions for Class 9 Maths Ganit Manjari Chapter 2 explain Introduction to Linear Polynomials from the new Ganita Manjari textbook, where students learn how algebraic expressions can represent changing quantities. The chapter introduces terms, variables, coefficients, constants, degree of a polynomial and linear polynomials.

The chapter moves from simple expressions to real-life linear situations such as bills, gym fees, plant growth, phone depreciation and temperature conversion. These exercise-wise solutions help students form equations, evaluate polynomials, understand linear growth and interpret graphs using slope and y-intercept.

NCERT Solutions for Class 9 Maths Chapter 2 PDF Download

The NCERT Solutions for Class 9 Maths Chapter 2 PDF CBSE 2026-27 gives students a ready reference for solving polynomial and linear relationship questions step by step. It helps students revise substitution, word problems, linear patterns, slope and graph-based questions without missing the logic behind each answer.

Students can use the PDF to revise:

1. Terms, variables, coefficients and constants
2. Degree of a polynomial
3. Linear, quadratic, cubic and constant polynomials
4. Evaluating polynomial values
5. Linear equations from word problems
6. Linear patterns
7. Linear growth and linear decay
8. Linear relationship in the form y = ax + b
9. Graphs of linear equations
10. Slope and y-intercept

Access Exercise-Wise NCERT Solutions for Class 9 Maths Chapter 2

Exercise	Main Focus	Solution Type
Exercise Set 2.1	Degree, coefficients and constants	Direct concept questions
Exercise Set 2.2	Polynomial values and linear equations	Substitution and word problems
Exercise Set 2.3	Linear patterns	Pattern-based expressions
Exercise Set 2.4	Linear growth and linear decay	Table and expression questions
Exercise Set 2.5	Linear relationships	Finding a and b in y = ax + b
Exercise Set 2.6	Graphs of linear relationships	Slope and y-intercept questions
End-of-Chapter Exercises	Mixed revision	Numerical, graph and reasoning questions

NCERT Solutions for Class 9 Maths Chapter 2 Exercise Set 2.1

Exercise Set 2.1 introduces students to polynomials, degree, coefficients and constant terms. Students learn how to identify the highest power of the variable and classify polynomials.

Question 1. Find the degrees of the following polynomials.

(i) 2x² – 5x + 3

Answer:
The highest power of x in 2x² – 5x + 3 is 2.

Final answer:
Degree = 2

(ii) y³ + 2y – 1

Answer:
The highest power of y in y³ + 2y – 1 is 3.

Final answer:
Degree = 3

(iii) -9

Answer:
-9 is a non-zero constant polynomial. A non-zero constant polynomial has degree 0.

Final answer:
Degree = 0

(iv) 4z – 3

Answer:
The highest power of z in 4z – 3 is 1.

Final answer:
Degree = 1

Question 2. Write polynomials of degrees 1, 2 and 3.

Answer:
Examples of polynomials are:

Degree 1 polynomial: 2x + 5
Degree 2 polynomial: x² – 3x + 1
Degree 3 polynomial: x³ + 2x² – 4

Final answer:
2x + 5, x² – 3x + 1, x³ + 2x² – 4

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

Answer:
Given polynomial:

x⁴ – 3x³ + 6x² – 2x + 7

Coefficient of x² = 6
Coefficient of x³ = -3

Final answer:
Coefficient of x² = 6, coefficient of x³ = -3

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

Answer:
The polynomial is:

4z³ + 5z² – 11

There is no z term in this polynomial. Therefore, the coefficient of z is 0.

Final answer:
Coefficient of z = 0

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

Answer:
The constant term is the term that does not contain the variable.

In 9x³ + 5x² – 8x – 10, the constant term is -10.

Final answer:
Constant term = -10

NCERT Solutions for Class 9 Maths Chapter 2 Exercise Set 2.2

Exercise Set 2.2 focuses on finding the value of polynomials by substituting given values. It also includes linear equation word problems based on ages, integers, coins, fences and rectangles.

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

(i) x = 0

Answer:
5x – 3 = 5(0) – 3
= -3

Final answer:
-3

(ii) x = -1

Answer:
5x – 3 = 5(-1) – 3
= -5 – 3
= -8

Final answer:
-8

(iii) x = 2

Answer:
5x – 3 = 5(2) – 3
= 10 – 3
= 7

Final answer:
7

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

(i) s = 0

Answer:
7s² – 4s + 6 = 7(0)² – 4(0) + 6
= 6

Final answer:
6

(ii) s = -3

Answer:
7s² – 4s + 6 = 7(-3)² – 4(-3) + 6
= 7(9) + 12 + 6
= 63 + 12 + 6
= 81

Final answer:
81

(iii) s = 4

Answer:
7s² – 4s + 6 = 7(4)² – 4(4) + 6
= 7(16) – 16 + 6
= 112 – 16 + 6
= 102

Final answer:
102

Question 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 be x years.

Then, Salil’s mother’s present age = 3x years.

After 5 years:

Salil’s age = x + 5
Mother’s age = 3x + 5

According to the question:

x + 5 + 3x + 5 = 70
4x + 10 = 70
4x = 60
x = 15

So, Salil’s present age = 15 years.
Mother’s present age = 3 × 15 = 45 years.

Final answer:
Salil is 15 years old, and his mother is 45 years old.

Question 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 two positive integers be 2x and 5x.

Their difference is 63.

5x – 2x = 63
3x = 63
x = 21

So:

2x = 2 × 21 = 42
5x = 5 × 21 = 105

Final answer:
The two integers are 42 and 105.

Question 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 be x.

Then, number of two-rupee coins = 3x.

Value of five-rupee coins = 5x
Value of two-rupee coins = 2 × 3x = 6x

Total value = ₹88

5x + 6x = 88
11x = 88
x = 8

So:

Five-rupee coins = 8
Two-rupee coins = 3 × 8 = 24

Final answer:
Ruby has 8 five-rupee coins and 24 two-rupee coins.

Question 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 shorter piece be x feet.

Then, the longer piece = 4x feet.

Total length = 300 feet

x + 4x = 300
5x = 300
x = 60

Shorter piece = 60 feet
Longer piece = 4 × 60 = 240 feet

Final answer:
The two pieces are 60 feet and 240 feet long.

Question 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 be x cm.

Then, length = 2x + 3 cm.

Perimeter = 24 cm

2(length + width) = 24
length + width = 12

Substitute:

2x + 3 + x = 12
3x + 3 = 12
3x = 9
x = 3

Width = 3 cm
Length = 2(3) + 3 = 9 cm

Final answer:
Length = 9 cm, width = 3 cm

NCERT Solutions for Class 9 Maths Chapter 2 Exercise Set 2.3

Exercise Set 2.3 focuses on linear patterns. Students learn how fixed increases or decreases can be represented using linear expressions.

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.

Answer:
Initial amount = ₹500
Pocket money received every month = ₹150

Amount after n months:

A(n) = 500 + 150n

At the end of the second month:

A(2) = 500 + 150(2)
= 500 + 300
= ₹800

Final answer:
Linear expression: A(n) = 500 + 150n

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.

Answer:
Initial number of members = 120
Members dropping out every hour = 9

After 1 hour:

120 – 9 = 111

After 2 hours:

120 – 18 = 102

After 3 hours:

120 – 27 = 93

After n hours:

M(n) = 120 – 9n

Final answer:
Members after n hours = 120 – 9n

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 of the rectangle.

Answer:
Length = 13 cm

Area = length × breadth
Area = 13 × 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

If breadth is b cm, then:

A = 13b

Final answer:
Areas are 156 cm², 130 cm² and 104 cm². Linear pattern: A = 13b

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.

Answer:
Length = 7 cm
Breadth = 11 cm

Volume = length × breadth × height
Volume = 7 × 11 × height
Volume = 77h

(i) Height = 5 cm

Volume = 77 × 5
= 385 cm³

(ii) Height = 9 cm

Volume = 77 × 9
= 693 cm³

(iii) Height = 13 cm

Volume = 77 × 13
= 1001 cm³

Final answer:
Volumes are 385 cm³, 693 cm³ and 1001 cm³. Linear pattern: V = 77h

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.

Answer:
Total pages = 500
Pages read per day = 20

Pages read in 15 days:

20 × 15 = 300

Pages left:

500 – 300 = 200

If n is the number of days, then pages left:

P(n) = 500 – 20n

Final answer:
Pages left after 15 days = 200. Linear pattern: P(n) = 500 – 20n

NCERT Solutions for Class 9 Maths Chapter 2 Exercise Set 2.4

Exercise Set 2.4 explains linear growth and linear decay. Linear growth means a quantity increases by a constant amount, while linear decay means a quantity decreases by a constant amount.

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.

Answer:
Initial height = 1.75 feet
Growth per month = 0.5 feet

Height after 7 months:

h = 1.75 + 0.5(7)
= 1.75 + 3.5
= 5.25 feet

Final answer:
5.25 feet

(ii) Make a table of values for t varying from 0 to 10 months and show how the height, h, increases every month.

Answer:

t months	0	1	2	3	4	5	6	7	8	9	10
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.

Answer:
The height after t months is:

h = 1.75 + 0.5t

This represents linear growth because the height increases by the fixed amount of 0.5 feet every month.

Final answer:
h = 1.75 + 0.5t

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.

Answer:
Initial value = ₹10,000
Decrease per year = ₹800

Decrease after 3 years:

800 × 3 = ₹2,400

Value after 3 years:

10,000 – 2,400 = ₹7,600

Final answer:
₹7,600

(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.

Answer:

t years	0	1	2	3	4	5	6	7	8
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.

Answer:
The value after t years is:

v = 10000 – 800t

This represents linear decay because the value decreases by the fixed amount of ₹800 every year.

Final answer:
v = 10000 – 800t

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.

Answer:
Initial population = 750
Increase per year = 50

Increase after 6 years:

50 × 6 = 300

Population after 6 years:

750 + 300 = 1050

Final answer:
1050 people

(ii) Make a table of values for t varying from 0 to 10 years and show how the population, P, increases every year.

Answer:

t years	0	1	2	3	4	5	6	7	8	9	10
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.

Answer:
The population after t years is:

P = 750 + 50t

This represents linear growth because the population increases by the fixed amount of 50 people every year.

Final answer:
P = 750 + 50t

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.

Answer:
Initial balance = ₹600
Decrease per day = ₹15

Remaining balance after x days:

b(x) = 600 – 15x

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

Final answer:
b(x) = 600 – 15x

(ii) After how many days will the balance run out?

Answer:
The balance runs out when b(x) = 0.

600 – 15x = 0
15x = 600
x = 40

Final answer:
The balance will run out after 40 days.

(iii) Make a table of values for x varying from 1 to 10 days and show how the balance b(x) reduces with time.

Answer:

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

NCERT Solutions for Class 9 Maths Chapter 2 Exercise Set 2.5

Exercise Set 2.5 focuses on linear relationships in the form y = ax + b. Students learn how to find the values of a and b using two given data points.

Question 1. A learning platform charges a fixed monthly fee and an additional cost per digital learning module accessed. A student observes 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:
Given relation:

y = ax + b

When x = 10, y = 400:

400 = 10a + b …(1)

When x = 14, y = 500:

500 = 14a + b …(2)

Subtract equation (1) from equation (2):

500 – 400 = 14a – 10a
100 = 4a
a = 25

Substitute a = 25 in equation (1):

400 = 10(25) + b
400 = 250 + b
b = 150

Final answer:
a = 25, b = 150

The linear relationship is:

y = 25x + 150

Question 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:
Given relation:

y = ax + b

When x = 10, y = 800:

800 = 10a + b …(1)

When x = 15, y = 1100:

1100 = 15a + b …(2)

Subtract equation (1) from equation (2):

1100 – 800 = 15a – 10a
300 = 5a
a = 60

Substitute a = 60 in equation (1):

800 = 10(60) + b
800 = 600 + b
b = 200

Final answer:
a = 60, b = 200

The linear relationship is:

y = 60x + 200

Question 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:
Given relation:

°C = a°F + b

When °C = 0 and °F = 32:

0 = 32a + b …(1)

When °C = 100 and °F = 212:

100 = 212a + b …(2)

Subtract equation (1) from equation (2):

100 – 0 = 212a – 32a
100 = 180a
a = 100/180
a = 5/9

Substitute a = 5/9 in equation (1):

0 = 32(5/9) + b
b = -160/9

Final answer:
a = 5/9, b = -160/9

The relationship is:

°C = (5/9)°F - 160/9

or

°C = (5/9)(°F - 32)

NCERT Solutions for Class 9 Maths Chapter 2 Exercise Set 2.6

Exercise Set 2.6 focuses on visualising linear relationships. Students draw graphs of straight lines and observe the role of slope and y-intercept.

Question 1. Draw the graphs of the following sets of lines. In each case, reflect on the role of a and b.

(i) y = 4x, y = 2x, y = x

Answer:
All three equations are of the form:

y = ax + b

Here, b = 0 for all three lines. Therefore, all lines pass through the origin.

Slopes are:

y = 4x has slope 4
y = 2x has slope 2
y = x has slope 1

Final answer:
All three lines pass through the origin. The larger the value of a, the steeper the line.

(ii) y = -6x, y = -3x, y = -x

Answer:
All three equations have b = 0, so all pass through the origin.

Slopes are:

y = -6x has slope -6
y = -3x has slope -3
y = -x has slope -1

Final answer:
All three lines pass through the origin and show negative slope. The more negative the value of a, the steeper the downward line.

(iii) y = 5x, y = -5x

Answer:
Both equations have b = 0, so both pass through the origin.

y = 5x has positive slope 5.
y = -5x has negative slope -5.

Final answer:
The two lines pass through the origin but slope in opposite directions.

(iv) y = 3x – 1, y = 3x, y = 3x + 1

Answer:
All three equations have the same slope:

a = 3

Their y-intercepts are:

y = 3x – 1 has y-intercept -1
y = 3x has y-intercept 0
y = 3x + 1 has y-intercept 1

Final answer:
The three lines are parallel because they have the same slope. They cut the y-axis at different points.

(v) y = -2x – 3, y = -2x, y = 2x + 3

Answer:
For y = -2x – 3:

slope = -2, y-intercept = -3

For y = -2x:

slope = -2, y-intercept = 0

For y = 2x + 3:

slope = 2, y-intercept = 3

Final answer:
The first two lines are parallel because they have the same slope -2. The third line is not parallel to them because its slope is 2.

NCERT Solutions for Class 9 Maths Chapter 2 End-of-Chapter Exercises

The end-of-chapter exercises include mixed questions on degree, polynomial values, linear equations, linear patterns, graphs, slope, y-intercepts, matchstick patterns and linear functions.

Question 1. Write a polynomial of degree 3 in the variable x, in which the coefficient of the x² term is -7.

Answer:
A polynomial of degree 3 must have the highest power of x as 3.

One such polynomial is:

x³ – 7x² + 2x + 1

Final answer:
x³ – 7x² + 2x + 1

Question 2. Find the values of the following polynomials at the indicated values of the variables.

(i) 5x² – 3x + 7 if x = 1

Answer:
Substitute x = 1:

5x² – 3x + 7
= 5(1)² – 3(1) + 7
= 5 – 3 + 7
= 9

Final answer:
9

(ii) 4t³ – t² + 6 if t = a

Answer:
Substitute t = a:

4t³ – t² + 6
= 4a³ – a² + 6

Final answer:
4a³ – a² + 6

Question 3. If we multiply a number by 5/2 and add 2/3 to the product, we get -7/12. Find the number.

Answer:
Let the number be x.

According to the question:

(5/2)x + 2/3 = -7/12

Move 2/3 to the right side:

(5/2)x = -7/12 - 2/3
= -7/12 - 8/12
= -15/12
= -5/4

Now multiply both sides by 2/5:

x = (-5/4) × (2/5)
x = -10/20
x = -1/2

Final answer:
The number is -1/2.

Question 4. A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?

Answer:
Let the smaller number be x.

Then, the larger number = 5x.

After adding 21 to both:

Smaller new number = x + 21
Larger new number = 5x + 21

According to the question:

5x + 21 = 2(x + 21)

5x + 21 = 2x + 42
3x = 21
x = 7

Smaller number = 7
Larger number = 5 × 7 = 35

Final answer:
The numbers are 7 and 35.

Question 5. If you have ₹800 and you save ₹250 every month, find the amount you have after (i) 6 months (ii) 2 years. Express this as a linear pattern.

Answer:
Initial amount = ₹800
Savings per month = ₹250

Amount after n months:

A(n) = 800 + 250n

(i) After 6 months

A(6) = 800 + 250(6)
= 800 + 1500
= ₹2300

(ii) After 2 years

2 years = 24 months

A(24) = 800 + 250(24)
= 800 + 6000
= ₹6800

Final answer:
After 6 months: ₹2300. After 2 years: ₹6800. Linear pattern: A(n) = 800 + 250n

Question 6. The digits of a two-digit number differ by 3. If the digits are interchanged, and the resulting number is added to the original number, we get 143. Find both the numbers.

Answer:
Let the tens digit be x and the ones digit be y.

Original number = 10x + y
Reversed number = 10y + x

Their sum is 143:

10x + y + 10y + x = 143
11x + 11y = 143
11(x + y) = 143
x + y = 13

The digits differ by 3:

x - y = 3 or y - x = 3

The two digits whose sum is 13 and difference is 3 are 8 and 5.

Therefore, the two possible numbers are:

85 and 58

Final answer:
The numbers are 85 and 58.

Question 7. Draw the graph of the following equations, and identify their slopes and y-intercepts. Also, find the coordinates of the points where these lines cut the y-axis. Are any of the lines parallel?

(i) y = -3x + 4

Answer:
Slope = -3
y-intercept = 4
Point where the line cuts y-axis = (0, 4)

(ii) 2y = 4x + 7

Answer:
Convert to y = ax + b form:

2y = 4x + 7
y = 2x + 7/2

Slope = 2
y-intercept = 7/2
Point where the line cuts y-axis = (0, 7/2)

(iii) 5y = 6x – 10

Answer:
Convert to y = ax + b form:

5y = 6x – 10
y = (6/5)x – 2

Slope = 6/5
y-intercept = -2
Point where the line cuts y-axis = (0, -2)

(iv) 3y = 6x – 11

Answer:
Convert to y = ax + b form:

3y = 6x – 11
y = 2x – 11/3

Slope = 2
y-intercept = -11/3
Point where the line cuts y-axis = (0, -11/3)

Parallel lines have equal slopes.

Final answer:
Lines 2y = 4x + 7 and 3y = 6x – 11 are parallel because both have slope 2.

Question 8. If the temperature of a liquid can be measured in Kelvin units as x K and in Fahrenheit units as y °F, the relation between the two systems of measurement of temperature is given by y = 9/5(x – 273) + 32.

(i) Find the temperature of the liquid in Fahrenheit if the temperature of the liquid is 313 K.

Answer:
Given:

x = 313

y = 9/5(x – 273) + 32
= 9/5(313 – 273) + 32
= 9/5(40) + 32
= 72 + 32
= 104

Final answer:
104 °F

(ii) If the temperature is 158 °F, then find the temperature in Kelvin.

Answer:
Given:

y = 158

158 = 9/5(x – 273) + 32
158 – 32 = 9/5(x – 273)
126 = 9/5(x – 273)

Multiply both sides by 5/9:

x – 273 = 126 × 5/9
x – 273 = 70
x = 343

Final answer:
343 K

Question 9. The work done by a body on the application of a constant force is the product of the constant force and the distance travelled by the body in the direction of the force. Express this in the form of a linear equation in two variables, work w and distance d, and draw its graph by taking the constant force as 3 units. What is the work done when the distance travelled is 2 units? Verify it by plotting it on the graph.

Answer:
Work = Force × Distance

Given force = 3 units

So:

w = 3d

This is a linear equation in two variables.

For d = 2:

w = 3(2)
w = 6

To draw the graph, use points such as:

d	0	1	2
w	0	3	6

Plot (0, 0), (1, 3) and (2, 6), and join them to get a straight line.

Final answer:
Equation: w = 3d. Work done when d = 2 is 6 units.

Question 10. The graph of a linear polynomial p(x) passes through the points (1, 5) and (3, 11).

(i) Find the polynomial p(x).

Answer:
Let:

p(x) = ax + b

The graph passes through (1, 5):

5 = a + b …(1)

The graph passes through (3, 11):

11 = 3a + b …(2)

Subtract equation (1) from equation (2):

6 = 2a
a = 3

Substitute a = 3 in equation (1):

5 = 3 + b
b = 2

Final answer:
p(x) = 3x + 2

(ii) Find the coordinates where the graph of p(x) cuts the axes.

Answer:
For y-axis, x = 0:

p(0) = 3(0) + 2
= 2

So, y-intercept point = (0, 2)

For x-axis, p(x) = 0:

3x + 2 = 0
3x = -2
x = -2/3

So, x-intercept point = (-2/3, 0)

Final answer:
Cuts y-axis at (0, 2) and x-axis at (-2/3, 0).

(iii) Draw the graph of p(x) and verify your answers.

Answer:
To draw the graph, plot the points (1, 5) and (3, 11). Join them with a straight line.

The line will cut the y-axis at (0, 2) and the x-axis at (-2/3, 0), verifying the answers.

Final answer:
The graph verifies the intercepts (0, 2) and (-2/3, 0).

Question 11. Let p(x) = ax + b and q(x) = cx + d be two linear polynomials such that p(0) = 5, p(x) – q(x) cuts the x-axis at (3, 0), and p(x) + q(x) is equal to 6x + 4 for all real x. Find p(x) and q(x).

Answer:
Given:

p(x) = ax + b
q(x) = cx + d

Since p(0) = 5:

b = 5

Also:

p(x) + q(x) = 6x + 4

So:

(a + c)x + (b + d) = 6x + 4

Therefore:

a + c = 6
b + d = 4

Since b = 5:

5 + d = 4
d = -1

Now:

p(x) – q(x) cuts the x-axis at (3, 0).
This means:

p(3) – q(3) = 0

p(3) = 3a + 5
q(3) = 3c - 1

So:

3a + 5 - (3c - 1) = 0
3a + 5 - 3c + 1 = 0
3(a - c) + 6 = 0
a - c = -2

Now solve:

a + c = 6
a - c = -2

Adding:

2a = 4
a = 2

Then:

c = 4

Final answer:
p(x) = 2x + 5, q(x) = 4x - 1

Question 12. Look at the first three stages of a growing pattern of hexagons made using matchsticks. A new hexagon gets added at every stage which shares a side with the last hexagon of the previous stage.

(i) Draw the next two stages of the pattern. How many matchsticks will be required at these stages?

Answer:
One hexagon needs 6 matchsticks.

When a new hexagon is added sharing one side, it needs 5 new matchsticks.

Stage 1 = 6 matchsticks
Stage 2 = 6 + 5 = 11 matchsticks
Stage 3 = 11 + 5 = 16 matchsticks
Stage 4 = 16 + 5 = 21 matchsticks
Stage 5 = 21 + 5 = 26 matchsticks

Final answer:
Stage 4 needs 21 matchsticks, and Stage 5 needs 26 matchsticks.

(ii) Complete the table.

Answer:

Stage Number	1	2	3	4	5	…	n
Number of matchsticks	6	11	16	21	26	…	5n + 1

(iii) Find a rule to determine the number of matchsticks required for the nth stage.

Answer:
The pattern starts at 6 and increases by 5 each time.

Number of matchsticks at nth stage:

M(n) = 6 + (n - 1)5
= 6 + 5n - 5
= 5n + 1

Final answer:
M(n) = 5n + 1

(iv) How many matchsticks will be required for the 15th stage of the pattern?

Answer:
M(n) = 5n + 1

For n = 15:

M(15) = 5(15) + 1
= 75 + 1
= 76

Final answer:
76 matchsticks

(v) Can 200 matchsticks form a stage in this pattern? Justify your answer.

Answer:
Use:

5n + 1 = 200

5n = 199
n = 199/5
n = 39.8

Since n is not a whole number, 200 matchsticks cannot form a complete stage in this pattern.

Final answer:
No, 200 matchsticks cannot form a stage in this pattern.

Question 13. Let p(x) = ax + b and q(x) = cx + d be two linear polynomials such that the graph of p(x) passes through (2, 3) and (6, 11), the graph of q(x) passes through (4, -1), and the graph of q(x) is parallel to the graph of p(x). Find p(x) and q(x). Also, find the coordinates of the point where these lines meet the x-axis.

Answer:
First find p(x).

Let:

p(x) = ax + b

It passes through (2, 3):

3 = 2a + b …(1)

It passes through (6, 11):

11 = 6a + b …(2)

Subtract equation (1) from equation (2):

8 = 4a
a = 2

Substitute a = 2 in equation (1):

3 = 2(2) + b
3 = 4 + b
b = -1

So:

p(x) = 2x - 1

Since q(x) is parallel to p(x), it has the same slope 2.

Let:

q(x) = 2x + d

It passes through (4, -1):

-1 = 2(4) + d
-1 = 8 + d
d = -9

So:

q(x) = 2x - 9

Now find x-intercepts.

For p(x):

2x - 1 = 0
2x = 1
x = 1/2

So p(x) meets x-axis at:

(1/2, 0)

For q(x):

2x - 9 = 0
2x = 9
x = 9/2

So q(x) meets x-axis at:

(9/2, 0)

Final answer:
p(x) = 2x - 1, q(x) = 2x - 9
x-intercepts: (1/2, 0) and (9/2, 0)

Question 14. What do all linear functions of the form f(x) = ax + a, a > 0, have in common?

Answer:
Given:

f(x) = ax + a

Factorise:

f(x) = a(x + 1)

To find the x-intercept, set f(x) = 0:

a(x + 1) = 0

Since a > 0, a cannot be 0.

So:

x + 1 = 0
x = -1

Therefore, all such functions pass through (-1, 0).

Also:

f(0) = a

So each function cuts the y-axis at (0, a), where a is positive.

Final answer:
All functions of the form f(x) = ax + a, a > 0, pass through (-1, 0). Their slope and y-intercept are both equal to a.

Topics Covered in NCERT Solutions for Class 9 Maths Chapter 2

1. Algebraic expressions
2. Terms, variables and coefficients
3. Constant term
4. One-variable polynomials
5. Degree of a polynomial
6. Linear polynomials
7. Quadratic polynomials
8. Cubic polynomials
9. Constant polynomials
10. Evaluating polynomials
11. Linear equations
12. Linear patterns
13. Linear growth
14. Linear decay
15. Linear relationships
16. Equation y = ax + b
17. Graphs of linear equations
18. Slope of a line
19. y-intercept of a line
20. Parallel lines

Important Formulas in NCERT Solutions for Class 9 Maths Chapter 2

Concept	Formula / Rule
Linear polynomial	ax + b
Degree of linear polynomial	1
Linear relationship	y = ax + b
Slope in y = ax + b	a
y-intercept in y = ax + b	b
Point where line cuts y-axis	(0, b)
Linear growth	Initial value + fixed increase × time
Linear decay	Initial value - fixed decrease × time
Work done with constant force	w = force × distance
Mid-value substitution	Replace variable with given value

NCERT Class 9 Maths Ganita Manjari 2026 Chapter Solutions

Chapter	Title
Chapter 1	Orienting Yourself: The Use of Coordinates
Chapter 2	Introduction to Linear Polynomials
Chapter 3	The World of Numbers
Chapter 4	Exploring Algebraic Identities
Chapter 5	I’m Up and Down, and Round and Round
Chapter 6	Measuring Space: Perimeter and Area
Chapter 7	The Mathematics of Maybe: Introduction to Probability
Chapter 8	Predicting What Comes Next: Exploring Sequences and Progressions