NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 2 Introduction to Linear Polynomials 2026-27

Think and Reflect (NCERT Textbook Page No. 17)

Question 1. Can you identify the terms, variables and coefficients of this algebraic expression?

Solution:
The given algebraic expression is 200l + 160w + 50lw.

Terms: 200l, 160w and 50lw
Variables: l and w
Coefficients: The coefficient of l is 200, the coefficient of w is 160, and the coefficient of lw is 50.

Question 2. How is it different from the algebraic expression in Example 1 ?

Solution:
The expression in Example 1 is 4x + 5y + 3, which is a linear polynomial in the variables x and y. The expression 200l + 160w + 50lw has variables l and w, but it is not a linear polynomial because it includes the product term lw.

Think and Reflect (NCERT Textbook Page No. 17)

Question 1. Can you identify the terms, variables and coefficients of this algebraic expression?

Solution:
The given algebraic expression is 10x - x².

Terms: 10x and -x²
Variable: x
Coefficients: The coefficient of x is 10, and the coefficient of x² is -1.

Question 2. Can you point out any similarity or difference between the algebraic expressions obtained in Examples 1 and 3?

Solution:
Similarity: Both expressions contain variables.

Difference: The expression in Example 1, 4x + 5y + 3, is a linear polynomial in x and y. The expression in Example 3, 10x - x², is not linear because it contains x². It is a quadratic expression in the variable x.

Think and Reflect (NCERT Textbook Page No. 19)

Find the perimeter of squares with sides 1 cm, 1.5 cm, 2 cm, 2.5 cm and 3 cm. What will happen to the perimeters if the sides increase by 0.5 cm?

Solution:
The perimeter of a square with side x cm is 4x cm.

For side 1 cm, perimeter = 4 × 1 = 4 cm
For side 1.5 cm, perimeter = 4 × 1.5 = 6 cm
For side 2 cm, perimeter = 4 × 2 = 8 cm
For side 2.5 cm, perimeter = 4 × 2.5 = 10 cm
For side 3 cm, perimeter = 4 × 3 = 12 cm

So, whenever the side increases by 0.5 cm, the perimeter increases by 2 cm. This shows a constant change, which is an important feature of a linear polynomial.

Think and Reflect (NCERT Textbook Page No. 19)

If a player paid ₹ 750, how many matches did he play?

Solution:
The given polynomial is 200 + 50m, where m is the number of matches played.

200 + 50m = 750
50m = 750 - 200
50m = 550
m = 11

Therefore, the player played 11 matches.

Think and Reflect (NCERT Textbook Page No. 20)

We have learnt that to evaluate the value of an algebraic expression, we substitute a value of the variable in the given expression. Consider Example 3, where the wire is bent to form a rectangle. Here, the area of the rectangle, 10x – x2, is a function of x. Can you interpret this as an input-output process? What value does the expression take when x = 6 cm?

Solution:
Yes, the expression 10x - x² can be understood as an input-output process. Here, x is the input, and the value of 10x - x² gives the output, which represents the area of the rectangle.

When x = 6 cm,

10x - x² = 10(6) - 6²
= 60 - 36
= 24

Therefore, the value of the expression is 24.

Think and Reflect (NCERT Textbook Page No. 22)

Predict the number of squares in the next three stages of the pattern and write the sequence of numbers up to Stage 7 of the pattern.

Solution:
Students should observe the given pattern carefully and continue it to find the next three stages. Then, they should write the number of squares from Stage 1 to Stage 7 based on the pattern.

Think and Reflect (NCERT Textbook Page No. 22)

Using the expression 2n – 1, can you find out how many tiles will be there in the 15th stage and the 26th stage of the pattern? Also, which stage will contain 21 tiles and 47 tiles?

Solution:
The number of tiles in the nth stage is given by 2n - 1.

For the 15th stage:
2(15) - 1 = 30 - 1 = 29
So, the 15th stage has 29 tiles.

For the 26th stage:
2(26) - 1 = 52 - 1 = 51
So, the 26th stage has 51 tiles.

For 21 tiles:
2n - 1 = 21
2n = 22
n = 11

So, 21 tiles will be in the 11th stage.

For 47 tiles:
2n - 1 = 47
2n = 48
n = 24

So, 47 tiles will be in the 24th stage.

Think and Reflect (NCERT Textbook Page No. 23)

What amount will be left on the 15th day? How many days will it take for the entire amount to be spent?

Solution:
The amount left on the nth day is given by 100 - 5n.

On the 15th day:
100 - 5(15) = 100 - 75 = 25

So, ₹25 will be left on the 15th day.

Since ₹5 is spent each day, the number of days required to spend ₹100 is:
100 ÷ 5 = 20

Therefore, the entire amount will be spent in 20 days.

Think and Reflect (NCERT Textbook Page No. 23)

For how many km will the fare be ₹ 130?

Solution:
The fare is given by 15n - 5, where n is the distance in kilometres.

15n - 5 = 130
15n = 135
n = 9

Therefore, the fare will be ₹130 for 9 km.

Think and Reflect (NCERT Textbook Page No. 24)

What is the cost for travelling 15 km? For how many kilometres will the cost of the journey be ₹ 700?

Solution:
The cost of the journey is given by C(d) = 100 + 60d, where d is the distance in kilometres.

For 15 km:
C(15) = 100 + 60(15)
= 100 + 900
= ₹1000

Now, if the cost is ₹700:

100 + 60d = 700
60d = 600
d = 10

Therefore, the cost for travelling 15 km is ₹1000, and the journey will cost ₹700 for 10 km.

Think and Reflect (NCERT Textbook Page No. 25)

What will be the height of the water at the end of 5 months?

Solution:
The height of water after t months is given by h(t) = 3 - 0.5t.

For t = 5:

h(5) = 3 - 0.5(5)
= 3 - 2.5
= 0.5 m

Therefore, the height of the water at the end of 5 months will be 0.5 m.

Think and Reflect (NCERT Textbook Page No. 27)

Can you guess what the numbers 20 and 150 in the equation y = 20x + 150 represent?

Solution:
In the equation y = 20x + 150, y represents the monthly bill and x represents the internet data used in GB.

Here, 20 represents the extra charge for each GB of data used, and 150 represents the fixed monthly charge.

Think and Reflect (NCERT Textbook Page No. 28)

Identify other points on the line by completing the following table.

Solution:
The equation of the straight line is y = 2x + 1.

When x = 2:
y = 2(2) + 1 = 5

When x = 5:
y = 2(5) + 1 = 11

When x = 9:
y = 2(9) + 1 = 19

When x = 12:
y = 2(12) + 1 = 25

When x = 20:
y = 2(20) + 1 = 41

So, the corresponding points are (2, 5), (5, 11), (9, 19), (12, 25), and (20, 41).

Think and Reflect (NCERT Textbook Page No. 33)

Differentiate between the graphs of the equations y = 3x + 1, and y = –3x + 1

Solution:
Both y = 3x + 1 and y = -3x + 1 represent straight lines, but their slopes are different.

For y = 3x + 1, the slope is positive. This means the line rises from left to right.

For y = -3x + 1, the slope is negative. This means the line falls from left to right.

Both lines have the same y-intercept, 1, because both pass through the point (0, 1). However, one line slopes upward, while the other slopes downward.

Think and Reflect (NCERT Textbook Page No. 35)

Does this help you to conclude anything about the linear equation y = ax + b when a is fixed but b varies?

Solution:
Yes. In the equation y = ax + b, if a remains fixed and b changes, the slope of the line remains the same. This means the lines are parallel to each other. Only their position changes because the value of b shifts the line up or down on the graph.

Exercise Set 2.1 Class 9 Maths Solutions

Question 1. Find the degrees of the following polynomials.

(i) 2x² – 5x + 3

Solution: In the polynomial 2x² – 5x + 3, the highest power of the variable x is 2. Therefore, the degree of the polynomial is 2.

(ii) y³ + 2y – 1

Solution: In the polynomial y³ + 2y – 1, the highest power of the variable y is 3. Therefore, the degree of the polynomial is 3.

(iii) – 9

Solution: The polynomial -9 is a constant polynomial. A non-zero constant polynomial has degree 0. Therefore, the degree of -9 is 0.

(iv) 4z – 3

Solution: In the polynomial 4z – 3, the highest power of the variable z is 1. Therefore, the degree of the polynomial is 1.

Question 2. Write the polynomials of degree 1, 2 and 3.

Solution:
Degree 1 polynomial: 4x + 2
Degree 2 polynomial: 3x² - 4x + 7
Degree 3 polynomial: x³ + 2x² - x + 10

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

Solution: In the polynomial x⁴ – 3x³ + 6x² – 2x + 7, the term containing x² is 6x². So, the coefficient of x² is 6. The term containing x³ is -3x³. So, the coefficient of x³ is -3.

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

Solution: The polynomial 4z³ + 5z² – 11 does not contain any z term. We can write it as 4z³ + 5z² + 0z – 11. Therefore, the coefficient of z is 0.

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

Solution: In the polynomial 9x³ + 5x² – 8x – 10, the term without any variable is -10. Therefore, the constant term is -10.

Exercise Set 2.2 Class 9 Maths Solutions

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

(i) x = 0

Solution:
Substitute x = 0 in the polynomial 5x – 3.

5x – 3 = 5(0) – 3
= 0 – 3
= -3

Therefore, the value of the polynomial is -3.

(ii) x = -1

Solution:
Substitute x = -1 in the polynomial 5x – 3.

5x – 3 = 5(-1) – 3
= -5 – 3
= -8

Therefore, the value of the polynomial is -8.

(iii) x = 2

Solution:
Substitute x = 2 in the polynomial 5x – 3.

5x – 3 = 5(2) – 3
= 10 – 3
= 7

Therefore, the value of the polynomial is 7.

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

(i) s = 0
(ii) s = -3
(iii) s = 4

Solution:

(i) When s = 0:

7s² - 4s + 6 = 7(0)² - 4(0) + 6
= 0 - 0 + 6
= 6

Therefore, the value of the polynomial is 6.

(ii) When s = -3:

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

Therefore, the value of the polynomial is 81.

(iii) When s = 4:

7s² - 4s + 6 = 7(4)² - 4(4) + 6
= 7(16) - 16 + 6
= 112 - 16 + 6
= 102

Therefore, the value of the polynomial is 102.

Question 3. The present age of Salihs 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 be x years.

Then, his 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 is 15 years.

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

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

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

Question 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 two positive integers be 2x and 5x.

Their difference is 63.

5x - 2x = 63
3x = 63
x = 21

First integer = 2 × 21 = 42
Second integer = 5 × 21 = 105

Therefore, 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 of ₹88, how many coins does she have of each type?

Solution:
Let the number of five-rupee coins be x.

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

Total value of five-rupee coins = 5x
Total value of two-rupee coins = 2(3x) = 6x

According to the question:

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

So, Ruby has 8 five-rupee coins.

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

Therefore, 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?

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

Then, the length of the longer piece = 4x feet.

Total length of the fence = 300 feet

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

So, the shorter piece is 60 feet long.

Longer piece = 4 × 60 = 240 feet

Therefore, 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?

Solution:
Let the width of the rectangle be x cm.

Then, the length of the rectangle = 2x + 3 cm.

Perimeter of a rectangle = 2(length + width)

2[(2x + 3) + x] = 24
2(3x + 3) = 24
6x + 6 = 24
6x = 18
x = 3

So, the width of the rectangle is 3 cm.

Length = 2(3) + 3 = 9 cm

Therefore, the dimensions of the rectangle are 9 cm and 3 cm.

Exercise Set 2.3 Class 9 Maths 

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.

Solution:
Initial amount in the savings account = ₹500
Pocket money received every month = ₹150

The amount will increase by ₹150 every month.

At the end of the nth month, the total amount can be written as:

A(n) = 500 + 150n

where n represents the number of months.

Therefore, the linear expression for the amount she will have in the nth month is ₹(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.

Solution:
Initial number of members = 120
Number of members leaving every hour = 9

After each hour, the number of members decreases by 9.

So, the number of members remaining after the nth hour can be represented as:

M(n) = 120 - 9n

where n represents the number of hours.

Therefore, the linear expression for the number of members left after the nth hour is 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.

Solution:
Length of the rectangle = 13 cm

Area of a rectangle = Length × Breadth

(i) When breadth = 12 cm:
Area = 13 × 12 = 156 cm²

(ii) When breadth = 10 cm:
Area = 13 × 10 = 130 cm²

(iii) When breadth = 8 cm:
Area = 13 × 8 = 104 cm²

If the breadth is b cm, then the area can be represented as:

A(b) = 13b

Therefore, the linear pattern for the area of the rectangle is A(b) = 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.

Solution:
Length of the rectangular box = 7 cm
Breadth of the rectangular box = 11 cm

Volume of a rectangular box = Length × Breadth × Height

So, Volume = 7 × 11 × height = 77 × height

(i) When height = 5 cm:
Volume = 77 × 5 = 385 cm³

(ii) When height = 9 cm:
Volume = 77 × 9 = 693 cm³

(iii) When height = 13 cm:
Volume = 77 × 13 = 1001 cm³

If the height is h cm, then the volume can be represented as:

V(h) = 77h

Therefore, the linear pattern for the volume of the rectangular box is V(h) = 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.

Solution:
Total number of pages in the book = 500
Number of pages read each day = 20

If Sarita reads for n days, the number of pages left can be written as:

L(n) = 500 - 20n

For 15 days:

L(15) = 500 - 20(15)
= 500 - 300
= 200

Therefore, 200 pages will be left after 15 days.

The linear pattern is L(n) = 500 - 20n.

Exercise Set 2.4 Class 9 Maths 

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.

Solution:
Initial height of the plant = 1.75 feet
Growth per month = 0.5 feet

Height after 7 months = Initial height + Growth in 7 months

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

Therefore, the height of the plant after 7 months will be 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.

Solution:

(iii) Find an expression that relates h and t, and explain why it represents linear growth.

Solution:
The height of the plant after t months can be written as:

h(t) = 1.75 + 0.5t

This represents linear growth because the plant’s height increases by the same amount, 0.5 feet, every month.

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.

Solution:
Initial value of the mobile phone = ₹10,000
Decrease in value every year = ₹800

Value after 3 years = Initial value - Depreciation for 3 years

v = 10000 - 800(3)
= 10000 - 2400
= ₹7600

Therefore, the value of the phone after 3 years will be ₹7600.

(ii) Make a table of values for; varying from 0 to 8 years and show how the value of the phone, v, depreciates with time.

Solution:

(iii) Find an expression that relates v and;, and explain why it represents linear decay.

Solution:
The value of the mobile phone after t years can be written as:

v(t) = 10000 - 800t

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

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.

Solution:
Initial population of the village = 750
Increase in population every year = 50

Population after 6 years = Initial population + Increase in 6 years

P = 750 + 50(6)
= 750 + 300
= 1050

Therefore, the population of the village after 6 years will be 1050.

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

Solution:

(iii) Find an expression that relates P and t, and explain why it represents linear growth.

Solution:
The population after t years can be written as:

P(t) = 750 + 50t

This represents linear growth because the population increases by a fixed number, 50 people, every year.

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 forx days. Explain why it represents linear decay.

Solution:
Initial prepaid balance = ₹600
Amount reduced each day = ₹15

The remaining balance after x days can be written as:

b(x) = 600 - 15x

This represents linear decay because the balance decreases by the same amount, ₹15, each day.

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

Solution:
The balance will run out when b(x) = 0.

600 - 15x = 0
15x = 600
x = 40

Therefore, 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.

Solution:

Exercise Set 2.5 Class 9 Maths

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.

Solution:
The monthly bill is given by the linear relation:

y = ax + b

Here, x is the number of modules accessed and y is the monthly bill.

When x = 10, y = 400:

400 = 10a + b …(i)

When x = 14, y = 500:

500 = 14a + b …(ii)

Subtract equation (i) from equation (ii):

500 - 400 = (14a + b) - (10a + b)
100 = 4a
a = 25

Now, substitute a = 25 in equation (i):

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

Therefore, a = 25 and b = 150.

So, the monthly bill can be written as:

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 billy 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:
The monthly bill is given by the linear relation:

y = ax + b

Here, x is the number of hours the badminton court is used and y is the monthly bill.

When x = 10, y = 800:

800 = 10a + b …(i)

When x = 15, y = 1100:

1100 = 15a + b …(ii)

Subtract equation (i) from equation (ii):

1100 - 800 = (15a + b) - (10a + b)
300 = 5a
a = 60

Now, substitute a = 60 in equation (i):

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

Therefore, a = 60 and b = 200.

So, the gym’s monthly bill can be written as:

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. (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:
The relation between Celsius and Fahrenheit is given as:

°C = a°F + b

When °C = 0 and °F = 32:

0 = 32a + b …(i)

When °C = 100 and °F = 212:

100 = 212a + b …(ii)

From equation (i):

b = -32a

Substitute b = -32a in equation (ii):

100 = 212a - 32a
100 = 180a
a = 100/180
a = 5/9

Now, substitute a = 5/9 in b = -32a:

b = -32 × 5/9
b = -160/9

Therefore, a = 5/9 and b = -160/9.

So, the linear relationship between Celsius and Fahrenheit is:

°C = (5/9)°F - 160/9

Exercise Set 2.6 Class 9 Maths 

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

Solution:

Image 1

The given equations are y = 4x, y = 2x, and y = x.

All three equations are in the form y = ax, where b = 0.

Observation:
All the lines pass through the origin (0, 0) because the value of b is 0. The value of a decides how steep the line will be. A greater value of a gives a steeper line. So, y = 4x is the steepest, followed by y = 2x and then y = x.

(ii) y = – 6x, y = – 3x, y = – x

Solution:

Image 2

The given equations are y = -6x, y = -3x, and y = -x.

All three equations are in the form y = ax, where b = 0 and a is negative.

Observation:
All the lines pass through the origin because the value of b is 0. Since the value of a is negative, all the lines slope downward from left to right. The greater the magnitude of a, the steeper the downward slope. So, y = -6x is the steepest, followed by y = -3x and then y = -x.

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

Solution:

Image 3

The given equations are y = 5x and y = -5x.

Both equations are in the form y = ax, where b = 0.

Observation:
Both lines pass through the origin because their y-intercept is 0. The line y = 5x slopes upward from left to right, while y = -5x slopes downward from left to right. Since the magnitude of a is the same in both equations, the two lines have equal steepness but in opposite directions.

(iv) y = 3x – 1, y = 3x, y = 3x + 1

Solution:

Image 4

The given equations are y = 3x - 1, y = 3x, and y = 3x + 1.

All three equations have the same value of a, which is 3.

Observation:
Since the slope is the same for all three lines, the lines are parallel to each other. The value of b decides where the line cuts the y-axis. For y = 3x - 1, the line cuts the y-axis at -1. For y = 3x, it passes through the origin. For y = 3x + 1, the line cuts the y-axis at 1. Thus, changing b shifts the line up or down without changing its slope.

(v) y = -2x – 3, y = -2x, y = 2x + 3

Solution:

Image 5

The given equations are y = -2x - 3, y = -2x, and y = 2x + 3.

Observation:
The equations y = -2x - 3 and y = -2x have the same slope, -2. So, their graphs are parallel lines. The only difference is the value of b, which changes the position of the line on the y-axis.

The equation y = 2x + 3 has a positive slope, so its graph rises from left to right. This line has a different direction from the first two lines.

Conclusion:

1. The value of a, the coefficient of x, decides the slope, steepness, and direction of the line.

2. The value of b, the constant term, decides the vertical shift and the point where the line cuts the y-axis.

Introduction to Linear Polynomials End of Chapter Exercises with Detailed Solutions

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

Solution:
A polynomial of degree 3 has the highest power of x as 3.

One example is:

p(x) = x³ + 2x² - 7x + 5

Here, the degree of the polynomial is 3, and the coefficient of the x term is -7.

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

(i) 5x² – 3x + 7 if x = 1

Solution:
Substitute x = 1 in the polynomial 5x² – 3x + 7.

5(1)² - 3(1) + 7
= 5 - 3 + 7
= 9

Therefore, the value of the polynomial is 9.

(ii) 4t³ – t² + 6 if t = a

Solution:
Substitute t = a in the polynomial 4t³ – t² + 6.

4(a)³ - (a)² + 6
= 4a³ - a² + 6

Therefore, the value of the polynomial is 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.

Solution:
Let the number be x.

According to the question:

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

Now, subtract 2/3 from both sides:

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

(5/2)x = -7/12 - 8/12

(5/2)x = -15/12

Now, multiply both sides by 2/5:

x = (-15/12) × (2/5)

x = -1/2

Therefore, the required 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?

Solution:
Let the smaller number be y.

Then, the larger number = 5y.

After adding 21 to both numbers:

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

According to the question:

5y + 21 = 2(y + 21)

5y + 21 = 2y + 42

3y = 21

y = 7

So, the smaller number is 7.

Larger number = 5 × 7 = 35

Therefore, the two numbers are 35 and 7.

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.

Solution:
Initial amount = ₹800
Amount saved every month = ₹250

The amount after x months can be represented as:

A(x) = 800 + 250x

(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

Therefore, the linear pattern is A(x) = 800 + 250x, where x is the number of months.

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.

Solution:
Let the two-digit number be 10a + b, where a is the tens digit and b is the ones digit.

The number formed by interchanging the digits is 10b + a.

According to the question:

(10a + b) + (10b + a) = 143

11a + 11b = 143

a + b = 13 …(i)

Also, the digits differ by 3:

a - b = 3 …(ii)

Add equations (i) and (ii):

a + b = 13
a - b = 3

2a = 16

a = 8

Now, substitute a = 8 in equation (i):

8 + b = 13

b = 5

So, the original number is 85.

The number formed by interchanging the digits is 58.

Therefore, the two 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.

(i) y = -3x + 4
(ii) 2y = 4x + 7
(iii) 5y = 6x – 10
(iv) 3y = 6x – 11

Are any of the lines parallel?

Solution:

(i) y = -3x + 4

This equation is already in the form y = ax + b.

Slope = -3
y-intercept = 4

When x = 0:

y = -3(0) + 4 = 4

So, the line cuts the y-axis at (0, 4).

(ii) 2y = 4x + 7

Divide both sides by 2:

y = 2x + 7/2

Slope = 2
y-intercept = 7/2

So, the line cuts the y-axis at (0, 7/2).

(iii) 5y = 6x - 10

Divide both sides by 5:

y = (6/5)x - 2

Slope = 6/5
y-intercept = -2

So, the line cuts the y-axis at (0, -2).

(iv) 3y = 6x - 11

Divide both sides by 3:

y = 2x - 11/3

Slope = 2
y-intercept = -11/3

So, the line cuts the y-axis at (0, -11/3).

Lines (ii) and (iv) are parallel because both have the same 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 the linear equation y = 9/5 (x – 273) + 32.

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

Solution:
The given equation is:

y = 9/5(x - 273) + 32

Substitute x = 313:

y = 9/5(313 - 273) + 32

y = 9/5(40) + 32

y = 72 + 32

y = 104

Therefore, the temperature of the liquid is 104°F.

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

Solution:
The given equation is:

y = 9/5(x - 273) + 32

Substitute y = 158:

158 = 9/5(x - 273) + 32

126 = 9/5(x - 273)

Multiply both sides by 5/9:

x - 273 = 70

x = 343

Therefore, the temperature of the liquid is 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.

Solution:
Work done = Force × Distance

Let work be w and distance be d.

Given force = 3 units.

So,

w = 3d

This is the required linear equation in two variables.

When d = 2:

w = 3 × 2
w = 6

Therefore, the work done is 6 units.

To draw the graph, take d on the x-axis and w on the y-axis. The equation w = 3d gives a straight line passing through the origin. The point (2, 6) lies on this line, which verifies the answer.

Image 6

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).

Solution:
Let the linear polynomial be:

p(x) = ax + b

Since the graph passes through (1, 5):

a + b = 5 …(i)

Since the graph also passes through (3, 11):

3a + b = 11 …(ii)

Subtract equation (i) from equation (ii):

3a + b - (a + b) = 11 - 5

2a = 6

a = 3

Now, substitute a = 3 in equation (i):

3 + b = 5

b = 2

Therefore, the polynomial is:

p(x) = 3x + 2

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

Solution:
The polynomial is:

p(x) = 3x + 2

To find the point where the graph cuts the y-axis, put x = 0:

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

So, the graph cuts the y-axis at (0, 2).

To find the point where the graph cuts the x-axis, put p(x) = 0:

3x + 2 = 0

3x = -2

x = -2/3

So, the graph cuts the x-axis at (-2/3, 0).

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

Solution:
To draw the graph of p(x) = 3x + 2, plot points such as (0, 2), (1, 5), and (3, 11), then join them to form a straight line.

The graph cuts the y-axis at (0, 2) and the x-axis at (-2/3, 0), which verifies the answers.

Image 7

Question 11. Let p(x) = ax + b and q(x) = cx + d be two linear polynomials such that:

(i) p(0) = 5.
(ii) The polynomial p(x) – q(x) cuts the x-axis at (3, 0).
(iii) The sum p(x) + q(x) is equal to 6x + 4 for all real x. Find the polynomials p(x) and q(x).

Solution:
Let:

p(x) = ax + b

q(x) = cx + d

From condition (i):

p(0) = 5

a(0) + b = 5

b = 5

So,

p(x) = ax + 5

From condition (iii):

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

(ax + 5) + (cx + d) = 6x + 4

(a + c)x + (5 + d) = 6x + 4

Comparing coefficients:

a + c = 6 …(1)

5 + d = 4

d = -1

So,

q(x) = cx - 1

From condition (ii), p(x) - q(x) cuts the x-axis at (3, 0). Therefore:

p(3) - q(3) = 0

p(3) = 3a + 5

q(3) = 3c - 1

So,

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

3a + 5 - 3c + 1 = 0

3a - 3c + 6 = 0

a - c = -2 …(2)

Now solve:

a + c = 6
a - c = -2

Adding both equations:

2a = 4

a = 2

Substitute a = 2 in equation (1):

2 + c = 6

c = 4

Therefore,

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.

Image 8

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

(ii) Complete the following table.

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

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

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

Solution:
One hexagon needs 6 matchsticks. When a new hexagon is added, it shares one side with the previous hexagon. So, only 5 new matchsticks are added at each new stage.

Image 9

The pattern is:

Stage 1 = 6 matchsticks
Stage 2 = 6 + 5 = 11 matchsticks
Stage 3 = 11 + 5 = 16 matchsticks

(i) The next two stages are:

Stage 4 = 16 + 5 = 21 matchsticks
Stage 5 = 21 + 5 = 26 matchsticks

Image 10

(ii) The completed table is:

(iii) The number of matchsticks increases by 5 at every stage.

For the nth stage:

Sₙ = 6 + (n - 1)5

Sₙ = 6 + 5n - 5

Sₙ = 5n + 1

Therefore, the rule is:

Sₙ = 5n + 1

(iv) For the 15th stage:

S₁₅ = 5(15) + 1

S₁₅ = 75 + 1

S₁₅ = 76

Therefore, 76 matchsticks are required for the 15th stage.

(v) To check whether 200 matchsticks can form a stage:

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.

Question 13. Let p(x) = ax + b and q(x) = cx + d be two linear polynomials such that:

(i) The graph of p(x) passes through the points (2, 3) and (6, 11).
(ii) The graph of q(x) passes through the point (4, -1).
(iii) The graph of q(x) is parallel to the graph of p(x). Find the polynomials p(x) and q(x). Also, find the coordinates of the point where these lines meet the x-axis.

Solution:
First, find p(x).

The graph of p(x) passes through (2, 3) and (6, 11).

Slope of p(x):

a = (11 - 3)/(6 - 2)

a = 8/4

a = 2

So,

p(x) = 2x + b

Using the point (2, 3):

3 = 2(2) + b

3 = 4 + b

b = -1

Therefore,

p(x) = 2x - 1

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

Thus,

q(x) = 2x + d

Since q(x) passes through (4, -1):

-1 = 2(4) + d

-1 = 8 + d

d = -9

Therefore,

q(x) = 2x - 9

Now, find the x-intercepts.

For p(x) = 2x - 1:

0 = 2x - 1

2x = 1

x = 1/2

So, p(x) meets the x-axis at (1/2, 0).

For q(x) = 2x - 9:

0 = 2x - 9

2x = 9

x = 9/2

So, q(x) meets the x-axis at (9/2, 0).

Therefore,

p(x) = 2x - 1 and q(x) = 2x - 9.

Their x-axis intercepts are (1/2, 0) and (9/2, 0), respectively.

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

Solution:
The given linear function is:

f(x) = ax + a, where a > 0

This can be written as:

f(x) = a(x + 1)

Common features:

1. Since a > 0, all the lines have a positive slope. So, they rise from left to right.

2. The y-intercept is found by putting x = 0:

f(0) = a

Since a > 0, every line cuts the y-axis above the origin.

3. The x-intercept is found by putting f(x) = 0:

ax + a = 0

a(x + 1) = 0

Since a > 0, a is not zero. So:

x + 1 = 0

x = -1

Therefore, all such lines pass through the fixed point (-1, 0).

Hence, all linear functions of the form f(x) = ax + a, a > 0, have a positive slope, cut the y-axis above the origin, and pass through the fixed point (-1, 0).

Why Class 9 Maths Chapter 2 Introduction to Linear Polynomials is Important?

Class 9 Maths Chapter 2 Introduction to Linear Polynomials is important because it helps students move from basic algebra to practical problem-solving. Students learn how to form expressions, evaluate polynomials, identify degree and coefficients, and understand linear equations through real-life examples like savings, population growth, fares, bills, and depreciation.

Prepare Better with Vedantu’s Class 9 Maths Chapter 2 NCERT Solutions

Vedantu’s NCERT Solutions for Class 9 Maths Chapter 2 Introduction to Linear Polynomials help students practise each exercise with clear steps. The solutions explain Exercise Set 2.1 to 2.6, including polynomial values, linear patterns, equations, graphs, slopes, and intercepts. These answers are useful for homework, revision, and CBSE exam preparation.

Access Exercise Wise NCERT Solutions for Chapter 2 Maths Class 9

CBSE Class 9 Maths Chapter 2 Introduction to Linear Polynomials Other Study Materials

Chapter-Specific NCERT Solutions for Class 9 Maths

Given below are the chapter-wise NCERT Solutions for Class 9 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

Additional Study Materials for Class 9 Maths