NCERT Solutions for Class 9 – Complete Chapter-wise Study Material
Introduction to Linear Polynomials is one of the most important chapters in the Class 9 Mathematics Ganita Manjari English NCERT Solutions curriculum. This chapter plays a significant role in helping students build a strong conceptual foundation while preparing for school examinations, class tests, unit tests, half-yearly examinations, annual examinations, and CBSE board assessments. The chapter has been carefully designed according to the latest NCERT syllabus, making it an essential part of every student's study plan.
The Introduction to Linear Polynomials - Class 9 Mathematics Ganita Manjari English NCERT Solutions available on ATP Education explain every question in a simple, accurate, and step-by-step manner. Each answer is prepared according to the latest CBSE guidelines so that students can understand the concepts clearly without confusion. Whether you are completing your homework, revising before examinations, or strengthening your understanding of the subject, these solutions provide reliable academic support throughout your learning journey.
One of the biggest advantages of studying Introduction to Linear Polynomials is that it helps students understand important concepts, definitions, examples, and textbook exercises in an organized way. Instead of memorizing answers, students learn how to develop logical thinking, improve analytical skills, and write well-structured answers in examinations. This chapter also helps improve problem-solving ability and encourages conceptual learning, which is essential for scoring higher marks in school and competitive examinations.
Our Class 9 Mathematics Ganita Manjari NCERT Solutions cover all textbook questions, important exercise questions, and chapter-wise explanations in English Medium. Every solution is written in easy-to-understand language, allowing students to revise the chapter quickly before examinations. Regular practice of these solutions improves confidence, strengthens subject knowledge, and reduces examination stress.
Students preparing for school assessments should carefully study Introduction to Linear Polynomials because questions from this chapter are frequently asked in objective questions, short answer questions, long answer questions, competency-based questions, and case-study questions. Understanding the concepts explained in this chapter also helps students connect related topics from other chapters, making overall learning more effective and meaningful.
At ATP Education, we continuously update our Class 9 Mathematics Ganita Manjari English NCERT Solutions according to the latest NCERT textbooks and CBSE curriculum. Students can confidently use these chapter-wise solutions for daily study, homework assistance, quick revision, examination preparation, and self-learning. By studying Introduction to Linear Polynomials thoroughly and practising every question regularly, students can strengthen their concepts, improve writing skills, and achieve better academic performance in both school and board examinations.
Introduction to Linear Polynomials - Class 9 Mathematics Ganita Manjari English NCERT Solutions
Introduction to Linear Polynomials
Exercise Set 2.1
Q1. Find the degrees of the following polynomials.
(i) 2x² – 5x + 3
Solution: Degree = 2
(ii) y³ + 2y – 1
Solution: Degree = 3
(iii) – 9
Solution: Degree = 0
(iv) 4z – 3
Solution: Degree = 1
Q2. Write polynomials of degrees 1, 2 and 3.
Polynomial of Degree 1 :
Solution: 2x + 5
Polynomial of Degree 2 :
Solution: x² + 3x + 1
Polynomial of Degree 2 :
Solution: x³ – 2x² + x + 4
Q3. What are the coefficients of x² and x³ in the polynomial x⁴ – 3x³ + 6x² – 2x + 7?
Solution:
Coefficient of x² = 6
Coefficient of x³ = –3
Q4. What is the coefficient of z in the polynomial 4z³ + 5z² – 11?
Solution:
There is no z term in the polynomial.
Therefore, coefficient of z = 0
Q5. What is the constant term of the polynomial 9x³ + 5x² – 8x – 10?
Solution:
The constant term is the term without variable.
Therefore, constant term = –10
Polynomials of degree 1 are called linear polynomials.
Introduction to Linear Polynomials
Exercise Set 2.2
Q1. Find the value of the linear polynomial 5x – 3 if:
(i) x = 0 (ii) x = –1 (iii) x = 2
Solutions:
P(x) = 5x - 3
(i) x = 0
P(0) = 5(0) - 3
= 0 - 3
= - 3 Ans
(ii) x = -1
P(-1) = 5(-1) - 3
= -5 - 3
= - 8 Ans
(iii) x = 2
P(2) = 5(2) - 3
= 10 - 3
= 7 Ans
Q2. Find the value of the quadratic polynomial 7s2 – 4s + 6 if:
(i) s = 0 (ii) s = –3 (iii) s = 4
Solution:
P(s) = 7s2 - 4s + 6
(i) s = 0
P(0) = 7(0)2 - 4(0) + 6
= 0 - 0 + 6
= 6 Ans
(ii) s = –3
P(–3) = 7(–3)2 - 4(–3) + 6
= 63 + 12 + 6
= 81 Ans
(iii) s = 4
P(4) = 7(4)2 - 4(4) + 6
= 112 - 16 + 6
= 118 - 16
= 102 Ans
Q3. 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 the Salil's present Age be x
Salil's Mother present age = 3x
After 5 years
Salil's age = x + 5
Salil's mother age = 3x + 5
So,
Therefore, Salil's Present Age = 15 years
Salil's Mother Present Age = 3 x 15 = 45 years
Q4. The difference between two positive integers is 63. The ratio of the two integers is 2:5. Find the two integers.
Solution:
Ratio of the two integers = 2 : 5
Let the first integer be 2x
And the second integer be 5x
So, 5x - 2x = 63
3x = 63
x = 63/3
x = 21
The first integer = 2 x 21 = 42
The second integer = 5 x 21 = 105
Q5. 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 the number of five-rupee coins be x.
Then, the number of two-rupee coins will be 3x.
Value of five-rupee coins = 5x
Value of two-rupee coins = 2(3x)
According to the question:
5x + 2(3x) = 88
Now solve:
5x + 6x = 88
11x = 88
x = 88/11
x = 8
So, number of five-rupee coins = 8
Number of two-rupee coins = 3 × 8 = 24
Therefore:
Five-rupee coins = 8
Two-rupee coins = 24
Q6. 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 will be 4x feet.
According to the question:
x + 4x = 300
Now solve:
5x = 300
x = 300/5
x = 60
So, length of the shorter piece = 60 feet
Length of the longer piece = 4 × 60
= 240 feet
Therefore:
Shorter piece = 60 feet
Longer piece = 240 feet
Q7. 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 will be 2x + 3 cm.
Perimeter of rectangle = 2(length + width)
According to the question:
2[(2x + 3) + x] = 24
Now solve:
2(3x + 3) = 24
6x + 6 = 24
6x = 24 − 6
6x = 18
x = 18/6
x = 3
So, width of the rectangle = 3 cm
Length of the rectangle = 2(3) + 3
= 6 + 3
= 9 cm
Therefore:
Width = 3 cm
Length = 9 cm
Introduction to Linear Polynomials
Exercise Set 2.3
Solve the following:
Q1. 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:
A student has ₹500 in her bank account and gets ₹150 every month as pocket money.
Amount at the end of each month
After 1 month:
500 + 150 = 650
After 2 months:
500 + 2(150) = 800
After 3 months:
500 + 3(150) = 950
Pattern:
650, 800, 950,…
Linear Expression
If n is the number of months, then:
A = 500 + 150n
So, the amount in the nth month is:
A = 500 + 150n
Q2. 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
Every hour, 9 members leave.
Members remaining
After 1 hour:
120 − 9 = 111
After 2 hours:
120 − 18 = 102
After 3 hours:
120 − 27 = 93
Pattern:
111, 102, 93,…
Linear Expression
If n is the number of hours:
M = 120 − 9n
So, the number of members after nnn hours is:
M = 120 − 9n
Q3. 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
(i) b = 12 cm
When Breadth = 12 cm
Area = Length x breadth
Area = 13 cm x 12 cm
= 156 cm2
(ii) b = 10 cm
When Breadth = 10 cm
Area = Length x breadth
Area = 13 cm x 10 cm
= 130 cm2
(ii) b = 8 cm
When Breadth = 8 cm
Area = Length x breadth
Area = 13 cm x 8 cm
= 104 cm2
Linear Pattern
If breadth =b
A=13b
So, the linear expression for area is:
A=13b
Q4. 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 rectangular box = Length × Breadth × Height
(i) h = 5 cm
When Height = 5 cm
Volume = Length × Breadth × Height
Volume = 7 cm × 11 cm × 5 cm
= 385 cm³
(ii) h = 9 cm
When Height = 9 cm
Volume = Length × Breadth × Height
Volume = 7 cm × 11 cm × 9 cm
= 693 cm³
(iii) h = 13 cm
When Height = 13 cm
Volume = Length × Breadth × Height
Volume = 7 cm × 11 cm × 13 cm
= 1001 cm³
Linear Pattern:
Volume = Length × Breadth × Height
V = 7 × 11 × h
V = 77h
Q5. 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 in the book = 500
Pages read every day = 20
Number of days = 15
Pages read in 15 days
= 20 × 15
= 300
Pages left after 15 days
= 500 − 300
= 200
So, 200 pages will be left after 15 days.
Linear Pattern:
Let the number of days be d.
Pages left = Total pages − Pages read
P = 500 − 20d
Introduction to Linear Polynomials
Exercise Set 2.4
Exercise Set 2.4
Q1. 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.
Q2. 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.
Q3. 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.
Q4. 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.
(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:
Q1. Suppose a plant has height 1.75 feet and it grows by 0.5 feet each month.
Solution:
Initial height of plant = 1.75 feet
Growth every month = 0.5 feet
(i) Height after 7 months
h = 1.75 + (0.5 × 7)
h = 1.75 + 3.5
h = 5.25 feet
So, the height after 7 months is 5.25 feet.
(ii) Table of values
t (months) h (feet)
0 1.75
1 2.25
2 2.75
3 3.25
4 3.75
5 4.25
6 4.75
7 5.25
8 5.75
9 6.25
10 6.75
(iii) Expression relating h and t
h = 1.75 + 0.5t
This represents linear growth because the height increases by the same amount every month.
Q2. A mobile phone is bought for ₹10,000. Its value decreases by ₹800 every year.
Solution:
Initial value of phone = ₹10,000
Decrease every year = ₹800
(i) Value after 3 years
v = 10000 − (800 × 3)
v = 10000 − 2400
v = ₹7600
So, the value after 3 years is ₹7600.
(ii) Table of values
t (years) v (₹)
0 10000
1 9200
2 8400
3 7600
4 6800
5 6000
6 5200
7 4400
8 3600
(iii) Expression relating v and t
v = 10000 − 800t
This represents linear decay because the value decreases by the same amount every year.
Q3. The initial population of a village is 750. Every year, 50 people move from a nearby city to the village.
Solution:
Initial population = 750
Increase every year = 50 people
(i) Population after 6 years
P = 750 + (50 × 6)
P = 750 + 300
P = 1050
So, the population after 6 years is 1050.
(ii) Table of values
t (years) P
0 750
1 800
2 850
3 900
4 950
5 1000
6 1050
7 1100
8 1150
9 1200
10 1250
(iii) Expression relating P and t
P = 750 + 50t
This represents linear growth because the population increases by the same amount every year.
Q4. A telecom company charges ₹600 for a certain recharge scheme. This prepaid balance is reduced by ₹15 each day after the recharge.
Solution:
Initial balance = ₹600
Reduction every day = ₹15
(i) Equation representing remaining balance
b(x) = 600 − 15x
This represents linear decay because the balance decreases by the same amount every day.
(ii) Number of days after which balance becomes zero
600 − 15x = 0
15x = 600
x = 600/15
x = 40
So, the balance will run out after 40 days.
(iii) Table of values
x (days) b(x)
1 585
2 570
3 555
4 540
5 525
6 510
7 495
8 480
9 465
10 450
Introduction to Linear Polynomials
Exercise Set 2.5
Exercise Set 2.5
Q1. 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:
Given relation:
y = ax + b
When x = 10, y = 400
So,
400 = 10a + b ......(1)
When x = 14, y = 500
So,
500 = 14a + b ......(2)
Subtract equation (1) from equation (2):
500 − 400 = 14a − 10a
100 = 4a
a = 100/4
a = 25
Now substitute a = 25 in equation (1):
400 = 10(25) + b
400 = 250 + b
b = 400 − 250
b = 150
Therefore:
a = 25
b = 150
Q2. 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 relation:
y = ax + b
When x = 10, y = 800
So,
800 = 10a + b ......(1)
When x = 15, y = 1100
So,
1100 = 15a + b ......(2)
Subtract equation (1) from equation (2):
1100 − 800 = 15a − 10a
300 = 5a
a = 300/5
a = 60
Now substitute a = 60 in equation (1):
800 = 10(60) + b
800 = 600 + b
b = 800 − 600
b = 200
Therefore:
a = 60
b = 200
Q3. 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.
Solution:
Given relation:
°C = a°F + b
When °C = 0 and °F = 32
So,
0 = 32a + b ......(1)
When °C = 100 and °F = 212
So,
100 = 212a + b ......(2)
Subtract equation (1) from equation (2):
100 − 0 = 212a − 32a
100 = 180a
a = 100/180
a = 5/9
Now substitute a = 5/9 in equation (1):
0 = 32(5/9) + b
0 = 160/9 + b
b = −160/9
Therefore:
a = 5/9
b = −160/9
Hence, the linear relationship between °C and °F is:
°C = (5/9)°F − 160/9
Introduction to Linear Polynomials
Exercise Set 2.6
Exercise Set 2.6
Q1. Draw the graphs of the following sets of lines. In each case, reflect on the role of ‘a’ and ‘b’.
Solution:
The general form of a linear equation is:
y = ax + b
where,
a = slope of the line
b = y-intercept
(i) y = 4x, y = 2x, y = x
Here,
a = 4, 2, 1
b = 0 in all cases
Observation:
As the value of a increases, the line becomes steeper.
Since b = 0, all lines pass through the origin (0, 0).
(ii) y = –6x, y = –3x, y = –x
Here,
a = –6, –3, –1
b = 0 in all cases
Observation:
Negative values of a give downward sloping lines.
Greater negative value means steeper downward slope.
Since b = 0, all lines pass through the origin.
(iii) y = 5x, y = –5x
Here,
a = 5 and –5
b = 0
Observation:
Positive a gives an upward sloping line.
Negative a gives a downward sloping line.
Both lines pass through the origin because b = 0.
(iv) y = 3x – 1, y = 3x, y = 3x + 1
Here,
a = 3 in all cases
b = –1, 0, 1
Observation:
All lines have the same slope, so they are parallel.
Changing b shifts the line upward or downward.
b determines where the line cuts the y-axis.
(v) y = –2x – 3, y = –2x, y = –2x + 3
Here,
a = –2 in all cases
b = –3, 0, 3
Observation:
All lines have the same negative slope, so they are parallel.
Changing b changes the y-intercept.
Negative b shifts the line downward and positive b shifts it upward.
Introduction to Linear Polynomials
End-of-Chapter Exercises
End-of-Chapter Exercises
Q1. Write a polynomial of degree 3 in the variable x, in which the coefficient of the x2 term is –7.
Q2. Find the values of the following polynomials at the indicated values of the variables.
(i) 5x2 – 3x + 7 if x = 1
(ii) 4t3 – t2 + 6 if t = a
Q3. If we multiply a number by 52 and add 23 to the product, we get –7 12. Find the number.
Q4. 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?
Q5. 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.
*Q6. 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.
*Q7. 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?
*Q8. 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 = 95
(x – 273) + 32.
(i) Find the temperature of the liquid in Fahrenheit if the temperature of the liquid is 313 K.
(ii) If the temperature is 158 °F, then find the temperature in Kelvin.
*Q9. 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 isthe work done when the distance travelled is 2 units? Verify it by
plotting it on the graph.
*Q10. The graph of a linear polynomial p(x) passes through the points (1, 5) and (3, 11).
(i) Find the polynomial p(x).
(ii) Find the coordinates where the graph of p(x) cuts the axes.
(iii) Draw the graph of p(x) and verify your answers.
*Q11. 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).
*Q12. 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
(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.
*Q13. 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.
*Q14. What do all linear functions of the form f(x) = ax + a, a > 0, have in common?
Mathematics Ganita Manjari
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