Important Questions Class 9 Maths Chapter 2 Introduction to Linear Polynomials

A linear polynomial is a polynomial of degree 1 in one variable and is written in the form ax + b, where a ≠ 0. Important questions class 9 maths chapter 2 cover algebraic expressions, degree, coefficients, linear patterns, linear growth, linear decay, linear relationships, slope, y-intercept, and graphs.

A taxi fare, monthly savings, plant growth, phone value, and internet bill can all be written using linear polynomials.

Class 9 Maths Chapter 2 helps students move from expressions to real-life patterns. The chapter begins with terms, coefficients, constants, and degree. It then shows how linear polynomials create straight-line relationships through growth, decay, slope, intercept, and graphs.

Key Takeaways

Class 9 Maths Ganita Manjari Part 1 Chapter List

Important Topics in Class 9 Maths Chapter 2

Class 9 maths chapter 2 important questions are mostly based on identifying degree, evaluating linear polynomials, forming equations from situations, recognising linear patterns, and reading graphs.

Students should first understand the structure of ax + b. After that, graph-based and word-problem questions become easier.

1. Algebraic expressions
2. Terms, variables, coefficients, and constants
3. Univariate polynomials
4. Degree of polynomial class 9
5. Constant, linear, quadratic, and cubic polynomials
6. Linear polynomial in the form ax + b
7. Evaluating a polynomial
8. Linear patterns with constant difference
9. Linear growth and decay class 9 maths
10. Linear relationships in the form y = ax + b
11. Slope and y-intercept
12. Graphs of linear polynomials
13. Parallel lines with the same slope
14. x-intercept and y-intercept
15. Real-life applications of linear polynomials

Important Questions Class 9 Maths Chapter 2 with Answers

Important questions class 9 maths chapter 2 should be revised topic-wise. Start with degree and coefficients, then move to evaluation, word problems, linear patterns, growth, decay, and graphs.

These class 9 maths chapter 2 questions and answers include short-answer, graph-based, application-based, and end-of-chapter style questions.

Very Short Answer Questions: Degree, Coefficients, and Classification

One-mark questions from this chapter usually test degree, coefficient, constant term, and classification.

Write only the exact answer with one supporting line.

Class 9 Maths Chapter 2 Important Questions

Q1. Find the degree of 2x² - 5x + 3.
Ans. The highest power of x is 2.

Degree = 2

So, it is a quadratic polynomial.

Q2. Find the degree of y³ + 2y - 1.
Ans. The highest power of y is 3.

Degree = 3

So, it is a cubic polynomial.

Q3. Find the degree of -9.
Ans. -9 is a non-zero constant polynomial.

It can be written as -9x⁰.

Degree = 0

Q4. Find the degree of 4z - 3.
Ans. The highest power of z is 1.

Degree = 1

So, it is a linear polynomial.

Q5. What is the coefficient of x² in x⁴ - 3x³ + 6x² - 2x + 7?
Ans. The coefficient of x² is 6.

Q6. What is the coefficient of x³ in x⁴ - 3x³ + 6x² - 2x + 7?
Ans. The coefficient of x³ is -3.

Q7. What is the coefficient of z in 4z³ + 5z² - 11?
Ans. There is no z term.

So, the coefficient of z is 0.

Q8. What is the constant term of 9x³ + 5x² - 8x - 10?
Ans. The constant term is -10.

Q9. Write a polynomial of degree 3 in variable x where the coefficient of x² is -7.
Ans. One possible polynomial is:

x³ - 7x² + 2x + 5

Q10. Is 3z + 7 a linear polynomial? Why?
Ans. Yes, 3z + 7 is a linear polynomial.

Its degree is 1 because the highest power of z is 1.

Short Answer Questions on Evaluating Linear Polynomials

Evaluating a polynomial means substituting the given value of the variable and simplifying.

These linear polynomials class 9 questions and answers are useful for 2-mark and 3-mark practice.

Introduction to Linear Polynomials Class 9 Important Questions

Q1. Find the value of 5x - 3 at x = 0, x = -1, and x = 2.
Ans.

At x = 0:

5(0) - 3 = -3

At x = -1:

5(-1) - 3 = -5 - 3 = -8

At x = 2:

5(2) - 3 = 10 - 3 = 7

So, the values are -3, -8, and 7.

Q2. A chess club charges ₹200 as joining fee plus ₹50 per match. Write the linear polynomial for total cost and find the cost for 11 matches.
Ans. Let m be the number of matches.

Total cost = 200 + 50m

For 11 matches:

200 + 50(11)

= 200 + 550

= ₹750

Q3. If a player paid ₹750 to the chess club, how many matches did he play?
Ans.

200 + 50m = 750

50m = 750 - 200

50m = 550

m = 11

So, he played 11 matches.

Q4. The sum of two numbers is 64. One number is 10 more than the other. Find both numbers.
Ans. Let the smaller number be x.

The larger number is x + 10.

x + x + 10 = 64

2x + 10 = 64

2x = 54

x = 27

So, the numbers are 27 and 37.

Q5. Salil’s mother is three times Salil’s age. After 5 years, their ages add up to 70. Find their present ages.
Ans. Let Salil’s present age be x.

Mother’s present age = 3x

After 5 years:

Salil’s age = x + 5

Mother’s age = 3x + 5

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

4x + 10 = 70

4x = 60

x = 15

Salil is 15 years old.

His mother is 45 years old.

Q6. The difference between two positive integers is 63. Their ratio is 2:5. Find both integers.
Ans. Let the integers be 2k and 5k.

5k - 2k = 63

3k = 63

k = 21

The numbers are:

2k = 42

5k = 105

So, the integers are 42 and 105.

Q7. Ruby has three times as many ₹2 coins as ₹5 coins. The total amount is ₹88. How many coins of each type does she have?
Ans. Let the number of ₹5 coins be x.

Number of ₹2 coins = 3x

Total value:

5x + 2(3x) = 88

5x + 6x = 88

11x = 88

x = 8

So, Ruby has 8 five-rupee coins and 24 two-rupee coins.

Q8. A farmer cuts a 300-foot fence into two pieces. The longer piece is four times the shorter piece. Find both lengths.
Ans. Let the shorter piece be x.

Longer piece = 4x

x + 4x = 300

5x = 300

x = 60

Shorter piece = 60 feet

Longer piece = 240 feet

Important Questions on Linear Patterns

A linear pattern is a sequence in which the difference between consecutive terms remains constant.

The constant difference helps form the rule for the nth term.

Linear Relationship Class 9 Maths Questions

Q1. A tile pattern follows the sequence 1, 3, 5, 7, 9, ... Express the number of tiles at Stage n. Find the tiles at Stage 15 and Stage 26.
Ans. The sequence increases by 2 each time.

Rule:

Number of tiles = 2n - 1

At Stage 15:

2(15) - 1 = 30 - 1 = 29

At Stage 26:

2(26) - 1 = 52 - 1 = 51

Q2. Which stage in the tile pattern has 21 tiles? Which stage has 47 tiles?
Ans.

For 21 tiles:

2n - 1 = 21

2n = 22

n = 11

So, Stage 11 has 21 tiles.

For 47 tiles:

2n - 1 = 47

2n = 48

n = 24

So, Stage 24 has 47 tiles.

Q3. A student has ₹500. She gets ₹150 every month. Find the linear expression for the amount in the nth month.
Ans. Month 1 amount = ₹500

Each month, ₹150 is added.

Amount in nth month:

500 + 150(n - 1)

= 500 + 150n - 150

= 350 + 150n

So, the expression is 350 + 150n.

Q4. A rally starts with 120 members. Nine members drop out every hour. Find the expression for members after n hours.
Ans. Initial members = 120

Members leaving every hour = 9

Members after n hours:

120 - 9n

This is a linear decay pattern.

Q5. Sarita reads a 500-page book at 20 pages per day. How many pages are left after 15 days? Write the expression.
Ans. Pages left after n days:

500 - 20n

After 15 days:

500 - 20(15)

= 500 - 300

= 200 pages

Q6. The perimeter of a square of side x is 4x. Find the perimeters for sides 1 cm, 1.5 cm, 2 cm, 2.5 cm, and 3 cm. What pattern do you observe?
Ans.

When the side increases by 0.5 cm, the perimeter increases by 2 cm.

This is a linear pattern.

Questions on Linear Growth and Linear Decay

Linear growth means a quantity increases by a fixed amount over equal intervals.

Linear decay means a quantity decreases by a fixed amount over equal intervals.

Linear Growth and Decay Class 9 Maths

Q1. A plant is 1.75 feet tall and grows 0.5 feet every month. Find its height after 7 months.
Ans. Let h(t) be the height after t months.

h(t) = 1.75 + 0.5t

After 7 months:

h = 1.75 + 0.5(7)

= 1.75 + 3.5

= 5.25 feet

This is linear growth because height increases by a fixed amount every month.

Q2. A mobile phone costs ₹10,000. Its value decreases by ₹800 every year. Find its value after 3 years.
Ans. Let v(t) be the value after t years.

v(t) = 10000 - 800t

After 3 years:

v = 10000 - 800(3)

= 10000 - 2400

= ₹7600

This is linear decay because the value decreases by a fixed amount each year.

Q3. A village has 750 people. Fifty people move in every year. Find the population after 6 years.
Ans. Let P(t) be the population after t years.

P(t) = 750 + 50t

After 6 years:

P = 750 + 50(6)

= 750 + 300

= 1050

This is linear growth.

Q4. A telecom recharge of ₹600 reduces by ₹15 every day. Write the balance after x days. After how many days will the balance become zero?
Ans. Balance after x days:

b(x) = 600 - 15x

For zero balance:

600 - 15x = 0

15x = 600

x = 40

The balance will become zero after 40 days.

Q5. Bela has ₹100. She spends ₹5 every day. After how many days will ₹40 be left?
Ans. Amount after n days:

100 - 5n

Set amount = ₹40:

100 - 5n = 40

5n = 60

n = 12

So, ₹40 will be left after 12 days.

Q6. The cost of a journey is C(d) = 100 + 60d. Find the cost for 15 km. For how many kilometres will the cost be ₹700?
Ans.

For 15 km:

C(15) = 100 + 60(15)

= 100 + 900

= ₹1000

For cost ₹700:

100 + 60d = 700

60d = 600

d = 10

So, ₹700 is charged for 10 km.

Questions on Linear Relationships in the Form y = ax + b

A linear relationship connects two variables using the form y = ax + b.

Here, a is the slope and b is the y-intercept.

Class 9 Maths Chapter 2 Questions and Answers

Q1. A telecom company charges a fixed monthly fee plus ₹20 per GB. At 10 GB, bill is ₹350. At 20 GB, bill is ₹550. Find a and b in y = ax + b.
Ans.

Using y = ax + b:

350 = 10a + b ... (i)

550 = 20a + b ... (ii)

Subtract (i) from (ii):

200 = 10a

a = 20

Using 350 = 10a + b:

350 = 10(20) + b

350 = 200 + b

b = 150

So, the relationship is:

y = 20x + 150

Here, ₹20 is the cost per GB and ₹150 is the fixed fee.

Q2. A learning platform charges a fixed fee plus cost per module. At 10 modules, bill is ₹400. At 14 modules, bill is ₹500. Find the relationship.
Ans.

Let y = ax + b.

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

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

Subtract:

100 = 4a

a = 25

Now:

400 = 10(25) + b

400 = 250 + b

b = 150

So, the relationship is:

y = 25x + 150

Q3. A gym charges a fixed fee plus ₹60 per hour for badminton. At 10 hours, bill is ₹800. At 15 hours, bill is ₹1100. Find a and b.
Ans.

Let y = ax + b.

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

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

Subtract:

300 = 5a

a = 60

Using equation (i):

800 = 10(60) + b

800 = 600 + b

b = 200

So, the relationship is:

y = 60x + 200

Q4. The Celsius-Fahrenheit relationship is °C = a(°F) + b. Ice melts at 0°C and 32°F. Water boils at 100°C and 212°F. Find a and b.
Ans.

Let C = aF + b.

For 0°C and 32°F:

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

For 100°C and 212°F:

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

Subtract:

100 = 180a

a = 5/9

Using equation (i):

0 = 32(5/9) + b

b = -160/9

So,

C = (5/9)F - 160/9

Or,

C = (5/9)(F - 32)

Graph-Based Questions from Class 9 Maths Chapter 2

Graph questions test slope, y-intercept, x-intercept, parallel lines, and steepness.

Always find at least two points before drawing a straight line.

Class 9 Maths Chapter 2 Graphs Questions

Q1. Plot y = 2x + 1. Identify two points on the line.
Ans.

At x = 0:

y = 2(0) + 1 = 1

Point = (0, 1)

At x = 3:

y = 2(3) + 1 = 7

Point = (3, 7)

Plot (0, 1) and (3, 7). Join them to get the line.

Q2. Points (-1, -3), (0, 0), (1, 3), (3, 9), and (4, 12) lie on the same line. Find the equation.
Ans. Check the relation:

For every point, y = 3x.

So, the equation of the line is:

y = 3x

Q3. What are the slope and y-intercept of y = 3x - 2?
Ans. Compare with y = ax + b.

Slope a = 3

y-intercept b = -2

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

Q4. Draw y = 2x - 1, y = 2x + 1, and y = 2x + 5 on the same axes. What do you observe?
Ans. All three lines have the same slope.

Slope = 2

Their y-intercepts are -1, 1, and 5.

So, the lines are parallel and cut the y-axis at different points.

Q5. In y = ax + b, what happens when a is fixed but b changes?
Ans. The slope remains the same.

The lines remain parallel.

Changing b shifts the line up or down.

Q6. In y = ax, what happens as a increases from 1/2 to 1 to 2?
Ans. All lines pass through the origin.

As a increases, the line becomes steeper.

Q7. What does a negative slope tell you about the graph?
Ans. A negative slope means the line falls from left to right.

It represents linear decay.

Q8. Draw the graphs of y = -(1/3)x, y = -x, and y = -3x. What changes and what stays the same?
Ans. All three lines pass through the origin.

All have negative slopes.

As the absolute value of the slope increases, the line becomes steeper.

So, y = -3x is steeper than y = -x and y = -(1/3)x.

Q9. The graph of a linear polynomial p(x) passes through (1, 5) and (3, 11). Find p(x). Find where it cuts both axes.
Ans.

Slope = (11 - 5) / (3 - 1)

= 6/2

= 3

So, p(x) = 3x + b

Use point (1, 5):

5 = 3(1) + b

b = 2

So,

p(x) = 3x + 2

It cuts the y-axis at:

x = 0

y = 2

Point = (0, 2)

It cuts the x-axis where:

3x + 2 = 0

x = -2/3

Point = (-2/3, 0)

Q10. Draw the graph of y = -3x + 4. Find its slope, y-intercept, and x-intercept.
Ans. Compare with y = ax + b.

Slope = -3

y-intercept = 4

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

For x-intercept:

0 = -3x + 4

3x = 4

x = 4/3

So, the line cuts the x-axis at (4/3, 0).

End-of-Chapter Practice Questions

These chapter 2 maths class 9 important questions cover higher-level algebra, real-life expressions, and linear relationships.

Solve each question with full steps.

Class 9 Maths Chapter 2 Extra Questions

Q1. If you have ₹800 and save ₹250 every month, find the amount after 6 months and 2 years. Express it as a linear pattern.
Ans. Amount after n months:

A(n) = 800 + 250n

After 6 months:

A(6) = 800 + 250(6)

= 800 + 1500

= ₹2300

After 2 years:

2 years = 24 months

A(24) = 800 + 250(24)

= 800 + 6000

= ₹6800

Q2. The digits of a two-digit number differ by 3. If the digits are interchanged and the new number is added to the original, the sum is 143. Find the numbers.
Ans. Let the tens digit be a and the units digit be b.

Given:

a - b = 3

Original number = 10a + b

Reversed number = 10b + a

Sum:

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

11a + 11b = 143

a + b = 13

Now solve:

a - b = 3

a + b = 13

Adding both:

2a = 16

a = 8

Then:

b = 5

Original number = 85

Reversed number = 58

Q3. Let p(x) = ax + b and q(x) = cx + d. Given p(0) = 5, p(x) - q(x) cuts the x-axis at (3, 0), and p(x) + q(x) = 6x + 4. Find p(x) and q(x).
Ans.

p(0) = 5

So, b = 5

p(x) = ax + 5

Now:

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

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

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

So,

a + c = 6

5 + d = 4

d = -1

Now:

q(x) = cx - 1

p(x) - q(x) = (ax + 5) - (cx - 1)

= (a - c)x + 6

It cuts the x-axis at (3, 0).

So,

0 = 3(a - c) + 6

3(a - c) = -6

a - c = -2

Now solve:

a + c = 6

a - c = -2

Adding:

2a = 4

a = 2

Then:

c = 4

So,

p(x) = 2x + 5

q(x) = 4x - 1

Q4. The temperature relation between Fahrenheit y and Kelvin x is y = (9/5)(x - 273) + 32. Find y when x = 313 K. Find x when y = 158°F.
Ans.

When x = 313:

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

= (9/5)(40) + 32

= 72 + 32

= 104°F

When y = 158:

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

126 = (9/5)(x - 273)

126 × 5/9 = x - 273

70 = x - 273

x = 343 K

Important Formulas and Rules from Class 9 Maths Chapter 2