Class 9 Maths Ganita Manjari Chapter 2 End of Chapter Exercises: Introduction to Linear Polynomials

Class 9 Maths Ganita Manjari Chapter 2 End of Chapter Exercises cover the complete revision of Introduction to Linear Polynomials. These exercises include polynomial degree, polynomial values, linear equations, linear patterns, graphing lines, slope, y-intercept and real-life applications of linear relationships.

By the end of Chapter 2, students are expected to connect three things clearly: polynomial basics, linear patterns, and straight-line graphs. The Class 9 Maths Ganita Manjari Chapter 2 End of Chapter Exercises bring all these ideas together through questions on polynomial values, linear equations, matchstick patterns, temperature conversion, slope, y-intercept and parallel lines.

These questions from Introduction to Linear Polynomials are not limited to direct formula use. Students also need to form equations from word problems, identify linear relationships in the form y = ax + b, and understand how a linear polynomial appears on a graph. The solutions below follow the textbook order and show each calculation in a clear, copy-friendly format.

Key Takeaways

Polynomial Basics: Degree, coefficient and substitution are revised together.
Linear Modelling: Word problems can be represented using linear equations.
Graph Skills: Slope, y-intercept and parallel lines are identified from equations.
Pattern Rules: Matchstick and savings patterns use linear expressions.

Class 9 Maths Ganita Manjari Chapter 2 End of Chapter Exercises Structure 2026

Class 9 Maths Ganita Manjari Chapter 2 End of Chapter Exercises

The End of Chapter Exercises of Chapter 2 include 14 questions. These Class 9 Maths linear polynomials answers help students revise polynomial basics, linear equations, word problems, temperature conversion, matchstick patterns and graphs of linear polynomials.

End of Chapter Exercises Question 1

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

Solution:

A polynomial of degree 3 must have the highest power of x equal to 3.

The coefficient of the x² term should be −7.

One possible polynomial is:

x³ − 7x² + 5x + 2

Answer: One such polynomial is x³ − 7x² + 5x + 2.

End of Chapter Exercises Question 2

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

(i) 5x² − 3x + 7 if x = 1

Solution:

Given polynomial:

5x² − 3x + 7

Substitute x = 1.

5x² − 3x + 7 = 5(1)² − 3(1) + 7

= 5 − 3 + 7

= 9

Answer: The value is 9.

(ii) 4t³ − t² + 6 if t = a

Solution:

Given polynomial:

4t³ − t² + 6

Substitute t = a.

4t³ − t² + 6 = 4a³ − a² + 6

Answer: The value is 4a³ − a² + 6.

End of Chapter Exercises 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

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

(5/2)x = −5/4

Now multiply both sides by 2/5.

x = (−5/4) × (2/5)

x = −1/2

Answer: The number is −1/2.

End of Chapter Exercises 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 positive number be x.

Then the larger number is 5x.

After adding 21 to both numbers:

Smaller new number = x + 21

Larger new number = 5x + 21

According to the question:

5x + 21 = 2(x + 21)

5x + 21 = 2x + 42

5x − 2x = 42 − 21

3x = 21

x = 7

So, the smaller number is 7.

Larger number:

5x = 5(7) = 35

Answer: The numbers are 7 and 35.

End of Chapter Exercises 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

After n months:

Amount = 800 + 250n

(i) After 6 months

Amount = 800 + 250(6)

= 800 + 1500

= 2300

(ii) After 2 years

2 years = 24 months

Amount = 800 + 250(24)

= 800 + 6000

= 6800

Answer: After 6 months, the amount is ₹2300. After 2 years, the amount is ₹6800. The linear pattern is 800 + 250n.

End of Chapter Exercises 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 tens digit be a and the ones digit be b.

Original number = 10a + b

Number after interchanging digits = 10b + a

Their sum is 143.

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

11a + 11b = 143

11(a + b) = 143

a + b = 13

Also, the digits differ by 3.

So, the digits are 8 and 5.

The possible numbers are:

85 and 58

Check:

85 + 58 = 143

Answer: The two numbers are 85 and 58.

End of Chapter Exercises 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

Solution:

This is in the form:

y = ax + b

So:

• Slope = −3
• y-intercept = 4
• Point where it cuts the y-axis = (0, 4)

(ii) 2y = 4x + 7

Solution:

Convert into y = ax + b form.

2y = 4x + 7

y = 2x + 7/2

So:

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

(iii) 5y = 6x − 10

Solution:

5y = 6x − 10

y = (6/5)x − 2

So:

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

(iv) 3y = 6x − 11

Solution:

3y = 6x − 11

y = 2x − 11/3

So:

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

Parallel lines

Lines with the same slope are parallel.

Equations (ii) and (iv) both have slope 2.

Answer: Lines 2y = 4x + 7 and 3y = 6x − 11 are parallel.

End of Chapter Exercises 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:

Given:

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

Answer: The temperature is 104°F.

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

Solution:

Given:

y = 158

Use:

y = (9/5)(x − 273) + 32

158 = (9/5)(x − 273) + 32

158 − 32 = (9/5)(x − 273)

126 = (9/5)(x − 273)

Multiply both sides by 5/9.

126 × 5/9 = x − 273

70 = x − 273

x = 343

Answer: The temperature is 343 K.

End of Chapter Exercises 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?

Solution:

Work done = Force × Distance

Given force = 3 units

So,

w = 3d

This is a linear equation in two variables.

To draw the graph, take some values of d and find w.

Plot the points (0, 0), (1, 3), (2, 6), (3, 9) and join them to get a straight line.

When d = 2:

w = 3(2)

w = 6

Answer: The linear equation is w = 3d. When distance is 2 units, work done is 6 units.

End of Chapter Exercises 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:

p(x) = ax + b

The graph passes through (1, 5).

So:

5 = a + b

The graph also passes through (3, 11).

So:

11 = 3a + b

Subtract the first equation from the second.

11 − 5 = (3a + b) − (a + b)

6 = 2a

a = 3

Now substitute a = 3 in:

5 = a + b

5 = 3 + b

b = 2

So:

p(x) = 3x + 2

Answer: 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

For the y-axis, put x = 0.

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

So, the line cuts the y-axis at:

(0, 2)

For the x-axis, put p(x) = 0.

3x + 2 = 0

3x = −2

x = −2/3

So, the line cuts the x-axis at:

(−2/3, 0)

Answer: The graph cuts the y-axis at (0, 2) and 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

Use two points:

• (0, 2)
• (1, 5)

Plot the points and join them. The graph is a straight line.

It passes through (1, 5) and (3, 11) and cuts the axes at (0, 2) and (−2/3, 0).

Answer: The graph verifies the answers.

End of Chapter Exercises 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) = 6x + 4 for all real x. Find p(x) and q(x).

Solution:

Given:

p(x) = ax + b

q(x) = cx + d

Since p(0) = 5:

b = 5

Also:

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

So:

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

(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) = (ax + 5) − (cx − 1)

p(x) − q(x) = (a − c)x + 6

This cuts the x-axis at (3, 0). So, when x = 3, the value is 0.

3(a − c) + 6 = 0

3(a − c) = −6

a − c = −2

Now solve:

a + c = 6

a − c = −2

Adding both equations:

2a = 4

a = 2

Then:

a + c = 6

2 + c = 6

c = 4

So:

p(x) = 2x + 5

q(x) = 4x − 1

Answer: p(x) = 2x + 5 and q(x) = 4x − 1.

End of Chapter Exercises 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?

Solution:

One hexagon needs 6 matchsticks.

When a new hexagon is added, it shares one side with the previous hexagon. So, each new hexagon adds 5 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

Answer: Stage 4 needs 21 matchsticks and Stage 5 needs 26 matchsticks.

(ii) Complete the following table.

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

Solution:

Stage 1 has 6 matchsticks.

Each new stage adds 5 matchsticks.

So, for stage n:

Number of matchsticks = 6 + 5(n − 1)

= 6 + 5n − 5

= 5n + 1

Answer: The rule is 5n + 1.

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

Solution:

Use:

5n + 1

For n = 15:

5(15) + 1 = 75 + 1

= 76

Answer: The 15th stage requires 76 matchsticks.

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

Solution:

Set:

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.

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

End of Chapter Exercises 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.

Solution:

Let:

p(x) = ax + b

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

Find slope:

a = (11 − 3) / (6 − 2)

a = 8 / 4

a = 2

So:

p(x) = 2x + b

Use point (2, 3).

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.

So:

q(x) = 2x + d

The graph of q(x) passes through (4, −1).

−1 = 2(4) + d

−1 = 8 + d

d = −9

So:

q(x) = 2x − 9

Now find x-axis intercepts.

For p(x):

2x − 1 = 0

2x = 1

x = 1/2

So, p(x) meets the x-axis at:

(1/2, 0)

For q(x):

2x − 9 = 0

2x = 9

x = 9/2

So, q(x) meets the x-axis at:

(9/2, 0)

Answer: p(x) = 2x − 1 and q(x) = 2x − 9. Their x-axis intercepts are (1/2, 0) and (9/2, 0).

End of Chapter Exercises Question 14

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

Solution:

Given:

f(x) = ax + a

Take a common.

f(x) = a(x + 1)

To find the x-intercept, put:

f(x) = 0

So:

a(x + 1) = 0

Since a > 0, a ≠ 0.

Therefore:

x + 1 = 0

x = −1

So, every graph of the form f(x) = ax + a, where a > 0, passes through:

(−1, 0)

Also:

f(0) = a

So, the y-intercept is positive.

Answer: All such linear functions pass through (−1, 0). They also have positive slope and positive y-intercept, both equal to a.

Final Answers for Class 9 Maths Ganita Manjari Chapter 2 End of Chapter Exercises

Concept Used in Introduction to Linear Polynomials End of Chapter Exercises

Introduction to Linear Polynomials End of Chapter Exercises revise polynomial values, linear equations, linear relationships and graph-based questions.

Important concepts include:

• degree and coefficient of a polynomial,
• value of a polynomial by substitution,
• linear equations from word problems,
• linear pattern and fixed change,
• graph of y = ax + b,
• slope and y-intercept Class 9,
• parallel lines,
• x-axis and y-axis intercepts,
• matchstick patterns using linear expressions.

These ideas complete the foundation of Class 9 Ganita Manjari linear polynomials solutions.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 2