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

Class 9 Ganita Manjari Chapter 2 Quick Links:

• Exercise Set 2.1
• Exercise Set 2.2
• Exercise Set 2.3
• Exercise Set 2.4
• Exercise Set 2.5
• Exercise Set 2.6
• End of Chapter Exercise

Class 9 Maths Ganita Manjari Chapter 2 Exercise Set 2.1 Solutions

(i) 2x² − 5x + 3
(ii) y³ + 2y − 1
(iii) −9
(iv) 4z − 3
Answer:
Degree of a polynomial is the highest power of the variable.
(i) 2x² − 5x + 3
Highest power of x = 2
∴ Degree = 2

(ii) y³ + 2y − 1
Highest power of y = 3
∴ Degree = 3

(iii) −9
This is a constant polynomial (no variable)
∴ Degree = 0

(iv) 4z − 3
Highest power of z = 1
∴ Degree = 1

Answer:
A polynomial of degree 1 (linear polynomial):
Example: 2x + 5

A polynomial of degree 2 (quadratic polynomial):
Example: x² + 3x + 1

A polynomial of degree 3 (cubic polynomial):
Example: x³ − 2x² + x + 4

Answer:
Given polynomial: x⁴ − 3x³ + 6x² − 2x + 7
Coefficient of x³ = −3
Coefficient of x² = 6

Answer:
The given polynomial is 4z³ + 5z² − 11.
There is no term containing z¹ (i.e., z).
Hence, the coefficient of z is 0.

Recall that polynomials of degree 1 are called linear polynomials. In this chapter, we shall study linear polynomials.
Answer:
The constant term is the term without any variable.
In the polynomial 9x³ + 5x² − 8x − 10, the constant term is −10.

Class 9 Maths Ganita Manjari Chapter 2 Exercise Set 2.2 Solutions

(i) x = 0
(ii) x = −1
(iii) x = 2
Answer:
Given polynomial: 5x − 3
(i) x = 0
= 5(0) − 3 = −3

(ii) x = −1
= 5(−1) − 3 = −5 − 3 = −8

(iii) x = 2
= 5(2) − 3 = 10 − 3 = 7

(i) s = 0
(ii) s = −3
(iii) s = 4
Answer:
(i) For s = 0
7s² − 4s + 6
= 7(0)² − 4(0) + 6 = 6

(ii) For s = −3
7s² − 4s + 6
= 7(−3)² − 4(−3) + 6
= 7(9) + 12 + 6
= 63 + 12 + 6 = 81

(iii) For s = 4
7s² − 4s + 6
= 7(4)² − 4(4) + 6
= 7(16) − 16 + 6
= 112 − 16 + 6 = 102

Answer:
Let Salil’s present age = x years
Mother’s present age = 3x years

After 5 years:
Salil’s age = x + 5
Mother’s age = 3x + 5

According to question: (x + 5) + (3x + 5) = 70
⇒ 4x + 10 = 70
⇒ 4x = 60
⇒ x = 15
Therefore, Salil’s age = 15 years
Mother’s age = 45 years

Answer:
Let the integers be 2x and 5x
Difference: 5x − 2x = 63
⇒ 3x = 63
⇒ x = 21

Integers: 2x = 42
⇒ 5x = 105
Required integers are 42 and 105.

Answer:
Let number of five-rupee coins = x
Number of two-rupee coins = 3x

Total value: 5x + 2(3x) = 88
⇒ 5x + 6x = 88
⇒ 11x = 88
⇒ x = 8
Hence:
Five-rupee coins = 8
Two-rupee coins = 24

Answer:
Let shorter piece = x feet
Longer piece = 4x feet

Total: x + 4x = 300
⇒ 5x = 300
⇒ x = 60

Therefore:
Shorter piece = 60 feet
Longer piece = 240 feet.

Answer:
Let width = x cm
Length = 2x + 3 cm

Perimeter = 2(length + width)
⇒ 2[(2x + 3) + x] = 24
⇒ 2(3x + 3) = 24
⇒ 6x + 6 = 24
⇒ 6x = 18
⇒ x = 3

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

Class 9 Maths Ganita Manjari Chapter 2 Exercise Set 2.3 Solutions

Find a linear expression to represent the amount she will have in the nᵗʰ month.
Answer:
Initial amount in bank = ₹500
Monthly pocket money = ₹150

At the end of:
2nd month = 500 + 2(150) = ₹800
3rd month = 500 + 3(150) = ₹950
4th month = 500 + 4(150) = ₹1100
and so on…

Let the amount in the nᵗʰ month be Aₙ
Then, Aₙ = 500 + 150n
Thus, the required linear expression is:
Aₙ = 150n + 500

Answer:
Initial members = 120
Members leaving each hour = 9
After:
1 hour = 120 − 9 = 111
2 hours = 120 − 18 = 102
3 hours = 120 − 27 = 93

Let number of members after n hours = Mₙ
Mₙ = 120 − 9n
Thus, required linear expression is Mₙ = 120 − 9n.

Answer:
Length = 13 cm
Area = Length × 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

Let breadth = x cm
Area = 13x
Thus, linear pattern is A = 13x.

Answer:
Length = 7 cm, Breadth = 11 cm
Volume = Length × Breadth × Height
= 7 × 11 × h = 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

Let height = h cm
Volume = 77h
Thus, linear pattern is V = 77h.

Answer:
Total pages = 500
Pages read per day = 20
Pages read in 15 days = 20 × 15 = 300
Pages left = 500 − 300 = 200
Let pages left after n days = Pₙ
So, Pₙ = 500 − 20n
Thus, linear pattern is Pₙ = 500 − 20n.

Class 9 Maths Ganita Manjari Chapter 2 Exercise Set 2.4 Solutions

(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.
Answer:
Given:
Initial height = 1.75 feet
Growth per month = 0.5 feet

(i) Height growth in each month = 0.5 feet
So, Height after 7 months:
h = 1.75 + (0.5 × 7)
= 1.75 + 3.5
= 5.25 feet

(ii) Table of values:

(iii) Expression:
Let height after t months = h
h = 1.75 + 0.5t
This represents linear growth because height increases by a constant amount (0.5 feet) every month.

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

(i) Decrease in value after 1 year = ₹800
So, Value after 3 years:
v = 10000 − (800 × 3)
= 10000 − 2400
= ₹7600

(ii) Table of values:

(iii) Expression:
Let value after t years = v
v = 10000 − 800t
This represents linear decay because the value decreases by a constant amount (₹800) every year.

(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.
Answer:
(i) Population after 6 years:
P = 750 + (50 × 6)
= 750 + 300
= 1050

(ii) Table of values:

(iii) Expression:
Let population after t years = P
P = 750 + 50t
This represents linear growth because the population increases by a constant number (50 people) every year.

(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.
Answer:
(i) Expression:
Let remaining balance after x days = b(x)
b(x) = 600 − 15x
This represents linear decay because the balance decreases by a constant amount (₹15) every day.

(ii) When balance runs out:
b(x) = 0
600 − 15x = 0
15x = 600
x = 40
So, the balance will run out after 40 days.

(iii) Table of values:

Class 9 Maths Ganita Manjari Chapter 2 Exercise Set 2.5 Solutions

Answer:
Given:
When x = 10, y = 400
When x = 14, y = 500
Using y = ax + b
For x = 10: 400 = 10a + b …(1)
For x = 14: 500 = 14a + b …(2)
Subtracting (1) from (2), we get:
500 − 400 = 14a − 10a
⇒ 100 = 4a
⇒ a = 25
Substituting a = 25 in (1), we get:
400 = 10(25) + b
⇒ 400 = 250 + b
⇒ b = 150
Therefore, a = 25 and b = 150.

Answer:
Given:
When x = 10, y = 800
When x = 15, y = 1100
Using y = ax + b
For x = 10: 800 = 10a + b …(1)
For x = 15: 1100 = 15a + b …(2)
Subtracting (1) from (2), we have:
1100 − 800 = 15a − 10a
⇒ 300 = 5a
⇒ a = 60
Substitute a = 60 in (1), we get:
800 = 10(60) + b
⇒ 800 = 600 + b
⇒ b = 200
Therefore, a = 60 and b = 200.

(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.)
Answer:
Given relation: °C = a°F + b
Using the point (°F, °C) = (32, 0), we have:
0 = 32a + b …(1)
Using the point (°F, °C) = (212, 100), we have:
100 = 212a + b …(2)
Subtracting (1) from (2), we get:
100 − 0 = 212a − 32a
⇒ 100 = 180a
⇒ a = 100/180
⇒ a = 5/9
Substitute a = 5/9 in (1), we get:
0 = 32(5/9) + b
⇒ 0 = 160/9 + b
⇒ b = −160/9
Therefore, a = 5/9 and b = −160/9.
Hence, the linear relationship is °C = (5/9)°F − 160/9.
It can also be written as °C = (5/9)(°F − 32).

Class 9 Maths Ganita Manjari Chapter 2 Exercise Set 2.6 Solutions

(i) y = 4x, y = 2x, y = x
Answer:
All lines are of the form y = ax (b = 0).
Observation:

• All lines pass through the origin (0,0).
• The value of ‘a’ (slope) determines steepness.
• Larger ‘a’ → steeper line.

(ii) y = −6x, y = −3x, y = −x
Answer:
All lines are of the form y = ax (b = 0).
Observation:

• All lines pass through the origin.
• Negative ‘a’ means lines slope downward.
• Larger magnitude of ‘a’ → steeper downward slope.

(iii) y = 5x, y = −5x
Answer:
Observation:

• Both lines pass through the origin.
• y = 5x slopes upward, y = −5x slopes downward.
• Same magnitude of ‘a’ → same steepness but opposite direction.

(iv) y = 3x − 1, y = 3x, y = 3x + 1
Answer:
All lines have same slope (a = 3).
Observation:

• Lines are parallel (same slope).
• Different values of ‘b’ shift the line up or down.
• b = −1 → line below origin
• b = 0 → passes through origin
• b = +1 → line above origin.

(v) y = −2x − 3, y = −2x, y = 2x + 3
Answer:
Observation:

• y = −2x − 3 and y = −2x have same slope (−2) → parallel lines.
• y = 2x + 3 has positive slope → different direction.
• ‘b’ changes vertical position of line.

Conclusion:

1. ‘a’ (coefficient of x) controls slope (steepness and direction).
2. ‘b’ (constant term) controls vertical shift (y-intercept).

Class 9 Maths Ganita Manjari Chapter 2 End of Chapter Exercises Solutions

Answer:
A polynomial of degree 3 has the general form: ax³ + bx² + cx + d
Given that coefficient of x² is −7, so b = −7
One such polynomial is x³ − 7x² + 2x + 1.
(Any polynomial of degree 3 with −7 as the coefficient of x² is correct)

(i) 5x² − 3x + 7 if x = 1
(ii) 4t³ − t² + 6 if t = a
Answer:
(i) 5x² − 3x + 7 if x = 1
Substitute x = 1:
= 5(1)² − 3(1) + 7
= 5 − 3 + 7
= 9

(ii) 4t³ − t² + 6 if t = a
Substitute t = a:
= 4a³ − a² + 6

Answer:
Let the number be x.
According to the question:
(5/2)x + 2/3 = -7/12
Subtract 2/3 from both sides, we get:
(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, we get:
x = (-5/4) × (2/5)
⇒ x = -1/2
Therefore, the number is -1/2.

Answer:
Let the smaller number be x.
Then the larger number = 5x
After adding 21 to both numbers: New numbers are x + 21 and 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 = 7 and the larger number = 5 × 7 = 35
Therefore, the two numbers are 7 and 35.

Answer:
Initial amount = ₹800
Saving every month = ₹250
Linear pattern: Amount after n months = 800 + 250n
(i) Amount after 6 months:
= 800 + 250(6)
= 800 + 1500
= ₹2300

(ii) Amount after 2 years: 2 years = 24 months
Amount after 24 months:
= 800 + 250(24)
= 800 + 6000
= ₹6800
Therefore:
Amount after 6 months = ₹2300
Amount after 2 years = ₹6800
Linear pattern: A = 800 + 250n, where n is the number of months.

Answer:
Let the tens digit be x and the units digit be y.
Then the original number = 10x + y
The interchanged number = 10y + x
According to the question: (10x + y) + (10y + x) = 143
⇒ 11x + 11y = 143
⇒ 11(x + y) = 143
⇒ x + y = 13 …(1)
The digits differ by 3.
So, x – y = 3 …(2)
Now adding (1) and (2), we get:
2x = 16
⇒ x = 8
Substitute x = 8 in (1), we have:
8 + y = 13
⇒ y = 5
Therefore, the original number = 85 and the interchanged number = 58
Hence, the two numbers are 85 and 58.

(i) y = -3x + 4
(ii) 2y = 4x + 7
(iii) 5y = 6x – 10
(iv) 3y = 6x – 11
Are any of the lines parallel?
Answer:
(i) y = -3x + 4
Comparing with y = ax + b:
Slope a = -3
y-intercept b = 4
So, the point, where the line cuts the y-axis, is (0, 4).

(ii) 2y = 4x + 7
Dividing both sides by 2: y = 2x + 7/2
Slope a = 2
y-intercept b = 7/2
So, the point, where the line cuts the y-axis, is (0, 7/2).

(iii) 5y = 6x – 10
Dividing both sides by 5: y = (6/5)x – 2
Slope a = 6/5
y-intercept b = -2
So, the point, where the line cuts the y-axis is (0, -2).

(iv) 3y = 6x – 11
Dividing both sides by 3: y = 2x – 11/3
Slope a = 2
y-intercept b = -11/3
So, the point, where the line cuts the y-axis is (0, -11/3).

Parallel lines:
Two lines are parallel if they have the same slope.
Equation (ii): slope = 2
Equation (iv): slope = 2
Therefore, lines (ii) and (iv) are parallel.

(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.
Answer:
(i) When x = 313 K, We have to find y:
Using equation y = (9/5)(x – 273) + 32, we have
y = (9/5)(313 – 273) + 32
⇒ y = (9/5)(40) + 32
⇒ y = 72 + 32
⇒ y = 104
Therefore, the temperature is 104 °F.

(ii) When y = 158 °F, we have to find x:
Using equation y = (9/5)(x – 273) + 32, we have
158 = (9/5)(x – 273) + 32
Subtracting 32 from both sides, we get:
158 – 32 = (9/5)(x – 273)
⇒ 126 = (9/5)(x – 273)
Multiply both sides by 5/9, we get:
126 × (5/9) = x – 273
⇒ 70 = x – 273
⇒ x = 343
Therefore, the temperature is 343 K.

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:
We know that:
Work done = Force × Distance
According to question, work done = w
Distance travelled = d
Constant force = 3 units
Therefore, w = 3d
This is the required linear equation in two variables.
If d = 2 units, force w = 3 × 2 = 6 units.
Taking w on y-axis and d on x-axis, we can plot the graph.

The point (2, 6) lie on a straight line, so, it is verified by graph.
So, the Work done when distance travelled is 2 units: w = 3 × 2
Hence, w = 6 units
Verification from graph:
The point corresponding to d = 2 is (2, 6), so the work done is 6 units.

(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.
Answer:
(i) Let p(x) = ax + b
Since the graph passes through (1, 5),
a(1) + b = 5
⇒ a + b = 5 …(1)
Since the graph passes through (3, 11),
a(3) + b = 11
⇒ 3a + b = 11 …(2)
Subtracting (1) from (2), we get:
3a + b – (a + b) = 11 – 5
⇒ 2a = 6
⇒ a = 3
Substituting a = 3 in (1), we have:
3 + b = 5
⇒ b = 2
Therefore, p(x) = 3x + 2.

(ii) Coordinates where the graph cuts the axes:
To find the y-axis intercept: Put x = 0
y = p(0) = 3(0) + 2 = 2
So, the graph cuts the y-axis at (0, 2).

To find the x-axis intercept: Put y = 0
3x + 2 = 0
⇒ 3x = -2
⇒ x = -2/3
So, the graph cuts the x-axis at (-2/3, 0).

(iii) From the graph, it verifies that the line passes through the given points and cuts:
y-axis at (0, 2)
x-axis at (-2/3, 0).

(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).
Answer:
Let p(x) = ax + b and q(x) = cx + d
Using the condition (i), we have p(0) = 5
⇒ a(0) + b = 5
⇒ b = 5
So, p(x) = ax + 5
Now, using condition (iii), we have
p(x) + q(x) = 6x + 4
⇒ (ax + 5) + (cx + d) = 6x + 4
⇒ (a + c)x + (5 + d) = 6x + 4
Comparing coefficients, we get a + c = 6 …(1)
5 + d = 4
⇒ d = –1
So, q(x) = cx – 1
Using condition (ii), p(x) − q(x) cuts x-axis at (3, 0)
⇒ p(3) − q(3) = 0
Here:
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)

Solving equations
From (1): a + c = 6
From (2): a − c = –2
Adding both: 2a = 4 ⇒ a = 2
Then: 2 + c = 6 ⇒ c = 4
Hence:
p(x) = 2x + 5
q(x) = 4x − 1.

(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.

Answer:
A single hexagon needs 6 matchsticks.
Since each new hexagon shares one side with the previous hexagon, only 5 new matchsticks are added at every new stage.
So the pattern is:
Stage 1 = 6
Stage 2 = 6 + 5 = 11
Stage 3 = 11 + 5 = 16

(i) Next two stages:
Stage 4: Number of matchsticks = 16 + 5 = 21
Stage 5: Number of matchsticks = 21 + 5 = 26
Therefore:

• Stage 4 requires 21 matchsticks
• Stage 5 requires 26 matchsticks

(ii) Complete table:

(iii) Rule for the nᵗʰ stage:
The number of matchsticks forms an arithmetic pattern: 6, 11, 16, 21, 26, …
First term = 6
Difference = 5
So, Number of matchsticks in the nᵗʰ stage
= 6 + (n – 1) × 5
= 6 + 5n – 5
= 5n + 1
Hence, the rule is Mₙ = 5n + 1

(iv) Matchsticks required for the 15th stage:
M₁₅ = 5(15) + 1
= 75 + 1
= 76
Therefore, 76 matchsticks will be required for the 15th stage.

(v) Can 200 matchsticks form a stage in this pattern?
For some stage n,
5n + 1 = 200
⇒ 5n = 199
⇒ n = 199/5
⇒ n = 39.8
Since n is not a whole number, 200 matchsticks cannot form any stage in this pattern.
Therefore, 200 matchsticks cannot form a stage in this pattern.

(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.
Answer:
Given:
p(x) = ax + b
q(x) = cx + d
First, to find p(x).
Since p(x) passes through (2, 3) and (6, 11), its slope is m = (11 – 3) / (6 – 2)
= 8/4 = 2
So, p(x) = 2x + b
Using the point (2, 3), we have:
3 = 2(2) + b
⇒ 3 = 4 + b
⇒ b = -1
Therefore, p(x) = 2x – 1.

Now to find q(x).
Since q(x) is parallel to p(x), it has the same slope.
So, slope of q(x) = 2
Hence, q(x) = 2x + d
Since q(x) passes through (4, -1), so -1 = 2(4) + d
⇒ -1 = 8 + d
⇒ d = -9
Therefore, q(x) = 2x – 9

Now we have to find where these lines meet the x-axis.
A line meets the x-axis where y = 0.
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).

Hence, p(x) = 2x – 1 and q(x) = 2x – 9.
The x-axis intercepts are:
For p(x): (1/2, 0)
For q(x): (9/2, 0)

Answer:
Given: f(x) = ax + a, where a > 0
We can write it as: f(x) = a(x + 1)
Common properties of all such linear functions are:

1. Slope:
   The slope is a, and since a > 0, all the lines have positive slope.
   So, all these lines rise from left to right.
2. y-intercept:
   Putting x = 0,
   f(0) = a
   So, the y-intercept is (0, a).
   Since a > 0, all the lines cut the y-axis above the origin.
3. x-intercept:
   To find where the line cuts the x-axis, put f(x) = 0
   ax + a = 0
   a(x + 1) = 0
   Since a > 0, a is not zero.
   So, x + 1 = 0
   or x = -1
   Thus, every line cuts the x-axis at the same point (-1, 0).

Therefore, all linear functions of the form f(x) = ax + a, a > 0, have the following in common:

• all have positive slope
• all cut the y-axis above the origin
• all pass through the fixed point (-1, 0).