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

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

Solution:

A degree 3 polynomial is of the form:
ax³ + bx² + cx + d
The coefficient of the x term must be -7.
So, one possible polynomial is:
P(x) = x³ + 2x² - 7x + 1
Answer: P(x) = x³ + 2x² - 7x + 1
(Answer may vary.)

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
For x = 1:
5(1)² - 3(1) + 7
= 5 - 3 + 7
= 9
Answer: 9

(ii) 4t³ – t² + 6 if t = a

Solution:

Given polynomial:
4t³ - t² + 6
For t = a:
4(a)³ - (a)² + 6
= 4a³ - a² + 6
Answer: 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. 12

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
Multiply both sides by 2/5:
x = (-5/4) × (2/5)
x = -1/2
Answer: The 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 x.
Then, the larger number = 5x.
After adding 21 to both numbers:
Smaller number = x + 21
Larger number = 5x + 21
According to the question:
5x + 21 = 2(x + 21)
5x + 21 = 2x + 42
5x - 2x = 42 - 21
3x = 21
x = 7
Therefore:
Smaller number = 7
Larger number = 5 × 7 = 35
Answer: The two numbers are 7 and 35.

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
(i) After 6 months:
Amount saved in 6 months:
₹250 × 6 = ₹1,500
Total amount:
₹800 + ₹1,500 = ₹2,300
Answer: After 6 months, the amount is ₹2,300.
(ii) After 2 years:
2 years = 24 months
Amount saved in 24 months:
₹250 × 24 = ₹6,000
Total amount:
₹800 + ₹6,000 = ₹6,800
Answer: After 2 years, the amount is ₹6,800.
If the number of months is represented by n, then the linear pattern is:
A = 800 + 250n
where A represents the total amount after n 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 tens digit be x and the units digit be y.
Since the digits differ by 3:
x - y = 3 ...(1)
The original number = 10x + y
The number obtained by interchanging the digits = 10y + x
According to the question:
(10x + y) + (10y + x) = 143
11x + 11y = 143
x + y = 13 ...(2)
Adding equations (1) and (2):
2x = 16
x = 8
Then:
y = 13 - 8 = 5
So, the original number is 85 and the number obtained by interchanging the digits is 58.
Answer: The two numbers are 85 and 58.

Question 7.
Draw the graph of the following equations, and identify their slopes andy-intercepts. Also, find the coordinates of the points where these lines cut the y-axis.

Solution:

(i) y = -3x + 4

Solution:

For the equation:

y = -3x + 4

Comparing it with the standard form:

y = ax + b

we get:

a = -3
b = 4

So:

Slope = -3

y-intercept = 4

The line cuts the y-axis when x = 0.

For x = 0:

y = -3(0) + 4
= 4

Therefore, the line cuts the y-axis at:

(0, 4)

A few points for plotting are:

(ii) 2y = 4x + 7

Solution:

For the equation:

2y = 4x + 7

Divide both sides by 2:

y = 2x + 7/2

Comparing with the standard form:

y = ax + b

we get:

a = 2
b = 7/2

So:

Slope = 2

y-intercept = 7/2 or 3.5

The line cuts the y-axis when x = 0.

For x = 0:

y = 2(0) + 7/2
= 7/2

Therefore, the line cuts the y-axis at:

(0, 7/2)

A few points for plotting are:

(iii) 5y = 6x – 10

Solution:

For the equation:

5y = 6x - 10

Divide both sides by 5:

y = (6/5)x - 2

Comparing with the standard form:

y = ax + b

we get:

a = 6/5
b = -2

So:

Slope = 6/5

y-intercept = -2

The line cuts the y-axis when x = 0.

For x = 0:

y = (6/5)(0) - 2
= -2

Therefore, the line cuts the y-axis at:

(0, -2)

A few points for plotting are:

(iv) 3y = 6x – 11

Solution:

For the equation:

3y = 6x - 11

Divide both sides by 3:

y = 2x - 11/3

Comparing with the standard form:

y = ax + b

we get:

a = 2
b = -11/3

So:

Slope = 2

y-intercept = -11/3

The line cuts the y-axis when x = 0.

For x = 0:

y = 2(0) - 11/3
= -11/3

Therefore, the line cuts the y-axis at:

(0, -11/3)

A few points for plotting are:

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
For x = 313 K:
y = (9/5)(313 - 273) + 32
= (9/5)(40) + 32
= 72 + 32
= 104
Answer: The temperature of the liquid is 104°F.

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

Solution:

Given:
y = (9/5)(x - 273) + 32
Here, y = 158°F.
So:
158 = (9/5)(x - 273) + 32
158 - 32 = (9/5)(x - 273)
126 = (9/5)(x - 273)
126 × 5/9 = x - 273
70 = x - 273
x = 343
Answer: The temperature 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 is given by:

Work = Force × Distance

Let:

w = work done
d = distance travelled

Since the constant force is 3 units:

w = 3d

So, the required linear equation is:

w = 3d

To draw the graph, take d on the x-axis and w on the y-axis.

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
The graph passes through (1, 5) and (3, 11).
So:
For x = 1, p(x) = 5:
a + b = 5 ...(1)
For x = 3, p(x) = 11:
3a + b = 11 ...(2)
Subtract equation (1) from equation (2):
3a + b - (a + b) = 11 - 5
2a = 6
a = 3
Substitute a = 3 in equation (1):
3 + b = 5
b = 2
Therefore:
p(x) = 3x + 2
Answer: p(x) = 3x + 2

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

Solution:

From part (i):
p(x) = 3x + 2
To find 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 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)

Answer:
x-axis: (-2/3, 0)
y-axis: (0, 2)

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

Solution:

From part (i), the polynomial is:

p(x) = 3x + 2

To draw the graph, take a few values of x:

Question 11.
Let p(x) = ax + b and q(x) = cx + d be two linear polynomials such that:
(i) m = 5.

Solution:

here appears to be a typo in the question. Condition (i) should be:

p(0) = 5

Given:

p(x) = ax + b

Substitute x = 0:

p(0) = a(0) + b

5 = b

Therefore:

b = 5

Hence,

p(x) = ax + 5

(ii) The polynomial p(x) – q(x) cuts the x-axis at (3, 0).

Solution:

Given:
p(x) = ax + b
q(x) = cx + d
From part (i):
b = 5
Therefore:
p(x) = ax + 5
Now,
p(x) - q(x) = ax + 5 - (cx + d)
= (a - c)x + (5 - d)
Since the graph of p(x) - q(x) cuts the x-axis at (3, 0), put x = 3 and p(x) - q(x) = 0:
3(a - c) + 5 - d = 0
Therefore:
3a - 3c + 5 - d = 0
Answer: 3(a - c) + 5 - d = 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:

Given:
p(x) = ax + b
q(x) = cx + d
From part (i):
b = 5
So:
p(x) = ax + 5
From part (iii):
p(x) + q(x) = 6x + 4
Therefore:
(a + c)x + (5 + d) = 6x + 4
Comparing coefficients:
a + c = 6 ...(1)
5 + d = 4
d = -1
So:
q(x) = cx - 1
From part (ii), p(x) - q(x) cuts the x-axis at (3, 0).
Therefore:
p(3) - q(3) = 0
3a + 5 - (3c - 1) = 0
3a - 3c + 6 = 0
a - c = -2 ...(2)
Now, solving equations (1) and (2):
a + c = 6
a - c = -2
Adding both equations:
2a = 4
a = 2
Therefore:
c = 4
Hence,
p(x) = 2x + 5
q(x) = 4x - 1
Answer: p(x) = 2x + 5 and 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.

Solution:

Each hexagon has 6 sides. When a new hexagon is added, it shares one side with the previous hexagon. Therefore, each new hexagon requires only 5 additional matchsticks.

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

Solution:

Stage 4 will contain 4 hexagons.
Number of matchsticks = 16 + 5 = 21
Stage 5 will contain 5 hexagons.
Number of matchsticks = 21 + 5 = 26
Answer:
Stage 4 = 21 matchsticks
Stage 5 = 26 matchsticks

(ii) Complete the following table.

Solution:

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

Solution:

Rule for the nth stage
At Stage 1, 6 matchsticks are required.
Each new stage requires 5 additional matchsticks.
Therefore:
Number of matchsticks = 6 + 5(n - 1)
= 6 + 5n - 5
= 5n + 1
Answer: The number of matchsticks required for the nth stage is 5n + 1.

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

Solution:

Matchsticks required for the 15th stage
Using the rule:
5n + 1
For n = 15:
5(15) + 1
= 75 + 1
= 76
Answer: 76 matchsticks are required for the 15th stage.

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

Solution:

Can 200 matchsticks form a stage in this pattern?
For a stage to have 200 matchsticks:
5n + 1 = 200
5n = 199
n = 199/5
n = 39.8
Since the stage number must be a whole number, 39.8 cannot represent a stage.
Answer: No, 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:

Solution:

Given:
p(x) = ax + b
q(x) = cx + d

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

Solution:

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

Since (2, 3) lies on p(x):

3 = 2a + b ...(1)

Since (6, 11) lies on p(x):

11 = 6a + b ...(2)

Subtract equation (1) from equation (2):

11 - 3 = 6a - 2a

8 = 4a

a = 2

Substitute a = 2 in equation (1):

3 = 2(2) + b

3 = 4 + b

b = -1

Therefore:

p(x) = 2x - 1

(ii) The graph of q(x) passes through the point (4, -1).

Solution:

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

Since:

q(x) = cx + d

Substitute x = 4 and q(x) = -1:

-1 = 4c + d

Therefore:

4c + d = -1

or,

d = -1 - 4c

So, from this condition alone, q(x) cannot be uniquely determined. Another condition is required to find the exact values of c and d.

Answer:
p(x) = 2x - 1
For q(x): 4c + d = -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:

From part (i):
p(x) = 2x - 1
Since q(x) is parallel to p(x), both lines have the same slope.
Therefore:
c = 2
So:
q(x) = 2x + d
The graph of q(x) passes through (4, -1).
Substitute x = 4 and q(x) = -1:
-1 = 2(4) + d
-1 = 8 + d
d = -9
Therefore:
q(x) = 2x - 9
Now, find where each line cuts the x-axis.
For p(x) = 2x - 1:
Set p(x) = 0:
2x - 1 = 0
2x = 1
x = 1/2
So, p(x) cuts the x-axis at:
(1/2, 0)
For q(x) = 2x - 9:
Set q(x) = 0:
2x - 9 = 0
2x = 9
x = 9/2
So, q(x) cuts the x-axis at:
(9/2, 0)
Answer:
p(x) = 2x - 1
q(x) = 2x - 9
p(x) meets the x-axis at (1/2, 0).
q(x) meets the x-axis at (9/2, 0).

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, where a > 0
Factor out a:
f(x) = a(x + 1)
For every such function, when x = -1:
f(-1) = a(-1 + 1)
= 0
So, all these linear functions pass through the common point:
(-1, 0)
Also, since a > 0, all the lines have a positive slope and rise from left to right.
Answer: All linear functions of the form f(x) = ax + a, a > 0, pass through the point (-1, 0) and have positive slopes.

Related Resource:

• • Exercise 2.1 – Degree, Coefficients and Constants
  • Exercise 2.2 – Polynomial Values and Linear Equations
  • Exercise 2.3 – Linear Patterns
  • Exercise 2.4 – Linear Growth and Linear Decay
  • Exercise 2.5 – Linear Relationships
  • Exercise 2.6 – Graphs of Linear Relationships