Introduction to Linear Polynomials Class 9 Extra Questions Maths Chapter 2 - #NCSOLVE 📚

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Class 9 Introduction to Linear Polynomials Extra Questions

Extra Questions of Introduction to Linear Polynomials

Class 9 Maths Chapter 2 Extra Questions

Introduction to Linear Polynomials Class 9 Short Question Answer

Question 1.
Write the numerical coefficient and degree of each term of \(\frac{x}{2}\) – 3x² + \(\frac{5}{2}\)x³ — 5x⁴
Solution:

Term	Numerical coefficient	Degree
\(\frac{x}{2}\)	\(\frac{1}{2}\)	1
-3x2	-3	2
\(\frac{5}{2}\)x3	\(\frac{5}{2}\)	3
-5x4	-5	4

Question 2.
If breadth of a rectangle is 13 cm and length is x cm, write the expression for the area of a rectangle. What is the area for x = 5 cm?
Solution:Given: Breadth of a rectangle is 13 cm and length is x cm.
So, area of rectangle = 13 × x = 13x cm² is the required equation for the area of rectangle.
And, the area of rectangle for x = 5 cm is 13x cm² = 13 × 5 = 65 cm²

Question 3.
For what value of c, the linear equation 2x + cy = 8 has equal values of x and v for its solution?
Solution:
The value of c for which the linear equation 2x + cy = 8 has equal values of x and y
i.e., x – y for its solution is
2x + cy = 8
⇒ 2x + cx = 8 [∵ y = x]
⇒ cx = 8 – 2x
∴ c = \(\frac{1}{2}\), x ≠ 0

Question 2.
Let y varies directly as x. If y = 12 when x = 4, then write a linear equation. What is the value of y, when x = 5?
Solution:
y varies directly as x.
⇒ y ∝ x,
∴ y = kx
Substituting y = 12 when x = 4, we get
12 = k × 4
⇒ k = \(\frac{12}{4}\) = 3
Hence, the required equation is y = 3x.
The value ofy when x = 5 is y = 3 × 5 = 15.

Introduction to Linear Polynomials Class 9 Long Question Answer

Question 1.
The taxi fare in a town is ₹ 15 for the first kilometre and ₹ 10 per kilometre for the subsequent distance.
(i) Taking the distance as ‘x’ km and total fare as ₹ y, write a linear equation for the above information.
(ii) Deepak hired a taxi for 10 km and paid ? 200 for it. He asked the-taxi-driver to donate the balance to an orphanage. What amount will the taxi-driver donate to the orphanage?
Solution:
(i) Total distance covered = x km
Total taxi fare for x km = ₹ y
Taxi fare for first km = ₹ 15
Remaining distance = (x – 1) km
Fare for (x – 1) km = ₹ 10 x (x – 1)
∴ Total fare = ₹ 10 (x – 1) + 15
⇒ 10(x – 1)+ 15 = y …(1)
which is the required equation representing the given problem.

(ii) Substituting x = 10 in the equation (1), we get
10(10 – 1)+ 15 = y
⇒ 100 – 10 + 15 = y
⇒ 90 + 15 = y
⇒ y = ₹ 105
Deepak paid ₹ 200.
∴ Balance amount = ₹ 200 – ₹ 105 = ₹ 95
∴ Donation to orphanage = ₹ 95

Introduction to Linear Polynomials Class 9 Case Based Questions

In a one-day cricket match, the organiser decided to donate as much money to an old-age home as the run scored by the first pair of batsman. Ratan and Naval were the opener batsman. They together scored ‘N’ runs such that for every 1 run of Ratan, Naval scored 3 runs.
(i) Set up a linear equation with x and y representing the number of runs of Ratan and Naval respectively. (ii) Find the number of runs scored by Ratan, if Naval scored 180 runs.
(iii) If Ratan’s score is 99, then what amount of money will be donated to the old-age home by the organiser?
OR
If Naval scored 126 runs, then how many runs have Ratan scored?
Solution:
(i) Number of runs scored by Ratan = x
∴ Number of runs scored by Naval = y = 3x
[For every 1 run of Ratan, Naval scored 3 runs]
Total number of runs scored together by the opener batsman = N
∴ x + y = N, where, y = 3x

(ii) Number of runs scored by Naval = 180
∴ y = 180
Since, y = 3x
⇒ 3x = 180
⇒ Runs scored by Ratan = 60

(iii) Number of runs scored by Ratan, x = 99
Number of runs scored by Naval, y = 3x
= 3 × 99 = 297
Now, x + y = N
N = 99 + 297 = 396
Organisers will donate ₹ 396 to the old age home.

OR

Naval scored 3 runs, when ratan scored 1 run
∴ Naval scored 126 runs, when Ratan scored
\(\frac{126}{3}\) = 42 runs.

Introduction to Linear Polynomials Class 9 Competency Based Questions

Question 1.
If the polynomial 2x + 5 is plotted, what will the line’s y-intercept be?
(a) 2
(b) 5
(c) 0
(d) -5
Solution:
Let y = 2x + 5
The y-intercept is the point where the line crosses the y-axis, i.e.,x = 0.
∴ By putting x = 0, in y = 2x + 5 y = 2 × 0 + 5 = 5
So, the y-intercept is 5.
Hence, (b) is the correct answer.

Question 2.
The cost of a journey is represented by the linear function C(d)= 100 + 60d, where d is the distance in km. What is the cost for 5 km?
(a) 300
(b) 400
(c) 500
(d) 600
Solution:
To find the cost for 5 km, substitute d = 5 into the given linear function:
C(d) = 100 + 60d
C(5) = 100 + 60 × 5 = 100 + 300 = 400
So, the cost for 5 km is 400.
Hence, option (b) is the correct answer.

Question 3.
Five years hence a man’s age will be three times his son’s age. Write a linear equation in two variables to represent this statement.
Solution:
Let the present age of the man and his son be x years and y years, respectively.
Five years hence, man’s age = (x + 5) years and son’s age = (y + 5) years
According to the question, we have
x + 5 = 3 (y + 5)
⇒ x + 5 = 3y + 15
⇒ x – 3y = 10
This is a linear equation in two variables which represents the given statement.

Question 4.
Seven times a given two digit number is equal to four times the number obtained by interchanging the digits. Write a linear equation in two variables to represent this statement.
Solution:
Let the digit in the units place be x and the digit in the tens place be y.
Then, the number = 10y + x .
The number obtained on interchanging the digits = 10x + y.
According to the given condition, we have
7 (10y + x) = 4(10x + y)
⇒ 70y + 7x = 40x + 4y
⇒ 66y – 33x = 0
Dividing both sides by 33, we get 2y – x = 0
This is a linear equation in two variables representing the given statement.

Question 5.
A professional photographer buys a high end DSLR camera for ₹ 45,000. Due to the release of newer models and wear and tear, the market value of the camera decreases at a constant rate of ₹ 3,500 per year.
(i) Find the market value of the camera exactly 4 years after the date of purchase.
Solution:
The initial value is ₹ 45,000. Each year it loses ₹ 3,500.
∴ Total decrease = 4 x 3,500 = ₹ 14,000
∴ Value after 4 years = 45,000 – 14,000 = ₹ 31,000

(ii) Create a table showing the value of the camera, v, for time t varying from 0 to 6 years.
Solution:

Time (t in years)	Value (v in ₹)
0 (Purchase)	45,000
1	41,500
2	38,000
3	34,500
4	31,000
5	27,500
6	24,000

(iii) Find a linear expression that relates the value v to the time 1 (in years).
Solution:
v = 45,000 – 3,5001, is the required linear expression that relates the value v to the time t.

(iv) Explain why this specific relationship represents linear decay.
Solution:
This represents linear decay because the value decreases by a fixed, constant amount (₹ 3,500) every year. In a graph, this would result in a straight line with a constant negative slope.

(v) After how many years will the camera’s value drop below ₹ 15,000?
Solution:
Since the camera value v < ₹ 15,000
⇒ 45,000 – 3,500t < 15,000
⇒ 30,000 < 3,500t
⇒ t > \(\frac{30000}{3500}\) = 8.57
Hence, the camera’s value drops below ₹ 15,000 during the 9th year.

Introduction to Linear Polynomials Extra Questions Class 9

Question 1.
For the polynomial \(\frac{x^3+2 x+1}{5}-\frac{7}{2} x^2-x^6\), find
(i) degree of the polynomial.
(ii) coefficient of x³.
(iii) coefficient of x⁶.
(iv) constant term.
Solution:
\(\frac{x^3+2 x+1}{5}-\frac{7}{2} x^2-x^6\) = \(\frac{1}{5} x^3+\frac{2}{5} x+\frac{1}{5}-\frac{7}{2} x^2-x^6\)
(i) 6
(ii) \(\frac {1}{5}\)
(iii) -1
(iv) \(\frac {1}{5}\)

Question 2.
Find the degree of the following polynomials.
(i) \(\frac{\sqrt{5}}{9}\)
(ii) 1 – x² – x³ + 2x⁷
(iii) (x – 1)(x + 1)
(iv) 4x⁴ + 0 . x³ + 0 . x⁵ + 5x + 7
(v) 2x – 3
Solution:
(i) 0
(ii) 7
(iii) 2
(iv) 4
(v) 1

Question 3.
Identify the types of the following polynomials based on terms.
(i) 20y³ + 3y + 8
(ii) 4x
(iii) x² + 5x
Solution:
(i) Trinomial
(ii) Monomial
(iii) Binomial

Question 4.
Identify the types of the following polynomials based on their degree.
(i) √5
(ii) 2x – 3
(iii) 2t³ – 3t² + 5t – 4
(iv) 3y² – 4y + 5
Solution:
(i) Constant
(ii) Linear
(iii) Cubic
(iv) Quadratic

Question 5.
Write each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, and c.
(i) 4x + 3y = 10
(ii) y + 2 = 3x
Solution:
(i) 4x + 3y = 10
4x + 3y – 10 = 0
On comparing with ax + by + c = 0,
we get a = 4, b = 3, and c = -10

(ii) y + 2 = 3x
-3x + y + 2 = 0
On comparing with ax + by + c = 0,
we get a = -3, b = 1 and c = 2

Question 6.
Write each of the following as an equation in two variables.
(i) 3x = -6
(ii) 4y = 8
(iii) x = 5
Solution:
(i) 3x = -6 can be written as 3x + 0y + 6 = 0
(ii) 4y = 8 can be written as 0x + 4y – 8 = 0
(iii) x = 5 can be written as 1x + 0y – 5 = 0

Question 7.
A rope of 250 m is cut into two pieces of different lengths. The longer piece is three times as long as the shorter piece. How long are the two pieces?
Solution:
Let the length of the shorter piece be x.
Then, the length of the longer piece is 3x.
According to the question,
the total length of the rope is 250 m.
⇒ x + 3x = 250
⇒ 4x = 250
⇒ x = 62.5
Thus, the length of the shorter piece, x = 62.5 m,
and the length of the longer piece,
3x = 3 × 62.5 = 187.5 m.
Hence, the lengths of the two pieces are 62.5 m and 187.5 m.

Question 8.
If the length of a rectangle is 5 more than twice its width and its perimeter is 34 cm, what are the dimensions of the rectangle?
Solution:
Let width = x cm and length = (2x + 5) cm.
According to the question,
the perimeter of the rectangle is 34 cm.
⇒ 2[(2x + 5) + x] = 34
⇒ 2(3x + 5) = 34
⇒ x = 4
∴ Width = 4 cm, Length = 13 cm.

Question 9.
Observe the following table.

Find the linear polynomial for this pattern and observe the change in y.
Solution:
Observe the change in y.
When x = 1, y = 5 = 3 × 1 + 2
When x = 2, y = 8 = 3 × 2 + 2 [8 – 5 = 3]
When x = 3, y = 11 = 3 × 3 + 2 [11 – 8 = 3]
When x = 4, y = 14 = 3 × 4 + 2 [14 – 11 = 3]
The value of y increases by 3.
So, the linear polynomial for this pattern is p(x) = 3x + 2.

Question 10.
The number of books read by a student over days is given below. Check whether the pattern is linear.

Solution:
Change from day 1 to 2: 5 – 2 = 3
Change from day 2 to 3, 9 – 5 = 4
Change from day 3 to 4, 14 – 9 = 5
The given pattern is not linear.

Question 11.
Suppose the length of a rectangle is 15 cm. Find the area if the breadth is (i) 14 cm, (i) 9 cm, (ii) 6 cm. Find the linear pattern representing the area of the rectangle.
Solution:
Given, length = 15 cm and let breadth = b cm
Now, area of rectangle = 15 × b
y = 15b
(i) b = 14 ⇒ y = 210
(ii) b = 9 ⇒ y = 135
(iii) b = 6 ⇒ y = 90

Question 12.
A book contains 420 pages. Aman reads 18 pages each day. Find the number of pages remaining after 10 days. Also, represent this situation by a linear expression.
Solution:
y = -18x + 420
For x = 10,
y = -18(10) + 420 = 240

Question 13.
Study the graph given below and answer the questions that follow.

(i) Find the point on the X-axis.
(ii) Find the point on the Y-axis.
(iii) Find the area of the triangle formed by the line and the axes.
Solution:
(i) (3, 0)
(ii) (0, 3)
(iii) \(\frac {9}{2}\) sq unit

Question 14.
Draw the graph of y = \(\frac {2}{3}\)x, y = x, and y = \(\frac {3}{2}\)x by selecting suitable points on these lines.
Solution:

Question 15.
The path of a train A is given by the equation 2x – 3y = 17 and the path of another train B is given by the linear equation 4x – 6y = 18. Represent the above situations graphically on the same Cartesian plane and write your observation.
Solution:

The two lines representing the paths of trains are parallel.

Question 16.
Find the equation of a line whose gradient is \(\frac{1}{\sqrt{3}}\) and y-intercept is 3.
Solution:
Put m = \(\frac{1}{\sqrt{3}}\) and c = 3 in the equation y = mx + c.
√3y = x + 3√3

Question 17.
Suppose a plant has height 2.5 feet and it grows by 0.6 feet each month.
(i) Find the height after 6 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.
Solution:
(i) h = 0.6t + 2.5
(ii)

(iii) This represents linear growth because the height increases by the same amount, 0.6 feet every month.

Question 18.
The initial amount of water in a tank is 500 L. Every hour, 40 L of water is added to the tank.
(i) Find the amount of water after 6 h.
(ii) Make a table of values for t varying from 0 to 10 h and show how the amount of water W increases every hour.
(iii) Find an expression that relates W and t, and explain why it represents linear growth.
Solution:
(i) W = 40t + 500
(ii)

(iii) This represents linear growth because the amount increases by the same quantity, 40 L, every hour.

Question 19.
A telecom company charges ₹ 800 for a certain recharge scheme. This prepaid balance is reduced by ₹ 20 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.
Solution:
(i) b(x) = -20x + 800
This represents linear decay because the balance decreases by the same amount, ₹ 20, every day.
(ii) 40 days
(iii)

Introduction to Linear Polynomials Class 9 Extra Questions for Practice

Multiple Choice Questions

Question 1.
What is the slope of the line represented by the equation y = 2x + 7?
(a) 7
(b) 2
(c) -2
(d) 0

Question 2.
Which of the following represents a linear growth pattern?
(a) y = 2x + 5
(b) y = x² + 5
(c) y = 5x³ + 3x + 2
(d) y = x²

Question 3.
Which of the following is the coefficient off2 in (4 + 4f2) (3/2 – 5)?
(a) -8
(b) 8
(c) 12
(d) 20

Assertion-Reason Questions

Two statements are given, one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer from the options (a), (b), (c), and (d) given below.
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(c) Assertion (A) is true, but Reason (R) is false.
(d) Assertion (A) is false, but Reason (R) is true.

Question 1.
Assertion (A): Linear decay describes a pattern where a quantity increases by a fixed amount over equal intervals.
Reason (R): In the equation y = ax + b, a negative value of ‘a’ represents a decreasing rate.

Question 2.
Assertion (A): The lines y = 3x + 2 and y = 3x + 5 are parallel to each other.
Reason (R): Two lines are parallel if they have the same slope a but different y-intercepts b.

Short Answer Type Questions

Question 1.
Consider the expression P(x) = x(x + 4) – x² + 7. A student claims this is a quadratic polynomial. Prove whether the student is correct or incorrect by simplifying the expression.

Question 2.
The length of a rectangle is (2x + 3) units and its breadth is 5 units. Express the perimeter as a linear polynomial in terms of x. If the perimeter is 46 units, find the value of x.

Question 3.
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?

Question 4.
In the algebraic expression 7x + 3y + 5z + 8. Identify the variables and the constant term. If x, y, and z all represent the same value k, simplify the expression into a univariate linear polynomial.

Question 5.
Draw the graphs of y = x, y = \(\frac{3}{4}\) x, y = 4x by selecting suitable points on these lines.

Question 6.
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) Find an expression that relates h and t, and explain why it represents linear growth.

Long Answer Type Questions

Question 1.
The following observed values of x andy are thought to satisfy a linear equation.

(i) Identify other points on the line by completing the given table.
(ii) Write the linear equation.
(iii) Draw the graph using the values of x, y given in the above table. At what points, the graph of the linear equation cuts the x-axis and the y-axis?

Question 2.
Draw the graphs of the following sets of lines. In each case, reflect on the role of ‘a’ and “b’.
(i) y = x, y = 2x, y = 3x
(ii) y = – 6x, y = – 3x, y = – x
(iii) y = 3x, y = -3x
(iv) y = 2x – 1, y = 2x, y = 2x + 1
(v) y = -2x – 5, y = -2x, y = 2x + 5

The post Introduction to Linear Polynomials Class 9 Extra Questions Maths Chapter 2 appeared first on Learn CBSE.

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