Students often refer to and Part 1 Class 9 Maths Chapter 2 Introduction to Linear Polynomials Solutions to verify their answers.

Introduction to Linear Polynomials Class 9 Solutions

Class 9 Ganita Manjari Chapter 2 Solutions

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

Think and Reflect (NCERT Textbook Page No. 17)

Question 1.
Can you identify the terms, variables and coefficients of this algebraic expression?
Solution:
Given algebraic expression = 200l + 160w + 50lw
Terms: 200l, 160w and 50lw
Variables: l and w
Coefficients: 200 is of variable l, 160 is of variable w, and 50 is of variable lw.

Question 2.
How is it different from the algebraic expression in Example 1?
Solution:
The algebraic expression given in the example is 4x + 5y + 3.
It is a linear polynomial in variables x and y.
While the algebraic expression given in Example 2 is 200l + 160w + 50lw, which has two variables, l and w, and the expression is not a linear polynomial because it contains a product term of two variables, lw.

Think and Reflect (NCERT Textbook Page No. 17)

Question 1.
Can you identify the terms, variables, and coefficients of this algebraic expression?
Solution:
Yes, the given algebraic expression is 10x – x²
Here, Terms: 10x and -x²
Variables: x
Coefficients: 10 of x and -1 of -x²

Question 2.
Can you point out any similarity or difference between the algebraic expressions obtained in Examples 1 and 3?
Solution:
Similarity: Both expressions use the same variable (x).
Difference: The algebraic expression given in Example 1 is 4x + 5y + 3, which is a linear polynomial with variables x and y.
While the algebraic expression given in Example 3 is 10x – x², which is a quadratic expression with variable x, because it contains the term x × x.

Think and Reflect (NCERT Textbook Page No. 19)

Find the perimeter of squares with sides 1 cm, 1.5 cm, 2 cm, 2.5 cm and 3 cm. What will happen to the perimeters if the sides increase by 0.5 cm?
Solution:
We know the perimeter of a square of side a cm = 4a cm
∴ Perimeter of square of side 1 cm = 4 × 1 = 4 cm
Perimeter of a square of side 1.5 cm = 4 × 1.5 = 6 cm
Perimeter of a square of side 2 cm = 4 × 2 = 8 cm
Perimeter of a square of side 2.5 cm = 4 × 2.5 = 10 cm
Perimeter of a square of side 3 cm = 4 × 3 = 12 cm
Observation: Every time the side increases by 0.5 cm, the perimeter increases by exactly 2 cm.
This constant rate of change a hallmark of linear polynomials.

Think and Reflect (NCERT Textbook Page No. 19)

If a player paid ₹ 750, how many matches did he play?
Solution:
Here, the polynomial given is 200 + 50m, where m is the number of matches.
If a player paid ₹ 750, we can find m using a simple linear equation as follows:
200 + 50m = 750
⇒ 50 m =750 – 200 = 550
⇒ m = \(\frac{550}{50}\) = 11 matches
Hence, the player played 11 matches.

Think and Reflect (NCERT Textbook Page No. 20)

Question 1.
We have learnt that to evaluate the value of an algebraic expression, we substitute a value of the variable in the given expression. Consider example 3, where the wire is bent to form a rectangle. Here, the area of the rectangle, 10x -x² is a function of x. Can you interpret this as an input-output process? What value does the expression take when x = 6 cm?
Solution:
Let f(x) = 10x – x².
We can interpret this as an input-output process.
The input is the value of x, which represents the side of the rectangle (in cm).
The output is the area of the rectangle, which depends on the value of x.
On putting x = 6, we get f(6) = 10(6) – (6)² = 60 – 36 = 24 cm²
Hence, the area of the rectangle, when x = 6 cm, is 24 cm².

Think and Reflect (NCERT Textbook Page No. 22)

Predict the number of squares in the next three stages of the pattern and write the sequence of numbers up to Stage 7 of the pattern.
Solution:
Do it Yourself

Think and Reflect (NCERT Textbook Page No. 22)

Using the expression 2n – 1, can you find out how many tiles will be there in the 15th stage and the 26th stage of the pattern? Also, which stage will contain 21 tiles and 47 tiles?
Solution:
Do it Yourself

Think and Reflect (NCERT Textbook Page No. 23)

What amount will be left on the 15th day? How many days will it take for the entire amount to be spent?
Solution:
Do it Yourself

Think and Reflect (NCERT Textbook Page No. 23)

For how many km will the fare be ₹ 130?
Solution:
Do it Yourself

Think and Reflect (NCERT Textbook Page No. 24)

What is the cost for travelling 15 km? For how many kilometres will the cost of the journey be ₹ 700?
Solution:
Do it Yourself

Think and Reflect (NCERT Textbook Page No. 25)

What will be the height of the water at the end of 5 months?
Solution:
Do it Yourself

Think and Reflect (NCERT Textbook Page No. 27)

Can you guess what the numbers 20 and 150 in the equation y = 20x + 150 represent?
Solution:
Do it Yourself

Think and Reflect (NCERT Textbook Page No. 28)

Identify other points on the line by completing the following table.

Solution:
Do it Yourself

Think and Reflect (NCERT Textbook Page No. 33)

Question 1.
Differentiate between the graphs of the equations y = 3x + 1 and y = -3x + 1.
Solution:
Given linear equations are y = 3x + 1 and y = -3x + 1.
For y = 3x + 1,
When x = 0, then y = 3(0) + 1 = 1
When x = 1, then y = 3(1) + 1 = 4
When x = 2, then y = 3 (2) + 1 = 7
So, we have the following table to draw the graph.
Here, we have three points A(0, 1), B (1, 4) and C(2, 7).
Now, plot these points on the graph and join them with a straight line.

Hence, AC is the required straight line.
For y = -3x + 1,
When x = 0, then y = -3(0) + 1 = 1
When x = 1, then y = -3(1) + 1 = -2
When x = 2, then y = -3(2) + 1 = -5
So, we have the following table to draw the graph.
Here, we have three points D(0, 1), E (1, -2) and F(2, -5).
Now, plot these on the graph and join them with a straight line.

Here, DF is the required straight line.

Think and Reflect (NCERT Textbook Page No. 35)

Does this help you to conclude anything about the linear equation y = ax + b when a is fixed but b varies?
Solution:
Do it Yourself

Ex 2.1 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 2.1 Solutions

Exercise 2.1 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 2.1 Solutions

Question 1.
Find the degrees of the following polynomials.
(i) 2x² – 5x + 3
Solution:
We have, 2x² – 5x + 3
Here, the highest power of x is 2, so its degree is 2.

(ii) y³ + 2y – 1
Solution:
We have, y³ + 2y – 1
Here, the highest power of y is 3, so its degree is 3.

(iii) -9
Solution:
We have -9
Here, the highest power of the variable is 0 (since it’s a constant), so its degree is 0.

(iv) 4z – 3
Solution:
We have, 4z-3
Here, the highest power of z is 1, so its degree is 1.

Question 2.
Write polynomials of degrees 1, 2 and 3.
Solution:
(i) A polynomial of degree 1,
2x + 3
This is a linear polynomial, where the highest power of x is 1.

(ii) A polynomial of degree 2,
x² – 4x + 5
This is a quadratic polynomial, where the highest power of x is 2.

(iii) A polynomial of degree 3, x³ + 2x² – 3x + 1
This is a cubic polynomial, where the highest power of x is 3.

Question 3.
What are the coefficients of x and x3 in the polynomial x⁴ – 3x³ + 6x² – 2x + 7?
Solution:
We have the polynomial is x⁴ – 3x³ + 6x² – 2x + 7.
Here, the coefficient of x² is 6 (the term associated with x²) and the coefficient of x³ is -3 (the term associated with x³).

Question 4.
What is the coefficient of z in the polynomial 4z³ + 5z² – 11?
Solution:
We have the polynomial is 4z³ + 5z² – 11. The term associated with z is not present in the polynomial, so the coefficient of z is 0.

Question 5.
What is the constant term of the polynomial 9x³ + 5x² – 8x – 10?
Solution:
We have the polynomial is 9x³ + 5x² – 8x – 10.
We know that the constant term is the term that does not involve any variable. In this case, it is -10.
Thus, the constant term is -10.

Ex 2.2 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 2.2 Solutions

Exercise 2.2 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 2.2 Solutions

Question 1.
Find the value of the linear polynomial 5x – 3, if
(i) x = 0
(ii) x = -1
(iii) x = 2
Solution:Given polynomial, p(x) = 5x – 3
(i) When x = 0,
p( 0) = 5(0) – 3 = 0 – 3 = -3
So, at x = 0, p(x) = -3

(ii) When x =-1,
p(-1) = 5(-1) – 3 = -5 – 3 = -8
So, at x = -1, p(x) = -8

(iii) When x = 2,
p(2) = 5(2) – 3 = 10 – 3 = 7
So, at x = 2, p(x) = 7

Question 2.
Find the value of the quadratic polynomial 7s² – 4s + 6, if
(i) s = 0
(ii) 5 = -3
(iii) s = 4
Solution:
Given polynomial, p(x) = 7s² – 4s + 6
(i) When s = 0,
p(0) = 7(0)² – 4(0) + 6 = 0 + 0 + 6 = 6
So, at s = 0, p(s) = 6

(ii) When s = -3,
p(-3) = 7(—3)² – 4(—3) + 6
= 7(9) + 12 + 6
= 63 + 12 + 6 = 81
So, at 5 = -3, p(s) = 81

(iii) When s = 4,
p(4) = 7(4)² – 4(4) + 6
= 7(16) – 16 + 6 = 112 – 16 + 6 = 102
So, at s = 4, p(s) = 102

Question 3.
The present age of Said’s mother is three times Said’s present age. After 5 yr, their ages will add upto 70 yr. Find their present ages.
Solution:
Let Salil’s present age be x.
Then, Salil’s mother’s present age = 3x.
After 5 years, Salil’s age = x + 5
and his mother’s age = 3x + 5.
Their total age after 5 years = (x + 5) + (3x + 5) = 70.
4x + 10 = 70
⇒ 4x = 60
⇒ x = 15
Thus, Salil’s present age is 15 years, and his mother’s present age is 3 × 15 = 45 years.

Question 4.
The difference between two positive integers is 63. The ratio of the two integers is 2 : 5. Find the two integers.
Solution:
Let the two integers be 2x and 5x, where x is a common factor.
According to the question,
5x – 2x = 63
3x = 63
x = \(\frac{63}{3}\) = 21
So, the first integer, 2x = 2 × 21 = 42
and the second integer, 5x = 5 × 21 = 105.
Hence, the two integers are 42 and 105.

Question 5.
Ruby has 3 times as many two rupee coins as she has five rupee coins. If she has a total of 188, how many coins does she have of each type?
Solution:
Let the number of five rupee coins be x.
Then, the number of two rupee coins is 3x.
According to the question,
2(3x) + 5x = 88
⇒ 6x + 5x = 88
⇒ 11x = 88
⇒ x = \(\frac{88}{11}\) = 8
Thus, the number of five-rupee coins is x = 8
and the number of two-rupee coins is 3x = 3 × 8 = 24.
Hence, Ruby has 8 five-rupee coins and 24 two-rupee coins.

Question 6.
A farmer cuts a 300 feet fence into two pieces of different size. The longer piece is four times as long as the shorter piece ? long are the two pieces?
Solution:
Let the length of the shorter piece be x.
Then, the length of the longer piece is 4x.
According to the question, the total length of the fence is 300 feet.
x + 4x = 300
⇒ 5x = 300
⇒ x = \(\frac{300}{5}\) = 60
Thus, the length of the shorter piece is x = 60 feet and the length of the longer piece is 4x = 4 × 60 = 240 feet.

Question 7.
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?
Solution:
Let the width of the rectangle be x.
Then, the length of the rectangle is 2x + 3.
We know that
The perimeter of rectangle = 2 (Length + Width)
On substituting the expression for length and width, we get
2 [(2x + 3) + x] = 24
⇒ 2(3x + 3) = 24
⇒ 6x + 6=24
⇒ 6x = 24 – 6
⇒ 6x = 18
So, the width of the rectangle is x =3 cm.
Now, substitute x =3 into the expression for the length.
Length = 2x + 3 = 2(3) + 3 = 6 + 3 = 9 cm
Hence, the dimensions of the rectangle are 3 cm by 9 cm.

Ex 2.3 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 2.3 Solutions

Exercise 2.3 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 2.3 Solutions

Question 1.
A student has ₹ 500 in her savings bank account. She gets ₹ 150 every month as pocket money. How much money will she have at the end of every month from the second month onwards? Find a linear expression to represent the amount she will have in the nth month.
Solution:
Given, initial amount (c) = ₹ 500
and rate of change (m) = 150 per month.
Now, linear expression
y = 150n + 500
Where, y is the total amount and n is the number of months.
To find how much money she will have from the second month onwards, we substitute the values of n.
At the end of the 2nd month (n =2)
y = 150(2) + 500 = 300 + 500 = ₹ 800
At the end of the 3rd month (n =3)
y = 150(3) + 500 = 450 + 500 = ₹ 950
At the end of the 4th month (n = 4)
y = 150(4) +500 = 600 + 500 = ₹ 1100

Question 2.
A rally starts with 120 members. Each hour, 9 members drop out of the group. How many members will remain after 1, 2, 3,… hours?
Find a linear expression to represent the number of members at the end of the nth h.
Solution:
Given, initial amount (c) =120 members and rate of change (m) = – 9 per hour.
Now, polynomial expression y =-9n + 120
Where, y is the remaining members and n is the number of hours.
Members remaining after each hour
After 1 h (n = 1)
y = -9(1) + 120 = 111 members After 2h (n = 2)
y = -9(2) + 120 = 102 members After 3h(n = 3)
y = -9(3) + 120 = 93 members

Question 3.
Suppose the length of a rectangle is 13 cm. Find the area if the breadth is (i) 12 cm, (ii) 10 cm and (iii) 8 cm. Find the linear pattern representing the area of the rectangle.
Solution:
Given, length (l) = 13 cm
We know that the area of a rectangle,
y = l × b
y = l3b, [∵ l = 13]
Where, y is the area and b is the breadth.
(i) For b = 12,
y = 13(12) = 156 cm²
(ii) For b = 10
y = 13(10) = 130 cm²
(iii) For b = 8
y = 13(8) = 104 cm²
Hence, the linear pattern is y = 13b.

Question 4.
Suppose the length of a rectangular box is 7 cm and breadth is 11 cm. Find the volume if the height is
(i) 5 cm, (ii) 9 cm and (iii) 13 cm.
Find the linear pattern representing the volume of the rectangular box.
Solution:
Given, length (l) = 7 cm and breadth (b) = 11 cm.
Now, volume of rectangular box,
V = l × b × h
⇒ V = 7 × 11 × h
⇒ V = 77h
Where, V is the volume and h is the height.
(i) For h =5,
V = 77(5) = 385 cm³

(ii) For h =9,
V = 77(9) = 693 cm³

(iii) For h = 13,
V = 77(13) = 1001 cm³
Hence, the linear pattern is V = 77h.

Question 5.
Sarita is reading a book of 500 pages. She reads 20 pages every day. How many pages will be left after 15 days? Express this as a linear pattern.
Solution:
Given, total pages (c) = 500
and rate of change (m) = -20 pages per day.
Now, polynomial expression y = -20x + 500
Where, y is the remaining pages and x is the number of days.
After 15 days (x= 15), the remaining pages,
y = -20(15) + 500 = -300 + 500 = 200
Hence, the pages left after 15 days are 200.

Ex 2.4 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 2.4 Solutions

Exercise 2.4 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 2.4 Solutions

Question 1.
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) 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:
Given, initial height (c) = 1.75 feet
and rate of growth (m) = 0.5 feet per month.
Now, polynomial expression
h = 0.51 + 1.75 …(i)
where h is the height and f is the number of months.

(i) After 7 months (t =7), the height,
h = 0.5(7) + 1.75 = 3.5 + 1.75 = 5.25 feet [using Eq. (i)]

(ii) From Eq. (i), for t = 0 to 10, the values are

(iii) The linear expression is h= 0.5f + 1.75 and it represents linear growth because the height increases by a constant amount of 0.5 feet every month.

Question 2.
A mobile phone is bought for ₹ 10000. Its value decreases by ₹ 800 every year.
(i) Find the value of the phone after 3 yr.
(ii) Make a table of values for t varying from 0 to 8 yr 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.
Solution:
Given, initial value (c) = ₹ 10000
and rate of change (m) = ₹ 800 per year.
Now, polynomial expression
v = -800f +10000 …(i)
Where, v is the value of the phone and t is the number of years.
(i) After 3 yr (f = 3), the value,
v = -800(3) + 10000 = -2400 + 10000 = 7600
[using Eq. (i)]

(iii) The linear expression is v = -8001 + 10000 and it represents linear decay because the value decreases by a constant amount of ₹ 800 every year.

Question 3.
The initial population of a village is 750. Every year, 50 people move from a nearby city to the village.
(i) Find the population of the village after 6 yr.
(ii) Make a table of values for 1 varying from 0 to 10 yr and show how the population, P increases every year.
(iii) Find an expression that relates P and 1, and explain why it represents linear growth.
Solution:
Given, initial population (c) = 750
and rate of change (m) = 50 people per year.
Now, polynomial expression
P = 501 + 750 …(i)
Where, P is the population and 1 is the number of years.

(i) After 6 yr (1 = 6), the population,
P = 50(6) + 750 = 300 + 750 = 1050 [using Eq. (i)]

(ii) From Eq. (i), for t = 0 to 10, the values are

(iii) The linear expression is P = 501 + 750 and it represents linear growth because the population increases by a constant number of 50 people every year.

Question 4.
A telecom company charges ₹ 600 for a certain recharge scheme. This prepaid balance is reduced by ₹ 15 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) Given, initial balance (c) = ₹ 600
and rate of change (m) = ₹ 15 per day.
Now, polynomial expression
b(x) =-15x + 600
Where, b(x) is the remaining balance and x is the number of days.
It represents linear decay because the balance decreases by a constant amount of ₹ 15 every day.

(ii) When the balance becomes zero.
Put b(x) = 0
⇒ —15x + 600 = 0
⇒ 15x = 600
⇒ x = \(\frac{600}{15}\) = 40
So, the balance runs out after 40 days.

(iii) For x = 1 to 10, the values arc

Hence, the linear expression is b(x) = -15x + 600 and it represents linear decay because the balance decreases by a constant amount of ₹ 15 every day.

Ex 2.5 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 2.5 Solutions

Exercise 2.5 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 2.5 Solutions

Question 1.
A learning platform charges a fixed monthly fee and an additional cost per digital learning module accessed. A student observes that when she accessed 10 modules, her bill was ? 400. When she accessed 14 modules, her bill was 1500. If the monthly bill y depends on the number of modules accessed, x, according to the relation y = ax +b, find the values of a and b.
Solution:
Given, monthly bill depends on x according to y = ax+ b.
Also, when x = 10, y = 400
10a + b = 400 … (i)
and when x = 14, y = 500
14a + b =500 …(ii)
From Eq. (i), we get
b = 400 – 10a …(iii)
On substituting b from Eq. (iii) in Eq. (ii),
we get 14a+ (400 – 10a) = 500
⇒ 14a + 400 – 10a = 500
⇒ 4a + 400 = 500
⇒ 4a = 100
⇒ a = 25
On substituting a = 25 in Eq. (iii), we get
b = 400-10(25)
⇒ b = 400 – 250 = 150
Hence, the values are a = 25 and b = 150.

Question 2.
A gym charges a fixed monthly fee and an additional cost per hour for using the badminton court. A student using the gym observed that when she used the badminton court for 10 hr, her bill was ₹ 800. When she used it for 15 hr, her bill was ₹ 1100. If the monthly bill y depends on the hours of use of the badminton court, x, according to the relation y=ax+b, find the values of a and b.
Solution:
Given, monthly bill depends on x according to y = ax+ b.
Also, when x = 10, y = 800
On substituting b from Eq. (iii) in Eq. (ii),
we get 15a + (800 – 10a) = 1100
⇒ 15a+ 800-10a =1100
⇒ 5a+ 800 =1100
⇒ 5a = 300
⇒ a = 60
On substituting a = 60 in Eq. (iii), we get
b = 800 – 10(60)
⇒ b = 800 – 600 = 200
Hence, the values are a = 60 and b = 200.

Question 3.
Consider the relationship between temperature measured in degrees Celsius (°C) and degrees Fahrenheit (°F), which is given by °C = a° F + b.
Find a and b, given that ice melts at 0 degrees Celsius and 32 degrees Fahrenheit, and water boils at 100 degrees Celsius and 212 degrees Fahrenheit.
[Hint When °C = 0, °F = 32 and when °C = 100, °F = 212 Use this information to find a and b, and thus linear relationship between °C and °F.]
Solution:
Given, the relation between Celsius and Fahrenheit °C =a°F + b ….(i)
Now, at °C = 0 and °F = 32 From Eq. (i), we get
0 = a × 32 +b
=> b = -32a …………(ii)
Now, at °C = 100 and °F = 212
From Eq. (i), we get
100 = a × 212 + (-32a)
[putting the value of b from Eq. (ii)]
⇒ 100 = 180a
⇒ a = \(\frac{100}{180}=\frac{5}{9}\)

On putting the value of a in Eq. (ii), we get
b = -32 × \(\frac{5}{9}=\frac{-160}{9}\)
a = \(\frac{5}{9}\) and b = \(\frac{-160}{9}\)

Hence, the linear relationship is
°C = \(\frac{5}{9}\) (°F – 32)

Ex 2.6 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 2.6 Solutions

Exercise 2.6 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 2.6 Solutions

Question 1.
Draw the graphs of the following sets of lines. In each case, reflect on the role of a and b.
(i) y = 4x, y = 2x, y = x
(ii) y = -6x, y = -3x, y = -x
(iii) y = 5x, y = -5x
(iv) y = 3x – 1, y = 3x, y = 3x + 1
(v) y = 2x – 1, y = -2x, y = 2x + 1
Solution:
(i) Given linear equations are y = 4x, y = 2x and y = x.
For y = 4x
When, x = 0 then y = 4(0) = 0
When, x = 1 then y – 4(1) = 4
When, x = 2 then y = 4(2) = 8
So, we have the following table to draw the graph.

Here, we have three points D(0, 1). E (1, -2) and F(2, -5).
Now, plot these points on the graph and join them by a straight line.
Thus, we get the straight line AC which represents the required graph.

For y = 2x
When, x = 0 then y = 2(0) = 0
When, x = 1 then y = 2(1) = 2
When, x = 2 then y = 2(2) = 4
So, we have the following table to draw the graph.

Here, we have three points A(0,0), E(1, 2) and F(2, 4).
Now, plot these points on the graph and join them by a straight line. Thus, we get the straight line AF, which represents the required graph.
For y = x
When, x = 0 then y = 0
When, x = 1 then y = 1
When, x = 2 then y = 2
So, we have the following table to draw the graph.

Here, we have three points A(0, 0), H(l, 1) and/(2,2).
Now, plot these points on the graph and join them by a straight line.
Thus, we get the straight line AI, which represents the required graph.

All the lines pass through the origin since b = 0. Also, the value of a changes the slope, so y = 4x is steepest then y =2x and then y = x.

(ii) Do same as Part (i).
All the lines pass through the origin, since b = 0. Also, the values of a are negative, so all the lines are decreasing. Further, larger |a| gives a steeper line, so y = -6x is steepest then y = -3x and then y = -x.

(iii) Do same as Part (i).
Both lines pass through the origin, since b = 0. Also, the values of a are equal in magnitude but opposite in sign, so one line is increasing (y = 5x) and the other is decreasing (y = -5x). Thus, the two lines are symmetric about the Y-axis.

(iv) Given linear equations are y = 3x -1, y = 3x and y =3x +1.
For y = 3x – 1
When, x = 0 then y = 3(0) – 1 = -1
When, x = 1 then y = 3 (1) – 1 = 2
When, x =2 then y = 3(2) – 1 = 5
So, we have the following table to draw the graph.

Here, we have three points A(0, -1), B (1, 2) and C (2,5).
Now, plot these points on the graph and join them by a straight line.
Thus, we get the straight line AC, which represents the required graph of given linear equation.
For y = 3x
When, x = 0 then y = 0
When, x = 1 then y = 3
When, x = 2 then y = 6
So, we have the following table to draw the graph.

Here, we have three points O (0, 0), E(1, 3) and F(2, 6).
Now, plot these points on the graph and join them by a straight line.
Thus, we get the straight line OF, which represents the required graph of given linear equation.
For y = 3x + 1
When, x = 0 then y = 1
When, x = 1 then y = 4
When, x = 2 then y = 7
So, we have the following table to draw the graph.

Here, we have three points G(0,1), H (1, 4) and I (2,7).
Now, plot these points on the graph and join them by a – straight line.
Thus, we get the straight line GI, which represents the required graph of given linear equation.

All the lines have the same slope (a = 3), so they are parallel to each other. The value of b changes the 7-intercept, so the lines shift upward or downward but remain parallel.

(v) Do same as Part (iv).
In the first two equations, the coefficient of x is the same
i. e. -2 and hence the slopes are equal. Therefore, these two lines are parallel to each other. Also, their constant b are different (-3 and 0), so they are distinct parallel lines. Now, in the third equation y = 2x + 3, the coefficient of x is different i.e. 2, so its slope is different from the first two lines. Hence, this line intersects both of them.

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

Introduction to Linear Polynomials End of Chapter Exercise Solutions

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 in x is of the form ax3 +bx2 +cx + d, wherea * 0.
Given that the coefficient of x2 is -7, so b = -7.
Takinga = 2, c = 5 and d = — lone such polynomial is 2×3 -lx2 +5x-l.

Question 2.
Find the values of the following polynomials at the indicated values of the variables.
(i) 5x² – 3x + 7 if x = 1
(ii) 4t³ – t² + 6 if t = a
Solution:
(i) Given, 5x² – 3x + 7 and x = 1
On substituting x = 1, we get
5(1)² – 3(1) + 7 = 5 – 3 + 7 = 9
Hence, the value is 9.

(ii) Given, 4t³ – t² + 6 and t = a
On substituting t = a, we get
4(a)³ – (a)² + 6 = 4a³ – a² + 6
Hence, the value is 4a³ – a² + 6.

Question 3.
If we multiply a number by \(\frac{5}{2}\) and add \(\frac{2}{3}\) to the product, we get \(\frac{-7}{12}\). Find the number.
Solution:
Let the number hex.
According to the question.
\(\frac{5}{2} x+\frac{2}{3}=-\frac{7}{12}\) …(i)
From Eq. (i), we get
\(\frac{5}{2} x=-\frac{7}{12}-\frac{2}{3}\) ..(ii)
On simplifying RHS, we get

Hence, \(\frac{1}{2}\) is the required number.

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 two numbers be x and y, where x = 5y.
ATQ, x + 21 = 2(y + 21)
Substituting x = 5y:
⇒ 5y + 21 =2(y + 21)
⇒ 5y + 21 = 2_y + 42
⇒ 3y = 21
⇒ y = 7
So, x = 5y = 35.
The numbers are 35 and 7.

Question 5.
If you have ₹ 800 and you save ₹ 250 every month, find the amount you have after (i) 6 months (ii) 2 yr. Express this as a linear pattern.
Solution:
Given, initial amount = 800
and saving per month = 250.
The amount after n months can be written as
A = 800 + 250n …(i)
For Eq. (i), when n = 6
A = 800 + 250 × 6
⇒ A = 800 + 1500 =2300
For Eq. (i), 2 yr = 24 months,
so when n = 24 A = 800 + 250 × 24
⇒ A = 800+ 6000 = 6800
Hence, the required amounts are 2300 and 6800, and the linear pattern is A = 800 + 250n.

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 bey.
Then, original number = 10x + y ….(i)
Given, digits differ by 3.
So, x – y = 3.

On interchanging the digits, we get
New number = 10y + x

According to the question, …(ii)
(10x + y) + (10y + x) = 143 …(iii)

From Eq. (iii), we get
11x + 11y = 143
⇒ x + y = 13

From Eq. (ii), we get …(iv)
x = y + 3 …(v)

On substituting x = y + 3 in Eq. (iv), we get
y + 3 + y = 13
⇒ 2y + 3 = 13
⇒ 2y = 10
⇒ y = 5
Now, on putting y = 5 in Eq. (v), we get x = 5 + 3 = 8
Now, original number = 10x + y = 10 × 8 + 5 = 85
and new number = 10y + x = 10 × 5 + 8 = 58.
Hence, the required numbers are 85 and 58.

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 7-axis.
(i) y = —3x + 4
(ii) 2y = 4x + 7
(iii) 6y = 6x – 10
(iv) 3y = 6x – 11
Are any of the lines parallel?
Solution:
(i) Given, equation of the line is
y = -3x + 4 …(i)
Plot the points A(0, 4) and B(2, -2) and join them to get the straight lineAB.

On comparing the given line with y = mx + c, we get
Slope, m = -3 and y-intercept, c = 4 and the line cuts the Y-axis at the point A(0, 4).

(ii) Given equation of the line is 2y = 4x + 7 …………. (i)
Now, 2y = 4x + 7 or y = 2x + \(\frac{7}{2}\)
Now, table for y = 2x + \(\frac{7}{2}\) is

Plot the points C(0, \(\frac{7}{2}\)) and D(2, \(\frac{15}{2}\)), and join them to get the straight line CD.
On comparing the given line with y = mx + c, we get
Slope, m = 2 and y-intercept, c = \(\frac{7}{2}\)
and the line cuts the 7-axis at the point A(0, \(\frac{7}{2}\))

(iii) Do same as Part (ii).
\(\frac{6}{5}\), -2 and (0, -2)

(iv) Do same as Part (ii).
Ans 2, \(\frac{-11}{3}\) and (0, \(\frac{-11}{3}\))
Since, the slopes of the lines are —3, 2, \(\frac{6}{5}\), 2, it is clear that the lines corresponding to parts (ii) and (iv) have equal slopes.
Hence, the lines 2y = 4x + 7 and 3y = 6x -11 are parallel.

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 = \(\frac{9}{5}\)(x – 273) + 32.
(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.
Solution:
Given relation between temperature,
y = \(\frac{9}{5}\)(x – 273) + 32 … (i)
Here, x represents temperature in Kelvin (K) and y represents temperature in Fahrenheit (°F).
For Eq. (i). when x =313k
y = \(\frac{9}{5}\)(313 – 273) + 32
y = \(\frac{9}{5}\)(40) + 32
y = 72 + 32 = 104
Thus, the temperature is 104 °F.
For Eq. (ii), when y =158
158 = \(\frac{9}{5}\)(x – 273) +32
158 – 32 = \(\frac{9}{5}\)(x – 273)
126 = \(\frac{9}{5}\)(x – 273)
126 x \(\frac{5}{9}\) = x – 273
⇒ 70 = x -273
⇒ x =70 + 273 = 343
Thus, 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:
Given the work done by a body is equal to the product of force and distance, therefore w =Fx d
Here, the constant force F = 3 units, hence w = 3d.
So, the linear equation in two variables is w = 3d
Now, to draw the graph, take some values of d and find corresponding w. .
When, d = 0 then w = 3(0) = 0
When, d = 1 then w = 3(1) = 3
When, d = 2 then w = 3(2) = 6
So, we have the following table to draw the graph.

Here, we have three points 0(0,0), A (1, 3) and 5(2, 6).
Now, plot these points on the graph and join them by a straight line.

Thus, we get the straight line OB, which represents the required graph of given linear equation. Also, from the graph, it is clear that, when d = 2 then w = 6.
Hence, the work done is 6 units, which is verified from the graph.

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).
(ii) Find the coordinates, where the graph of p(x) cuts the axes.
(iii) Draw the graph of p(x) and verify your answers.
Solution:
Given linear polynomial can be written as p(x) = ax + b.
Since, the graph passes through the points (1, 5) and (3, 11).
For x = 1, a(1) + b =5
⇒ a + b = 5 …(i)
For x=3, a(3) + b = 11
⇒ 3a + b = 11 ……….(ii)

From Eq. (i), we get b = 5 – a
On substituting the value of b in Eq. (ii), we get
3a + (5 – a) =11
⇒3a + 5 – a =11
⇒ 2a + 5 =11
⇒ 2a = 6
⇒ a = 3
Now, b – 5 = a
= 5 – 3 = 2
Hence, p(x) = 3x + 2
When, x= 0 then p(x) = 2, so the graph cuts Y-axis at (0,2).
For X-axis, p(x) =0
0 = 3x + 2 ⇒ x = –\(\frac{2}{3}\)
So, the graph cuts X-axis at (-\(\frac{2}{3}\), 0)

Now, to draw the graph, take some values.
When, x = 0 then p(x) = 2
When, x = 1 then p (x) = 5
When, x = 3 then p(x) = 11
So, we have the following table

Here, we have three points A (0, 2), B (1, 5) and C (3, 11).
Now, plot these points on the graph and join them by a straight line.

Thus, we get the straight line AC, which represents the graph of p(x) = 3x + 2. Also, from the graph, it is clear that it cuts Y axis at (0, 2) and X-axis at (-\(\frac{2}{3}\),0) hence the results are verified.

Question 11.
Let p(x) = ax +b and q{x) = cx + d be two linear polynomials such that
(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).
Solution:
Given p(x) = ax + b …(i)
and q(x) = cx + d … (ii)

(i) Given, p(0) =5
From Eq. (i), we get
a(0) + b = 5
⇒ b = 5 …(iii)

(ii) Given, p(x) – q(x) = (a – c)x + (b – d)
Since, it cuts X-axis at (3, 0), p(3) – q(3) = 0
So, (a – c)3 + (b – d) =0
⇒ 3(a – c) + (5 – (-1)) = 0 [•■• b = 5]
⇒ 3(a – c) + 6 = 0
⇒ a – c = -2 … (iv)

(iii) Given, p(x) + q(x) =6x + 4
(ax + b) + (cx + d) = 6x + 4 [using Eqs. (i) and (ii)]
⇒ (a + c)x + (b + d) = 6x + 4
On comparing both sides we get
a + c = 6 …….(v)
b + d= 4 …….(vi)
From Eqs. (iv) and (v), we get
a – (6 – a) = 2
a – 6 + a = -2
⇒ 2a – 6 = -2
⇒ 2a = 4
⇒ a = 2
Now, c = 6 – a = 6 – 2 = 4
Now, from Eqs. (iii) and (vi), we get
5 + d =4
⇒ d = —1
Hence, 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.

(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.
Solution:
(i) In this pattern, first hexagon requires 6 matchsticks.
Also, each new hexagon shares one side with the previous hexagon, so each new hexagon requires 5 more matchsticks.
For Stage 1, number of matchsticks = 6
For Stage 2, number of matchsticks = 6 + 5 = 11
For Stage 3, number of matchsticks = 11 + 5 = 16

(ii) The completed table is

(iii) Since, the first hexagon needs 6 matchsticks and every new hexagon adds 5 matchitics,
Number of matchstics at nth stage = 6 + (n – 1)5
= 6 + 5n – 5
= 5n + 1
Hence, number of matchsticks required for the n th stage is 5n + 1.

(iv) For the 15th stage,
5n + 1 = 5(15) + 1 = 75 + 1 = 76
Hence, 76 matchsticks are required for the 15th stage.

(v) For 200 matchsticks,
5n + 1 = 200
5n = 199
n = \(\frac{199}{5}\) = 39.8
Since, n is not a natural number, so 200 matchsticks cannot form a stage in this pattern.

Question 13.
Let p(x) = ax +b and q(x) = cx + d be two linear polynomials such that
(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 meet X-axis.
Solution:
Given, p(x) = ax + b and q(x) = cx + d
(i) The graph of p(x) passes through (2, 3) and (6, 11).
For x = 2,
2a + b = 3 …(i)
For x = 6,
6a + b = 11 ……..(ii)
From Eq. (i), we get
b = 3 – 2a

On substituting in Eq. (ii), we get
6a + (3 – 2a) = 11
⇒ 4a + 3 = 11
⇒ 4a = 8
⇒ a = 2
Now, b = 3 – 2a = 3 – 2(2) = -1
Hence, p(x) = 2x – 1 and q(x) = 2x + d.

(ii) Given the graph of q(x) passes through (4, -1) then
-1 = c (4) + d
– 1 = 4c + d

(iii) Since, the graph of q(x) is parallel to the graph of p(x), so the coefficient of x will be same.
So, -1 = 2 × 4 + d
⇒ d=- 9
Hence, q (x) = 2x – 9
Now, for X-axis, p(x) = 0
2x – 1 = 0
2x = 1
x = \(\frac{1}{2}\)

So, p(x) meets the X-axis at (\(\frac{1}{2}\), 0)

For X-axis, q(x) = 0
2x – 9 = 0
2x = 9
⇒ x = \(\frac{9}{2}\)
So, q(x) meets the X-axis at (\(\frac{9}{2}\), 0)

Question 14.
What do all linear functions of the form f(x) = ax + a, a > 0 have in common?
Solution:
Given linear function f(x) = ax + a, where a > 0
Now, f(x) = a(x + 1)
For x-intercept, f(x) = 0
So, a(x + 1) = 0
Since, a ≠ 0,
x + 1 = 0
x = -1
Hence, all such linear functions pass through the fixed point (-1, 0) on the X-axis.
Also, since a > 0 the graph is an increasing straight line.
Hence, all linear functions of the form f(x) = ax + a have a common point (-1, 0) and all are increasing lines.