Introduction to Linear Polynomials Class 9 Questions and Answers PDF Free Download

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NCERT Solutions for Class 9 Maths Chapter 2 Introduction to Linear Polynomials: Exercise Questions

NCERT Class 9 Maths Chapter 2 Introduction to Linear Polynomials question answers with detailed explanations are provided below.

Think and Reflect (Page 17)

Question 1. Can you identify the terms, variables and coefficients of this algebraic expression?

Expression 2: $200 l+160 w+50 l w$

$\textbf{Answer:}$

Given: $200 l+160 w+50 l w$
In this algebraic expression, the terms are: $200l$, $160w$, and $50lw$
Variable of this algebraic expression are $l$ and $w$.
The coefficient of $l$ is 200.
The coefficient of $w$ is 160.
The coefficient of $lw$ is 50.

Question 2. How is it different from the algebraic expression in Example 1?
Expression 1: $4 x+5 y+3$

$\textbf{Answer:}$

Expression 1: $4 x+5 y+3$
Expression 2: $200 l+160 w+50 l w$

There are many differences between these two expressions.

• Expression 1 has two variable terms and one constant term.
• Expression 2 has three variable terms and no constant term.
• Expression 1 is a linear algebraic expression.
• Expression 2 is not a linear algebraic expression, as there is $lw$.

Think and Reflect (Page 17)

Question 1. Can you identify the terms, variables and coefficients of this algebraic expression?

Expression 3: $10 x-x^2$

$\textbf{Answer:}$

Given: $10 x-x^2$
In this algebraic expression, the terms are: $10x$ and $-x^2$
Variable of this algebraic expression is $x$.
The coefficient of $x$ is 10.
The coefficient of $x^2$ is -1.

Question 2. Can you point out any similarity or difference between the algebraic expressions obtained in Examples 1 and 3?

$\textbf{Answer:}$

Similarity:

• Both are algebraic expressions.
• Both contain variables and coefficients.
• Both have more than one term.

Difference:

• Expression 1 has two variables, i.e., x and y.
• Expression 3 has only one variable, i.e., x.
• Expression 1 is linear because the highest power of the variables is 1.
• Expression 3 is quadratic because the highest power of x is 2.
• Expression 1 contains a constant term of 3, but Expression 3 does not contain a constant term.

Question 1. Find the degrees of the following polynomials:
(i) $2 x^2-5 x+3$
(ii) $y^3+2 y-1$
(iii) -9
(iv) $4 z-3$

$\textbf{Answer:}$

We know that the highest power of the variable of the polynomial is the degree of the polynomial.

(i) The highest power of x is 2. So, Degree of the polynomial = 2
(ii) The highest power of y is 3. So, Degree of the polynomial = 3
(iii) This is a non-zero constant polynomial. So, Degree of the polynomial = 0
(iv) The highest power of z is 1. So, Degree of the polynomial = 1

Question 2. Write polynomials of degrees 1,2 and 3.

$\textbf{Answer:}$

Degree 1: $2 x+5$
Degree 2: $y^2-4 y+7$
Degree 3: $z^3+2 z^2-z+1$

Question 3. What are the coefficients of $x^2$ and $x^3$ in the polynomial $x^4-3 x^3+6 x^2-2 x+7$?

$\textbf{Answer:}$
The coefficients of $x^2$ is +6.

The coefficients of $x^3$ is -3,

Question 4. What is the coefficient of $z$ in the polynomial $4 z^3+5 z^2-11$?

$\textbf{Answer:}$
In the given polynomial, there is no z term.
So, the coefficient of z = 0

Question 5. What is the constant term of the polynomial $9 x^3+5 x^2-8 x-10$?

$\textbf{Answer:}$

The constant term of the polynomial is -10.

Think and Reflect (Page 19)

Question: 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?

$\textbf{Answer:}$

We know that the perimeter of a square = 4 × Length of the side

When length = 1 cm,
Perimeter of a square = 4 × 1 = 4 cm

When length = 1.5 cm,
Perimeter of a square = 4 × 1.5 = 6 cm

When length = 2 cm,
Perimeter of a square = 4 × 2 = 8 cm

When length = 2.5 cm,
Perimeter of a square = 4 × 2.5 = 10 cm

When length = 3 cm,
Perimeter of a square = 4 × 3 = 12 cm

If the side increases by 0.5, then the perimeter will increase by $4\times 0.5=2$ cm
Hence, the perimeter increases by 2 cm whenever the side increases by 0.5 cm.

Think and Reflect (Page 19)

A chess club charges a joining fee of `200 plus `50 for every match played. The following table shows the amount a player will have to pay as the number of matches varies.

If a player paid `750, how many matches did he play?

$\textbf{Answer:}$

Here, the total amount paid is 750.
According to the question,
$200+50m=750$, where $m$ is the number of matches
$⇒50m=550$
$\therefore m=11$

Hence, the player played 11 matches to get paid 750.

Think and Reflect (Page 20)

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, $10 x-x^2$, is a function of $x$. Can you interpret this as an input-output process? What value does the expression take when $x=6 \mathrm{~cm}$?

$\textbf{Answer:}$

Given: The area of the rectangle, $10 x-x^2$
This means if we input the value of variable $x$, we will get the area of the rectangle as an output.

When $x=6$ cm, the area of the rectangle $=10\times6-6^2=24 \ \text{cm}^2$

Question 1. Find the value of the linear polynomial $5 x-3$ if:
(i) $x=0$
(ii) $ x=-1$
(iii) $x=2$

$\textbf{Answer:}$
Given polynomial $=5 x-3$
(i) When $x=0$,
$5\times0-3=-3$
So, the value will be -3.

(ii) When $x=-1$,
$5\times(-1)-3=-8$
So, the value will be -8.

(iii) When $x=2$,
$5\times2-3=7$
So, the value will be 7.

Question 2. Find the value of the quadratic polynomial $7 s^2-4 s+6$ if:
(i) $s=0$
(ii) $s=-3$
(iii) $s=4$

$\textbf{Answer:}$

Given polynomial $=7 s^2-4 s+6$
(i) When $s=0$,
$7(0)^2-4(0)+6=6$
So, the value will be 6.

(ii) When $s=-3$,
$7(-3)^2-4(-3)+6 $
$ =7(9)+12+6 =63+12+6=81$
So, the value will be 81.

(iii) When $s=4$,
$ 7(4)^2-4(4)+6 $
$ =7(16)-16+6$
$= 112-16+6=102$
So, the value will be 102.

Question 3. The present age of Salil's mother is three times Salil's present age. After 5 years, their ages will add up to 70 years. Find their present ages.

$\textbf{Answer:}$

Let the present age of Salil be $x$.
So, his mother's present age is $3x$.
According to the question,
$(x+5)+(3x+5)=70$
$⇒4x=60$
$\therefore x= 15$
So, Salil's present age = 15 years.
His mother's present age $=3x=3\times15=45$ years

Question 4. The difference between two positive integers is 63. The ratio of the two integers is 2:5. Find the two integers.

$\textbf{Answer:}$

The ratio of the two integers is 2:5
Let the integers be $2x$ and $5x$.
According to the question,
$5x-2x=63$
$⇒3x=63$
$\therefore x=21$

So, the integers are: $2x=2\times21=42$ and $5x=5\times21=105$

Question 5. Ruby has 3 times as many two-rupee coins as she has five-rupee coins. If she has a total ₹88, how many coins does she have of each type?

$\textbf{Answer:}$

Let the amount of 2 rupee coins Rubi has be $x$.
So, she has $3x$ amount of 3-rupee coins.

Total value of Rubi's 2 rupee coins = $2x$
and total 3-rupee coins value = $3x\times3=9x$
According to the question,
$2x+9x=88$
$\therefore x= 8$

So, she has 8 amounts of 2-rupee coins and $3\times8=24$ amounts of 3-rupee coins.

Question 6. A farmer cuts a 300 feet fence into two pieces of different sizes. The longer piece is four times as long as the shorter piece. How long are the two pieces?

$\textbf{Answer:}$

Let the shorter piece be $x$ feet.
So, the longer piece =$4x$
According to the question,
$x+4x=300$
$\therefore x=60$

So, the shorter piece is 60 feet.
and the longer piece is $300-60=240$ 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?

$\textbf{Answer:}$

Let the width of the rectangle be $x$ cm.
So, the length = $2x+3$
We know that the perimeter of a rectangle = 2 (Length + Width)
$⇒ 24 = 2 (2x+3+x)$
$⇒6x+6=24$
$\therefore x =3$

So, the width of the rectangle is 3 cm,
and length is $2x+3=2\times3+3=9$ cm

Think and Reflect (Page 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.

$\textbf{Answer:}$

Stage 1: 1

Stage 2: 1 + 2 = 3

Stage 3: 1 + 4 = 5

Stage 4: 1 + 6 = 7

Stage 5: 1 + 8 = 9

Stage 6: 1 + 10 = 11

Stage 7: 1 + 12 = 13

Solve the following:
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 $n^{\text {th }}$ month.

$\textbf{Answer:}$

Initial amount in savings account = ₹500
Pocket money received every month = ₹150
Amount after 1st month: 500 + 150 = 650
Amount after 2nd month: 500 + 2 × (150) = 800
Amount after 3rd month: 500 + 3 × (150) = 950
.
.
.

Amount after nth month: 500 + n × (150) = 500 + 150n
$\therefore$ The required linear expression is: (500 + 150n).

Question 2. A rally starts with 120 members. Each hour, 9 members drop out of the group. How many members will remain after $1,2,3, \ldots$ hours? Find a linear expression to represent the number of members at the end of the $n^{\text {th }}$ hour.

$\textbf{Answer:}$

Initial number of members = 120
Members dropping out every hour = 9
After 1 hour, the number of members: 120 - 9 = 111
After 2 hours, the number of members: 120 - (2 × 9) = 102
After 3 hours, the number of members: 120 - (3 × 9) = 93
.
.
.
.

After n hours, the number of members: 120 - n × 9 = 120 - 9n
$\therefore$ The required linear expression is: (120 - 9n)

Question 3. Suppose the length of a rectangle is 13 cm. Find the area if the breadth is
(i) 12 cm,
(ii) 10 cm,
(iii) 8 cm.
Find the linear pattern representing the area of the rectangle.

$\textbf{Answer:}$

Length of rectangle = 13 cm
Area of a rectangle = Length × Breadth

(i) When breadth = 12 cm
Area of the rectangle = 13 × 12 = 156 $\text{cm}^2$

(ii) When breadth = 10 cm
Area of the rectangle = 13 × 10 = 130 $\text{cm}^2$

(iii) When breadth = 8 cm
Area of the rectangle = 13 × 8 = 104 $\text{cm}^2$

So, if the breadth is $x$ cm, the area of the rectangle = $13x\ \text{cm}^2$
Hence, the linear pattern representing the area is: $13x$

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, (iii) 13 cm.
Find the linear pattern representing the volume of the rectangular box.

$\textbf{Answer:}$

Length of rectangular box = 7 cm
Breadth = 11 cm
We know that the volume of a cuboid = Length × Breadth × Height
Let the height be $h$ cm.
So, volume of a cuboid = 7 × 11 × $h$ = $77h\ \text{cm}^3$

(i) When height = 5 cm
Volume of the cuboid = $77\times5=385 \ \text{cm}^3$

(i) When height = 9 cm
Volume of the cuboid = $77\times9=693 \ \text{cm}^3$

(i) When height = 13 cm
Volume of the cuboid = $77\times13=1001 \ \text{cm}^3$

Hence, the linear pattern representing the volume is: $77h$

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.

$\textbf{Answer:}$

Total pages in the book = 500
Pages read every day by Sarita = 20
So, Pages read in 15 days by Sarita = 20 × 15 = 300
$\therefore$ Pages left after 15 days $=500-300=200$

If $d$ represents the number of days, Pages left after $d$ days: $500- 20d$
Hence, the linear pattern is: $500- 20d$

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.

$\textbf{Answer:}$

Initial height of plant = 1.75 feet
Growth every month = 0.5 feet
(i) Height after 7 months = $1.75 + 7\times0.5=1.75+3.5=5.25$ feet
Hence, the height after 7 months is 5.25 feet.

(ii)

(iii)
Height (h) = 1.75 + 0.5t
Since the height increases by a constant amount of 0.5 every month, it represents linear growth.

Question 2. A mobile phone is bought for ₹ 10,000. Its value decreases by ₹800 every year.
(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.

$\textbf{Answer:}$

Initial value of mobile phone = ₹10000
Depreciation every year = ₹800

(i) Value of the phone after 3 years = 10000 - (3 × 800) = 7600
Hence, the value after 3 years is ₹7600.

(ii)

(iii)
Value (v) = 10000 - 800t
Since the value decreases by a constant amount of ₹800 every year, it represents linear decay.

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 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.

$\textbf{Answer:}$

Initial population of village = 750
Increase every year = 50

(i) Population after 6 years in the village = 750 + (50 × 6) = 1050
Hence, the population after 6 years is 1050.

(ii)

(iii)
Population (P) = 750 + 50t
Since the population increases by a constant amount of 50 every year, it represents linear growth.

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.

$\textbf{Answer:}$

Initial prepaid balance = ₹600
Reduction every day = ₹15

(i) Equation for the remaining balance, $b(x)=600-15x$
Since the balance decreases by a constant amount of ₹15 every day, it represents linear decay.

(ii)
The balance reaches zero, meaning it runs out.
According to the question,
$600-15x=0$
$\therefore x=40$ days
Hence, the balance will run out after 40 days.

(iii)

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 ₹500. If the monthly bill $y$ depends on the number of modules accessed, $x$, according to the relation $y=a x+b$, find the values of $a$ and $b$.

$\textbf{Answer:}$

Let the charges per module be $a$, and the fixed monthly charge of the platform be $b$.
According to the question,
10a + b = 400------------(1)
14a + b = 500------------(2)
Subtracting equation 1 from equation 2, we get,
4a = 100
$\therefore$ a = 25
Putting the value of $a$ in equation 1, we get,
10 × 25 + b = 400
$\therefore$ b = 400 - 250 = 150

Hence, the values of $a$ and $b$ are 25 and 150, respectively.
Also, the required relation is: 25x + 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 hours, her bill was ₹800. When she used it for 15 hours, her bill was ₹1100. If the monthly bill $y$ depends on the hours of the use of the badminton court, $x$, according to the relation $y=a x+b$, find the values of $a$ and $b$.

$\textbf{Answer:}$

Let the charges per hour be $a$, and the fixed monthly charge of the Gym be $b$.
According to the question,
10a + b = 800------------(1)
15a + b = 1100------------(2)
Subtracting equation 1 from equation 2, we get,
5a = 300
$\therefore$ a = 60
Putting the value of $a$ in equation 1, we get,
10 × 60 + b = 800
$\therefore$ b = 800 - 650 = 200

Hence, the values of $a$ and $b$ are 60 and 200, respectively.
Also, the required relation is: 60x + 200

Question 3. Consider the relationship between temperature measured in degrees Celsius ( ${ }^{\circ} \mathrm{C}$ ) and degrees Fahrenheit ( ${ }^{\circ} \mathrm{F}$ ), which is given by ${ }^{\circ} \mathrm{C}=a^{\circ} \mathrm{F}+b$. 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 ${ }^{\circ} \mathrm{C}=0,{ }^{\circ} \mathrm{F}=32$ and when ${ }^{\circ} \mathrm{C}=100,{ }^{\circ} \mathrm{F}=212$. Use this information to find $a$ and $b$, and thus, the linear relationship between ${ }^{\circ} \mathrm{C}$ and ${ }^{\circ} \mathrm{F}$.)

$\textbf{Answer:}$

Given: ${ }^{\circ} \mathrm{C}=a^{\circ} \mathrm{F}+b$
We know that ice melts at 0 degrees Celsius and 32 degrees Fahrenheit, and water boils at 100 degrees Celsius and 212 degrees Fahrenheit.
So,
0 = 32a + b------------(1)
100 = 212a + b------------(2)
Subtracting equation 1 from equation 2, we get,
180a = 100
$\therefore$ a = $\frac59$
Putting the value of $a$ in equation 1, we get,
$0=32\times \frac59+b$
$\therefore b=-\frac{160}9$
$\therefore$ b = 800 - 650 = 200

Hence, the values of $a$ and $b$ are $\frac59$ and $-\frac{160}9$, respectively.
Also, the required relation is:
${ }^{\circ} C=\frac{5}{9}^{\circ} F-\frac{160}{9}$
$⇒\frac{{ }^{\circ} C}5=\frac{^{\circ}F-32}9$

Question 1. Draw the graphs of the following sets of lines. In each case, reflect on the role of ' $a$ ' and ' $b$'.

(i) $y=4 x, y=2 x, y=x$

(ii) $y=-6 x, y=-3 x, y=-x$

(iii) $y=5 x, y=-5 x$

(iv) $y=3 x-1, y=3 x, y=3 x+1$

(v) $y=-2 x-3, y=-2 x, y=2 x+3$

$\textbf{Answer:}$

(i)
Given: $y=4 x, y=2 x, y=x$

Here, the slopes are: 4, 2, and 1

y-intercept points are: 0, 0, and 0

Observation:

• All lines pass through the origin.
• A greater value of "a" gives a steeper line.
• Since the value of "b" is 0, all lines cut the y-axis at (0,0).

(ii)
Given: $y=-6 x, y=-3 x, y=-x$

Here, the slopes are: -6, -3, and -1

y-intercept points are: 0, 0, and 0

Observation:

• All lines pass through the origin.
• Negative value of "a" gives a downward-sloping line.
• A larger negative value makes the line steeper.

(iii)
Given: $y=5 x, y=-5 x$

Here, the slopes are: 5 and -5

y-intercept points are: 0 and 0

Observation:

• Both lines pass through the origin.
• Positive slope rises from left to right.
• Negative slope falls from left to right.
• The same magnitude of slope gives equal steepness.

(iv)
Given: $y=3 x-1, y=3 x, y=3 x+1$

Here, the slopes are: 3, 3, and 3

y-intercept points are: -1, 0, and 1

Observation:

• All lines are parallel because slopes are equal.
• Changing "b" shifts the line up or down.
• Positive "b" shifts upward.
• Negative "b" shifts downward.

(v)
Given: $y=-2 x-3, y=-2 x, y=2 x+3$

Here, the slopes are: -2, -2, and -2

y-intercept points are: -3, 0, and 3

Observation:

• All lines are parallel because slopes are equal.
• Negative slope means lines fall from left to right.
• Changing "b" changes the y-intercept and shifts the graph vertically.

Question 1. Write a polynomial of degree 3 in the variable $x$, in which the coefficient of the $x^2$ term is -7.

$\textbf{Answer:}$

A polynomial of degree 3 must have the highest power $x^3$.
So, the required polynomial is: $3x^3-7x^2+x+2$

Question 2. Find the values of the following polynomials at the indicated values of the variables.
(i) $5 x^2-3 x+7$ if $x=1$
(ii) $4 t^3-t^2+6$ if $t=a$

$\textbf{Answer:}$

(i)
Given: $5 x^2-3 x+7$
When $x=1$, the value of the polynomial is:
$(5\times1^2)-(3\times1)+7=9$

(ii)
Given: $4 t^3-t^2+6$
When $t=a$, the value of the polynomial is:
$4a^3-a^2+6$

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.

$\textbf{Answer:}$

Let the number be $x$.
According to the question,
$x\times\frac52+\frac23=-\frac7{12}$
$⇒\frac{5x}{2}=-\frac{15}{12}$
$\therefore x=-\frac12$

Hence, the required number is $-\frac12$.

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?

$\textbf{Answer:}$

Let the smaller number be $x$.
Then the larger number is: $5x$
According to the question,
$(x+21)\times2=5x+21$
$⇒2x+42=5x+21$
$⇒3x=21$
$\therefore x=7$
So, the larger number $=5x=5\times7=35$

Hence, the 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.

$\textbf{Answer:}$

Initial amount = ₹800
Monthly savings = ₹250

(i)
After 6 months, I will have:
800 + (6 × 250) = 2300
Hence, the amount after 6 months is ₹2300.

(ii)
After 2 years, i.e., 24 months, I will have:
800 + (24 × 250) = 6800

Hence, the linear pattern is: $800+250n$, where $n$ is the number of 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.

$\textbf{Answer:}$

Let the two-digit number be $10x+y$, where $x$ is the tens digit, and $y$ is the units digit.
Given: $x-y=3----------(1)$
If the digits are interchanged, the number becomes $10y+x$
According to the question,
$10x+y+10y+x=143$
$⇒11x+11y=143$
$⇒x+y=13-------------(2)$
Adding equation 1 to equation 2, we get,
$2x=10$
$\therefore x=8$
Putting the value $x$ in equation 1, we get $y=5$
Hence, the original number $=10x+y=10\times8+5=85$
And the number after digit interchange is 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 y -axis.
(i) $y=-3 x+4$
(ii) $2 y=4 x+7$
(iii) $5 y=6 x-10$
(iv) $3 y=6 x-11$

Are any of the lines parallel?

$\textbf{Answer:}$

We know that the Slope-Intercept Form of linear equations is the $y=mx+c$, where $m$ is the slope, and $ c$ is the y-intercept, where the line crosses the vertical axis

(i)
Given: $y=-3 x+4$
Comparing it with $y=mx+c$, we get, $m=-3$ and $c=4$
$\therefore$ The coordinates of the points where these lines cut the y-axis are (0, 4).

(ii)
Given: $2 y=4 x+7$
$⇒y=2x+\frac72$
Comparing it with $y=mx+c$, we get, $m=2$ and $c=\frac72$
$\therefore$ The coordinates of the points where these lines cut the y-axis are $(0, \frac72)$.

(iii)
Given: $5 y=6 x-10$
$⇒y=\frac65 x-2$
Comparing it with $y=mx+c$, we get, $m=\frac65$ and $c=-2$
$\therefore$ The coordinates of the points where these lines cut the y-axis are $(0, -2)$.

(iv)
Given: $3 y=6 x-11$
$⇒y=2x-\frac{11}2$
Comparing it with $y=mx+c$, we get, $m=2$ and $c=-\frac{11}2$
$\therefore$ The coordinates of the points where these lines cut the y-axis are $(0, -\frac{11}2)$.

Since equations (ii) and (iv) have the same slope, $m=2$, those lines are parallel.
Hence, the parallel lines are: $2 y=4 x+7$ and $3 y=6 x-11$

*Question 8. If the temperature of a liquid can be measured in Kelvin units as $x \mathrm{~K}$ and in Fahrenheit units as $y^{\circ} \mathrm{F}$, 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 $158^{\circ} \mathrm{F}$, then find the temperature in Kelvin.

$\textbf{Answer:}$

Given linear equation: $y=\frac{9}{5}(x-273)+32$

(i)
When the temperature of the liquid is 313 K, i.e., $x=313$,
$y=\frac{9}{5}(313-273)+32=72+32=104$ F

Hence, the temperature is 102 F.

(ii)
When the temperature of the liquid is $158^{\circ} \mathrm{F}$, i.e., $y=158$
$158=\frac{9}{5}(x-273)+32$
$⇒158-32=\frac{9}{5}(x-273)$
$⇒126=\frac{9}{5}(x-273)$
$⇒630=9(x-273)$
$⇒x-273=70$
$\therefore x=343$ K

Hence, the temperature in Kelvin is 343K.

*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.

$\textbf{Answer:}$

We know that:
Work done (w) = Force × Distance (d)
Constant force $=3$ units
Therefore, $w=3d$
Hence, the linear equation is $w=3d$

If $\mathrm{d}=2$ units, $\mathrm{w}=3 \times 2=6$ units.
Hence, the work done is 6 units.

Taking w on the y-axis and d on the x-axis, we can plot the graph.

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

* 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.

$\textbf{Answer:}$

(i)
Let the linear polynomial be $p(x)=a x+b$.
Given: The graph passes through points $(1,5)$ and $(3,11)$.
Putting these values in the equation, we get,
$5=a(1)+b $
$\Rightarrow a+b=5----------(1)$
$ 11=a(3)+b$
$\Rightarrow 3 a+b=11---------(2)$
Subtracting the first equation from the second:

$
\begin{aligned}
& \Rightarrow(3 a+b)-(a+b)=11-5 \\
& \Rightarrow 2 a=6 \\
& \Rightarrow a=3
\end{aligned}
$
Substituting $a=3$ into $a+b=5$ :
$
\begin{aligned}
& \Rightarrow 3+b=5 \\
& \Rightarrow b=2
\end{aligned}
$
Substituting the values of $a$ and $b$ into $p(x)$, we get $p(x)=3 x+2$

Hence, the polynomial is $p(x)=3 x+2$.

(ii)

To find the $y$-intercept, lets take $x=0$ :

$
\begin{aligned}
& \Rightarrow p(0)=3(0)+2 \\
& \Rightarrow p(0)=2
\end{aligned}
$
The graph cuts the $y$-axis at $(0,2)$.
To find the $x$-intercept, lets take $p(x)=0$ :
$
\begin{aligned}
& \Rightarrow 3 x+2=0 \\
& \Rightarrow 3 x=-2 \\
& \Rightarrow x=-\frac{2}{3}
\end{aligned}
$
The graph cuts the $x$-axis at $\left(-\frac{2}{3}, 0\right)$.
Hence, the graph cuts the axes at $(0,2)$ and $\left(-\frac{2}{3}, 0\right)$.

(iii)
For $x=1$:
$ p(1)=3(1)+2=5$ (Verified)
For $x=3$:
$p(3)=3(3)+2=11$ (Verified)
Hence, the coordinates and the polynomial are verified.

*Question 11. Let $p(x)=a x+b$ and $q(x)=c x+d$ 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 $6 x+4$ for all real $x$.

Find the polynomials $p(x)$ and $q(x)$.

$\textbf{Answer:}$

Let $p(x)=a x+b$ and $q(x)=c x+d$.
Given $p(0)=5$ [Condition (i)]
$
\begin{aligned}
& \Rightarrow a(0)+b=5 \\
& \Rightarrow b=5
\end{aligned}
$
Given the sum $p(x)+q(x)=6 x+4$ [Condition (iii)]
$
\begin{aligned}
& \Rightarrow(a x+b)+(c x+d)=6 x+4 \\
& \Rightarrow(a+c) x+(b+d)=6 x+4
\end{aligned}
$
Comparing the constant terms on both sides:
$
\begin{aligned}
& \Rightarrow b+d=4 \\
& \Rightarrow 5+d=4 \\
& \Rightarrow d=-1
\end{aligned}
$
Comparing the coefficients of $x$ on both sides:
$
\Rightarrow a+c=6
$
The polynomial difference
$=p(x)-q(x)$
$=(a x+b)-(c x+d)$
$=(a-c) x+(b-d)$
Substituting $b=5$ and $d=-1$:
$\Rightarrow p(x)-q(x)=(a-c) x+(5-(-1))=(a-c) x+6$
Since $p(x)-q(x)$ cuts the $x$-axis at $(3,0)$, [Condition (ii)]
$
\begin{aligned}
& \Rightarrow(a-c)(3)+6=0 \\
& \Rightarrow 3(a-c)=-6 \\
& \Rightarrow a-c=-2
\end{aligned}
$
We now have a system of two linear equations: $a+c=6$ and $a-c=-2$.
Adding the two equations:
$
\begin{aligned}
& \Rightarrow(a+c)+(a-c)=6+(-2) \\
& \Rightarrow 2 a=4 \\
& \Rightarrow a=2
\end{aligned}
$
Substituting $a=2$ into $a+c=6$, we get,
$
\begin{aligned}
& \Rightarrow 2+c=6 \\
& \Rightarrow c=4
\end{aligned}
$
Substituting the values into $p(x)$ and $q(x)$ :
$
\begin{aligned}
& \Rightarrow p(x)=2 x+5 \\
& \Rightarrow q(x)=4 x-1
\end{aligned}
$
Hence, the polynomials are $p(x)=2 x+5$ and $q(x)=4 x-1$.

*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 $n^{\text {th }}$ 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.

$\textbf{Answer:}$

(i)

It takes 6 matchsticks to make a single hexagon, as it has six sides.
Since each new hexagon shares one common side with the previous hexagon, only 5 new matchsticks are added at every new stage.
So the pattern is:
1st Hexagon = 6 sticks
two hexagons = 6 + 5 = 11 sticks
three hexagons = 6 + 5 + 5 = 16 sticks
four hexagons = 6 + 5 + 5 + 5 = 21 sticks
five hexagons = 6 + 5 + 5 + 5 + 5 = 26 sticks

This is an arithmetic pattern, where the first term is 6, and the common difference is 5.
Hence,
$y=6+5(x-1)$, where $x$ is the number of stages
$⇒y=5x+1$
Hence, the linear equation is $y=5x+1$.

(ii)

As we know, the pattern is: $y=5x+1$, where $x$ is the number of stages

(iii)

This is an arithmetic pattern, where the first term is 6, and the common difference is 5.

Hence,

$y=6+5(n-1)$, where $n$ is the number of stages

$⇒y=5n+1$

Hence, the linear equation is $y=5n+1$.

(iv)

We know,

$y=5n+1$ where $n$ is the number of stages

If $n=15$, $y=5\times15+1=76$

Hence, the required matchsticks for the 15th stage are 76.

(v)

We know,

$y=5n+1$ where $n$ is the number of stages

We have to check if $y=200$, then the value of $n$ is a whole number or not.

$200=5n+1$

$⇒5n=199$

$⇒n=39.8$, which is not a whole number

Hence, 200 matchsticks cannot form a stage in this pattern.

*Question 13. Let $p(x)=a x+b$ and $q(x)=c x+d$ 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 lines meet the x -axis.

$\textbf{Answer:}$

Let $p(x)=a x+b$.
Given: The graph passes through $(2,3)$ and $(6,11)$.
So, $ 3=a(2)+b$
$ \Rightarrow 2 a+b=3 -------------(1)$
Also, $11=a(6)+b $
$\Rightarrow 6 a+b=11-----------(2)$
Subtracting equation 1 from equation 2, we get,
$
\begin{aligned}
& \Rightarrow(6 a+b)-(2 a+b)=11-3 \\
& \Rightarrow 4 a=8 \\
& \Rightarrow a=2
\end{aligned}
$
Substituting $a=2$ into equation 1, we get,
$
\begin{aligned}
& \Rightarrow 2(2)+b=3 \\
& \Rightarrow 4+b=3 \\
& \Rightarrow b=-1
\end{aligned}
$
Substituting the values of $a$ and $b$ into $p(x)$
$
\Rightarrow p(x)=2 x-1
$
Since the graph of $q(x)=c x+d$ is parallel to the graph of $p(x)$, their slopes must be equal.
$
\begin{aligned}
& \Rightarrow c=a \\
& \Rightarrow c=2
\end{aligned}
$
The graph of $q(x)$ passes through $(4,-1)$. [Given]
$\begin{aligned}
& \Rightarrow-1=2(4)+d \\
& \Rightarrow-1=8+d \\
& \Rightarrow d=-9
\end{aligned}
$
Substituting the values of $c$ and $d$ into $q(x)$ :
$
\Rightarrow q(x)=2 x-9
$
Taking $p(x)=0$, we get,
$
\begin{aligned}
& \Rightarrow 2 x-1=0 \\
& \Rightarrow 2 x=1 \\
& \Rightarrow x=\frac{1}{2}
\end{aligned}
$
Taking $q(x)=0$, we get,
$
\begin{aligned}
& \Rightarrow 2 x-9=0 \\
& \Rightarrow 2 x=9 \\
& \Rightarrow x=\frac{9}{2}
\end{aligned}
$
Hence, the polynomials are $p(x)=2 x-1$ and $q(x)=2 x-9$, and they meet the x -axis at ( $\frac{1}{2}, 0$ ) and $\left(\frac{9}{2}, 0\right)$ respectively.

*Question 14. What do all linear functions of the form $f(x)=a x+a, a>0$, have in common?

$\textbf{Answer:}$

Let the linear function be $f(x)=a x+a$.

$\Rightarrow f(x)=a(x+1)$

Taking $f(x)=0$, we get,

$\Rightarrow a(x+1)=0$

Since $a>0$, we can divide both sides by $a$ :

$ x+1=0$
$\Rightarrow x=-1$

The coordinates of this point on the graph are $(-1,0)$.

This result is completely independent of the value of $a$.

So, every such line passes through the fixed point $(-1,0)$.

Additionally, because the slope is $a$ and it is given that $a>0$ :

⇒ All such lines have a positive slope and strictly increase from left to right.

Hence, all these linear functions have a common $x$-intercept at $(-1,0)$ and a positive slope.

Introduction to Linear Polynomials Class 9 Chapter 2: Topics

Students will explore the following topics in NCERT Class 9 Maths Chapter 2 Introduction to Linear Polynomials:

• 2.1 Introduction
• 2.2 Linear Polynomials
• 2.3 Exploring linear patterns
• 2.4. Linear growth and linear decay
• 2.5 Linear Relationships
• 2.6 Visualising linear relationships

Class 9 Maths NCERT Chapter 2 Solutions: Extra Questions

Question 1: Find the value of $y$ in the equation $ 2x + 3y = 12 $ when $x = 3 $.

$\textbf{Answer:}$

First, substitute $x = 3 $ into the equation:

We get, $2(3) + 3y = 12 \Rightarrow 6 + 3y = 12$

$3y = 12 - 6 = 6 \Rightarrow y = \frac{6}{3} = 2$

Thus, the value of $y$ is 2.

Question 2:

If two mixers and one T.V cost Rs. 700, while two T.Vs and one mixer cost Rs. 980. The value of one T.V is:

$\textbf{Answer:}$

Let the T.V be denoted by T, and the mixer be denoted by M.
2M + T = 700
2T + M = 980
Adding both equations:
2T + M + (2M + T) = 980 + 700
⇒ T + M = $\frac{1680}{3}$ = 560
Again,
2T + M = 980
⇒ T + T + M = 980
⇒ T + 560 = 980
$\therefore$ T = 420
$\therefore$ Value of one TV is Rs. 420.
Hence, the correct answer is Rs. 420.

Question 3:

Student A got 25 marks more than student B, and the marks of A were equal to 60% of the sum of their marks. What are the marks obtained by A and B, respectively?

$\textbf{Answer:}$

Let the marks of student A be p and the marks of student B be q.
From the problem, we have two equations:
Since A got 25 marks more than B
⇒ p = q + 25 ...(1)
Since the marks of A were equal to 60% of the sum of their marks
⇒ p = 0.6 × (p + q) ...(2)
Substituting equation 1 into equation 2, we get:
⇒ q + 25 = 0.6 × (q + 25 + q)
⇒ q + 25 = 1.2q + 15
⇒ 0.2q = 10
⇒ q = 50
$\therefore$ p = 50 + 25 = 75
Hence, the correct answer is 75 and 50.

Question 4:

The cost of 3 pieces of bread and 1 packet of juice at a certain store is Rs. 323.50. At the same store, the cost of 5 pieces of bread and 1 packet of juice is Rs. 433.50. What will be the cost of 2 pieces of bread and 2 packets of juice at this store?

$\textbf{Answer:}$

Let the cost of one bread be b, and the cost of one packet of juice be j.
The cost of 3 breads and 1 packet of juice.
⇒ 3b + j = 323.50---(1)
The cost of 5 breads and 1 packet of juice.
⇒ 5b + j = 433.50---(2)
Subtracting the first equation from the second,
⇒ 2b = 433.50 - 323.50 = 110
⇒ b = $\frac{110}{2}$ = 55
Substituting b = 55 into the first equation,
⇒ 3 × 55 + j = 323.50
⇒ j = 323.50 - 165 = 158.50
So, the cost of one bread is Rs. 55, and the cost of one packet of juice is Rs. 158.50.
Therefore, the cost of 2 breads and 2 packets of juice is,
⇒ 2b + 2j = 2 × 55 + 2 × 158.50 = 110 + 317 = 427
Hence, the correct answer is Rs. 427.

Question 5:

The cost of 12 apples and 15 oranges is INR 660. The cost of 15 apples and 21 oranges is INR 870. Find the cost of one apple.

$\textbf{Answer:}$

Let the cost of one apple be $x$, and one orange be $y$.
According to the question,
$12x+15y=660$ .......(1)
And, $15x+21y=870$ ........(2)
Multiplying equation (1) by 5 and equation (2) by 4, we get,
$60x+75y = 3300$ and $60x+84y=3480$
Subtracting these equations, we get,
$75y-84y=3300-3480$
⇒ $9y=180$
⇒ $y=20$
Putting this value in (1), we get,
$12x+15(20)=660$
⇒ $12x+300=660$
⇒ $12x = 360$
⇒ $x=30$
⇒ The cost of one apple is INR 30
Hence, the correct answer is INR 30.

NCERT Solutions for Class 9 Maths Chapter Wise

We at Careers360 compiled all the NCERT Class 9 Maths solutions in one place for easy student reference. The following links will allow you to access them.

NCERT Class 9 Books & Syllabus

Careers360 provides links to the Class 9 NCERT Books and the latest NCERT Syllabus which students can easily access. Students can access them through the links below.

• NCERT Books Class 9 Maths
• NCERT Books for Class 9 Science
• NCERT Syllabus Class 9 Maths
• NCERT Books Class 9
• NCERT Syllabus Class 9