---
title: "NCERT Solutions Class 9 Maths Introduction to Linear Polynomials"
url: https://www.swavid.com/maths/class/9/chapter/introduction-to-linear-polynomials/ncert-solutions
dateModified: 2026-10-07T15:41:27+00:00
---

# NCERT Solutions Class 9 Maths Introduction to Linear Polynomials

This chapter's questions cover the fundamentals of polynomials, linear equations, linear patterns, growth and decay, and linear relationships. Students practice finding degrees, evaluating polynomials, solving word problems, and graphing lines.

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## Exercise Set 2.1

### Question 1

*4 marks · Short answer*

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

**Part (i)**

1. The highest power of the variable x in $2x^2 - 5x + 3$ is 2.
2. Therefore, the degree of the polynomial is 2.

Answer (i): 2

**Part (ii)**

1. The highest power of the variable y in $y^3 + 2y - 1$ is 3.
2. Therefore, the degree of the polynomial is 3.

Answer (ii): 3

**Part (iii)**

1. The constant polynomial $-9$ can be written as $-9x^0$.
2. Therefore, the degree of the polynomial is 0.

Answer (iii): 0

**Part (iv)**

1. The highest power of the variable z in $4z - 3$ is 1.
2. Therefore, the degree of the polynomial is 1.

Answer (iv): 1

**Answer:** The degrees are (i) 2, (ii) 3, (iii) 0, (iv) 1.

> Common mistake: Confusing the coefficient of the leading term with the degree of the polynomial.

### Question 2

*3 marks · Short answer*

Write polynomials of degrees 1, 2 and 3.

**Solution**

1. A polynomial of degree 1 is a linear polynomial, for example, $x + 1$.
2. A polynomial of degree 2 is a quadratic polynomial, for example, $x^2 + 2x + 1$.
3. A polynomial of degree 3 is a cubic polynomial, for example, $x^3 + 1$. (Answers may vary as any valid polynomial of the respective degree is correct).

**Answer:** Degree 1: $x + 1$, Degree 2: $x^2 + 2x + 1$, Degree 3: $x^3 + 1$

> Common mistake: Writing an expression that is not a polynomial, such as having a variable in the denominator or with a fractional exponent.

### Question 3

*3 marks · Short answer*

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

**Solution**

1. The given polynomial is $x^4 - 3x^3 + 6x^2 - 2x + 7$.
2. The term containing $x^2$ is $6x^2$, so its coefficient is 6.
3. The term containing $x^3$ is $-3x^3$, so its coefficient is $-3$.

**Answer:** Coefficient of $x^2$ is 6 and coefficient of $x^3$ is $-3$.

> Common mistake: Forgetting to include the negative sign when writing the coefficient of $x^3$.

### Question 4

*2 marks · Very short answer*

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

**Solution**

1. The given polynomial is $4z^3 + 5z^2 - 11$.
2. Since the term with variable $z$ (i.e., $z^1$) is absent, it can be written as $0z$.
3. Therefore, the coefficient of $z$ is 0.

**Answer:** 0

> Common mistake: Writing the coefficient of $z^2$ or $z^3$ instead of $z$.

### Question 5

*2 marks · Very short answer*

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

**Solution**

1. The given polynomial is $9x^3 + 5x^2 - 8x - 10$.
2. The constant term is the term independent of the variable x.
3. Therefore, the constant term is $-10$.

**Answer:** -10

> Common mistake: Omitting the negative sign from the constant term.

## Exercise Set 2.2

### Question 1

*3 marks · Short answer*

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

**Part (i)**

1. Substitute $x = 0$ in the polynomial $5x - 3$.
2. We get $5(0) - 3 = 0 - 3 = -3$.

Answer (i): $-3$

**Part (ii)**

1. Substitute $x = -1$ in the polynomial $5x - 3$.
2. We get $5(-1) - 3 = -5 - 3 = -8$.

Answer (ii): $-8$

**Part (iii)**

1. Substitute $x = 2$ in the polynomial $5x - 3$.
2. We get $5(2) - 3 = 10 - 3 = 7$.

Answer (iii): $7$

**Answer:** (i) $-3$, (ii) $-8$, (iii) $7$

> Common mistake: Making sign errors while substituting negative values for $x$.

### Question 2

*3 marks · Short answer*

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

**Part (i)**

1. Substitute $s = 0$ in the polynomial $7s^2 - 4s + 6$.
2. $7(0)^2 - 4(0) + 6 = 0 - 0 + 6 = 6$

Answer (i): $6$

**Part (ii)**

1. Substitute $s = -3$ in the polynomial $7s^2 - 4s + 6$.
2. $7(-3)^2 - 4(-3) + 6 = 7(9) + 12 + 6 = 63 + 12 + 6 = 81$

Answer (ii): $81$

**Part (iii)**

1. Substitute $s = 4$ in the polynomial $7s^2 - 4s + 6$.
2. $7(4)^2 - 4(4) + 6 = 7(16) - 16 + 6 = 112 - 16 + 6 = 102$

Answer (iii): $102$

**Answer:** (i) $6$, (ii) $81$, (iii) $102$

> Common mistake: Sign errors while substituting negative numbers, such as writing $(-3)^2$ as $-9$ instead of $9$.

### Question 3

*3 marks · Short answer*

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.

**Solution**

1. Let Salil's present age be $x$ years.
2. His mother's present age is $3x$ years.
3. After 5 years, Salil's age will be $x + 5$ and his mother's age will be $3x + 5$.
4. According to the question, $(x + 5) + (3x + 5) = 70$.
5. Simplify the equation: $4x + 10 = 70$, which gives $4x = 60$, so $x = 15$.
6. Salil's present age is 15 years and his mother's present age is $3 \times 15 = 45$ years.

**Answer:** Salil's present age is 15 years, and his mother's present age is 45 years.

> Common mistake: Forgetting to add 5 to both ages for the future condition.

### Question 4

*3 marks · Short answer*

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

**Solution**

1. Let the two positive integers be $2x$ and $5x$ based on the given ratio $2:5$.
2. The difference between the two integers is given as 63, so $5x - 2x = 63$.
3. Simplify the equation: $3x = 63$, which gives $x = 21$.
4. Calculate the first integer: $2 \times 21 = 42$.
5. Calculate the second integer: $5 \times 21 = 105$.

**Answer:** The two integers are 42 and 105.

> Common mistake: Subtracting in the wrong order leading to negative values.

### Question 5

*3 marks · Short answer*

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?

**Solution**

1. Let the number of five-rupee coins be $x$.
2. The number of two-rupee coins is $3x$.
3. The total value of five-rupee coins is $5x$ and the total value of two-rupee coins is $2 \times 3x = 6x$.
4. The total amount is given as ₹88, so $5x + 6x = 88$.
5. Simplify to get $11x = 88$, which gives $x = 8$.
6. Number of five-rupee coins is 8 and number of two-rupee coins is $3 \times 8 = 24$.

**Answer:** Ruby has 24 two-rupee coins and 8 five-rupee coins.

> Common mistake: Multiplying the coin count incorrectly with their face values.

### Question 6

*3 marks · Short answer*

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?

**Solution**

1. Let the length of the shorter piece of the fence be $x$ feet.
2. The length of the longer piece is $4x$ feet.
3. The total length of the fence is 300 feet, so $x + 4x = 300$.
4. Simplify the equation: $5x = 300$, which gives $x = 60$.
5. The shorter piece is 60 feet and the longer piece is $4 \times 60 = 240$ feet.

**Answer:** The two pieces are 60 feet and 240 feet long.

> Common mistake: Confusing which piece is 4 times the other.

### Question 7

*3 marks · Short answer*

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**

1. Let the width of the rectangle be $x$ cm.
2. The length of the rectangle is given as $2x + 3$ cm.
3. The perimeter of the rectangle is given by the formula $2(\text{length} + \text{width}) = 24$.
4. Substitute the expressions for length and width into the perimeter formula: $2((2x + 3) + x) = 24$.
5. Simplify the equation: $2(3x + 3) = 24$, which gives $6x + 6 = 24$.
6. Solve for $x$: $6x = 18$, so $x = 3$.
7. Therefore, the width is $3$ cm and the length is $2(3) + 3 = 9$ cm.

**Answer:** Width = $3$ cm, Length = $9$ cm

> Common mistake: Forgetting to multiply the entire sum of length and width by 2 when using the perimeter formula.

## Exercise Set 2.3

### Question 1

*3 marks · Short answer*

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.

**Part (i)**

1. At the end of the 1st month, the amount is ₹$(500 + 150 \times 1) = \text{₹}650$.
2. At the end of the 2nd month, the amount is ₹$(500 + 150 \times 2) = \text{₹}800$.
3. At the end of the 3rd month, the amount is ₹$(500 + 150 \times 3) = \text{₹}950$.

Answer (i): ₹650, ₹800, and ₹950

**Part (ii)**

1. Initial amount = ₹500 and monthly addition = ₹150.
2. For $n$ months, the total addition is ₹$150n$.
3. Thus, the linear expression representing the amount in the $n^{\text{th}}$ month is ₹$(500 + 150n)$.

Answer (ii): ₹$(500 + 150n)$

**Answer:** The amount at the end of month $n$ is ₹$(500 + 150n)$.

> Common mistake: Including the initial amount multiplied by $n$ or missing the initial fixed amount.

### Question 2

*3 marks · Short answer*

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

**Part (i)**

1. After 1 hour, remaining members = $120 - 9 \times 1 = 111$.
2. After 2 hours, remaining members = $120 - 9 \times 2 = 102$.
3. After 3 hours, remaining members = $120 - 9 \times 3 = 93$.

Answer (i): 111, 102, and 93 members

**Part (ii)**

1. Initial members = 120 and members leaving every hour = 9.
2. After $n$ hours, total members leaving = $9n$.
3. Thus, the linear expression for the remaining members is $120 - 9n$.

Answer (ii): $120 - 9n$

**Answer:** The number of members remaining after $n$ hours is $120 - 9n$.

> Common mistake: Adding the rate of decrease instead of subtracting it.

### Question 3

*3 marks · Short answer*

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.

**Part (i)**

1. Length = $13\text{ cm}$ and breadth = $12\text{ cm}$.
2. Area = $\text{length} \times \text{breadth} = 13 \times 12 = 156\text{ cm}^2$.

Answer (i): $156\text{ cm}^2$

**Part (ii)**

1. Length = $13\text{ cm}$ and breadth = $10\text{ cm}$.
2. Area = $13 \times 10 = 130\text{ cm}^2$.

Answer (ii): $130\text{ cm}^2$

**Part (iii)**

1. Length = $13\text{ cm}$ and breadth = $8\text{ cm}$. Area = $13 \times 8 = 104\text{ cm}^2$.
2. If breadth is $x\text{ cm}$, the linear pattern representing the area is $13x$.

Answer (iii): $104\text{ cm}^2$, linear pattern: $13x$

**Answer:** The areas are $156\text{ cm}^2$, $130\text{ cm}^2$, and $104\text{ cm}^2$, and the linear pattern is $13x$.

> Common mistake: Using perimeter formula instead of area.

### Question 4

*3 marks · Short answer*

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.

**Part (i)**

1. Length = $7\text{ cm}$, breadth = $11\text{ cm}$, and height = $5\text{ cm}$.
2. Volume = $\text{length} \times \text{breadth} \times \text{height} = 7 \times 11 \times 5 = 385\text{ cm}^3$.

Answer (i): $385\text{ cm}^3$

**Part (ii)**

1. Length = $7\text{ cm}$, breadth = $11\text{ cm}$, and height = $9\text{ cm}$.
2. Volume = $7 \times 11 \times 9 = 693\text{ cm}^3$.

Answer (ii): $693\text{ cm}^3$

**Part (iii)**

1. Length = $7\text{ cm}$, breadth = $11\text{ cm}$, and height = $13\text{ cm}$. Volume = $7 \times 11 \times 13 = 1001\text{ cm}^3$.
2. If height is $h\text{ cm}$, the linear pattern representing the volume is $77h$.

Answer (iii): $1001\text{ cm}^3$, linear pattern: $77h$

**Answer:** The volumes are $385\text{ cm}^3$, $693\text{ cm}^3$, and $1001\text{ cm}^3$, and the linear pattern is $77h$.

> Common mistake: Multiplying only two dimensions or adding them.

### Question 5

*3 marks · Short answer*

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**

1. Total pages in the book = 500.
2. Pages read per day = 20, so after 15 days, total pages read = $20 \times 15 = 300$.
3. Pages left after 15 days = $500 - 300 = 200$.
4. If $d$ represents the number of days, the linear pattern for the pages left is $500 - 20d$.

**Answer:** 200 pages will be left, expressed as the linear pattern $500 - 20d$.

> Common mistake: Adding the read pages to the total instead of subtracting.

## Exercise Set 2.4

### Question 1

*3 marks · Short answer*

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.

**Part (i)**

1. Initial height = $1.75$ feet and growth rate = $0.5$ feet per month.
2. Height after $7$ months $= 1.75 + 7 \times 0.5 = 1.75 + 3.5 = 5.25$ feet.

Answer (i): $5.25$ feet

**Part (ii)**

1. For $t = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10$, the height $h$ is $1.75, 2.25, 2.75, 3.25, 3.75, 4.25, 4.75, 5.25, 5.75, 6.25, 6.75$ feet respectively.

Answer (ii): Table of values showing $h$ increasing by $0.5$ each month

**Part (iii)**

1. The expression relating $h$ and $t$ is $h = 1.75 + 0.5t$.
2. This represents linear growth because the height increases by a constant amount ($0.5$ feet) over equal intervals of time ($1$ month).

Answer (iii): $h = 1.75 + 0.5t$

**Answer:** The height after 7 months is 5.25 feet, the expression relating $h$ and $t$ is $h = 1.75 + 0.5t$, representing linear growth.

> Common mistake: Adding the initial height multiple times or multiplying incorrectly.

### Question 2

*3 marks · Short answer*

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.

**Part (i)**

1. Initial value = ₹10,000 and annual depreciation = ₹800.
2. Value after $3$ years $= 10000 - 3 \times 800 = 10000 - 2400 = ₹7600$.

Answer (i): ₹7600

**Part (ii)**

1. For $t = 0, 1, 2, 3, 4, 5, 6, 7, 8$, the value $v$ is ₹$10000, 9200, 8400, 7600, 6800, 6000, 5200, 4400, 3600$ respectively.

Answer (ii): Table of values showing $v$ decreasing by ₹800 each year

**Part (iii)**

1. The expression relating $v$ and $t$ is $v = 10000 - 800t$.
2. This represents linear decay because the value decreases by a constant amount (₹800) over equal intervals of time ($1$ year).

Answer (iii): $v = 10000 - 800t$

**Answer:** The value of the phone after 3 years is ₹7600, the expression relating $v$ and $t$ is $v = 10000 - 800t$, representing linear decay.

> Common mistake: Adding the depreciation instead of subtracting.

### Question 3

*3 marks · Short answer*

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.

**Part (i)**

1. Initial population = $750$ and annual increase = $50$ people.
2. Population after $6$ years $= 750 + 6 \times 50 = 750 + 300 = 1050$.

Answer (i): $1050$

**Part (ii)**

1. For $t = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10$, the population $P$ is $750, 800, 850, 900, 950, 1000, 1050, 1100, 1150, 1200, 1250$ respectively.

Answer (ii): Table of values showing $P$ increasing by $50$ each year

**Part (iii)**

1. The expression relating $P$ and $t$ is $P = 750 + 50t$.
2. This represents linear growth because the population increases by a constant amount ($50$) over equal intervals of time ($1$ year).

Answer (iii): $P = 750 + 50t$

**Answer:** The population after 6 years is 1050, the expression relating $P$ and $t$ is $P = 750 + 50t$, representing linear growth.

> Common mistake: Multiplying the initial population by the number of years.

### Question 4

*3 marks · Short answer*

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.

**Part (i)**

1. Initial balance = ₹600 and daily reduction = ₹15.
2. The remaining balance after $x$ days is given by the linear equation $b(x) = 600 - 15x$.
3. This represents linear decay because the balance decreases by a constant amount (₹15) over equal intervals of time ($1$ day).

Answer (i): $b(x) = 600 - 15x$

**Part (ii)**

1. Set the remaining balance to $0$: $600 - 15x = 0$.
2. $15x = 600$, which gives $x = \frac{600}{15} = 40$ days.

Answer (ii): $40$ days

**Part (iii)**

1. For $x = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10$, the balance $b(x)$ is $585, 570, 555, 540, 525, 510, 495, 480, 465, 450$ respectively.

Answer (iii): Table of values showing $b(x)$ reducing by ₹15 each day

**Answer:** The balance is modelled by $b(x) = 600 - 15x$, it runs out after 40 days, and decreases by ₹15 daily.

> Common mistake: Forgetting to equate the balance to zero when finding when it runs out.

## Exercise Set 2.5

### Question 1

*3 marks · Short answer*

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 $y = ax + b$, find the values of $a$ and $b$.

**Solution**

1. We are given the linear relation $y = ax + b$.
2. When $x = 10$, $y = 400$, so $400 = 10a + b$.
3. When $x = 14$, $y = 500$, so $500 = 14a + b$.
4. From the first equation, $b = 400 - 10a$. Substituting this in the second equation gives $500 = 14a + (400 - 10a)$, which simplifies to $100 = 4a$, so $a = 25$.
5. Substituting $a = 25$ gives $b = 400 - 10(25) = 150$.
6. Thus, $a = 25$ and $b = 150$.

**Answer:** $a = 25$ and $b = 150$

> Common mistake: Errors in simultaneous linear equations substitution or signs.

### Question 2

*3 marks · Short answer*

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 $y = ax + b$, find the values of $a$ and $b$.

**Solution**

1. We are given the linear relation $y = ax + b$.
2. When $x = 10$, $y = 800$, so $800 = 10a + b$.
3. When $x = 15$, $y = 1100$, so $1100 = 15a + b$.
4. From the first equation, $b = 800 - 10a$. Substituting this into the second equation gives $1100 = 15a + (800 - 10a)$, which yields $300 = 5a$, so $a = 60$.
5. Substituting $a = 60$ gives $b = 800 - 10(60) = 200$.
6. Thus, $a = 60$ and $b = 200$.

**Answer:** $a = 60$ and $b = 200$

> Common mistake: Wrong assignment of variables $x$ and $y$.

### Question 3

*3 marks · Short answer*

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, the linear relationship between °C and °F.)

**Solution**

1. We are given the linear relation $\text{\textdegree}C = a \text{\textdegree}F + b$.
2. When $\text{\textdegree}C = 0$, $\text{\textdegree}F = 32$, so $0 = 32a + b$.
3. When $\text{\textdegree}C = 100$, $\text{\textdegree}F = 212$, so $100 = 212a + b$.
4. From the first equation, $b = -32a$. Substituting this into the second equation gives $100 = 212a - 32a$, which simplifies to $100 = 180a$, so $a = \frac{100}{180} = \frac{5}{9}$.
5. Substituting $a = \frac{5}{9}$ gives $b = -32 \left(\frac{5}{9}\right) = -\frac{160}{9}$.
6. Thus, $a = \frac{5}{9}$ and $b = -\frac{160}{9}$.

**Answer:** $a = \frac{5}{9}$ and $b = -\frac{160}{9}$

> Common mistake: Confusing the independent and dependent variables as specified in the relation.

## Exercise Set 2.6

### Question 1

*3 marks · Short answer*

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) $y = –2x – 3, y = –2x, y = 2x + 3$

**Part (i)**

1. Find points for $y = 4x$, $y = 2x$, and $y = x$, such as $(0, 0)$ and $(1, 4)$, $(1, 2)$, $(1, 1)$.
2. Plot the lines passing through the origin with positive slopes $4, 2, 1$.
3. Observe that as $a$ increases, the line becomes steeper.

Answer (i): All lines pass through the origin; larger $a$ gives a steeper line.

**Part (ii)**

1. Find points for $y = -6x$, $y = -3x$, and $y = -x$, such as $(0, 0)$ and $(1, -6)$, $(1, -3)$, $(1, -1)$.
2. Plot the lines passing through the origin with negative slopes $-6, -3, -1$.
3. Observe that negative slopes represent linear decay and larger magnitude makes the line steeper downwards.

Answer (ii): All lines pass through the origin with negative slope representing linear decay.

**Part (iii)**

1. Plot $y = 5x$ passing through $(0, 0)$ and $(1, 5)$ with a positive slope representing linear growth.
2. Plot $y = -5x$ passing through $(0, 0)$ and $(1, -5)$ with a negative slope representing linear decay.
3. Observe that equal magnitude of slope with opposite signs results in reflections across the x-axis.

Answer (iii): One line shows linear growth and the other shows linear decay with equal steepness.

**Answer:** Graphs drawn and role of slope $a$ and y-intercept $b$ analysed.

> Common mistake: Confusing the steepness of lines when comparing slopes greater than 1 and less than 1.

## End-of-Chapter Exercises

### Question 1

*3 marks · Short answer*

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

**Solution**

1. A polynomial of degree 3 in the variable $x$ has the general form $ax^3 + bx^2 + cx + d$, where $a \neq 0$.
2. We are given that the coefficient of the $x^2$ term is $-7$, so $b = -7$.
3. Choosing $a = 1$, $c = 0$, and $d = 0$ as an example, a polynomial of degree 3 with coefficient of $x^2$ as $-7$ is $x^3 - 7x^2$.

**Answer:** $x^3 - 7x^2$ (or any polynomial of the form $ax^3 - 7x^2 + cx + d$ where $a \neq 0$)

> Common mistake: Confusing the degree of the polynomial with the coefficient of a specific term.

### Question 2

*3 marks · Short answer*

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

**Part (i)**

1. Let $p(x) = 5x^2 - 3x + 7$.
2. Substitute $x = 1$ into the polynomial to get $p(1) = 5(1)^2 - 3(1) + 7$.
3. Simplify the expression to obtain $5 - 3 + 7 = 9$.

Answer (i): $9$

**Part (ii)**

1. Let $p(t) = 4t^3 - t^2 + 6$.
2. Substitute $t = a$ into the polynomial to get $p(a) = 4(a)^3 - (a)^2 + 6$.
3. Simplify to obtain $4a^3 - a^2 + 6$.

Answer (ii): $4a^3 - a^2 + 6$

**Answer:** (i) $9$, (ii) $4a^3 - a^2 + 6$

> Common mistake: Arithmetic errors while substituting values.

### Question 3

*3 marks · Short answer*

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**

1. Let the required number be $x$.
2. According to the question, multiplying the number by $\frac{5}{2}$ and adding $\frac{2}{3}$ gives $\frac{-7}{12}$, so $\frac{5}{2}x + \frac{2}{3} = \frac{-7}{12}$.
3. Subtract $\frac{2}{3}$ from both sides to get $\frac{5}{2}x = \frac{-7}{12} - \frac{2}{3} = \frac{-7 - 8}{12} = \frac{-15}{12} = \frac{-5}{4}$.
4. Multiply both sides by $\frac{2}{5}$ to solve for $x$, giving $x = \frac{-5}{4} \times \frac{2}{5} = \frac{-1}{2}$.st

**Answer:** $\frac{-1}{2}$

> Common mistake: Errors in fraction addition and subtraction.

### Question 4

*3 marks · Short answer*

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**

1. Let the another number be $x$. Then the positive number is $5x$.
2. When 21 is added to both numbers, the new numbers become $x + 21$ and $5x + 21$.
3. According to the condition, one of the new numbers becomes twice the other, so $5x + 21 = 2(x + 21)$.
4. Expand and solve the equation: $5x + 21 = 2x + 42$, which gives $3x = 21$, or $x = 7$.
5. The numbers are $7$ and $5(7) = 35$.

**Answer:** $7$ and $35$

> Common mistake: Multiplying the wrong side by 2 in the equation.

### Question 5

*3 marks · Short answer*

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.

**Part (i)**

1. The initial amount is ₹800 and the monthly saving is ₹250.
2. The amount after $n$ months is given by the linear expression $800 + 250n$.
3. After 6 months ($n = 6$), the amount is $800 + 250(6) = 800 + 1500 = \text{₹}2300$.

Answer (i): ₹2300

**Part (ii)**

1. 2 years is equal to 24 months ($n = 24$).
2. Substitute $n = 24$ into the expression $800 + 250n$ to get $800 + 250(24) = 800 + 6000 = \text{₹}6800$.

Answer (ii): ₹6800

**Part (iii)**

1. Express the total amount after $n$ months as the linear pattern $800 + 250n$.

Answer (iii): $800 + 250n$

**Answer:** (i) ₹2300, (ii) ₹6800, Linear pattern: $800 + 250n$

> Common mistake: Forgetting to convert years to months in part (ii).

### Question 6

*3 marks · Short answer*

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**

1. Let the units digit be $y$ and the tens digit be $x$. The two-digit number can be written as $10x + y$.
2. Since the digits differ by 3, we have $x - y = 3$ or $y - x = 3$. Let us assume $x - y = 3$, so $x = y + 3$.
3. The original number is $10(y + 3) + y = 11y + 30$. The number obtained by interchanging digits is $10y + (y + 3) = 11y + 3$.
4. The sum of the original number and the interchanged number is given as 143, so $(11y + 30) + (11y + 3) = 143$.
5. Simplify to get $22y + 33 = 143$, which gives $22y = 110$, or $y = 5$.
6. Then $x = 5 + 3 = 8$, so the original number is 85 and the other possible number is 58.

**Answer:** $85$ and $58$

> Common mistake: Ignoring the case where the digits can be in reverse order (e.g., tens digit smaller than units digit).

### Question 7

*5 marks · Case-based*

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) $y = –3x + 4$
(ii) $2y = 4x + 7$
(iii) $5y = 6x – 10$
(iv) $3y = 6x – 11$
Are any of the lines parallel?

**Part (i)**

1. Given equation is $y = -3x + 4$.
2. Comparing with $y = ax + b$, the slope is $a = -3$ and the y-intercept is $b = 4$.
3. The line cuts the y-axis at $(0, 4)$.

Answer (i): Slope = $-3$, y-intercept = $4$, y-axis point = $(0, 4)$

**Part (ii)**

1. Given equation is $2y = 4x + 7$, which can be written as $y = 2x + \frac{7}{2}$.
2. Comparing with $y = ax + b$, the slope is $a = 2$ and the y-intercept is $b = \frac{7}{2}$.
3. The line cuts the y-axis at $\left(0, \frac{7}{2}\right)$.

Answer (ii): Slope = $2$, y-intercept = $\frac{7}{2}$, y-axis point = $\left(0, \frac{7}{2}\right)$

**Part (iii)**

1. Given equation is $5y = 6x - 10$, which can be written as $y = \frac{6}{5}x - 2$.
2. Comparing with $y = ax + b$, the slope is $a = \frac{6}{5}$ and the y-intercept is $b = -2$.
3. The line cuts the y-axis at $(0, -2)$.

Answer (iii): Slope = $\frac{6}{5}$, y-intercept = $-2$, y-axis point = $(0, -2)$

**Part (iv)**

1. Given equation is $3y = 6x - 11$, which can be written as $y = 2x - \frac{11}{3}$.
2. Comparing with $y = ax + b$, the slope is $a = 2$ and the y-intercept is $b = -\frac{11}{3}$.
3. The line cuts the y-axis at $\left(0, -\frac{11}{3}\right)$.

Answer (iv): Slope = $2$, y-intercept = $-\frac{11}{3}$, y-axis point = $\left(0, -\frac{11}{3}\right)$

**Part (v)**

1. Lines with equal slopes are parallel to each other.
2. Equation (ii) and equation (iv) both have the same slope $a = 2$ but different y-intercepts.

Answer (v): Yes, the lines in (ii) and (iv) are parallel.

**Answer:** The slopes, y-intercepts, points of intersection with the y-axis, and parallel lines are determined for each equation.

> Common mistake: Forgetting to express the equation in the standard form $y = ax + b$ before identifying the slope and y-intercept.

### Question 8

*4 marks · Case-based*

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.

**Part (i)**

1. Given equation is $y = \frac{9}{5}(x - 273) + 32$.
2. Substitute $x = 313$ into the equation.
3. $y = \frac{9}{5}(313 - 273) + 32 = \frac{9}{5}(40) + 32$.
4. $y = 9 \times 8 + 32 = 72 + 32 = 104$.

Answer (i): $104$ °F

**Part (ii)**

1. Given equation is $y = \frac{9}{5}(x - 273) + 32$.
2. Substitute $y = 158$ into the equation.
3. $158 = \frac{9}{5}(x - 273) + 32$.
4. $126 = \frac{9}{5}(x - 273)$.
5. $126 \times \frac{5}{9} = x - 273$, which gives $70 = x - 273$.
6. $x = 70 + 273 = 343$.

Answer (ii): $343$ K

**Answer:** The required temperatures are 104 °F and 343 K.

> Common mistake: Errors in arithmetic calculation while transposing numbers.

### Question 9

*4 marks · Case-based*

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.

**Part (i)**

1. Work done $w$ is the product of constant force and distance $d$.
2. Given constant force is 3 units.
3. Thus, the linear equation is $w = 3d$.

Answer (i): $w = 3d$

**Part (ii)**

1. Substitute $d = 2$ in the equation $w = 3d$.
2. $w = 3 \times 2 = 6$ units.

Answer (ii): $6$ units

**Part (iii)**

1. Plot the points such as $(0, 0)$, $(1, 3)$, and $(2, 6)$ on the graph.
2. Join the points to get a straight line passing through the origin representing $w = 3d$.
3. The point $(2, 6)$ lies on the line, verifying the calculated work done.

Answer (iii): Graph drawn and point $(2, 6)$ verified.

**Answer:** Linear equation is $w = 3d$, work done for 2 units distance is 6 units.

> Common mistake: Confusing the dependent and independent variables while plotting.

### Question 10

*5 marks · Case-based*

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.

**Part (i)**

1. Let the linear polynomial be $p(x) = ax + b$.
2. Since the graph passes through $(1, 5)$, we get $a(1) + b = 5$, or $a + b = 5$.
3. Since it passes through $(3, 11)$, we get $a(3) + b = 11$, or $3a + b = 11$.
4. Subtracting the first equation from the second gives $2a = 6$, so $a = 3$.
5. Substituting $a = 3$ in $a + b = 5$ gives $b = 2$.
6. Thus, the polynomial is $p(x) = 3x + 2$.

Answer (i): $p(x) = 3x + 2$

**Part (ii)**

1. To find where it cuts the y-axis, put $x = 0$, giving $y = 3(0) + 2 = 2$. Point is $(0, 2)$.
2. To find where it cuts the x-axis, put $y = 0$, giving $3x + 2 = 0$, so $x = -\frac{2}{3}$. Point is $\left(-\frac{2}{3}, 0\right)$.

Answer (ii): y-axis intercept: $(0, 2)$; x-axis intercept: $\left(-\frac{2}{3}, 0\right)$

**Part (iii)**

1. Plot the points $(1, 5)$ and $(3, 11)$ on a graph paper and join them with a straight line.
2. Observe that the line cuts the y-axis at $(0, 2)$ and the x-axis at $\left(-\frac{2}{3}, 0\right)$, verifying our answers.

Answer (iii): Graph drawn and intercepts verified.

**Answer:** The polynomial is $p(x) = 3x + 2$, cutting the axes at $(0, 2)$ and $\left(-\frac{2}{3}, 0\right)$.

> Common mistake: Arithmetic errors when solving simultaneous equations for $a$ and $b$.

### Question 11

*3 marks · Short answer*

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**

1. Let $p(x) = ax + b$ and $q(x) = cx + d$.
2. From condition (i), $p(0) = a(0) + b = 5$, which gives $b = 5$.
3. From condition (iii), $p(x) + q(x) = (ax + b) + (cx + d) = (a + c)x + (b + d) = 6x + 4$.
4. Comparing coefficients, we get $a + c = 6$ and $b + d = 4$.
5. Since $b = 5$, we have $5 + d = 4$, which gives $d = -1$.
6. From condition (ii), $p(x) - q(x) = (ax + b) - (cx + d) = (a - c)x + (b - d)$.
7. The polynomial $p(x) - q(x)$ cuts the x-axis at $(3, 0)$, meaning $x = 3$ is its zero.
8. Therefore, $(a - c)(3) + (5 - (-1)) = 0$, which gives $3(a - c) + 6 = 0$, so $a - c = -2$.
9. We have the system: $a + c = 6$ and $a - c = -2$.
10. Adding both equations gives $2a = 4$, so $a = 2$. Then $c = 4$.
11. Thus, the polynomials are $p(x) = 2x + 5$ and $q(x) = 4x - 1$.

**Answer:** $p(x) = 2x + 5$ and $q(x) = 4x - 1$

> Common mistake: Failing to use the condition that a polynomial cutting the x-axis at $(3, 0)$ means its value is zero at $x = 3$.

### Question 12

*5 marks · Case-based*

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.

**Part (i)**

1. Draw stage 4 and stage 5 by adding a new hexagon sharing a side with the previous stage.
2. Stage 1 has 6 matchsticks, Stage 2 has 10, Stage 3 has 14.
3. Stage 4 requires $14 + 4 = 18$ matchsticks and Stage 5 requires $18 + 4 = 22$ matchsticks.

Answer (i): Stage 4 has 18 matchsticks and Stage 5 has 22 matchsticks.

**Part (ii)**

1. Count matchsticks for each stage: Stage 1 = 6, Stage 2 = 10, Stage 3 = 14, Stage 4 = 18, Stage 5 = 22.
2. The table values for $n$ are $6, 10, 14, 18, 22, \dots, 4n + 2$.

Answer (ii): Table completed with values 6, 10, 14, 18, 22, ..., $4n + 2$

**Part (iii)**

1. The number of matchsticks forms an arithmetic sequence with first term 6 and common difference 4.
2. The rule for the $n^{\text{th}}$ stage is given by $6 + (n - 1)4 = 4n + 2$.

Answer (iii): $4n + 2$

**Part (iv)**

1. Substitute $n = 15$ in the rule $4n + 2$.
2. $4(15) + 2 = 60 + 2 = 62$.

Answer (iv): $62$ matchsticks

**Part (v)**

1. Equate the rule to 200: $4n + 2 = 200$.
2. $4n = 198$, which gives $n = \frac{198}{4} = 49.5$.
3. Since $n$ must be a positive integer (stage number), 200 matchsticks cannot form a complete stage.

Answer (v): No, because $n$ is not a positive integer.

**Answer:** The rule is $4n + 2$, 15th stage has 62 matchsticks, and 200 matchsticks cannot form a stage.

> Common mistake: Assuming the first stage has 4 matchsticks instead of counting all 6 sides of the hexagon.

### Question 13

*5 marks · Long answer*

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 lines meet the x-axis.

**Part (i)**

1. Given $p(x) = ax + b$ passes through $(2, 3)$ and $(6, 11)$, substitute the coordinates to get $2a + b = 3$ and $6a + b = 11$.
2. Subtract the first equation from the second to get $4a = 8$, which gives $a = 2$.
3. Substitute $a = 2$ into $2a + b = 3$ to get $4 + b = 3$, so $b = -1$.
4. Thus, the polynomial is $p(x) = 2x - 1$.

Answer (i): $p(x) = 2x - 1$

**Part (ii)**

1. Since the graph of $q(x) = cx + d$ is parallel to $p(x) = 2x - 1$, their slopes are equal, so $c = 2$.
2. Thus, $q(x) = 2x + d$.
3. Since $q(x)$ passes through $(4, -1)$, substitute $x = 4$ and $y = -1$ to get $-1 = 2(4) + d$.
4. Solving for $d$ gives $d = -1 - 8 = -9$.
5. Thus, the polynomial is $q(x) = 2x - 9$.

Answer (ii): $q(x) = 2x - 9$

**Part (iii)**

1. To find where the line $y = 2x - 1$ meets the x-axis, put $y = 0$ to get $0 = 2x - 1$, which gives $x = \frac{1}{2}$.
2. To find where the line $y = 2x - 9$ meets the x-axis, put $y = 0$ to get $0 = 2x - 9$, which gives $x = \frac{9}{2}$.
3. Thus, the lines meet the x-axis at $(\frac{1}{2}, 0)$ and $(\frac{9}{2}, 0)$ respectively.

Answer (iii): The points are $(\frac{1}{2}, 0)$ and $(\frac{9}{2}, 0)$.

**Answer:** $p(x) = 2x - 1$, $q(x) = 2x - 9$, and the lines do not meet the x-axis at the same point (they meet at $(\frac{1}{2}, 0)$ and $(\frac{9}{2}, 0)$ respectively).

> Common mistake: Confusing parallel lines as having different slopes or making arithmetic errors while solving simultaneous linear equations.

### Question 14

*3 marks · Short answer*

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

**Solution**

1. Compare the given function $f(x) = ax + a$ with the standard linear equation form $y = ax + b$, where the slope is $a$ and the y-intercept is $a$.
2. Since $a > 0$, all such linear functions have a positive slope $a$ and a positive y-intercept $a$.
3. Substitute $x = -1$ into the function to get $f(-1) = a(-1) + a = 0$, which shows that the graph of every such function passes through the fixed point $(-1, 0)$.

**Answer:** All such linear functions have the same slope and y-intercept (both equal to $a$), and all their graphs pass through the fixed point $(-1, 0)$.

> Common mistake: Forgetting to check the common x-intercept by substituting $x = -1$.

## Frequently asked questions

### How many exercises and questions are there in NCERT Solutions for Class 9 Maths Chapter 2?

This chapter features 6 exercise sets along with end-of-chapter exercises, containing a total of 39 questions covering various types like very short answer, short answer, long answer, and case-based problems. You can find SwaVid's free PDF and step-by-step solutions for all these questions right on this page.

### Which topics do the questions in this chapter cover?

The questions cover essential concepts such as coefficients and constant terms of polynomials, degree of a polynomial, evaluating linear and quadratic polynomials, and linear equations in one variable. Other topics include word problems on coin denominations, ratios, linear decay and growth patterns, and graphing linear equations to understand the role of slope and y-intercept.

### What are the hardest question types in this chapter and how should I approach them?

The most challenging questions are typically the end-of-chapter problems involving case studies and long answers, such as finding linear polynomials from given points and slopes or analyzing linear equations and work done graphs. To approach these, carefully identify the given conditions, apply the standard form $f(x) = ax + b$, and break down the steps logically.

### How can I write answers for full marks in Class 9 Maths Chapter 2 exams?

To secure full marks, you should clearly state the given information, write down the relevant algebraic formulas, and show each step of your calculation sequentially. Referring to SwaVid's step-by-step solutions available on this page will help you understand the ideal presentation format expected by examiners.

### Is the free PDF for these NCERT solutions available for the current academic session?

Yes, the comprehensive solutions are tailored according to the new NCERT book based on the NCF 2023 for the 2026-27 session. You can easily access and download SwaVid's free PDF and step-by-step solutions directly on this page to aid your exam preparation.

## Related pages

- [Exercise 2.1 solutions](https://www.swavid.com/maths/class/9/chapter/introduction-to-linear-polynomials/ncert-solutions/exercise-2-1)
- [Exercise 2.2 solutions](https://www.swavid.com/maths/class/9/chapter/introduction-to-linear-polynomials/ncert-solutions/exercise-2-2)
- [Exercise 2.3 solutions](https://www.swavid.com/maths/class/9/chapter/introduction-to-linear-polynomials/ncert-solutions/exercise-2-3)
- [Exercise 2.4 solutions](https://www.swavid.com/maths/class/9/chapter/introduction-to-linear-polynomials/ncert-solutions/exercise-2-4)
- [Exercise 2.5 solutions](https://www.swavid.com/maths/class/9/chapter/introduction-to-linear-polynomials/ncert-solutions/exercise-2-5)
- [Exercise 2.6 solutions](https://www.swavid.com/maths/class/9/chapter/introduction-to-linear-polynomials/ncert-solutions/exercise-2-6)
- [Class 9 Maths chapters](https://www.swavid.com/maths/class/9)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student