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

Q: What is Class 9 Maths Ganita Manjari Chapter 2 about?

It covers polynomials and degree, linear polynomials, linear patterns, linear growth and decay, linear relationships y = ax + b, and plotting straight-line graphs.

Q: What is a linear polynomial?

A linear polynomial is a polynomial of degree 1, such as 2x + 3 or 5 − 4y.

Q: What do a and b mean in y = ax + b?

a is the slope of the line and b is the y-intercept (where the line cuts the y-axis).

Q: Are these Class 9 Maths Ganita Manjari Chapter 2 solutions free?

Yes. All solutions are free and follow the official NCERT Ganita Manjari textbook for 2026–27.

These Class 9 Maths Ganita Manjari Chapter 2 solutions cover Introduction to Linear Polynomials from the new NCF-2023 textbook (2026–27). Checked against NCERT Ganita Manjari, Grade 9, Part I, first edition April 2026, pages 16–40. The guide follows Exercise Sets 2.1–2.6 and all 14 end-of-chapter questions. Sample choices and the limits of real-world models are stated explicitly.

Class: 9 Subject: Mathematics Book: Ganita Manjari (Part 1) Chapter: 2 Exercises: 2.1–2.6, End-of-Chapter Session: 2026–27

On this page

Chapter overview
Key concepts & definitions
Key facts & formulas
Exercise Set 2.1
Exercise Set 2.2
Exercise Set 2.3
Exercise Set 2.4
Exercise Set 2.5
Exercise Set 2.6
End-of-Chapter Exercises (Q1–Q14)
Common mistakes to avoid
Practice MCQs & Assertion–Reason
Quick revision summary
FAQs

Chapter 2 Overview

Chapter 2 of Ganita Manjari, Introduction to Linear Polynomials, builds on algebraic expressions to define polynomials in one variable and their degree, then focuses on linear polynomials (degree 1). It shows how linear expressions model real-life linear patterns, linear growth and decay, and linear relationships of the form $y=ax+b$ , and how to plot these as straight lines. Use the graphs to compare slopes and intercepts, and reproduce the constructions on graph paper.

Key Concepts & Definitions

Polynomial (in one variable): an algebraic expression like $x^{2}+5x+3$ made of terms with whole-number powers of the variable.

Degree: the highest power with a non-zero coefficient after simplifying. Degree 1 = linear, 2 = quadratic, 3 = cubic. A non-zero constant has degree 0; the zero polynomial has no defined degree.

Linear polynomial: a polynomial ax + b with $a\ne 0$ , of degree 1; for example, 2x + 3 or 5 − 4y.

Linear pattern: a sequence in which the difference between consecutive terms is constant.

Linear growth / decay: a quantity that increases (growth) or decreases (decay) by a fixed amount over equal intervals.

Linear relationship: $y=ax+b$ , a straight line with slope a and y-intercept b; when $b=0$ the line passes through the origin.

Key Facts & Formulas

Value of a polynomial: substitute the given value of the variable and simplify.

Linear relationship: $y=ax+b$ — a = slope (rate of change), b = y-intercept (value when $x=0$ ).

Plotting a line: find two distinct points (e.g. $x=0$ and one other value), plot them, join with a ruler and extend in both directions. For a physical model, display only the meaningful domain.

Parallel lines: have the same slope a (only b differs). Growth $\Rightarrow a>0$ ; decay $\Rightarrow a<0$ .

Think and Reflect — introductory expressions (page 17)

In 200l + 160w + 50lw, the terms are 200l, 160w and 50lw; the variables are l and w, with numerical coefficients 200, 160 and 50. Unlike 4x + 5y + 3, this expression contains a product of variables (total degree 2), and no non-zero constant term.

In $10x-x^{2}$ the terms are 10x and $-x^{2}$ ; the coefficients are 10 and −1. It uses one variable and has degree 2, whereas 4x + 5y + 3 uses two variables with terms of degree at most 1. Both expressions are sums of terms with numerical coefficients.

Exercise Set 2.1

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

SOLUTION (i) Highest power = 2 → degree 2. (ii) Highest power = 3 → degree 3. (iii) −9 is a constant → degree 0. (iv) Highest power = 1 → degree 1.

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

SOLUTION Sample answers. Degree 1: 2x + 5; Degree 2: $x^{2}-3x+1$ ; Degree 3: $4x^{3}+x^{2}-2$ . (Other correct answers are possible.)

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

SOLUTION Coefficient of

x^{2}=6

; coefficient of

x^{3}=-3

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

SOLUTIONThere is no z term, so the coefficient of z is 0.

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

SOLUTIONThe constant term is −10.

Think and Reflect 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? Answer. Perimeter = 4 × side: 4, 6, 8, 10, 12 cm. Each time the side increases by 0.5 cm, the perimeter increases by a constant 2 cm (a linear pattern).

Think and Reflect A chess club charges ₹200 joining fee plus ₹50 per match. If a player paid ₹750, how many matches did he play? Answer.

\begin{aligned}200+50m&=750\\\Rightarrow 50m&=550\\\Rightarrow m&=11\end{aligned}

matches.

Exercise Set 2.2

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

SOLUTION (i)

5\left(0\right)-3=-3

. (ii)

5\left(-1\right)-3=-8

. (iii)

5\left(2\right)-3=7

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

SOLUTION (i)

7\left(0\right)-4\left(0\right)+6=6

. (ii)

\begin{aligned}7\left(9\right)-4\left(-3\right)+6&=63+12+6\\&=81\text{.}\end{aligned}

(iii)

\begin{aligned}7\left(16\right)-4\left(4\right)+6&=112-16+6\\&=102\text{.}\end{aligned}

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.

SOLUTION Let Salil’s age = x, so mother’s age = 3x. After 5 years:

\begin{aligned}\left(x+5\right)+\left(3x+5\right)&=70\\\Rightarrow 4x+10&=70\\\Rightarrow 4x&=60\\\Rightarrow x&=15\text{.}\end{aligned}

∴ Salil is 15 years and his mother is 45 years.

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 integers be 2k and 5k. Then

\begin{aligned}5k-2k&=63\\\Rightarrow 3k&=63\\\Rightarrow k&=21\text{.}\end{aligned}

∴ the integers are 42 and 105.

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?

SOLUTION Let five-rupee coins = x, so two-rupee coins = 3x. Value:

\begin{aligned}5x+2\left(3x\right)&=88\\\Rightarrow 5x+6x&=88\\\Rightarrow 11x&=88\\\Rightarrow x&=8\text{.}\end{aligned}

∴ 8 five-rupee coins and 24 two-rupee coins (32 coins in all).

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?

SOLUTION Let the shorter piece = x, so the longer = 4x. Then

\begin{aligned}x+4x&=300\\\Rightarrow 5x&=300\\\Rightarrow x&=60\text{.}\end{aligned}

∴ the pieces are 60 feet and 240 feet.

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 width = w, so length = 2w + 3. Perimeter:

\begin{aligned}2\left(w+2w+3\right)&=24\\\Rightarrow 2\left(3w+3\right)&=24\\\Rightarrow 3w+3&=12\\\Rightarrow w&=3\text{.}\end{aligned}

∴ width = 3 cm, length =

2\left(3\right)+3=9\,\mathrm{cm}

Think and Reflect The area of the rectangle,

10x-x^{2}

, is a function of x. What value does the expression take when

x=6\,\mathrm{cm}

? Answer. Input x → calculate width 10 − x → multiply x(10 − x) → output area. A non-degenerate rectangle requires

0<x<10\,\mathrm{cm}

. At

x=6

, width = 4 cm and area =

6\times 4=24\,\mathrm{cm}^{2}

Exercise Set 2.3

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^th month.

SOLUTION Amount after n months = 500 + 150n. Month 2 → 500 + 300 = ₹800, month 3 → ₹950, month 4 → ₹1100, … Linear expression: 500 + 150n.

Assumption: ₹500 is the starting balance (month 0), all pocket money is saved, and there is no interest or withdrawal. Month 1 has ₹650.

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 n^th hour.

SOLUTION After 1 hr → 111, 2 hr → 102, 3 hr → 93, … Linear expression: 120 − 9n.

The formula applies at whole hours $n=0$ , 1, …, 13: after 13 hours, 3 members remain. Continuing to $n=14$ would give −6, which is impossible. The stated pattern cannot continue unchanged once fewer than 9 people remain.

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.

SOLUTION Area = 13 × breadth. (i)

13\times 12=156\,\mathrm{cm}^{2}

; (ii)

13\times 10=130\,\mathrm{cm}^{2}

; (iii)

13\times 8=104\,\mathrm{cm}^{2}

. Linear pattern: area = 13b (it falls by

26\,\mathrm{cm}^{2}

for every 2 cm fall in breadth).

If the breadth continues as 12, 10, 8, …, its n-th value is 14 − 2n. The corresponding area sequence is $A_{n}=182-26n\,\mathrm{cm}^{2}$ , for positive breadth. The general area relation is $A=13b$ .

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.

SOLUTION Volume =

7\times 11

× height = 77h. (i)

77\times 5=385\,\mathrm{cm}^{3}

; (ii)

77\times 9=693\,\mathrm{cm}^{3}

; (iii)

77\times 13=1001\,\mathrm{cm}^{3}

. Linear pattern: volume = 77h.

For heights 5, 9, 13, …, $h_{n}=4n+1$ and $V_{n}=308n+77\,\mathrm{cm}^{3}$ . The common increase is $308\,\mathrm{cm}^{3}$ per stage.

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 Pages left after n days = 500 − 20n. After 15 days:

\begin{aligned}500-20\left(15\right)&=500-300\\&=200\end{aligned}

pages.

The model applies for whole days $0\le n\le 25$ . The book is finished on day 25; the number of unread pages cannot become negative.

Think and Reflect Using 2n − 1, how many tiles are there in the 15th and 26th stages? Which stage contains 21 tiles and 47 tiles? Answer. 15th:

2\left(15\right)-1=29

; 26th:

2\left(26\right)-1=51

2n-1=21\Rightarrow n=11

;

2n-1=47\Rightarrow n=24

Further Think and Reflect answers (pages 22–25)

Tile stages 1–7 contain 1, 3, 5, 7, 9, 11, 13 tiles; the next three stages after stage 4 are 9, 11 and 13.

Bela: 100 − 5(15) = ₹25 on day 15; $100-5n=0$ gives 20 days to spend the full amount.

Auto-rickshaw: $15n-5=130$ gives $15n=135$ , so 9 km. Use this expression only for $n\ge 2$ ; the initial 2 km cost ₹25.

Water height: $h\left(5\right)=3-0.5\left(5\right)=0.5m$ . In this ideal model the tank becomes empty at $t=6$ months; negative water heights are not physical.

Exercise Set 2.4

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

SOLUTION (i)

\begin{aligned}h&=1.75+0.5\left(7\right)\\&=1.75+3.5\\&=5.25\end{aligned}

feet. (ii) t: 0,1,2,…,10 → h: 1.75, 2.25, 2.75, 3.25, 3.75, 4.25, 4.75, 5.25, 5.75, 6.25, 6.75. (iii) $h=1.75+0.5t$ . The height increases by a fixed 0.5 ft each month (slope > 0), so it is linear growth.

Time (months)	Height (ft)
0	1.75
1	2.25
2	2.75
3	3.25
4	3.75
5	4.25
6	4.75
7	5.25
8	5.75
9	6.25
10	6.75

2. A mobile phone is bought for ₹10,000. Its value decreases by ₹800 every year. (i) Find the value after 3 years. (ii) Make a table of values for t from 0 to 8 years. (iii) Find an expression relating v and t, and explain why it represents linear decay.

SOLUTION (i)

v=10000-800\left(3\right)

= ₹7600. (ii) t: 0–8 → v: 10000, 9200, 8400, 7600, 6800, 6000, 5200, 4400, 3600. (iii) $v=10000-800t$ . The value falls by a fixed ₹800 each year (slope < 0), so it is linear decay.

Time (years)	Value (₹)
0	10000
1	9200
2	8400
3	7600
4	6800
5	6000
6	5200
7	4400
8	3600

This is the textbook’s fixed depreciation model, not a prediction of resale prices. It must not be extended to negative monetary values.

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 after 6 years. (ii) Make a table of values for t from 0 to 10 years. (iii) Find an expression relating P and t, and explain why it represents linear growth.

SOLUTION (i)

P=750+50\left(6\right)=1050

. (ii) t: 0–10 → P: 750, 800, 850, 900, 950, 1000, 1050, 1100, 1150, 1200, 1250. (iii) $P=750+50t$ . The population rises by a fixed 50 each year (slope > 0) ⇒ linear growth.

Time (years)	Population (people)
0	750
1	800
2	850
3	900
4	950
5	1000
6	1050
7	1100
8	1150
9	1200
10	1250

This model assumes no other population changes such as births, deaths or departures.

4. A telecom company charges ₹600 for a recharge scheme. The prepaid balance is reduced by ₹15 each day after the recharge. (i) Write an equation for the remaining balance b(x) after 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 from 1 to 10 days.

SOLUTION (i) $b\left(x\right)=600-15x$ . The balance falls by a fixed ₹15 per day (slope < 0) ⇒ linear decay. (ii)

600-15x=0\Rightarrow x=40

days. (iii) x: 1–10 → b: 585, 570, 555, 540, 525, 510, 495, 480, 465, 450.

Time (days)	Balance (₹)
1	585
2	570
3	555
4	540
5	525
6	510
7	495
8	480
9	465
10	450

The balance model has domain $0\le x\le 40$ days before exhaustion; after day 40 a recharge or a new rule is needed.

Think and Reflect For the cost

C\left(d\right)=100+60d

: what is the cost for 15 km? For how many km will the cost be ₹700? Answer.

C\left(15\right)=100+60\left(15\right)

= ₹1000.

\begin{aligned}700&=100+60d\\\Rightarrow 60d&=600\\\Rightarrow d&=10\,\mathrm{km}\text{.}\end{aligned}

Exercise Set 2.5

1. A learning platform charges a fixed monthly fee plus a cost per module. When 10 modules were accessed the bill was ₹400; for 14 modules it was ₹500. If $y=ax+b$ , find a and b.

SOLUTION

400=10a+b

and

500=14a+b

. Subtract:

100=4a\Rightarrow a=25

b=400-10\left(25\right)=150

. ∴ $a=25$ , $b=150$

\left(y=25x+150\right)

2. A gym charges a fixed monthly fee plus a cost per hour of badminton. For 10 hours the bill was ₹800; for 15 hours it was ₹1100. If $y=ax+b$ , find a and b.

SOLUTION

800=10a+b

and

1100=15a+b

. Subtract:

300=5a\Rightarrow a=60

b=800-10\left(60\right)=200

. ∴ $a=60$ , $b=200$

\left(y=60x+200\right)

3. The relation between Celsius and Fahrenheit is ${}^{\circ}\mathrm{C}=a\cdot {}^{\circ}\mathrm{F}+b$ . Given ice melts at 0°C / 32°F and water boils at 100°C / 212°F, find a and b.

SOLUTION

0=32a+b

and

100=212a+b

. Subtract:

\begin{aligned}100&=180a\\\Rightarrow a&=\frac{100}{180}=\frac{5}{9}\text{.}\end{aligned}

\begin{aligned}b&=-32a\\&=-32\left(\frac{5}{9}\right)\\&=-\frac{160}{9}\text{.}\end{aligned}

\begin{aligned}\therefore {}^{\circ}\mathrm{C}&=\left(\frac{5}{9}\right){}^{\circ}\mathrm{F}-\frac{160}{9}\\&=\left(\frac{5}{9}\right)\left({}^{\circ}\mathrm{F}-32\right)\text{.}\end{aligned}

Think and Reflect In

y=20x+150

, what do the numbers 20 and 150 represent? Answer. 20 is the slope — the cost per GB of data; 150 is the y-intercept — the fixed monthly fee charged even when

x=0

Think and Reflect — slopes and intercepts

The graphs $y=3x+1$ and $y=-3x+1$ both pass through (0, 1). The first rises and the second falls; they have equal steepness and are reflections in the y-axis.

For $y=-ax$ with $a>0$ , all lines pass through the origin and fall. Larger a means greater steepness: $a>1$ is steeper than $y=-x$ , while $0<a<1$ is less steep.

With slope fixed and b changing in $y=ax+b$ , distinct values of b give parallel lines. The y-intercept b is signed; the distance of (0, b) from the origin is |b|.

Exercise Set 2.6

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

SOLUTION Plot each using two points, then observe the role of a (slope) and b (y-intercept): (i)

y=4x\to \left(0,0\right)

,(1,4);

y=2x\to \left(1,2\right)

;

y=x\to \left(1,1\right)

. All pass through the origin

\left(b=0\right)

; the line gets steeper as a increases. (ii)

y=-6x

,−3x,−x → (1,−6),(1,−3),(1,−1). All through the origin with negative slope (fall left-to-right); steeper as |a| increases. (iii) y = 5x → (1,5);

y=-5x\to \left(1,-5\right)

. Same steepness, opposite signs — mirror images in the x-axis. (iv)

y=3x-1

, 3x, 3x+1 → same slope 3, y-intercepts −1, 0, 1. They are parallel (a fixed, b changes). (v)

y=-2x-3

and

y=-2x

are parallel (slope −2);

y=2x+3

has slope +2, so it is not parallel to them. Conclusion: a controls slope/direction, b controls the y-intercept; equal a ⇒ parallel lines.

Exercise 2.6, part 1. Equal scales: each grid interval is 1 unit. Lines continue beyond the displayed window.

Exercise 2.6, part 2. Equal scales: each grid interval is 1 unit. Lines continue beyond the displayed window.

Exercise 2.6, part 3. Equal scales: each grid interval is 1 unit. Lines continue beyond the displayed window.

Exercise 2.6, part 4. Equal scales: each grid interval is 1 unit. Lines continue beyond the displayed window.

Exercise 2.6, part 5. Equal scales: each grid interval is 1 unit. Lines continue beyond the displayed window.

Think and Reflect Complete the table for

y=2x+1

x=1

, 2, 5, 7, 9, 12, 20. Answer.

y=2x+1\to x

: 1,2,5,7,9,12,20 give y: 3, 5, 11, 15, 19, 25, 41.

End-of-Chapter Exercises — Questions 1–14

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

SOLUTIONOne example: $x^{3}-7x^{2}+2x+1$ (any degree-3 polynomial with the

x^{2}

coefficient equal to −7).

2. 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 $t=a$

SOLUTION (i)

\begin{aligned}5\left(1\right)-3\left(1\right)+7&=5-3+7\\&=9\text{.}\end{aligned}

(ii) Substitute

t=a

: $4a^{3}-a^{2}+6$ .

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 be x:

\left(\frac{5}{2}\right)x+\frac{2}{3}=-\frac{7}{12}

\begin{aligned}\left(\frac{5}{2}\right)x&=-\frac{7}{12}-\frac{2}{3}\\&=-\frac{7}{12}-\frac{8}{12}\\&=-\frac{15}{12}\\&=-\frac{5}{4}\text{.}\end{aligned}

\begin{aligned}x&=\left(-\frac{5}{4}\right)\times \left(\frac{2}{5}\right)\\&=-\frac{10}{20}\\&=-\frac{1}{2}\text{.}\end{aligned}

4. A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?

SOLUTION Let the smaller number = x, larger = 5x. After adding 21: (5x + 21) and (x + 21).

\begin{aligned}5x+21&=2\left(x+21\right)\\\Rightarrow 5x+21&=2x+42\\\Rightarrow 3x&=21\\\Rightarrow x&=7\text{.}\end{aligned}

∴ the numbers are 7 and 35 (check:

56=2\times 28

5. If you have ₹800 and you save ₹250 every month, find the amount you have after (i) 6 months (ii) 2 years. Express this as a linear pattern.

SOLUTION Amount after n months = 800 + 250n. (i) 800 + 250(6) = ₹2300. (ii) 2 years = 24 months: 800 + 250(24) = ₹6800.

*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 digits be a (tens) and b (units). Number = 10a + b, reversed = 10b + a. Sum =

\begin{aligned}11\left(a+b\right)&=143\\\Rightarrow a+b&=13\text{.}\end{aligned}

Also

|a-b|=3

. Solving:

a=8

b=5

(or

a=5

b=8

). ∴ the numbers are 85 and 58.

*7. Draw the graph of the following equations, and identify their slopes and y-intercepts. Also find 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?

SOLUTION Write each as

y=ax+b

: (i)

y=-3x+4

→ slope −3, y-intercept 4, cuts y-axis at (0, 4). (ii)

y=2x+\frac{7}{2}

→ slope 2, y-intercept 3.5, cuts (0, 3.5). (iii)

y=\left(\frac{6}{5}\right)x-2

→ slope 1.2, y-intercept −2, cuts (0, −2). (iv)

y=2x-\frac{11}{3}

→ slope 2, y-intercept

-\frac{11}{3}

, cuts

\left(0,-\frac{11}{3}\right)

. Lines (ii) and (iv) are parallel (both have slope 2).

End-of-chapter Question 7: four lines. Equal scales: each grid interval is 1 unit. Lines continue beyond the displayed window.

*8. The relation between Kelvin (x K) and Fahrenheit (y °F) is $y=\left(\frac{9}{5}\right)\left(x-273\right)+32$ . (i) Find y if $x=313K$ . (ii) If $y=158{}^{\circ}\mathrm{F}$ , find x.

SOLUTION (i)

\begin{aligned}y&=\left(\frac{9}{5}\right)\left(313-273\right)+32\\&=\left(\frac{9}{5}\right)\left(40\right)+32\\&=72+32\\&=104{}^{\circ}\mathrm{F}\text{.}\end{aligned}

(ii)

\begin{aligned}158&=\left(\frac{9}{5}\right)\left(x-273\right)+32\\\Rightarrow 126&=\left(\frac{9}{5}\right)\left(x-273\right)\\\Rightarrow x-273&=70\\\Rightarrow x&=343K\text{.}\end{aligned}

These answers use the textbook’s stated approximation with 273. Do not silently replace its given equation when solving this exercise.

*9. Work done = constant force × distance. Express this as a linear equation in two variables (work w and distance d), and draw its graph for a constant force of 3 units. What is the work done when the distance travelled is 2 units?

SOLUTION With force = 3: $w=3d$ . This is a straight line through the origin with slope 3 (points (0,0), (1,3), (2,6)). When

d=2

w=3\left(2\right)=6

units.

The 6 units are units of work (force unit × distance unit). If force is in newtons and distance in metres, the answer is 6 joules. For this distance model use $d\ge 0$ .

Work and distance: horizontal

x=d

; vertical

y=w

. Equal scales: each grid interval is 1 unit. Lines continue beyond the displayed window.

The full supporting line is shown for comparison; only its portion with $d\ge 0$ models the stated travel distance.

*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 where its graph cuts the axes. (iii) Draw the graph and verify.

SOLUTION (i) Slope

a=\frac{11-5}{3-1}=3

;

b=5-3\left(1\right)=2

⇒ $p\left(x\right)=3x+2$ . (ii) y-axis:

x=0\Rightarrow \left(0,2\right)

. x-axis:

\begin{aligned}3x+2&=0\\\Rightarrow x&=-\frac{2}{3}\\\Rightarrow \left(-\frac{2}{3},0\right)\text{.}\end{aligned}

(iii) Plot (0, 2), (1, 5), (3, 11) and $\left(-\frac{2}{3},0\right)$ , then join them. Substitution confirms all four points satisfy $y=3x+2$ .

Question 10: intercepts of

p\left(x\right)=3x+2

. Equal scales: each grid interval is 1 unit. Lines continue beyond the displayed window.

*11. Let $p\left(x\right)=ax+b$ and $q\left(x\right)=cx+d$ be linear polynomials such that: (i) $p\left(0\right)=5$ ; (ii) p(x) – q(x) cuts the x-axis at (3, 0); (iii) $p\left(x\right)+q\left(x\right)=6x+4$ for all real x. Find p(x) and q(x).

SOLUTION (i)

p\left(0\right)=b=5

. (iii)

a+c=6

and

b+d=4\Rightarrow d=-1

. (ii)

\begin{aligned}p\left(x\right)-q\left(x\right)&=\left(a-c\right)x+\left(b-d\right)\\&=\left(a-c\right)x+6\text{;}\end{aligned}

x=3

\begin{aligned}3\left(a-c\right)+6&=0\\\Rightarrow a-c&=-2\text{.}\end{aligned}

With

a+c=6

a=2

c=4

. ∴ $p\left(x\right)=2x+5$ and $q\left(x\right)=4x-1$ .

*12. Hexagons are made with matchsticks; a new hexagon shares a side with the previous one. (i) Draw the next two stages — how many matchsticks at each? (ii) Complete the table. (iii) Find the rule for the n^th stage. (iv) Matchsticks for the 15th stage? (v) Can 200 matchsticks form a stage?

SOLUTION Stage 1 needs 6 matchsticks; each new hexagon adds 5 (one shared side). So 1→6, 2→11, 3→16. (i) Stage

4=21

, Stage

5=26

matchsticks. (ii)/(iii) Rule for the n^th stage: 5n + 1. (iv) 15th stage:

5\left(15\right)+1=76

. (v)

\begin{aligned}5n+1&=200\\\Rightarrow 5n&=199\\\Rightarrow n&=39.8\text{,}\end{aligned}

not a whole number ⇒ No, 200 matchsticks cannot form a stage.

Stages 4 and 5: each new hexagon shares one complete side. Schematic; not to scale.

Stage	Matchsticks
1	6
2	11
3	16
4	21
5	26
n	5n + 1

*13. $p\left(x\right)=ax+b$ and $q\left(x\right)=cx+d$ such that: (i) p(x) passes through (2, 3) and (6, 11); (ii) q(x) passes through (4, –1); (iii) q(x) is parallel to p(x). Find p(x) and q(x), and where each meets the x-axis.

SOLUTION (i) Slope of

p=\frac{11-3}{6-2}=2

;

b=3-2\left(2\right)=-1

⇒ $p\left(x\right)=2x-1$ . (iii) Parallel ⇒ slope of q is also 2. (ii) Through (4, −1):

-1=2\left(4\right)+d\Rightarrow d=-9

⇒ $q\left(x\right)=2x-9$ . x-axis:

p\left(x\right)=0\Rightarrow x=\frac{1}{2}

, i.e.

\left(\frac{1}{2},0\right)

;

q\left(x\right)=0\Rightarrow x=\frac{9}{2}

, i.e.

\left(\frac{9}{2},0\right)

. (Being parallel, the two lines never meet each other.)

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

SOLUTION

f\left(x\right)=ax+a=a\left(x+1\right)

. At

x=-1

f\left(-1\right)=0

for every a. So all such lines pass through the point (−1, 0) (their common x-intercept), and being

a>0

they all rise from left to right.

Common Mistakes to Avoid

Watch out for these

Confusing a coefficient (the number) with the term or the variable.
Forgetting that a non-zero constant has degree 0, and that a missing term means coefficient 0.
Swapping slope (a) and y-intercept (b) in $y=ax+b$ .
Sign slips when transposing terms while solving a linear equation.
Substituting negatives wrongly: $\left(-3\right)^{2}=9$ , not −9.
Misreading word problems (which quantity is ‘twice’ or ‘5 times’ the other).

Practice MCQs & Assertion–Reason

Additional ClearStudy practice, separate from the NCERT exercises.

1. The degree of the polynomial 7x − 4 is:

(a) 0 (b) 1 (c) 2 (d) 7

2. The value of 3x − 5 at $x=2$ is:

(a) 1 (b) −1 (c) 11 (d) 6

3. A polynomial of degree 1 is called a:

(a) constant polynomial (b) linear polynomial (c) quadratic polynomial (d) cubic polynomial

4. In $y=ax+b$ , the letter ‘a’ represents the:

(a) y-intercept (b) slope (c) constant term (d) degree

5. The y-intercept of the line $y=2x+5$ is:

(a) 2 (b) 5 (c) −5 (d) 0

6. The constant term of $4x^{2}-3x+9$ is:

(a) 4 (b) −3 (c) 9 (d) 0

7. Two distinct non-vertical lines are parallel if they have the same:

(a) y-intercept (b) slope (c) x-intercept (d) degree

8. The line $y=3x$ passes through the point:

(a) (0, 3) (b) (3, 0) (c) (0, 0) (d) (1, 0)

9. If $2x+1=9$ , then x =

(a) 3 (b) 4 (c) 5 (d) 8

10. The coefficient of x in $5x^{2}-7x+2$ is:

(a) 5 (b) −7 (c) 2 (d) 7

Answer key: 1-(b), 2-(a), 3-(b), 4-(b), 5-(b), 6-(c), 7-(b), 8-(c), 9-(b), 10-(b).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: The degree of a non-zero constant polynomial is 0.

Reason: A constant c can be written as $c\cdot x^{0}$ .

A-R 2. Assertion: The lines $y=2x+3$ and $y=2x-4$ are parallel.

Reason: Distinct non-vertical lines with equal slopes are parallel.

A-R 3. Assertion: The graph of $y=ax$ $\left(b=0\right)$ passes through the origin.

Reason: When $x=0$ , $y=0$ for the equation $y=ax$ .

A-R 4. Assertion: $3x^{2}+2x+1$ is a linear polynomial.

Reason: A linear polynomial has degree 1.

A-R 5. Assertion: In linear decay the slope is negative.

Reason: Linear decay means the quantity decreases by a fixed amount over equal intervals.

Answer key: 1-(A), 2-(A), 3-(A), 4-(D), 5-(A).

Quick Revision Summary

A non-zero polynomial’s degree is its highest power with a non-zero coefficient; degree 1 = linear, 2 = quadratic, 3 = cubic. Non-zero constants have degree 0.
To find a polynomial’s value, substitute the variable and simplify.
A linear pattern has a constant difference between consecutive terms.
Linear growth: increases by a fixed amount (slope $a>0$ ); linear decay: decreases by a fixed amount (slope $a<0$ ).
A linear relationship is $y=ax+b$ : a = slope, $b=y$ -intercept; if $b=0$ the line passes through the origin.
Parallel lines share the same slope a (only b differs).

Presenting clear mathematical working

State the degree/coefficient/constant clearly, show each step when solving linear equations, and for word problems always define the variable first (“Let $\ldots =x$ “). For graph questions, write the line as $y=ax+b$ , find two neat points, and label the slope and y-intercept.

Frequently Asked Questions

What is Class 9 Maths Ganita Manjari Chapter 2 about?

It covers polynomials and degree, linear polynomials, linear patterns, linear growth and decay, linear relationships $y=ax+b$ , and plotting straight-line graphs.

What is a linear polynomial?

A linear polynomial is a polynomial of degree 1, such as 2x + 3 or 5 − 4y.

What do a and b mean in

y=ax+b

a is the slope of the line and b is the y-intercept (where the line cuts the y-axis).

Are these Class 9 Maths Ganita Manjari Chapter 2 solutions free?

Yes. All solutions are free and follow the official NCERT Ganita Manjari textbook for 2026–27.

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