Class 9 · MathsNCERT 2026 Revised Syllabus
Chapter 2: Introduction to Linear Polynomials (0904CH02)
Ganita Manjari Part I · 39 questions · 7 exercises
Chapter code: 0904ch02 · Book: Ganita Manjari Part I · Class 9 Maths
Find complete NCERT solutions for Class 9 Maths, Ganita Manjari Part I Chapter 2. Every exercise has a clear, step-by-step answer for chapter code 0904CH02.
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Exercise Set 2.1(5 questions)
1Find the degrees of the following polynomials:(i) 2x2 – 5x + 3(ii) y3 + 2y – 1(iii) – 9(iv) 4z – 3
School Level Solution
i) Degree = 2 Ans.ii) Degree = 3 Ans.iii) Degree = 0 Ans.iv) Degree = 1 Ans.
2Write polynomials of degrees 1, 2 and 3.
School Level Solution
degree 1 = 7x + 2degree 2 = 7x² + 6x + 2degree 3 = x³ + 30x² + x + 2
3What are the coefficients of x2 and x3 in the polynomial x4 – 3x3 + 6x2 – 2x + 7?
School Level Solution
The coefficient of x² is 6 while the coefficient of x³ is −3.
4What is the coefficient of z in the polynomial 4z3 + 5z2 – 11?
School Level Solution
The coefficient of z in 4z³ + 5z² − 11 is 0.
5What is the constant term of the polynomial 9x3 + 5x2 – 8x –10?
School Level Solution
Constant term of the polynomial 9x³ + 5x² − 8x − 10 is −10.
Exercise Set 2.2(7 questions)
1Find the value of the linear polynomial 5x – 3 if:(i) x=0(ii) x = –1(iii) x = 2
School Level Solution
i) ∵ x = 0∴ 5x − 3 = 5×0 − 3 = −3 Ans.ii) ∵ x = −1∴ 5x − 3 = 5×(−1) − 3 = −8 Ans.iii) ∵ x = 2∴ 5x − 3 = 5×2 − 3 = 7 Ans.
2Find the value of the quadratic polynomial 7s2 – 4s + 6 if:(i) s=0(ii) s = –3(iii) s = 4
School Level Solution
i) ∵ s = 07s² − 4s + 6 = 7×0² − 4×0 + 6 = 6 Ans.ii) ∵ s = −37s² − 4s + 6 = 7×(−3)² − 4×(−3) + 6= 63 + 12 + 6 = 81 Ans.iii) ∵ s = 47s² − 4s + 6 = 7×4² − 4×4 + 6= 112 − 16 + 6 = 102 Ans.
3The 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.
School Level Solution
Let Salil’s present age be x years.Salil’s mother’s present age = 3x years.After 5 years,(x+5) + (3x+5) = 70⇒ 4x + 10 = 70⇒ x = 15 years∴ Salil’s present age = 15 years.∴ Salil’s mother’s present age = 45 years.
4The difference between two positive integers is 63. The ratio of the two integers is 2:5. Find the two integers.
School Level Solution
Let the two positive integers be 2x and 5x.Their difference = 5x − 2x = 63⇒ 3x = 63⇒ x = 21∴ The two required numbers are 42 and 105.
5Ruby 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?
School Level Solution
No. of ₹5 coins = xNo. of ₹2 coins = 3xAmount contributed by ₹5 coins = 5xAmount contributed by ₹2 coins = 2(3x) = 6x∴ 5x + 6x = 88⇒ 11x = 88⇒ x = 8∴ ₹5 coins = 8 and ₹2 coins = 24.
6A 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?
School Level Solution
Let the shorter piece be x feet.The longer piece = 4x feet.x + 4x = 300⇒ 5x = 300⇒ x = 60∴ Length of the shorter piece = 60 ft and longer piece = 240 ft.
7If 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?
School Level Solution
Let breadth = x cm.Length = (2x + 3) cm.Perimeter = 2(l+b)24 = 2[(2x+3)+x]⇒ 24 = 6x + 6⇒ x = 3∴ Breadth = 3 cm and length = 9 cm.
Exercise Set 2.3(5 questions)
1A student has ₹500 in her savings bank account. She gets ₹150 every month as pocket money. How much money will she have at the end of every month from the second month onwards? Find a linear expression to represent the amount she will have in the nth month.
School Level Solution
Initial savings = ₹500Pocket money received every month = ₹150Amount after 1 month = 500 + 150 = ₹650Amount after 2 months = 500 + 300 = ₹800Amount after 3 months = 500 + 450 = ₹950∴ Linear expression for amount after n months = 150n + 500 Ans.
2A rally starts with 120 members. Each hour, 9 members drop out of the group. How many members will remain after 1, 2, 3, … hours? Find a linear expression to represent the number of members at the end of the nth hour.
School Level Solution
Initial members = 120Members dropping out every hour = 9People remaining after 1 hour = 120 − 9 = 111People remaining after 2 hours = 120 − 18 = 102People remaining after 3 hours = 120 − 27 = 93∴ Linear expression = 120 − 9n Ans.
3Suppose 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.
School Level Solution
Length of rectangle = 13 cm.i) Breadth = 12 cm ⇒ Area = 13×12 = 156 cm².ii) Breadth = 10 cm ⇒ Area = 13×10 = 130 cm².iii) Breadth = 8 cm ⇒ Area = 13×8 = 104 cm².The areas decrease by 26 cm² each time.∴ Linear pattern = 182 − 26n = 13(14 − 2n).
4Suppose 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.
School Level Solution
Length = 7 cm and breadth = 11 cm.i) h = 5 cm: V = 7×11×5 = 385 cm³.ii) h = 9 cm: V = 7×11×9 = 693 cm³.iii) h = 13 cm: V = 7×11×13 = 1001 cm³.∴ Linear pattern = 77(4n+1).
5Sarita 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.
School Level Solution
Initial pages = 500Pages read per day = 20Pages left after n days = 500 − 20nAfter 15 days = 500 − 20×15 = 200 pages Ans.
Exercise Set 2.4(4 questions)
1Suppose 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.
School Level Solution
Initial height = 1.75 ftGrowth each month = 0.5 fti) Height after 7 months = 1.75 + 7×0.5 = 5.25 ft.ii) Table:
iii) h = 1.75 + 0.5t = 0.5t + 1.75. It represents linear growth because h changes by a constant 0.5 ft per month.
| t (months) | h (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 |
2A 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.
School Level Solution
Initial value of mobile phone = ₹10,000Decrease each year = ₹800i) Value after 3 years = 10000 − 3×800 = ₹7600.ii) Table:
iii) v = 10000 − 800t. Since the value decreases by a constant amount every year, it is linear decay.
| t (years) | v (₹) |
|---|---|
| 0 | 10000 |
| 1 | 9200 |
| 2 | 8400 |
| 3 | 7600 |
| 4 | 6800 |
| 5 | 6000 |
| 6 | 5200 |
| 7 | 4400 |
| 8 | 3600 |
3The 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.
School Level Solution
Initial population = 750Increase every year = 50i) Population after 6 years = 750 + 6×50 = 1050.ii) Table:
iii) P = 750 + 50t. The increase is constant, so it represents linear growth.
| t (years) | P |
|---|---|
| 0 | 750 |
| 1 | 800 |
| 2 | 850 |
| 3 | 900 |
| 4 | 950 |
| 5 | 1000 |
| 6 | 1050 |
| 7 | 1100 |
| 8 | 1150 |
| 9 | 1200 |
| 10 | 1250 |
4A 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.
School Level Solution
Let x be the number of days before the stock runs out. From the table, the stock decreases linearly by 50 kg per day, so the linear equation is w = 500 − 50x.
The stock becomes zero when 0 = 500 − 50x, giving x = 10 days.
Thus the completed table is: x (days) = 0, 2, 4, 6, 8, 10 and w (kg) = 500, 400, 300, 200, 100, 0.
Exercise Set 2.5(3 questions)
1A 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.
School Level Solution
Let y = ax + b.When x = 10, y = 400:10a + b = 400 ...(1)When x = 14, y = 500:14a + b = 500 ...(2)Subtracting (1) from (2):4a = 100 ⇒ a = 25.Putting a = 25 in (1): 250 + b = 400 ⇒ b = 150.∴ y = 25x + 150 Ans.
2A 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.
School Level Solution
Let y = ax + b.When x = 10, y = 800:10a + b = 800 ...(1)When x = 15, y = 1100:15a + b = 1100 ...(2)Subtracting, 5a = 300 ⇒ a = 60.Putting a = 60: 600 + b = 800 ⇒ b = 200.∴ y = 60x + 200 Ans.
3Consider 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.)
School Level Solution
Given, when °F = 32, °C = 0:0 = 32a + b ...(1)When °F = 212, °C = 100:100 = 212a + b ...(2)Subtracting, 100 = 180a⇒ a = 59From (1), b = −32×59 = −1609.∴ °C = (59)°F − 1609.
Exercise Set 2.6(1 questions)
1Draw 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
School Level Solution
i) y = 4x, y = 2x, y = x have y-intercept 0; their slopes are 4, 2 and 1 respectively.ii) y = −6x, y = −3x, y = −x have slopes −6, −3 and −1; all pass through the origin.iii) y = 5x and y = −5x have slopes 5 and −5; they are mirror lines about the y-axis.iv) y = 3x − 1, 3x, 3x + 1 have the same slope 3 and different y-intercepts −1, 0, 1, so they are parallel.v) y = −2x − 3, y = −2x, y = 2x + 3 have slopes −2, −2, 2; the first two are parallel.
End-of-Chapter Exercises(14 questions)
1Write a polynomial of degree 3 in the variable x, in which the coefficient of the x2 term is –7.
School Level Solution
One polynomial of degree 3 in x with coefficient of x² = −7 is x³ − 7x² + 2x + 1 Ans.
2Find the values of the following polynomials at the indicated values of the variables.(i) 5x2 – 3x + 7 if x = 1(ii) 4t3 – t2 + 6 if t = a
School Level Solution
i) For x = 1: 5x² − 3x + 7 = 5 − 3 + 7 = 9 Ans.ii) For t = a: 4t³ − t² + 6 = 4a³ − a² + 6 Ans.
3If we multiply a number by 2 and add 3 to the product, we get –712. Find the number.
School Level Solution
Let the number be x.2x + 3 = −7122x = −4312∴ x = −4324 Ans.
4A 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?
School Level Solution
Let the smaller number be x and the larger number be 5x.After adding 21: x+21 and 5x+21.The larger becomes twice the smaller:5x + 21 = 2(x+21)⇒ 5x + 21 = 2x + 42⇒ 3x = 21 ⇒ x = 7∴ numbers = 7 and 35.
5If 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.
School Level Solution
Initial amount = ₹800, saving every month = ₹250.i) After 6 months = 800 + 6×250 = ₹2300.ii) After 2 years = 24 months = 800 + 24×250 = ₹6800.Linear pattern = 800 + 250n.
★6The 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.
School Level Solution
Let one of the digits of the two digit no. be x.
The other digit = x − 3
The two digit no. = 10x + (x − 3)
= 10x + x − 3
= 11x − 3
The no. formed by inter-changing the digits =
10(x − 3) + x
= 10x − 30 + x
= 11x − 30
ATQ,
11x − 3 + 11x − 30 = 143
22x − 33 = 143
22x = 143 + 33
22x = 176
x = 17622
⇒ x = 8
The other digit = 8 − 3 = 5
The required nos. are 85 and 58 Ans.
★7Draw 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?
School Level Solution
(i) y = −3x + 4
Slope = −3
y-intercept = 4
Point where it cuts y-axis = (0, 4)
(ii) 2y = 4x + 7
y = 2x + 72
Slope = 2
y-intercept = 72
Point where it cuts y-axis = (0, 72)
(iii) 5y = 6x − 10
y = 65x − 2
Slope = 65
y-intercept = −2
Point where it cuts y-axis = (0, −2)
(iv) 3y = 6x − 11
y = 2x − 113
Slope = 2
y-intercept = −113
Point where it cuts y-axis = (0, −113)
(ii) and (iv) are parallel because their slopes are equal.
★8If 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 = 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.
School Level Solution
Given, y = 95(x − 273) + 32
x is Kelvin.
y is Fahrenheit.
i) x = 313 K
y = 95(313 − 273) + 32
= 95 × 40 + 32
= 72 + 32
= 104°F Ans.
Temperature of the liquid in Fahrenheit scale = 104°F
ii) Given, temperature is 158°F.
⇒ y = 158
158 = 95(x − 273) + 32
158 − 32 = 95(x − 273)
95(x − 273) = 126
x − 273 = 126 × 59
x − 273 = 70
x = 273 + 70 = 343 K Ans.
158°F in Kelvin is 343 K.
★9The 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.
School Level Solution
Work done = distance × force
Constant force = 3 units
w = d × 3
= 3d
When distance travelled = 2 units,
w = 3d
w = 3 × 2
w = 6 units
Verified by the point (2, 6) on the graph.
★10The 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.
School Level Solution
Let p(x) = ax + b.
Given, p(1) = 5
a + b = 5 ...(1)
Also, p(3) = 11
3a + b = 11 ...(2)
2a = 6
a = 3
From (1), b = 5 − 3 = 2
p(x) = 3x + 2
Now,
For x-axis, y = 0
0 = 3x + 2
3x = −2
x = −23
∴ Point = (−23, 0)
For y-axis, x = 0
y = 2
∴ Point = (0,2)
Hence verified.
★11Let 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).
School Level Solution
Let p(x) = ax + b and q(x) = cx + d.
i) p(0) = 5
a×0 + b = 5
∴ b = 5
iii) p(x) + q(x) = 6x + 4
(a+c)x + (b+d) = 6x + 4
a+c = 6 and b+d = 4
Since b=5, d = −1.
ii) p(x) − q(x) cuts x-axis at (3,0).
p(3) − q(3) = 0
(3a+5) − (3c−1) = 0
3a − 3c + 6 = 0
a − c = −2
Now, a+c=6
⇒ a=2, c=4
p(x) = 2x+5
q(x) = 4x−1
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 table for stages 1, 2, 3, 4, 5, …, n. (iii) Find a rule for the number of matchsticks required for the nth stage. (iv) How many matchsticks are required at the 15th stage? (v) How many matchsticks are required to make 200 matchsticks in the pattern?
School Level Solution
Stage 4 = 21 matchsticks.
∴ Stage 5 = 26 matchsticks.
iii) Rule to determining the no. of matchsticks in the nth stage =
[6 + (n−1)5] Ans.
= 6 + 5(n−1)
= 6 + 5n − 5
= 5n+1 Ans.
iv) No. of matchsticks required in the 15th stage = 5n+1
= 5×15+1
= 75+1
= 76 Ans.
v) To check if 200 matchsticks form a stage, let 5n+1=200
5n = 199
n = 1995 = 3945
n is not a natural no.
200 matchsticks do not form a stage.
| Stage no. | No. of matchsticks |
|---|---|
| 1 | 6 |
| 2 | 11 |
| 3 | 16 |
| 4 | 21 |
| 5 | 26 |
| … | … |
| n | 6+(n−1)5 |
★13Let 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.
School Level Solution
i) Let p(x) = ax+b.
Given, p(2)=3 and p(6)=11
2a+b=3 ...(1)
6a+b=11 ...(2)
4a=8
a=2
From (1), b=−1
p(x)=2x−1
ii) Given, q(x)=cx+d.
Since q(x) is parallel to p(x),
∴ slope of q(x) = slope of p(x)
∴ c=2
q(x) passes through (4,−1)
−1=2×4+d
d=−9
q(x)=2x−9
Now, p(x) meets x-axis when y=0:
0=2x−1 ⇒ x=12
∴ p(x) cuts x-axis at (12,0).
q(x) meets x-axis:
0=2x−9 ⇒ x=92=412
∴ q(x) cuts x-axis at (412,0).
★14What do all linear functions of the form f(x) = ax + a, a > 0, have in common?
School Level Solution
Given, f(x)=ax+a, a>0
⇒ y=ax+a
Here, slope = a and y-intercept = a.
a>0
slope > 0
it shows linear growth
f(x) will pass through positive y-axis.
For y=0,
0=ax+a
a(x+1)=0
x+1=0
x=−1
point of (x,y)=(−1,0) common.