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

NCERT Solutions Class 9 Maths Chapter 2 Introduction to Linear Polynomials

Exercise 2.1

1. Find the degrees of the following polynomials:

Solution:

(i) 2x² − 5x + 3
The highest power of x is 2.
Degree = 2

(ii) y³ + 2y − 1
The highest power of y is 3.
Degree = 3

(iii) −9
This is a constant, which can be written as −9x⁰.
Degree = 0

(iv) 4z − 3
The highest power of z is 1.
Degree = 1

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

Solution:

(Answers may vary; any correct example is acceptable.)

• Degree 1 (linear polynomial): 3x + 5
• Degree 2 (quadratic polynomial): x² + 4x − 7
• Degree 3 (cubic polynomial): 2x³ − x² + x − 9

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3. What are the coefficients of x² and x³ in the polynomial x⁴ − 3x³ + 6x² − 2x + 7?

Solution:

Coefficient of x³ = −3
Coefficient of x² = 6

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4. What is the coefficient of z in the polynomial 4z³ + 5z² − 11?

Solution:

The term containing z (i.e., z¹) is missing in this polynomial, so its coefficient is 0.

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5. What is the constant term of the polynomial 9x³ + 5x² − 8x − 10?

Solution:

The constant term (the term with no variable) is −10.

Exercise 2.2

1. Find the value of the linear polynomial 5x − 3 if:

Solution:

(i) x = 0
5(0) − 3 = 0 − 3 = −3

(ii) x = −1
5(−1) − 3 = −5 − 3 = −8

(iii) x = 2
5(2) − 3 = 10 − 3 = 7

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2. Find the value of the quadratic polynomial 7s² − 4s + 6 if:

Solution:

(i) s = 0
7(0)² − 4(0) + 6 = 0 − 0 + 6 = 6

(ii) s = −3
7(−3)² − 4(−3) + 6 = 7(9) + 12 + 6 = 63 + 12 + 6 = 81

(iii) s = 4
7(4)² − 4(4) + 6 = 7(16) − 16 + 6 = 112 − 16 + 6 = 102

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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 present age = x years.
Then his mother’s present age = 3x years.

After 5 years:
Salil’s age = (x + 5)
Mother’s age = (3x + 5)

According to the question:
(x + 5) + (3x + 5) = 70
4x + 10 = 70
4x = 60
x = 15

Salil’s present age = 15 years
Mother’s present age = 3 × 15 = 45 years

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

Solution:

Let the two integers be 2k and 5k.

5k − 2k = 63
3k = 63
k = 21

The two integers are 2 × 21 = 42 and 5 × 21 = 105.

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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 the number of five-rupee coins = x
Then the number of two-rupee coins = 3x

Total value:
5(x) + 2(3x) = 88
5x + 6x = 88
11x = 88
x = 8

Number of five-rupee coins = 8
Number of two-rupee coins = 3 × 8 = 24
Total number of coins = 8 + 24 = 32

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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 feet
Then the longer piece = 4x feet

x + 4x = 300
5x = 300
x = 60

Shorter piece = 60 feet
Longer piece = 4 × 60 = 240 feet

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7. If the length of a rectangle is three more than twice its width and its perimeter is 24 cm, what are the dimensions of the rectangle?

Solution:

Let the width = w cm
Then the length = (2w + 3) cm

Perimeter = 2(length + width) = 24
length + width = 12
(2w + 3) + w = 12
3w + 3 = 12
3w = 9
w = 3

Width = 3 cm
Length = 2(3) + 3 = 9 cm

Exercise 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:

At the end of the n^th month, the amount = ₹500 + ₹150 × n.

Linear expression: 500 + 150n

(End of month 2 = ₹800, end of month 3 = ₹950, and so on.)

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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:

Linear expression: 120 − 9n

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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 = length × breadth = 13 × b

(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

Linear pattern: Area, A = 13b (where b is the breadth)

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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 = length × breadth × height = 7 × 11 × h = 77h

(i) Height = 5 cm ⇒ Volume = 77 × 5 = 385 cm³
(ii) Height = 9 cm ⇒ Volume = 77 × 9 = 693 cm³
(iii) Height = 13 cm ⇒ Volume = 77 × 13 = 1001 cm³

Linear pattern: Volume, V = 77h (where h is the height)

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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 d days = 500 − 20d

After 15 days: 500 − 20(15) = 500 − 300 = 200 pages

Linear pattern: Pages left = 500 − 20d

Exercise 2.4

1. Suppose a plant has height 1.75 feet and it grows by 0.5 feet each month.

Solution:

(i) Find the height after 7 months.
h = 1.75 + 0.5(7) = 1.75 + 3.5 = 5.25 feet

(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.

h = 1.75 + 0.5t

This represents linear growth because as t increases by 1 (one month), the value of h always increases by a fixed amount, 0.5 feet. The plant’s height grows by a constant amount over equal intervals of time.

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

Solution:

(i) Find the value of the phone after 3 years.
v = 10000 − 800(3) = 10000 − 2400 = ₹7600

(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.

v = 10000 − 800t

This represents linear decay because as t increases by 1 (one year), the value v always decreases by a fixed amount, ₹800. The value decreases by a constant amount over equal time intervals.

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3. The initial population of a village is 750. Every year, 50 people move from a nearby city to the village.

Solution:

(i) Find the population of the village after 6 years.
P = 750 + 50(6) = 750 + 300 = 1050

(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.

P = 750 + 50t

This represents linear growth because for every increase of 1 in t (one year), the population P increases by a fixed amount of 50. The quantity increases by a constant value over equal intervals.

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4. A telecom company charges ₹600 for a certain recharge scheme. This prepaid balance is reduced by ₹15 each day after the recharge.

Solution:

(i) Write an equation that models the remaining balance b(x) after using the scheme for x days. Explain why it represents linear decay.

b(x) = 600 − 15x

This represents linear decay because as x increases by 1 day, the balance decreases by a fixed amount of ₹15 every time — a constant decrease over equal intervals.

(ii) After how many days will the balance run out?

600 − 15x = 0
15x = 600
x = 40

The balance will run out after 40 days.

(iii) Make a table of values for x varying from 1 to 10 days and show how the balance b(x), reduces with time.

Exercise 2.5

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

Solution:

When x = 10, y = 400: 400 = 10a + b …(1)
When x = 14, y = 500: 500 = 14a + b …(2)

Subtracting (1) from (2):
100 = 4a
a = 25

Substituting in (1):
400 = 10(25) + b
400 = 250 + b
b = 150

a = 25, b = 150 (so y = 25x + 150)

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

Solution:

When x = 10, y = 800: 800 = 10a + b …(1)
When x = 15, y = 1100: 1100 = 15a + b …(2)

Subtracting (1) from (2):
300 = 5a
a = 60

Substituting in (1):
800 = 10(60) + b
800 = 600 + b
b = 200

a = 60, b = 200 (so y = 60x + 200)

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3. Consider the relationship between temperature measured in degrees Celsius (°C) and degrees Fahrenheit (°F), which is given by °C = a °F + b. Find a and b, given that ice melts at 0 degrees Celsius and 32 degrees Fahrenheit, and water boils at 100 degrees Celsius and 212 degrees Fahrenheit.

Solution:

When °C = 0, °F = 32: 0 = 32a + b …(1)
When °C = 100, °F = 212: 100 = 212a + b …(2)

Subtracting (1) from (2):
100 = 180a
a = 100/180 = 5/9

Substituting in (1):
0 = 32(5/9) + b
b = −160/9

a = 5/9, b = −160/9

So the relationship is °C = (5/9)°F − 160/9, which is equivalent to the familiar formula °C = (5/9)(°F − 32).

Exercise 2.6

1. Draw the graphs of the following sets of lines. In each case, reflect on the role of ‘a’ and ‘b’.

Solution:

For all parts below, plot each line using two points: (0, 0) and one more point obtained by substituting a convenient value of x.

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

Observation: All three lines pass through the origin (0, 0), since b = 0 in every case. As the value of ‘a’ increases (1 → 2 → 4), the line becomes steeper, i.e., it rises more quickly. The line y = 4x is the steepest, and y = x is the least steep among the three.

(ii) y = −6x, y = −3x, y = −x

Observation: All three lines pass through the origin and slope downward (from left to right) since ‘a’ is negative in each case. As the magnitude of ‘a’ increases, the line becomes steeper in the downward direction. y = −6x is the steepest, y = −x is the least steep.

(iii) y = 5x, y = −5x

Observation: Both lines pass through the origin and have the same steepness (since |a| = 5 for both), but they slope in opposite directions — y = 5x rises from left to right (positive slope/linear growth), while y = −5x falls from left to right (negative slope/linear decay). The two lines are mirror images of each other about the x-axis.

(iv) y = 3x − 1, y = 3x, y = 3x + 1

Observation: Here, ‘a’ is fixed at 3 for all three lines, so all three lines have the same slope (same steepness/inclination) and are parallel to each other. Only the y-intercept ‘b’ differs (−1, 0, 1), so the lines are shifted vertically — up or down — but never meet.

(v) y = −2x − 3, y = −2x, y = 2x + 3

Observation: The first two lines (y = −2x − 3 and y = −2x) have the same slope (a = −2), so they are parallel to each other, differing only in y-intercept. The third line, y = 2x + 3, has a different (positive) slope, so it is not parallel to the other two; it slopes upward instead of downward.

End-of-Chapter Exercises

1. Write a polynomial of degree 3 in the variable x, in which the coefficient of the x² term is −7.

Solution:

(Answers may vary.) One example:

2x³ − 7x² + 4x − 1

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2. Find the values of the following polynomials at the indicated values of the variables.

Solution:

(i) 5x² − 3x + 7 if x = 1

5(1)² − 3(1) + 7 = 5 − 3 + 7 = 9

(ii) 4t³ − t² + 6 if t = a

4(a)³ − (a)² + 6 = 4a³ − a² + 6

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3. If we multiply a number by 5/2 and add 2/3 to the product, we get −7/12. Find the number.

Solution:

Let the number be x.

(5/2)x + 2/3 = −7/12

(5/2)x = −7/12 − 2/3 = −7/12 − 8/12 = −15/12 = −5/4

x = (−5/4) × (2/5) = −10/20 = −1/2

The number is −1/2.

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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
Then the larger (positive) number = 5x

After adding 21 to both:
New numbers are (x + 21) and (5x + 21)

Since one becomes twice the other, and 5x + 21 > x + 21:
5x + 21 = 2(x + 21)
5x + 21 = 2x + 42
3x = 21
x = 7

The numbers are 7 and 5 × 7 = 35.

Check: 7 + 21 = 28, 35 + 21 = 56, and 56 = 2 × 28. ✓

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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:

Linear pattern: Amount, A = 800 + 250n (where n is the number of months)

(i) After 6 months:
A = 800 + 250(6) = 800 + 1500 = ₹2300

(ii) After 2 years (24 months):
A = 800 + 250(24) = 800 + 6000 = ₹6800

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*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 digit) and b (units digit), with a > b, so that a − b = 3.

Original number = 10a + b
Number after interchanging digits = 10b + a

Sum: (10a + b) + (10b + a) = 143
11a + 11b = 143
11(a + b) = 143
a + b = 13

Now we solve:
a − b = 3
a + b = 13

Adding: 2a = 16 ⇒ a = 8
Then b = 13 − 8 = 5

The original number is 85 (digits 8 and 5), and the number obtained after interchanging the digits is 58.

Check: 85 + 58 = 143. ✓ Also, 8 − 5 = 3. ✓

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*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. Are any of the lines parallel?

Solution:

(i) y = −3x + 4

This is already in the form y = ax + b.
Slope (a) = −3, y-intercept (b) = 4
Point where it cuts the y-axis: (0, 4)

Two points to plot: (0, 4) and (1, 1)

(ii) 2y = 4x + 7

Dividing throughout by 2: y = 2x + 7/2
Slope (a) = 2, y-intercept (b) = 7/2 = 3.5
Point where it cuts the y-axis: (0, 3.5)

Two points to plot: (0, 3.5) and (1, 5.5)

(iii) 5y = 6x − 10

Dividing throughout by 5: y = (6/5)x − 2
Slope (a) = 6/5 = 1.2, y-intercept (b) = −2
Point where it cuts the y-axis: (0, −2)

Two points to plot: (0, −2) and (5, 4)

(iv) 3y = 6x − 11

Dividing throughout by 3: y = 2x − 11/3
Slope (a) = 2, y-intercept (b) = −11/3
Point where it cuts the y-axis: (0, −11/3)

Two points to plot: (0, −11/3) and (2, 1/3)

Are any of the lines parallel?

Lines (ii) [y = 2x + 7/2] and (iv) [y = 2x − 11/3] both have slope 2.

Yes, lines (ii) and (iv) are parallel, since they have the same slope (a = 2) but different y-intercepts.

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*8. If the temperature of a liquid can be measured in Kelvin units as x K and in Fahrenheit units as y °F, the relation between the two systems of measurement of temperature is given by the linear equation y = (9/5)(x − 273) + 32.

Solution:

(i) Find the temperature of the liquid in Fahrenheit if the temperature of the liquid is 313 K.

y = (9/5)(313 − 273) + 32
y = (9/5)(40) + 32
y = 72 + 32
y = 104°F

(ii) If the temperature is 158 °F, then find the temperature in Kelvin.

158 = (9/5)(x − 273) + 32
158 − 32 = (9/5)(x − 273)
126 = (9/5)(x − 273)
126 × (5/9) = x − 273
70 = x − 273
x = 343

The temperature is 343 K.

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*9. The work done by a body on the application of a constant force is the product of the constant force and the distance travelled by the body in the direction of the force. Express this in the form of a linear equation in two variables (work w and distance d), and draw its graph by taking the constant force as 3 units. What is the work done when the distance travelled is 2 units? Verify it by plotting it on the graph.

Solution:

Work = Force × Distance

Taking the constant force as 3 units:

w = 3d

This is a linear equation in the variables w and d, passing through the origin (0, 0), with slope 3.

To plot the graph, we can use the points (0, 0) and (1, 3).

Work done when distance travelled is 2 units:
w = 3(2) = 6 units

This can be verified on the graph: the point (2, 6) lies on the line w = 3d.

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*10. The graph of a linear polynomial p(x) passes through the points (1, 5) and (3, 11).

Solution:

(i) Find the polynomial p(x).

Let p(x) = ax + b.

Since the graph passes through (1, 5): a + b = 5 …(1)
Since the graph passes through (3, 11): 3a + b = 11 …(2)

Subtracting (1) from (2):
2a = 6
a = 3

Substituting in (1): 3 + b = 5 ⇒ b = 2

p(x) = 3x + 2

(ii) Find the coordinates where the graph of p(x) cuts the axes.

y-axis (set x = 0): p(0) = 3(0) + 2 = 2 ⇒ point (0, 2)

x-axis (set p(x) = 0):
3x + 2 = 0
x = −2/3 ⇒ point (−2/3, 0)

(iii) Draw the graph of p(x) and verify your answers.

Plot the points (1, 5) and (3, 11) (or, more simply, (0, 2) and (−2/3, 0)), join them with a straight line, and extend in both directions. The line should pass through (0, 2) on the y-axis and (−2/3, 0) on the x-axis, confirming the answers found above.

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*11. Let p(x) = ax + b and q(x) = cx + d be two linear polynomials such that: (i) p(0) = 5. (ii) The polynomial p(x) − q(x) cuts the x-axis at (3, 0). (iii) The sum p(x) + q(x) is equal to 6x + 4 for all real x. Find the polynomials p(x) and q(x).

Solution:

From condition (i):
p(0) = b = 5, so b = 5

From condition (iii):
p(x) + q(x) = (ax + b) + (cx + d) = (a + c)x + (b + d) = 6x + 4
So, a + c = 6 and b + d = 4.
Since b = 5, we get d = 4 − 5 = −1

From condition (ii):
p(x) − q(x) = (ax + b) − (cx + d) = (a − c)x + (b − d)
This cuts the x-axis at (3, 0), meaning when x = 3, p(x) − q(x) = 0:
(a − c)(3) + (b − d) = 0
b − d = 5 − (−1) = 6
3(a − c) + 6 = 0
a − c = −2

Now solve:
a + c = 6
a − c = −2

Adding: 2a = 4 ⇒ a = 2
Then c = 6 − 2 = 4

p(x) = 2x + 5
q(x) = 4x − 1

Check: p(x) + q(x) = (2x + 5) + (4x − 1) = 6x + 4 ✓
p(x) − q(x) = (2x + 5) − (4x − 1) = −2x + 6, which equals 0 when x = 3 ✓

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*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.

Solution:

(i) Draw the next two stages of the pattern. How many matchsticks will be required at these stages?

A single hexagon uses 6 matchsticks. Each time a new hexagon is added sharing one side with the previous hexagon, only 5 new matchsticks are needed (since one side is shared, that stick is not repeated).

Stage 1: 6 matchsticks
Stage 2: 6 + 5 = 11 matchsticks
Stage 3: 11 + 5 = 16 matchsticks
Stage 4: 16 + 5 = 21 matchsticks
Stage 5: 21 + 5 = 26 matchsticks

(ii) Complete the following table.

(iii) Find a rule to determine the number of matchsticks required for the n^th stage.

Each new stage adds 5 matchsticks to the previous stage (a linear pattern with common difference 5), starting from 6 at Stage 1.

Number of matchsticks at Stage n = 5n + 1

(iv) How many matchsticks will be required for the 15th stage of the pattern?

5(15) + 1 = 75 + 1 = 76 matchsticks

(v) Can 200 matchsticks form a stage in this pattern? Justify your answer.

We need 5n + 1 = 200
5n = 199
n = 199/5 = 39.8

Since n must be a whole number (a stage number), and 39.8 is not a whole number, 200 matchsticks cannot form a stage in this pattern.

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*13. Let p(x) = ax + b and q(x) = cx + d be two linear polynomials such that: (i) The graph of p(x) passes through the points (2, 3) and (6, 11). (ii) The graph of q(x) passes through the point (4, −1). (iii) The graph of q(x) is parallel to the graph of p(x). Find the polynomials p(x) and q(x). Also, find the coordinates of the point where these lines meet the x-axis.

Solution:

Finding p(x):

Since the graph passes through (2, 3): 2a + b = 3 …(1)
Since the graph passes through (6, 11): 6a + b = 11 …(2)

Subtracting (1) from (2):
4a = 8
a = 2

Substituting in (1): 2(2) + b = 3 ⇒ b = −1

p(x) = 2x − 1

Finding q(x):

Since q(x) is parallel to p(x), they have the same slope: c = 2.

Since the graph of q(x) passes through (4, −1):
2(4) + d = −1
8 + d = −1
d = −9

q(x) = 2x − 9

Points where the lines meet the x-axis:

For p(x) = 0: 2x − 1 = 0 ⇒ x = 1/2 ⇒ point (1/2, 0)

For q(x) = 0: 2x − 9 = 0 ⇒ x = 9/2 ⇒ point (9/2, 0)

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

Solution:

For f(x) = ax + a, the y-intercept is also ‘a’ (since b = a here), and the slope is also ‘a’.

Setting f(x) = 0: ax + a = 0 ⇒ a(x + 1) = 0. Since a > 0 (a ≠ 0), we get x = −1.

All such lines pass through the point (−1, 0) on the x-axis, regardless of the value of a (as long as a > 0). This is because f(−1) = a(−1) + a = −a + a = 0 for every value of a. So every line of the form y = ax + a has the same x-intercept, (−1, 0), and all such lines pass through this common point, only differing in their slope/steepness.