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Class 9 · Mathematics · New NCERT

NCERT Solutions for Class 9 Mathematics Chapter 2: Introduction to Linear Polynomials

Chapter 2 introduces polynomials in one variable and focuses on linear polynomials (degree 1). Starting from everyday situations such as buying boxes of pens, fencing a garden and fixed-plus-per-use charges, the chapter explains terms, coefficients, variables and degree, treats a polynomial as an input-output process, and uses linear patterns to model linear growth and linear decay. It then shows how to find the linear relationship y = ax + b from two data points and how to draw its graph, where a is the slope and b is the y-intercept.

Chapter overview

  • Algebraic expressions, terms, coefficients and variables
  • Polynomials in one variable and their degree (constant, linear, quadratic, cubic)
  • Linear polynomials and linear equations
  • Polynomials as input-output processes (functions)
  • Linear patterns and the nth term
  • Linear growth and linear decay
  • Finding the linear relationship y = ax + b from two points
  • Graphing linear relationships: slope, y-intercept and parallel lines

Introduction to Linear Polynomials

This chapter introduces linear polynomials — polynomials of degree 1 — and builds up from the vocabulary of polynomials (degree, coefficients, constant term) to evaluating them, writing linear patterns, modelling linear growth and decay, finding the constants a and b in the relation y = ax + b, and graphing lines to see how a (slope) and b (y-intercept) control their position. The solutions below cover Exercise Sets 2.1 to 2.6 and the End-of-Chapter Exercises.

Exercise Set 2.1

Q1. Find the degrees of the following polynomials:

Q1(i). 2x² − 5x + 3

Answer

The degree of a polynomial is the highest power of the variable.

Steps

  1. Highest power of x = 2
  2. ∴ Degree = 2
Degree=2\text{Degree} = 2

Q1(ii). y³ + 2y − 1

Answer

Steps

  1. Highest power of y = 3
  2. ∴ Degree = 3
Degree=3\text{Degree} = 3

Q1(iii). −9

Answer

This is a constant polynomial (no variable).

∴ Degree=0\therefore\ \text{Degree} = 0

Q1(iv). 4z − 3

Answer

Steps

  1. Highest power of z = 1
  2. ∴ Degree = 1
Degree=1\text{Degree} = 1

Q2. Write polynomials of degrees 1, 2 and 3.

Answer

A polynomial of degree 1 (linear polynomial). Example:

2x+52x + 5

A polynomial of degree 2 (quadratic polynomial). Example:

x2+3x+1x^2 + 3x + 1

A polynomial of degree 3 (cubic polynomial). Example:

x3−2x2+x+4x^3 - 2x^2 + x + 4

Q3. What are the coefficients of x² and x³ in the polynomial x⁴ − 3x³ + 6x² − 2x + 7?

Answer

Given polynomial: x⁴ − 3x³ + 6x² − 2x + 7

Steps

  1. Coefficient of x³ = −3
  2. Coefficient of x² = 6
Coefficient of x3=−3,  Coefficient of x2=6\text{Coefficient of } x^3 = -3,\ \ \text{Coefficient of } x^2 = 6

Q4. What is the coefficient of z in the polynomial 4z³ + 5z² − 11?

Answer

The given polynomial is 4z³ + 5z² − 11. There is no term containing z¹ (i.e., z).

Hence, the coefficient of z is 0\text{Hence, the coefficient of } z \text{ is } 0

Q5. What is the constant term of the polynomial 9x³ + 5x² − 8x − 10? Recall that polynomials of degree 1 are called linear polynomials. In this chapter, we shall study linear polynomials.

Answer

The constant term is the term without any variable. In the polynomial 9x³ + 5x² − 8x − 10, the constant term is −10.

Constant term=−10\text{Constant term} = -10

Exercise Set 2.2

Q1. Find the value of the linear polynomial 5x − 3 if:

Q1(i). x = 0

Answer

Given polynomial: 5x − 3

Steps

  1. 5(0) − 3 = −3
5x−3=−3 when x=05x - 3 = -3 \text{ when } x = 0

Q1(ii). x = −1

Answer

Steps

  1. 5(−1) − 3 = −5 − 3 = −8
5x−3=−8 when x=−15x - 3 = -8 \text{ when } x = -1

Q1(iii). x = 2

Answer

Steps

  1. 5(2) − 3 = 10 − 3 = 7
5x−3=7 when x=25x - 3 = 7 \text{ when } x = 2

Q2. Find the value of the quadratic polynomial 7s² − 4s + 6 if:

Q2(i). s = 0

Answer

Steps

  1. 7s² − 4s + 6
  2. = 7(0)² − 4(0) + 6 = 6
7s2−4s+6=6 when s=07s^2 - 4s + 6 = 6 \text{ when } s = 0

Q2(ii). s = −3

Answer

Steps

  1. 7s² − 4s + 6
  2. = 7(−3)² − 4(−3) + 6
  3. = 7(9) + 12 + 6
  4. = 63 + 12 + 6 = 81
7s2−4s+6=81 when s=−37s^2 - 4s + 6 = 81 \text{ when } s = -3

Q2(iii). s = 4

Answer

Steps

  1. 7s² − 4s + 6
  2. = 7(4)² − 4(4) + 6
  3. = 7(16) − 16 + 6
  4. = 112 − 16 + 6 = 102
7s2−4s+6=102 when s=47s^2 - 4s + 6 = 102 \text{ when } s = 4

Q3. 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.

Answer

Let Salil's present age = x years. Then mother's present age = 3x years. After 5 years, Salil's age = x + 5 and mother's age = 3x + 5.

According to the question:

Steps

  1. (x + 5) + (3x + 5) = 70
  2. ⇒ 4x + 10 = 70
  3. ⇒ 4x = 60
  4. ⇒ x = 15

Therefore, Salil's age = 15 years and mother's age = 45 years.

Salil=15 years,  Mother=45 years\text{Salil} = 15 \text{ years},\ \ \text{Mother} = 45 \text{ years}

Q4. The difference between two positive integers is 63. The ratio of the two integers is 2:5. Find the two integers.

Answer

Let the integers be 2x and 5x. Their difference is 63:

Steps

  1. 5x − 2x = 63
  2. ⇒ 3x = 63
  3. ⇒ x = 21
  4. Integers: 2x = 42 and 5x = 105
The required integers are 42 and 105\text{The required integers are } 42 \text{ and } 105

Q5. 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?

Answer

Let the number of five-rupee coins = x. Then the number of two-rupee coins = 3x. Total value:

Steps

  1. 5x + 2(3x) = 88
  2. ⇒ 5x + 6x = 88
  3. ⇒ 11x = 88
  4. ⇒ x = 8

Hence:

  • Five-rupee coins = 8
  • Two-rupee coins = 24

Q6. 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?

Answer

Let the shorter piece = x feet. Then the longer piece = 4x feet. Total:

Steps

  1. x + 4x = 300
  2. ⇒ 5x = 300
  3. ⇒ x = 60

Therefore:

  • Shorter piece = 60 feet
  • Longer piece = 240 feet

Q7. 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?

Answer

Let width = x cm. Then length = 2x + 3 cm. Perimeter = 2(length + width):

Steps

  1. 2[(2x + 3) + x] = 24
  2. ⇒ 2(3x + 3) = 24
  3. ⇒ 6x + 6 = 24
  4. ⇒ 6x = 18
  5. ⇒ x = 3

Therefore:

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

Exercise Set 2.3

Q1. 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 nth month.

Answer

Initial amount in bank = ₹500. Monthly pocket money = ₹150. At the end of:

Steps

  1. 2nd month = 500 + 2(150) = ₹800
  2. 3rd month = 500 + 3(150) = ₹950
  3. 4th month = 500 + 4(150) = ₹1100
  4. and so on…

Let the amount in the nth month be Aₙ. Then Aₙ = 500 + 150n. Thus, the required linear expression is:

An=150n+500A_n = 150n + 500

Q2. 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 nth hour.

Answer

Initial members = 120. Members leaving each hour = 9. After:

Steps

  1. 1 hour = 120 − 9 = 111
  2. 2 hours = 120 − 18 = 102
  3. 3 hours = 120 − 27 = 93

Let the number of members after n hours = Mₙ. Then Mₙ = 120 − 9n. Thus, the required linear expression is:

Mn=120−9nM_n = 120 - 9n

Q3. 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.

Answer

Length = 13 cm and Area = Length × Breadth.

Steps

  1. (i) Breadth = 12 cm: Area = 13 × 12 = 156 cm²
  2. (ii) Breadth = 10 cm: Area = 13 × 10 = 130 cm²
  3. (iii) Breadth = 8 cm: Area = 13 × 8 = 104 cm²

Let breadth = x cm. Then Area = 13x. Thus, the linear pattern is:

A=13xA = 13x

Q4. 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.

Answer

Length = 7 cm, Breadth = 11 cm. Volume = Length × Breadth × Height = 7 × 11 × h = 77h.

Steps

  1. (i) Height = 5 cm: Volume = 77 × 5 = 385 cm³
  2. (ii) Height = 9 cm: Volume = 77 × 9 = 693 cm³
  3. (iii) Height = 13 cm: Volume = 77 × 13 = 1001 cm³

Let height = h cm. Then Volume = 77h. Thus, the linear pattern is:

V=77hV = 77h

Q5. 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.

Answer

Steps

  1. Total pages = 500
  2. Pages read per day = 20
  3. Pages read in 15 days = 20 × 15 = 300
  4. Pages left = 500 − 300 = 200

Let the pages left after n days = Pₙ. So Pₙ = 500 − 20n. Thus, the linear pattern is:

Pn=500−20nP_n = 500 - 20n

Exercise Set 2.4

Q1. Suppose a plant has height 1.75 feet and it grows by 0.5 feet each month.

Q1(i). Find the height after 7 months.

Answer

Given: initial height = 1.75 feet; growth per month = 0.5 feet. Height growth in each month = 0.5 feet, so the height after 7 months is:

Steps

  1. h = 1.75 + (0.5 × 7)
  2. = 1.75 + 3.5
  3. = 5.25 feet

Q1(ii). Make a table of values for t varying from 0 to 10 months and show how the height, h, increases every month.

Answer

t (months)h (feet)
01.75
12.25
22.75
33.25
43.75
54.25
64.75
75.25
85.75
96.25
106.75

Q1(iii). Find an expression that relates h and t, and explain why it represents linear growth.

Answer

Let the height after t months = h. Then:

h=1.75+0.5th = 1.75 + 0.5t

This represents linear growth because the height increases by a constant amount (0.5 feet) every month.

Q2. A mobile phone is bought for ₹10,000. Its value decreases by ₹800 every year.

Q2(i). Find the value of the phone after 3 years.

Answer

Given: initial value = ₹10,000; decrease per year = ₹800. Decrease in value after 1 year = ₹800, so the value after 3 years is:

Steps

  1. v = 10000 − (800 × 3)
  2. = 10000 − 2400
  3. = ₹7600

Q2(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.

Answer

t (years)v (₹)
010000
19200
28400
37600
46800
56000
65200
74400
83600

Q2(iii). Find an expression that relates v and t, and explain why it represents linear decay.

Answer

Let the value after t years = v. Then:

v=10000−800tv = 10000 - 800t

This represents linear decay because the value decreases by a constant amount (₹800) every year.

Q3. The initial population of a village is 750. Every year, 50 people move from a nearby city to the village.

Q3(i). Find the population of the village after 6 years.

Answer

Steps

  1. P = 750 + (50 × 6)
  2. = 750 + 300
  3. = 1050

Q3(ii). Make a table of values for t varying from 0 to 10 years and show how the population, P, increases every year.

Answer

t (years)P (population)
0750
1800
2850
3900
4950
51000
61050
71100
81150
91200
101250

Q3(iii). Find an expression that relates P and t, and explain why it represents linear growth.

Answer

Let the population after t years = P. Then:

P=750+50tP = 750 + 50t

This represents linear growth because the population increases by a constant number (50 people) every year.

Q4. A telecom company charges ₹600 for a certain recharge scheme. This prepaid balance is reduced by ₹15 each day after recharge.

Q4(i). Write an equation that models the remaining balance b(x) after using the scheme for x days. Explain why it represents linear decay.

Answer

Let the remaining balance after x days = b(x). Then:

b(x)=600−15xb(x) = 600 - 15x

This represents linear decay because the balance decreases by a constant amount (₹15) every day.

Q4(ii). After how many days will the balance run out?

Answer

The balance runs out when b(x) = 0:

Steps

  1. 600 − 15x = 0
  2. 15x = 600
  3. x = 40

So, the balance will run out after 40 days.

Q4(iii). Make a table of values for x varying from 1 to 10 days and show how the balance b(x) reduces with time.

Answer

x (days)b(x) (₹)
1585
2570
3555
4540
5525
6510
7495
8480
9465
10450

Exercise Set 2.5

Q1. A learning platform charges a fixed monthly fee and an additional cost per digital learning module accessed. A student observed 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.

Answer

Given: when x = 10, y = 400; when x = 14, y = 500. Using y = ax + b:

Steps

  1. For x = 10: 400 = 10a + b …(1)
  2. For x = 14: 500 = 14a + b …(2)

Subtracting (1) from (2), we get:

Steps

  1. 500 − 400 = 14a − 10a
  2. ⇒ 100 = 4a
  3. ⇒ a = 25

Substituting a = 25 in (1), we get:

Steps

  1. 400 = 10(25) + b
  2. ⇒ 400 = 250 + b
  3. ⇒ b = 150
a=25 and b=150a = 25 \text{ and } b = 150

Q2. 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.

Answer

Given: when x = 10, y = 800; when x = 15, y = 1100. Using y = ax + b:

Steps

  1. For x = 10: 800 = 10a + b …(1)
  2. For x = 15: 1100 = 15a + b …(2)

Subtracting (1) from (2), we have:

Steps

  1. 1100 − 800 = 15a − 10a
  2. ⇒ 300 = 5a
  3. ⇒ a = 60

Substituting a = 60 in (1), we get:

Steps

  1. 800 = 10(60) + b
  2. ⇒ 800 = 600 + b
  3. ⇒ b = 200
a=60 and b=200a = 60 \text{ and } b = 200

Q3. 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.)

Answer

Given relation: °C = a°F + b. Using the point (°F, °C) = (32, 0):

Steps

  1. 0 = 32a + b …(1)

Using the point (°F, °C) = (212, 100):

Steps

  1. 100 = 212a + b …(2)

Subtracting (1) from (2), we get:

Steps

  1. 100 − 0 = 212a − 32a
  2. ⇒ 100 = 180a
  3. ⇒ a = 100/180
  4. ⇒ a = 5/9

Substituting a = 5/9 in (1), we get:

Steps

  1. 0 = 32(5/9) + b
  2. ⇒ 0 = 160/9 + b
  3. ⇒ b = −160/9

Therefore, a = 5/9 and b = −160/9. Hence, the linear relationship is:

∘C=59 ∘F−1609^\circ C = \tfrac{5}{9}\,^\circ F - \tfrac{160}{9}

It can also be written as:

∘C=59 (∘F−32)^\circ C = \tfrac{5}{9}\,(^\circ F - 32)

Exercise Set 2.6

Q1. Draw the graphs of the following sets of lines. In each case, reflect on the role of ‘a’ and ‘b’.

Q1(i). y = 4x, y = 2x, y = x

Answer

All lines are of the form y = ax (b = 0).

Observation:

  • All lines pass through the origin (0, 0).
  • The value of ‘a’ (slope) determines steepness.
  • Larger ‘a’ → steeper line.
Graph of y = 4x (red), y = 2x (green) and y = x (blue): three rising straight lines through the origin, with the labelled points (1, 4) and (2, 8) on y = 4x, (1, 2) and (2, 4) on y = 2x, and (1, 1) and (2, 2) on y = x. The larger the value of a, the steeper the line.

Image unavailable. Open original

Graph of y = 4x (red), y = 2x (green) and y = x (blue): three rising straight lines through the origin, with the labelled points (1, 4) and (2, 8) on y = 4x, (1, 2) and (2, 4) on y = 2x, and (1, 1) and (2, 2) on y = x. The larger the value of a, the steeper the line.

Q1(ii). y = −6x, y = −3x, y = −x

Answer

All lines are of the form y = ax (b = 0).

Observation:

  • All lines pass through the origin.
  • Negative ‘a’ means lines slope downward.
  • Larger magnitude of ‘a’ → steeper downward slope.
Graph of y = −6x (red), y = −3x (green) and y = −x (blue): three falling straight lines through the origin, with the labelled points (1, −6) and (2, −12) on y = −6x, (1, −3) and (2, −6) on y = −3x, and (−1, 1) and (1, −1) on y = −x. A stray label (−2, −2) with no plotted point also appears in the lower-left part of the grid.

Image unavailable. Open original

Graph of y = −6x (red), y = −3x (green) and y = −x (blue): three falling straight lines through the origin, with the labelled points (1, −6) and (2, −12) on y = −6x, (1, −3) and (2, −6) on y = −3x, and (−1, 1) and (1, −1) on y = −x. A stray label (−2, −2) with no plotted point also appears in the lower-left part of the grid.

Q1(iii). y = 5x, y = −5x

Answer

Observation:

  • Both lines pass through the origin.
  • y = 5x slopes upward, y = −5x slopes downward.
  • Same magnitude of ‘a’ → same steepness but opposite direction.
Graph of y = 5x (red, rising) and y = −5x (green, falling) through the origin, with the labelled points (1, 5) and (2, 10) on y = 5x and (1, −5) and (2, −10) on y = −5x. The two lines have the same steepness but slope in opposite directions; some y-axis tick labels are missing and the plotted dots sit slightly off the exact grid positions.

Image unavailable. Open original

Graph of y = 5x (red, rising) and y = −5x (green, falling) through the origin, with the labelled points (1, 5) and (2, 10) on y = 5x and (1, −5) and (2, −10) on y = −5x. The two lines have the same steepness but slope in opposite directions; some y-axis tick labels are missing and the plotted dots sit slightly off the exact grid positions.

Q1(iv). y = 3x − 1, y = 3x, y = 3x + 1

Answer

All lines have the same slope (a = 3).

Observation:

  • Lines are parallel (same slope).
  • Different values of ‘b’ shift the line up or down.
  • b = −1 → line below origin
  • b = 0 → passes through origin
  • b = +1 → line above origin.
Graph of three parallel rising lines of slope 3: y = 3x − 1 (red, through (0, −1), with labelled points (0, −1), (1, 2) and (2, 5)), y = 3x (green, through the origin, with labelled point (2, 6)) and y = 3x + 1 (blue, through (0, 1), with labelled point (2, 7)). In the figure, the blue line’s point at (1, 4) is mislabelled “(1, 3)”.

Image unavailable. Open original

Graph of three parallel rising lines of slope 3: y = 3x − 1 (red, through (0, −1), with labelled points (0, −1), (1, 2) and (2, 5)), y = 3x (green, through the origin, with labelled point (2, 6)) and y = 3x + 1 (blue, through (0, 1), with labelled point (2, 7)). In the figure, the blue line’s point at (1, 4) is mislabelled “(1, 3)”.

Q1(v). y = −2x − 3, y = −2x, y = 2x + 3

Answer

Observation:

  • y = −2x − 3 and y = −2x have the same slope (−2) → parallel lines.
  • y = 2x + 3 has positive slope → different direction.
  • ‘b’ changes the vertical position of the line.

Conclusion:

Steps

  1. ‘a’ (coefficient of x) controls slope (steepness and direction).
  2. ‘b’ (constant term) controls vertical shift (y-intercept).
Graph of y = 2x + 3 (blue, rising, labelled points (0, 3), (1, 5) and (2, 7)), y = −2x − 3 (red, falling, labelled points (0, −3), (1, −5) and (2, −7)) and y = −2x (green, falling, labelled points (1, −2) and (2, −4)). The figure also carries an extra label “(2, −10)” beside a red dot that does not match y = −2x − 3, and the plotted dots sit slightly off their labelled grid positions.

Image unavailable. Open original

Graph of y = 2x + 3 (blue, rising, labelled points (0, 3), (1, 5) and (2, 7)), y = −2x − 3 (red, falling, labelled points (0, −3), (1, −5) and (2, −7)) and y = −2x (green, falling, labelled points (1, −2) and (2, −4)). The figure also carries an extra label “(2, −10)” beside a red dot that does not match y = −2x − 3, and the plotted dots sit slightly off their labelled grid positions.

End-of-Chapter Exercises

Q1. Write a polynomial of degree 3 in the variable x, in which the coefficient of the x² term is −7.

Answer

A polynomial of degree 3 has the general form ax³ + bx² + cx + d. Given that the coefficient of x² is −7, we have b = −7. One such polynomial is:

x3−7x2+2x+1x^3 - 7x^2 + 2x + 1

(Any polynomial of degree 3 with −7 as the coefficient of x² is correct.)

Q2. Find the values of the following polynomials at the indicated values of the variables.

Q2(i). 5x² − 3x + 7 if x = 1

Answer

Substitute x = 1:

Steps

  1. = 5(1)² − 3(1) + 7
  2. = 5 − 3 + 7
  3. = 9
5x2−3x+7=9 when x=15x^2 - 3x + 7 = 9 \text{ when } x = 1

Q2(ii). 4t³ − t² + 6 if t = a

Answer

Substitute t = a:

4t3−t2+6=4a3−a2+64t^3 - t^2 + 6 = 4a^3 - a^2 + 6

Q3. If we multiply a number by 5/2 and add 2/3 to the product, we get −7/12. Find the number.

Answer

Let the number be x. According to the question:

Steps

  1. (5/2)x + 2/3 = −7/12

Subtract 2/3 from both sides:

Steps

  1. (5/2)x = −7/12 − 2/3
  2. ⇒ (5/2)x = −7/12 − 8/12
  3. ⇒ (5/2)x = −15/12
  4. ⇒ (5/2)x = −5/4

Now multiply both sides by 2/5:

Steps

  1. x = (−5/4) × (2/5)
  2. ⇒ x = −1/2

Therefore, the number is −1/2.

Q4. 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?

Answer

Let the smaller number be x. Then the larger number = 5x. After adding 21 to both numbers, the new numbers are x + 21 and 5x + 21. According to the question:

Steps

  1. 5x + 21 = 2(x + 21)
  2. ⇒ 5x + 21 = 2x + 42
  3. ⇒ 5x − 2x = 42 − 21
  4. ⇒ 3x = 21
  5. ⇒ x = 7

So, the smaller number = 7 and the larger number = 5 × 7 = 35. Therefore, the two numbers are 7 and 35.

Q5. 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.

Answer

Initial amount = ₹800. Saving every month = ₹250. Linear pattern: Amount after n months = 800 + 250n.

(i) Amount after 6 months:

Steps

  1. = 800 + 250(6)
  2. = 800 + 1500
  3. = ₹2300

(ii) Amount after 2 years (2 years = 24 months):

Steps

  1. = 800 + 250(24)
  2. = 800 + 6000
  3. = ₹6800

Therefore, the amount after 6 months = ₹2300 and the amount after 2 years = ₹6800.

A=800+250n, where n is the number of monthsA = 800 + 250n, \text{ where } n \text{ is the number of months}

Q6. 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.

Answer

Let the tens digit be x and the units digit be y. Then the original number = 10x + y and the interchanged number = 10y + x. According to the question:

Steps

  1. (10x + y) + (10y + x) = 143
  2. ⇒ 11x + 11y = 143
  3. ⇒ 11(x + y) = 143
  4. ⇒ x + y = 13 …(1)

The digits differ by 3. So:

Steps

  1. x − y = 3 …(2)

Adding (1) and (2), we get:

Steps

  1. 2x = 16
  2. ⇒ x = 8

Substituting x = 8 in (1), we have:

Steps

  1. 8 + y = 13
  2. ⇒ y = 5

Therefore, the original number = 85 and the interchanged number = 58. Hence, the two numbers are 85 and 58.

Q7. 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.

Q7(i). y = −3x + 4

Answer

Comparing with y = ax + b:

Steps

  1. Slope a = −3
  2. y-intercept b = 4

So, the point where the line cuts the y-axis is (0, 4).

Q7(ii). 2y = 4x + 7

Answer

Dividing both sides by 2: y = 2x + 7/2.

Steps

  1. Slope a = 2
  2. y-intercept b = 7/2

So, the point where the line cuts the y-axis is (0, 7/2).

Q7(iii). 5y = 6x − 10

Answer

Dividing both sides by 5: y = (6/5)x − 2.

Steps

  1. Slope a = 6/5
  2. y-intercept b = −2

So, the point where the line cuts the y-axis is (0, −2).

Q7(iv). 3y = 6x − 11

Answer

Dividing both sides by 3: y = 2x − 11/3.

Steps

  1. Slope a = 2
  2. y-intercept b = −11/3

So, the point where the line cuts the y-axis is (0, −11/3).

Q7(v). Are any of the lines parallel?

Answer

Two lines are parallel if they have the same slope.

Steps

  1. Equation (ii): slope = 2
  2. Equation (iv): slope = 2

Therefore, lines (ii) and (iv) are parallel.

Graph of the four lines from Question 7: y = −3x + 4 (green, falling, labelled points (0, 4) and (1, 1)), 2y = 4x + 7 (red, rising, labelled points (1, 5.5) and (2, 7.5)), 5y = 6x − 10 (purple, rising, labelled points (0, −2) and (4, 5)) and 3y = 6x − 11 (cyan, with labelled points (0, 3.5), (1, −5/3) and (0, −11/3)). In the figure, the cyan line is drawn as a vertical line along the y-axis instead of a slanted line of slope 2, and the purple point labelled (4, 5) is drawn at (5, 4).

Image unavailable. Open original

Graph of the four lines from Question 7: y = −3x + 4 (green, falling, labelled points (0, 4) and (1, 1)), 2y = 4x + 7 (red, rising, labelled points (1, 5.5) and (2, 7.5)), 5y = 6x − 10 (purple, rising, labelled points (0, −2) and (4, 5)) and 3y = 6x − 11 (cyan, with labelled points (0, 3.5), (1, −5/3) and (0, −11/3)). In the figure, the cyan line is drawn as a vertical line along the y-axis instead of a slanted line of slope 2, and the purple point labelled (4, 5) is drawn at (5, 4).

Q8. 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.

Q8(i). Find the temperature of the liquid in Fahrenheit if the temperature of the liquid is 313 K.

Answer

When x = 313 K, we have to find y. Using the equation y = (9/5)(x − 273) + 32:

Steps

  1. y = (9/5)(313 − 273) + 32
  2. ⇒ y = (9/5)(40) + 32
  3. ⇒ y = 72 + 32
  4. ⇒ y = 104

Therefore, the temperature is 104 °F.

Q8(ii). If the temperature is 158 °F, then find the temperature in Kelvin.

Answer

When y = 158 °F, we have to find x. Using the equation y = (9/5)(x − 273) + 32:

Steps

  1. 158 = (9/5)(x − 273) + 32

Subtracting 32 from both sides, we get:

Steps

  1. 158 − 32 = (9/5)(x − 273)
  2. ⇒ 126 = (9/5)(x − 273)

Multiplying both sides by 5/9, we get:

Steps

  1. 126 × (5/9) = x − 273
  2. ⇒ 70 = x − 273
  3. ⇒ x = 343

Therefore, the temperature is 343 K.

Q9. 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.

Answer

We know that Work done = Force × Distance. According to the question, work done = w, distance travelled = d and constant force = 3 units. Therefore:

w=3dw = 3d

This is the required linear equation in two variables. If d = 2 units, the work done is w = 3 × 2 = 6 units. Taking w on the y-axis and d on the x-axis, we can plot the graph.

Graph of the straight line y = 3x through the origin, representing w = 3d, with the point (2, 6) marked and labelled and a second point plotted at (1, 3). The axes are labelled X and Y rather than d and w.

Image unavailable. Open original

Graph of the straight line y = 3x through the origin, representing w = 3d, with the point (2, 6) marked and labelled and a second point plotted at (1, 3). The axes are labelled X and Y rather than d and w.

The point (2, 6) lies on the straight line, so it is verified by the graph. Hence, when the distance travelled is 2 units, w = 3 × 2 = 6 units.

Verification from the graph: the point corresponding to d = 2 is (2, 6), so the work done is 6 units.

Q10. The graph of a linear polynomial p(x) passes through the points (1, 5) and (3, 11).

Q10(i). Find the polynomial p(x).

Answer

Let p(x) = ax + b. Since the graph passes through (1, 5):

Steps

  1. a(1) + b = 5
  2. ⇒ a + b = 5 …(1)

Since the graph passes through (3, 11):

Steps

  1. a(3) + b = 11
  2. ⇒ 3a + b = 11 …(2)

Subtracting (1) from (2), we get:

Steps

  1. 3a + b − (a + b) = 11 − 5
  2. ⇒ 2a = 6
  3. ⇒ a = 3

Substituting a = 3 in (1), we have:

Steps

  1. 3 + b = 5
  2. ⇒ b = 2
p(x)=3x+2p(x) = 3x + 2

Q10(ii). Find the coordinates where the graph of p(x) cuts the axes.

Answer

To find the y-axis intercept, put x = 0:

Steps

  1. y = p(0) = 3(0) + 2 = 2

So, the graph cuts the y-axis at (0, 2).

To find the x-axis intercept, put y = 0:

Steps

  1. 3x + 2 = 0
  2. ⇒ 3x = −2
  3. ⇒ x = −2/3

So, the graph cuts the x-axis at (−2/3, 0).

Q10(iii). Draw the graph of p(x) and verify your answers.

Answer

From the graph, it verifies that the line passes through the given points and cuts the y-axis at (0, 2) and the x-axis at (−2/3, 0).

Graph of the line y = 3x + 2 with labelled points (−2/3, 0) on the x-axis (marked with an arrow), (0, 2), (1, 5) and (3, 11). The plotted dots are drawn slightly off their exact grid positions, and some y-axis tick labels are missing.

Image unavailable. Open original

Graph of the line y = 3x + 2 with labelled points (−2/3, 0) on the x-axis (marked with an arrow), (0, 2), (1, 5) and (3, 11). The plotted dots are drawn slightly off their exact grid positions, and some y-axis tick labels are missing.

Q11. 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).

Answer

Let p(x) = ax + b and q(x) = cx + d. Using condition (i), p(0) = 5:

Steps

  1. a(0) + b = 5
  2. ⇒ b = 5

So, p(x) = ax + 5. Now, using condition (iii):

Steps

  1. p(x) + q(x) = 6x + 4
  2. ⇒ (ax + 5) + (cx + d) = 6x + 4
  3. ⇒ (a + c)x + (5 + d) = 6x + 4

Comparing coefficients, we get a + c = 6 …(1), and:

Steps

  1. 5 + d = 4
  2. ⇒ d = −1

So, q(x) = cx − 1. Using condition (ii), p(x) − q(x) cuts the x-axis at (3, 0), so p(3) − q(3) = 0. Here p(3) = 3a + 5 and q(3) = 3c − 1. So:

Steps

  1. (3a + 5) − (3c − 1) = 0
  2. ⇒ 3a + 5 − 3c + 1 = 0
  3. ⇒ 3a − 3c + 6 = 0
  4. ⇒ a − c = −2 …(2)

Solving equations (1) and (2): from (1), a + c = 6; from (2), a − c = −2. Adding both:

Steps

  1. 2a = 4 ⇒ a = 2
  2. Then 2 + c = 6 ⇒ c = 4

Hence:

p(x)=2x+5p(x) = 2x + 5q(x)=4x−1q(x) = 4x - 1

Q12. 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.

Stages 1 to 3 of the hexagon matchstick pattern: Stage 1 is a single hexagon, Stage 2 is two hexagons sharing one side, and Stage 3 is three hexagons in which two upper hexagons each share a side with a lower middle hexagon.

Image unavailable. Open original

Stages 1 to 3 of the hexagon matchstick pattern: Stage 1 is a single hexagon, Stage 2 is two hexagons sharing one side, and Stage 3 is three hexagons in which two upper hexagons each share a side with a lower middle hexagon.

Q12(i). Draw the next two stages of the pattern. How many matchsticks will be required at these stages?

Answer

A single hexagon needs 6 matchsticks. Since each new hexagon shares one side with the previous hexagon, only 5 new matchsticks are added at every new stage. So the pattern is:

Steps

  1. Stage 1 = 6
  2. Stage 2 = 6 + 5 = 11
  3. Stage 3 = 11 + 5 = 16

Next two stages:

Steps

  1. Stage 4: number of matchsticks = 16 + 5 = 21
  2. Stage 5: number of matchsticks = 21 + 5 = 26
Stages 1 to 5 of the hexagon matchstick pattern. Stages 1 to 3 repeat the given figures; Stage 4 shows four hexagons (three arranged around a central hexagon), and Stage 5 adds a fifth hexagon attached to the lower hexagon on its lower-right side.

Image unavailable. Open original

Stages 1 to 5 of the hexagon matchstick pattern. Stages 1 to 3 repeat the given figures; Stage 4 shows four hexagons (three arranged around a central hexagon), and Stage 5 adds a fifth hexagon attached to the lower hexagon on its lower-right side.

Therefore:

  • Stage 4 requires 21 matchsticks
  • Stage 5 requires 26 matchsticks

Q12(ii). Complete the following table.

A blank table with the row “Stage Number” showing 1, 2, 3, 4, 5, …, n and an empty row “Number of matchsticks” to be completed.

Image unavailable. Open original

A blank table with the row “Stage Number” showing 1, 2, 3, 4, 5, …, n and an empty row “Number of matchsticks” to be completed.

Answer

The completed table: Stage Number 1, 2, 3, 4, 5 has Number of matchsticks 6, 11, 16, 21, 26; the “…” column is left blank; and Stage Number n has Number of matchsticks 5n + 1.

Image unavailable. Open original

The completed table: Stage Number 1, 2, 3, 4, 5 has Number of matchsticks 6, 11, 16, 21, 26; the “…” column is left blank; and Stage Number n has Number of matchsticks 5n + 1.

Q12(iii). Find a rule to determine the number of matchsticks required for the nth stage.

Answer

The number of matchsticks forms an arithmetic pattern: 6, 11, 16, 21, 26, … The first term = 6 and the difference = 5. So, the number of matchsticks in the nth stage:

Steps

  1. = 6 + (n − 1) × 5
  2. = 6 + 5n − 5
  3. = 5n + 1

Hence, the rule is:

Mn=5n+1M_n = 5n + 1

Q12(iv). How many matchsticks will be required for the 15th stage of the pattern?

Answer

Steps

  1. M₁₅ = 5(15) + 1
  2. = 75 + 1
  3. = 76

Therefore, 76 matchsticks will be required for the 15th stage.

Q12(v). Can 200 matchsticks form a stage in this pattern? Justify your answer.

Answer

For some stage n:

Steps

  1. 5n + 1 = 200
  2. ⇒ 5n = 199
  3. ⇒ n = 199/5
  4. ⇒ n = 39.8

Since n is not a whole number, 200 matchsticks cannot form any stage in this pattern.

Q13. 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.

Answer

Given: p(x) = ax + b and q(x) = cx + d. First, to find p(x). Since p(x) passes through (2, 3) and (6, 11), its slope is:

Steps

  1. m = (11 − 3)/(6 − 2) = 8/4 = 2

So, p(x) = 2x + b. Using the point (2, 3):

Steps

  1. 3 = 2(2) + b
  2. ⇒ 3 = 4 + b
  3. ⇒ b = −1

Therefore, p(x) = 2x − 1.

Now to find q(x). Since q(x) is parallel to p(x), it has the same slope, so the slope of q(x) = 2 and q(x) = 2x + d. Since q(x) passes through (4, −1):

Steps

  1. −1 = 2(4) + d
  2. ⇒ −1 = 8 + d
  3. ⇒ d = −9

Therefore, q(x) = 2x − 9.

Now we find where these lines meet the x-axis. A line meets the x-axis where y = 0. For p(x) = 2x − 1:

Steps

  1. 0 = 2x − 1
  2. ⇒ 2x = 1
  3. ⇒ x = 1/2

So, p(x) meets the x-axis at (1/2, 0). For q(x) = 2x − 9:

Steps

  1. 0 = 2x − 9
  2. ⇒ 2x = 9
  3. ⇒ x = 9/2

So, q(x) meets the x-axis at (9/2, 0).

Hence, p(x) = 2x − 1 and q(x) = 2x − 9. The x-axis intercepts are:

  • For p(x): (1/2, 0)
  • For q(x): (9/2, 0)

Q14. What do all linear functions of the form f(x) = ax + a, a > 0, have in common?

Answer

Given: f(x) = ax + a, where a > 0. We can write it as f(x) = a(x + 1). The common properties of all such linear functions are:

1. Slope: The slope is a, and since a > 0, all the lines have positive slope. So, all these lines rise from left to right.

2. y-intercept: Putting x = 0, f(0) = a. So, the y-intercept is (0, a). Since a > 0, all the lines cut the y-axis above the origin.

3. x-intercept: To find where the line cuts the x-axis, put f(x) = 0:

Steps

  1. ax + a = 0
  2. a(x + 1) = 0
  3. Since a > 0, a is not zero. So, x + 1 = 0, or x = −1

Thus, every line cuts the x-axis at the same point (−1, 0).

Therefore, all linear functions of the form f(x) = ax + a, a > 0, have the following in common:

  • all have positive slope
  • all cut the y-axis above the origin
  • all pass through the fixed point (−1, 0).
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