NCERT Solutions for Class 9 Maths Chapter 2 Introduction to Linear Polynomials - Ganita Manjari
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NCERT Solutions for Chapter 2 Introduction to Linear Polynomials Class 9 Maths
Page No. 17
Think and Reflect
1. Can you identify the terms, variables and coefficients of this algebraic expression?
200l + 160w + 50lw
Answer
The algebraic expression is 200l + 160w + 50lw.
Terms: 200l, 160w, 50lw
Variables: l and w
Coefficients:
- Coefficient of l = 200
- Coefficient of w = 160
- Coefficient of lw = 50
There is no constant term in this expression.
2. How is it different from the algebraic expression in Example 1?
Answer
The algebraic expression in Example 1 is 4x + 5y + 3, whereas the expression in this example is 200l + 160w + 50lw.
- Example 1 contains two variable terms and one constant term (3).
- This expression has no constant term.
- Example 1 contains only single-variable terms (x and y), whereas this expression also contains the product of two variables (lw).
1. Can you identify the terms, variables and coefficients of this algebraic expression?
Answer
The algebraic expression is 10x − x².
Terms: 10x, −x²
Variable: x
Coefficients:
- Coefficient of x² = −1
- Coefficient of x = 10
There is no constant term in this expression.
2. Can you point out any similarity or difference between the algebraic expressions obtained in Examples 1 and 3?
Answer
Similarity:
- Both are algebraic expressions.
- Both contain variable terms with numerical coefficients.
Difference:
- Example 1, 4x + 5y + 3, has two variables (x and y) and includes a constant term.
- Example 3, 10x − x², has only one variable (x) and does not contain a constant term.
- Example 1 is a linear algebraic expression in two variables, whereas Example 3 contains the term x², making it a quadratic expression.
Page No. 18
Exercise Set 2.1
1. Find the degrees of the following polynomials:
(i) 2x² − 5x + 3
(ii) y³ + 2y − 1
(iii) −9
(iv) 4z − 3
Answer
(i) The highest power of x is 2.
Degree = 2
(ii) The highest power of y is 3.
Degree = 3
(iii) −9 is a constant polynomial.
Degree = 0
(iv) The highest power of z is 1.
Degree = 1
2. Write polynomials of degrees 1, 2 and 3.
Answer
One possible set of polynomials is:
- Degree 1: 3x + 7
- Degree 2: x² + 5x + 1
- Degree 3: 5y³ + y² + 2y − 1
3. What are the coefficients of x² and x³ in the polynomial
x⁴ − 3x³ + 6x² − 2x + 7?
Answer
Coefficient of x² = 6
Coefficient of x³ = −3
4. What is the coefficient of z in the polynomial 4z³ + 5z² − 11?
Answer
The given polynomial is 4z³ + 5z² − 11.
Since there is no z-term in the polynomial, the coefficient of z is 0.
Coefficient of z = 0
5. What is the constant term of the polynomial 9x³ + 5x² − 8x −10?
Answer
The constant term is the term that does not contain any variable.
In the polynomial 9x³ + 5x² − 8x − 10, the constant term is −10.
Constant term = −10
Page No. 19
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
We know that the perimeter of a square = 4 × Side.
| Side (cm) | Perimeter (cm) |
|---|---|
| 1 | 4 |
| 1.5 | 6 |
| 2 | 8 |
| 2.5 | 10 |
| 3 | 12 |
When the side increases by 0.5 cm, the perimeter increases by 2 cm each time.
Thus, the perimeters form the linear pattern: 4, 6, 8, 10, 12, …
If a player paid ₹750, how many matches did he play?
Answer
According to the given information,
Total amount paid = ₹200 + ₹50 × (Number of matches)
Let the number of matches played be m.
Then,
₹200 + 50m = ₹750
50m = 750 − 200
50m = 550
m = 550 ÷ 50 = 11
Therefore, the player played 11 matches.
We have learnt that to evaluate the value of an algebraic expression, we substitute a value of the variable in the given expression. Consider Example 3, where the wire is bent to form a rectangle. Here, the area of the rectangle, 10x − x², is a function of x. Can you interpret this as an input-output process? What value does the expression take when x = 6 cm?
Answer
Yes. The expression 10x − x² can be interpreted as an input-output process.
- The input is the value of x (length of the rectangle).
- The expression 10x − x² processes the input.
- The output is the area of the rectangle.
When x = 6 cm,
Area = 10 × 6 − 6²
= 60 − 36
= 24 cm²
Therefore, when x = 6 cm, the area of the rectangle is 24 cm².
Page No. 21
Exercise Set 2.2
1. Find the value of the linear polynomial 5x − 3 if:
(i) x = 0
(ii) x = −1
(iii) x = 2
Answer
(i) Substitute x = 0 in the polynomial 5x − 3.
5(0) − 3 = 0 − 3 = −3
Therefore, the value of the polynomial is −3.
(ii) Substitute x = −1 in the polynomial 5x − 3.
5(−1) − 3 = −5 − 3 = −8
Therefore, the value of the polynomial is −8.
(iii) Substitute x = 2 in the polynomial 5x − 3.
5(2) − 3 = 10 − 3 = 7
Therefore, the value of the polynomial is 7.
2. Find the value of the quadratic polynomial 7s² − 4s + 6 if:
(i) s = 0
Answer
Substitute s = 0 in the polynomial 7s² − 4s + 6.
7(0)² − 4(0) + 6 = 6
Therefore, the value of the polynomial is 6.
(ii) s = −3
Answer
Substitute s = −3 in the polynomial 7s² − 4s + 6.
7(−3)² − 4(−3) + 6
= 7 × 9 + 12 + 6
= 63 + 12 + 6
= 81
Therefore, the value of the polynomial is 81.
(iii) s = 4
Answer
Substitute s = 4 in the polynomial 7s² − 4s + 6.
7(4)² − 4(4) + 6
= 7 × 16 − 16 + 6
= 112 − 16 + 6
= 102
Therefore, the value of the polynomial is 102.
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.
Answer
Let Salil's present age be x years.
Then, his mother's present age = 3x years.
After 5 years,
Salil's age = (x + 5) years
Mother's age = (3x + 5) years
According to the question,
(x + 5) + (3x + 5) = 70
⇒ 4x + 10 = 70
⇒ 4x = 60
⇒ x = 15
Therefore, Salil's present age = 15 years.
Mother's present age = 3 × 15 = 45 years.
Hence, the present ages are:
- Salil = 15 years
- Salil's mother = 45 years
4. The difference between two positive integers is 63. The ratio of the two integers is 2:5. Find the two integers.
Answer
Let the two integers be 2x and 5x.
According to the question,
5x − 2x = 63
⇒ 3x = 63
⇒ x = 21
Therefore,
First integer = 2 × 21 = 42
Second integer = 5 × 21 = 105
Hence, the two 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?
Answer
Let the number of five-rupee coins be x.
Then, the number of two-rupee coins = 3x.
According to the question,
5x + 2(3x) = 88
⇒ 5x + 6x = 88
⇒ 11x = 88
⇒ x = 8
Therefore,
Number of five-rupee coins = 8
Number of two-rupee coins = 3 × 8 = 24
Verification:
Value of five-rupee coins = 8 × ₹5 = ₹40
Value of two-rupee coins = 24 × ₹2 = ₹48
Total value = ₹40 + ₹48 = ₹88
Hence, Ruby has 8 five-rupee coins and 24 two-rupee coins.
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?
Answer
Let the length of the shorter piece be x feet.
Then, the length of the longer piece = 4x feet.
According to the question,
x + 4x = 300
⇒ 5x = 300
⇒ x = 60
Therefore,
Length of the shorter piece = 60 feet
Length of the longer piece = 4 × 60 = 240 feet
Verification:
60 + 240 = 300
Hence, the two pieces are 60 feet and 240 feet long.
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?
Answer
Let the width of the rectangle be x cm.
Then, the length of the rectangle = (2x + 3) cm.
The perimeter of a rectangle is given by:
Perimeter = 2 × (Length + Width)
According to the question,
2[(2x + 3) + x] = 24
⇒ 2(3x + 3) = 24
⇒ 6x + 6 = 24
⇒ 6x = 18
⇒ x = 3
Therefore,
Width = 3 cm
Length = 2 × 3 + 3 = 9 cm
Verification:
Perimeter = 2 × (9 + 3) = 2 × 12 = 24 cm
Hence, the dimensions of the rectangle are 9 cm × 3 cm.
Page No. 22
Think and Reflect
Predict the number of squares in the next three stages of the pattern and write the sequence of numbers up to Stage 7 of the pattern.
Answer
Each new stage is formed by adding 2 more square tiles than the previous stage.
The first four stages have 1, 3, 5, 7 square tiles.
Therefore, the next three stages will have:
- Stage 5 = 9 squares
- Stage 6 = 11 squares
- Stage 7 = 13 squares
Hence, the sequence of the number of square tiles up to Stage 7 is: 1, 3, 5, 7, 9, 11, 13.
Using the expression 2n − 1, can you find out how many tiles will be there in the 15th stage and the 26th stage of the pattern? Also, which stage will contain 21 tiles and 47 tiles?
Answer
The number of tiles in the nth stage is given by:
Number of tiles = 2n − 1
For the 15th stage:
Number of tiles = 2 × 15 − 1 = 30 − 1 = 29
Therefore, the 15th stage has 29 tiles.
For the 26th stage:
Number of tiles = 2 × 26 − 1 = 52 − 1 = 51
Therefore, the 26th stage has 51 tiles.
To find the stage containing 21 tiles:
2n − 1 = 21
⇒ 2n = 22
⇒ n = 11
Therefore, Stage 11 contains 21 tiles.
To find the stage containing 47 tiles:
2n − 1 = 47
⇒ 2n = 48
⇒ n = 24
Therefore, Stage 24 contains 47 tiles.
What amount will be left on the 15th day? How many days will it take for the entire amount to be spent?
Answer
The amount left after n days is given by:
Amount left = ₹(100 − 5n)
Amount left on the 15th day:
= ₹(100 − 5 × 15)
= ₹(100 − 75)
= ₹25
Therefore, the amount left on the 15th day is ₹25.
Time taken for the entire amount to be spent:
When the entire amount is spent, the amount left = ₹0.
100 − 5n = 0
5n = 100
n = 20
Therefore, it will take 20 days for the entire amount to be spent.
For how many km will the fare be ₹130?
Answer
For a journey of n km (n ≥ 2), the fare is given by:
Fare = 15n − 5
Given that the fare is ₹130,
15n − 5 = 130
15n = 135
n = 9
Therefore, the fare will be ₹130 for a journey of 9 km.
Page No. 23
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 nth month.
Answer
Initially, the student has ₹500 in her savings bank account.
She receives ₹150 as pocket money every month.
The amount at the end of each month is as follows:
| Month | Amount (₹) |
|---|---|
| 2 | 500 + 2 × 150 = 800 |
| 3 | 500 + 3 × 150 = 950 |
| 4 | 500 + 4 × 150 = 1100 |
| 5 | 500 + 5 × 150 = 1250 |
Therefore, the amount in the nth month is given by:
Let the month number be n.
Since the student already has ₹500 at the beginning, the total amount after n months is:
Total Amount = Initial Amount + Pocket Money received in n months
⇒ Total Amount = 500 + 150n
⇒ A = 150n + 500
Therefore, the required linear expression is A = 150n + 500.
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 nth hour.
Answer
The number of members remaining is as follows:
| Hour | Members Remaining |
|---|---|
| 1 | 120 − 9 = 111 |
| 2 | 120 − 18 = 102 |
| 3 | 120 − 27 = 93 |
Initial number of members = 120
Number of members dropping out every hour = 9
Let the number of hours be n.
Members who drop out in n hours = 9n
Therefore,
Members Remaining = Initial Members − Members Dropped Out
⇒ Members Remaining = 120 − 9n
⇒ M = 120 − 9n
Therefore, the required linear expression is M = 120 − 9n.
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.
Answer
Area of a rectangle = Length × Breadth
Given, Length = 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 form the pattern: 156, 130, 104, …
We know that,
Area of a Rectangle = Length × Breadth
Let the breadth of the rectangle be b cm.
Therefore,
⇒ Area = 13 × b
⇒ A = 13b
Thus, the required linear expression representing the area of the rectangle 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.
Answer
Volume of a rectangular box = Length × Breadth × Height
Given, Length = 7 cm and Breadth = 11 cm.
(i) Height = 5 cm
Volume = 7 × 11 × 5 = 385 cm³
(ii) Height = 9 cm
Volume = 7 × 11 × 9 = 693 cm³
(iii) Height = 13 cm
Volume = 7 × 11 × 13 = 1001 cm³
We know that,
Volume of a Rectangular Box = Length × Breadth × Height
Let the height of the box be h cm.
Therefore,
⇒ Volume = 7 × 11 × h
⇒ Volume = 77h
⇒ V = 77h
Thus, the required linear expression representing the volume of the rectangular box is V = 77h.
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.
Answer
Total number of pages in the book = 500
Number of pages read every day = 20
Let the number of days be n.
Pages read in n days = 20n
Therefore,
Pages Left = Total Pages − Pages Read
⇒ Pages Left = 500 − 20n
⇒ P = 500 − 20n
Thus, the required linear pattern is P = 500 − 20n.
Pages left after 15 days:
⇒ P = 500 − 20 × 15
⇒ P = 500 − 300
= 200 pages
Think and Reflect
What is the cost for travelling 15 km? For how many kilometres will the cost of the journey be ₹700?
Answer
The cost of the journey is given by:
C(d) = 100 + 60d
Cost for travelling 15 km:
C(15) = 100 + 60 × 15
= 100 + 900
= ₹1000
Therefore, the cost for travelling 15 km is ₹1000.
For the cost to be ₹700:
100 + 60d = 700
60d = 600
d = 10
Therefore, the cost of the journey will be ₹700 for travelling 10 km.
What will be the height of the water at the end of 5 months?
Answer
The height of water is given by:
h(t) = 3 − 0.5t
Substitute t = 5.
h(5) = 3 − 0.5 × 5
= 3 − 2.5
= 0.5 m
Therefore, the height of the water at the end of 5 months is 0.5 m.
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.
Answer
Initial height of the plant = 1.75 feet
Growth every month = 0.5 feet
Growth in 7 months = 7 × 0.5 = 3.5 feet
Height after 7 months = 1.75 + 3.5 = 5.25 feet
Therefore, the height of the plant after 7 months is 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.
Answer
| Month, t | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| Height, h (feet) | 1.75 | 2.25 | 2.75 | 3.25 | 3.75 | 4.25 | 4.75 | 5.25 | 5.75 | 6.25 | 6.75 |
(iii) Find an expression that relates h and t, and explain why it represents linear growth.
Answer
The initial height of the plant is 1.75 feet.
The plant grows by 0.5 feet every month.
Therefore, the relation between the height h and time t is:
h = 1.75 + 0.5t
This represents linear growth because the height increases by the same amount, 0.5 feet, every month.
2. A mobile phone is bought for ₹10,000. Its value decreases by ₹800 every year.
(i) Find the value of the phone after 3 years.
Answer
Initial value of the phone = ₹10,000
Decrease in value every year = ₹800
Decrease in 3 years = 3 × 800 = ₹2,400
Value after 3 years = 10,000 − 2,400 = ₹7,600
Therefore, the value of the phone after 3 years is ₹7,600.
(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
| Year, t | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|---|
| Value, v (₹) | 10000 | 9200 | 8400 | 7600 | 6800 | 6000 | 5200 | 4400 | 3600 |
(iii) Find an expression that relates v and t, and explain why it represents linear decay.
Answer
The initial value of the phone is ₹10,000.
Its value decreases by ₹800 every year.
Therefore, the relation between the value v and time t is:
v = 10000 − 800t
This represents linear decay because the value decreases by the same amount, ₹800, every year.
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 of the village after 6 years.
Answer
Initial population of the village = 750
Increase in population every year = 50
Increase in 6 years = 6 × 50 = 300
Population after 6 years = 750 + 300 = 1050
Therefore, the population of the village after 6 years is 1050.
(ii) Make a table of values for t varying from 0 to 10 years and show how the population, P, increases every year.
Answer
| Year, t | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| Population, P | 750 | 800 | 850 | 900 | 950 | 1000 | 1050 | 1100 | 1150 | 1200 | 1250 |
(iii) Find an expression that relates P and t, and explain why it represents linear growth.
Answer
Initial population of the village = 750
Number of people moving to the village every year = 50
Let the number of years be t.
Increase in population in t years = 50t
Therefore,
Population after t years = Initial Population + Increase in Population
⇒ P = 750 + 50t
Thus, the required expression is P = 750 + 50t.
4. A telecom company charges ₹600 for a certain recharge scheme. This prepaid balance is reduced by ₹15 each day after the recharge.
(i) Write an equation that models the remaining balance b(x) after using the scheme for x days. Explain why it represents linear decay.
Answer
Initial prepaid balance = ₹600
Balance reduced every day = ₹15
Let the number of days be x.
Balance reduced in x days = ₹15x
Therefore,
Remaining Balance = Initial Balance − Balance Used
⇒ b(x) = 600 − 15x
Thus, the required equation is b(x) = 600 − 15x.
This represents linear decay because the balance decreases by the same amount, ₹15, every day.
(ii) After how many days will the balance run out?
Answer
The balance runs out when the remaining balance becomes ₹0.
⇒ b(x) = 0
⇒ 600 − 15x = 0
⇒ 15x = 600
⇒ x = 40
Therefore, the prepaid balance will run out after 40 days.
Verification:
Remaining balance after 40 days
⇒ b(40) = 600 − 15 × 40
⇒ 600 − 600
⇒ ₹0
Therefore, 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.
Answer
| Days, x | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| Balance, b(x) (₹) | 585 | 570 | 555 | 540 | 525 | 510 | 495 | 480 | 465 | 450 |
Think and Reflect
Can you guess what the numbers 20 and 150 in the equation y = 20x + 150 represent?
Answer
In the equation y = 20x + 150:
- 20 represents the cost per GB of internet data used.
- 150 represents the fixed monthly fee charged by the telecom company.
Thus, the monthly bill consists of a fixed charge of ₹150 and an additional charge of ₹20 for each GB of data used.
Page No. 25
Exercise Set 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.
Answer
Given,
When x = 10, y = 400
When x = 14, y = 500
Using y = ax + b, we get:
400 = 10a + b ...(1)
500 = 14a + b ...(2)
Subtracting equation (1) from equation (2),
100 = 4a
⇒ a = 25
Substituting a = 25 in equation (1),
400 = 10 × 25 + b
⇒ 400 = 250 + b
⇒ b = 150
Therefore, a = 25 and b = 150.
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.
Answer
Given,
When x = 10, y = 800
When x = 15, y = 1100
Using y = ax + b, we get:
800 = 10a + b ...(1)
1100 = 15a + b ...(2)
Subtracting equation (1) from equation (2),
300 = 5a
⇒ a = 60
Substituting a = 60 in equation (1),
800 = 10 × 60 + b
⇒ 800 = 600 + b
⇒ b = 200
Therefore, a = 60 and b = 200.
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.
(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,
When °F = 32, °C = 0
When °F = 212, °C = 100
Using °C = a°F + b, we get:
0 = 32a + b ...(1)
100 = 212a + b ...(2)
Subtracting equation (1) from equation (2),
100 = 180a
⇒ a = 5/9
Substituting a = 5/9 in equation (1),
0 = 32 × (5/9) + b
⇒ b = −160/9
Therefore,
a = 5/9
b = −160/9
Hence, the required linear relationship is:
°C = (5/9)°F − 160/9
Page No. 28
Think and Reflect
Identify other points on the line by completing the following table.
Answer
The given points are (1, 3) and (7, 15).
The linear relationship is:
y = 2x + 1
Using this relation, we get:
| x | 1 | 2 | 5 | 7 | 9 | 12 | 20 |
|---|---|---|---|---|---|---|---|
| y | 3 | 5 | 11 | 15 | 19 | 25 | 41 |
Hence, the other points on the line are (2, 5), (5, 11), (9, 19), (12, 25) and (20, 41).
Differentiate between the graphs of the equations y = 3x + 1, and y = −3x + 1.
Answer
| Equation | Nature of Graph |
|---|---|
| y = 3x + 1 | The graph is a rising straight line. As the value of x increases, the value of y also increases. The line has a positive slope. |
| y = −3x + 1 | The graph is a falling straight line. As the value of x increases, the value of y decreases. The line has a negative slope. |
Both graphs intersect the y-axis at y = 1, but they slope in opposite directions.
Does this help you to conclude anything about the linear equation y = ax + b when a is fixed but b varies?
Answer
Yes. When a is fixed and b varies, the graphs of the equations are parallel straight lines because they have the same slope.
Changing the value of b only changes the y-intercept. The graph shifts upward if b increases and downward if b decreases, but its slope remains unchanged.
Therefore, when a is fixed and b varies, the graphs are parallel to each other and differ only in their y-intercepts.
Page No. 36
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
Answer
Role of a and b:
- Here, the values of a are 4, 2 and 1, while b = 0 in all three equations.
- Since a is positive, all the lines rise from left to right.
- As the value of a increases, the line becomes steeper.
- Since b = 0, all the lines pass through the origin (0, 0).
(ii) y = −6x, y = −3x, y = −x
Answer
Role of a and b:
- Here, the values of a are −6, −3 and −1, while b = 0.
- Since a is negative, all the lines fall from left to right.
- The greater the numerical value of |a|, the steeper the line.
- Since b = 0, all the lines pass through the origin.
(iii) y = 5x, y = −5x
Answer
Role of a and b:
- Here, |a| = 5 for both equations, but the signs are different.
- Positive a gives a rising line, while negative a gives a falling line.
- Since b = 0, both lines pass through the origin.
(iv) y = 3x − 1, y = 3x, y = 3x + 1
Answer
Role of a and b:
- Here, a = 3 is the same for all three equations.
- Since a is fixed, all the lines have the same slope and are parallel.
- The values of b are −1, 0 and 1.
- Changing b changes only the y-intercept and shifts the line upward or downward without changing its slope.
(v) y = −2x − 3, y = −2x, y = −2x + 3
Answer
Role of a and b:
- Here, a = −2 is the same for all three equations.
- Since a is negative, all the lines slope downward from left to right.
- The values of b are −3, 0 and 3.
- Changing b changes only the y-intercept, so the lines remain parallel but shift vertically.
Conclusion:
- The coefficient a determines the slope (steepness and direction) of the line.
- The constant b determines the point where the line cuts the y-axis (y-intercept).
1. 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 must have the highest power of x equal to 3.
The coefficient of the x² term must be −7.
One such polynomial is
\(p(x)=2x^3-7x^2+5x-4\)
Many different answers are possible. Any polynomial whose highest power is 3 and whose \(x^2\) term has coefficient −7 is correct.
2. Find the values of the following polynomials at the indicated values of the variables.
(i) 5x² − 3x + 7 if x = 1
Answer
Substitute x = 1 in the polynomial.
5(1)² − 3(1) + 7
⇒ 5 − 3 + 7
⇒ 9
Therefore, the value of the polynomial is 9.
(ii) 4t³ − t² + 6 if t = a
Answer
Substitute t = a in the polynomial.
4(a)³ − (a)² + 6
⇒ 4a³ − a² + 6
Therefore, the value of the polynomial is 4a³ − a² + 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.
Answer
Let the required number be x.
According to the question,
\(\frac{5}{2}x+\frac{2}{3}=-\frac{7}{12}\)
⇒ \(\frac{5}{2}x=-\frac{7}{12}-\frac{2}{3}\)
⇒ \(\frac{5}{2}x=-\frac{7}{12}-\frac{8}{12}\)
⇒ \(\frac{5}{2}x=-\frac{15}{12}\)
⇒ \(\frac{5}{2}x=-\frac{5}{4}\)
⇒ \(x=-\frac{5}{4}\times\frac{2}{5}\)
⇒ \(x=-\frac{1}{2}\)
Therefore, the required number is \(-\frac{1}{2}\).
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?
Answer
Let the smaller number be x.
Then, the larger number = 5x.
After adding 21 to both the numbers,
Smaller number = x + 21
Larger number = 5x + 21
According to the question,
5x + 21 = 2(x + 21)
⇒ 5x + 21 = 2x + 42
⇒ 3x = 21
⇒ x = 7
Therefore,
Smaller number = 7
Larger number = 5 × 7 = 35
Verification:
After adding 21, the numbers become 28 and 56.
Since 56 = 2 × 28, the condition is verified.
Hence, the required numbers are 7 and 35.
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.
Answer
Initial amount = ₹800
Amount saved every month = ₹250
Let the number of months be n.
Amount saved in n months = ₹250n
Therefore,
Total Amount = Initial Amount + Savings
⇒ Total Amount = 800 + 250n
⇒ A = 800 + 250n
Thus, the required linear pattern is A = 800 + 250n.
(i) Amount after 6 months:
Substitute n = 6 in the expression.
⇒ A = 800 + 250 × 6
⇒ A = 800 + 1500
⇒ ₹2300
(ii) Amount after 2 years:
2 years = 24 months
Substitute n = 24 in the expression.
⇒ A = 800 + 250 × 24
⇒ A = 800 + 6000
⇒ ₹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.
Answer
Let the tens digit be x.
Since the digits differ by 3, let the units digit be x − 3.
Therefore,
Original Number = 10 × (Tens Digit) + Units Digit
⇒ Original Number = 10x + (x − 3)
⇒ Original Number = 11x − 3
After interchanging the digits,
New Number = 10 × (Units Digit) + Tens Digit
⇒ New Number = 10(x − 3) + x
⇒ New Number = 11x − 30
According to the question,
Original Number + New Number = 143
⇒ (11x − 3) + (11x − 30) = 143
⇒ 22x − 33 = 143
⇒ 22x = 176
⇒ x = 8
Therefore,
Tens digit = 8
Units digit = 8 − 3 = 5
Original Number = 85
Interchanged Number = 58
Verification:
Digits differ by 3.
⇒ 8 − 5 = 3
Sum of the two numbers:
⇒ 85 + 58 = 143
Hence, the required numbers are 85 and 58.
For digit problems, first define the digits as variables. Then write the number using the place value rule:
Two-digit Number = 10 × (Tens Digit) + Units Digit. This makes it easy to form the required equation.
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.
(i) y = −3x + 4
Answer
The equation is already in the form y = ax + b.
- Slope, a = −3
- y-intercept, b = 4
- The line cuts the y-axis at (0, 4).
(ii) 2y = 4x + 7
Answer
Write the equation in the form y = ax + b.
\(2y = 4x + 7\)
⇒ \(y = 2x + \frac{7}{2}\)
- Slope, \(a = 2\)
- y-intercept, \(b = \frac{7}{2}\)
- The line cuts the y-axis at \(\left(0,\frac{7}{2}\right)\).
(iii) 5y = 6x − 10
Answer
Write the equation in the form y = ax + b.
\(5y = 6x - 10\)
⇒ \(y = \frac{6}{5}x - 2\)
- Slope, \(a = \frac{6}{5}\)
- y-intercept, \(b = -2\)
- The line cuts the y-axis at (0, −2).
(iv) 3y = 6x − 11
Answer
Write the equation in the form y = ax + b.
\(3y = 6x - 11\)
⇒ \(y = 2x - \frac{11}{3}\)
- Slope, \(a = 2\)
- y-intercept, \(b = -\frac{11}{3}\)
- The line cuts the y-axis at \(\left(0,-\frac{11}{3}\right)\).
Are any of the lines parallel?
Answer
Two lines are parallel if they have the same slope.
The slopes of the given lines are:
- (i) −3
- (ii) 2
- (iii) \(\frac{6}{5}\)
- (iv) 2
Since equations (ii) and (iv) have the same slope (2) but different y-intercepts, their graphs are parallel.
Hence, the graphs of equations (ii) and (iv) are parallel.
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 \(y=\frac{9}{5}(x-273)+32\).
(i) Find the temperature of the liquid in Fahrenheit if the temperature of the liquid is 313 K.
Answer
(i) The given relation between Kelvin and Fahrenheit is
\[ y=\frac{9}{5}(x-273)+32 \]
where,
- x = Temperature in Kelvin (K)
- y = Temperature in Fahrenheit (°F)
Given,
Temperature of the liquid = 313 K
⇒ x = 313
Substitute x = 313 in the given equation.
\[ \begin{aligned} y&=\frac{9}{5}(313-273)+32\\ &=\frac{9}{5}(40)+32\\ &=72+32\\ &=104 \end{aligned} \]
Therefore, the temperature of the liquid is 104°F.
(ii) If the temperature is 158 °F, then find the temperature in Kelvin.
Answer
Given,
\(y=158^\circ\text{F}\)
Substitute \(y = 158\) in the given equation.
\(158=\frac{9}{5}(x-273)+32\)
⇒ \(126=\frac{9}{5}(x-273)\)
⇒ \(70=x-273\)
⇒ \(x=343\)
Therefore, the temperature of the liquid is 343 K.
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.
Answer
We know that,
Work Done = Force × Distance
Let
- Work done = w
- Distance travelled = d
- Constant force = 3 units
Therefore, w = 3d
This is the required linear equation.
Table of values:
| d | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| w | 0 | 3 | 6 | 9 | 12 |
When the distance travelled is 2 units,
w = 3 × 2
⇒ w = 6 units
The point (2, 6) lies on the graph, which verifies the answer.
Therefore, the work done is 6 units.
10. The graph of a linear polynomial p(x) passes through the points (1, 5) and (3, 11).
(i) Find the polynomial p(x).
Answer
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 equation (1) from equation (2),
⇒ 2a = 6
⇒ a = 3
Substituting a = 3 in equation (1),
⇒ 3 + b = 5
⇒ b = 2
Therefore,
p(x) = 3x + 2
(ii) Find the coordinates where the graph of p(x) cuts the axes.
Answer
y-axis:
At x = 0,
p(0) = 3(0) + 2
⇒ p(0) = 2
Therefore, the graph cuts the y-axis at (0, 2).
x-axis:
At the x-axis, p(x) = 0.
3x + 2 = 0
⇒ 3x = −2
⇒ \(x=-\frac{2}{3}\)
Therefore, the graph cuts the x-axis at \(\left(-\frac{2}{3},0\right)\).
(iii) Draw the graph of p(x) and verify your answers.
Answer
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 \(q(x)\).
Answer
Step 1: Find the polynomial \(p(x)\)
Let
\(p(x)=ax+b\)
The graph passes through the points \((2,3)\) and \((6,11)\).
First, find the slope.
\[ m=\frac{11-3}{6-2}=\frac{8}{4}=2 \]
Therefore,
⇒ \(a=2\)
Substitute the point \((2,3)\) into \(y=2x+b\).
⇒ \(3=2\times2+b\)
⇒ \(3=4+b\)
⇒ \(b=-1\)
Hence,
\(p(x)=2x-1\)
Step 2: Find the polynomial \(q(x)\)
Since the graph of \(q(x)\) is parallel to the graph of \(p(x)\), both lines have the same slope.
Therefore,
⇒ \(c=2\)
Let
\(q(x)=2x+d\)
The graph passes through the point \((4,-1)\).
⇒ \(-1=2\times4+d\)
⇒ \(-1=8+d\)
⇒ \(d=-9\)
Hence,
\(q(x)=2x-9\)
Step 3: Find where each graph cuts the x-axis
At the x-axis, \(y=0\).
For \(p(x)\):
⇒ \(2x-1=0\)
⇒ \(2x=1\)
⇒ \(\displaystyle x=\frac12\)
Therefore, \(p(x)\) cuts the x-axis at
\(\left(\frac12,0\right)\).
For \(q(x)\):
⇒ \(2x-9=0\)
⇒ \(2x=9\)
⇒ \(\displaystyle x=\frac92\)
Therefore, \(q(x)\) cuts the x-axis at
\(\left(\frac92,0\right)\).
Verification:
For \(p(x)\):
⇒ \(p(2)=2(2)-1=3\)
⇒ \(p(6)=2(6)-1=11\)
For \(q(x)\):
⇒ \(q(4)=2(4)-9=-1\)
Final Answer:
- \(\boxed{p(x)=2x-1}\)
- \(\boxed{q(x)=2x-9}\)
- \(p(x)\) cuts the x-axis at \(\left(\frac12,0\right)\).
- \(q(x)\) cuts the x-axis at \(\left(\frac92,0\right)\).
For two parallel lines, the slopes are always equal, but their y-intercepts are different. To find the equation of a line:
- Find the slope.
- Use one given point to determine the y-intercept.
- Write the equation in the form \(y=mx+c\).
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?
Answer
Each new hexagon shares one side with the previous hexagon.
Therefore, every new hexagon requires 5 additional matchsticks.
Stage 4:
Number of matchsticks = 16 + 5 = 21
Stage 5:
Number of matchsticks = 21 + 5 = 26
Hence, Stage 4 requires 21 matchsticks and Stage 5 requires 26 matchsticks.
(ii) Complete the following table.
Answer
| Stage Number | 1 | 2 | 3 | 4 | 5 | … | n |
|---|---|---|---|---|---|---|---|
| Number of Matchsticks | 6 | 11 | 16 | 21 | 26 | … | 5n + 1 |
(iii) Find a rule to determine the number of matchsticks required for the nth stage.
Answer
The number of matchsticks forms the pattern:
6, 11, 16, 21, 26, …
Each stage requires 5 more matchsticks than the previous stage.
Therefore, the rule is:
Number of matchsticks = 6 + (n − 1) × 5
⇒ Number of matchsticks = 6 + 5n − 5
⇒ Number of matchsticks = 5n + 1
Hence, the number of matchsticks required for the nth stage is 5n + 1.
(iv) How many matchsticks will be required for the 15th stage of the pattern?
Answer
From part (iii), the number of matchsticks required for the nth stage is
Number of matchsticks = 5n + 1
For the 15th stage,
Number of matchsticks = 5 × 15 + 1
⇒ 75 + 1
⇒ 76
Therefore, 76 matchsticks are required for the 15th stage.
(v) Can 200 matchsticks form a stage in this pattern? Justify.
Answer
For a stage to be formed, the number of matchsticks must satisfy
5n + 1 = 200
⇒ 5n = 199
⇒ \(\displaystyle n=\frac{199}{5}=39.8\)
Since n is not a whole number, 200 matchsticks cannot correspond to any stage of the pattern.
Hence, 200 matchsticks cannot form a stage in this pattern.
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.
Answer
Step 1: Find \(p(x)\)
Let
\(p(x)=ax+b\)
Since the graph passes through (2, 3),
\(2a+b=3\) … (1)
Since the graph passes through (6, 11),
\(6a+b=11\) … (2)
Subtracting equation (1) from equation (2),
⇒ \(4a=8\)
⇒ \(a=2\)
Substituting \(a=2\) in equation (1),
⇒ \(4+b=3\)
⇒ \(b=-1\)
Therefore,
\(\boxed{p(x)=2x-1}\)
Step 2: Find \(q(x)\)
Since the graph of \(q(x)\) is parallel to the graph of \(p(x)\), both have the same slope.
Hence,
\(q(x)=2x+d\)
Since the graph passes through (4, −1),
\(-1=2(4)+d\)
⇒ \(-1=8+d\)
⇒ \(d=-9\)
Therefore,
\(\boxed{q(x)=2x-9}\)
Step 3: Find where the graphs cut the x-axis
For \(p(x)\):
At the x-axis, \(p(x)=0\).
\(2x-1=0\)
⇒ \(2x=1\)
⇒ \(\displaystyle x=\frac{1}{2}\)
Therefore, the graph of \(p(x)\) cuts the x-axis at
\(\left(\frac{1}{2},0\right)\).
For \(q(x)\):
At the x-axis, \(q(x)=0\).
\(2x-9=0\)
⇒ \(2x=9\)
⇒ \(\displaystyle x=\frac{9}{2}\)
Therefore, the graph of \(q(x)\) cuts the x-axis at
\(\left(\frac{9}{2},0\right)\).
Hence,
- \(\boxed{p(x)=2x-1}\)
- \(\boxed{q(x)=2x-9}\)
- \(p(x)\) cuts the x-axis at \(\left(\frac{1}{2},0\right)\).
- \(q(x)\) cuts the x-axis at \(\left(\frac{9}{2},0\right)\).
14. What do all linear functions of the form \(f(x)=ax+a,\; a>0,\) have in common?
Answer
For the linear function
\(f(x)=ax+a\)
we have:
- The slope is \(a\).
- The y-intercept is also \(a\).
Since \(a>0\), all such graphs:
- are rising straight lines,
- cut the y-axis at the point \((0,a)\), and
- have the same numerical value for the slope and the y-intercept.
Hence, all linear functions of the form \(f(x)=ax+a,\; a>0,\) are increasing straight lines whose slope and y-intercept are equal.
