GANITA MANJARI CLASS 9, Chapter 2, Introduction to Linear Polynomials

 Introduction to Linear Polynomials 

EXERCISE SET 2.1

1. Find the degrees of the following polynomials:

(i) 2x² − 5x + 3

Degree = 2

(ii) y³ + 2y − 1

Degree = 3

(iii) −9

Degree = 0

(iv) 4z − 3

Degree = 1

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

Degree 1: x + 2

Degree 2: x² + 3x + 1

Degree 3: x³ + 2x² + x + 5

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

Coefficient of x² = 6

Coefficient of x³ = −3

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

Coefficient of z = 0

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

Constant term = −10

EXERCISE SET 2.2

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

(i) x = 0

5(0) − 3 = −3

Answer = −3

(ii) x = −1

5(−1) − 3 = −8

Answer = −8

(iii) x = 2

5(2) − 3 = 7

Answer = 7

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

(i) s = 0

7(0)² − 4(0) + 6 = 6

Answer = 6

(ii) s = −3

7(−3)² − 4(−3) + 6

= 63 + 12 + 6

= 81

Answer = 81

(iii) s = 4

7(4)² − 4(4) + 6

= 112 − 16 + 6

= 102

Answer = 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.

Let Salil's age = x years

Mother's age = 3x years

(x + 5) + (3x + 5) = 70

4x + 10 = 70

4x = 60

x = 15

Salil's present age = 15 years

Mother's present age = 45 years

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

Let the integers be 2x and 5x.

5x − 2x = 63

3x = 63

x = 21

Integers = 42 and 105

5. Ruby has 3 times as many two-rupee coins as five-rupee coins. If she has a total ₹88, how many coins does she have of each type ?

Let the number of ₹5 coins = x

Number of ₹2 coins = 3x

5x + 2(3x) = 88

11x = 88

x = 8

₹5 coins = 8

₹2 coins = 24

6. A farmer cuts a 300-feet fence into two pieces. The longer piece is four times the shorter piece.How long are the two pieces ?

Let the shorter piece = x ft

Longer piece = 4x ft

x + 4x = 300

5x = 300

x = 60

Shorter piece = 60 ft

Longer piece = 240 ft

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?

Let width = x cm

Length = 2x + 3 cm

2(L + W) = 24

L + W = 12

(2x + 3) + x = 12

3x + 3 = 12

3x = 9

x = 3

Width = 3 cm

Length = 9 cm

Answer: Length = 9 cm, Width = 3 cm.

EXERCISE SET 2.3

1. A student has ₹500 in her savings bank account. She gets ₹150 every month as pocket money.How much money will she have at the end of every month from the second month onwards ? Find a linear expression to represent the amount she will have in the nth month.

Solution :

Amount after 1 month = ₹500 + ₹150 = ₹650

Amount after 2 months = ₹500 + ₹300 = ₹800

Amount after 3 months = ₹500 + ₹450 = ₹950

Amount after 4 months = ₹500 + ₹600 = ₹1100

Linear expression for the amount after n months:

A = 500 + 150n

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.

Solution : 

After 1 hour = 120 − 9 = 111

After 2 hours = 120 − 18 = 102

After 3 hours = 120 − 27 = 93

After 4 hours = 120 − 36 = 84

Linear expression for the number of members after n hours:

M = 120 − 9n

3. Suppose the length of a rectangle is 13 cm. Find the area of the breadth is (i) 12 cm , (ii) 10 cm, ( iii ) 8 cm. Find the linear pattern representing the area of the rectangle.

Solution : 

Length of a rectangle = 13 cm.

(i) Breadth = 12 cm

Area = 13 × 12 = 156 cm²

(ii) Breadth = 10 cm

Area = 13 × 10 = 130 cm²

(iii) Breadth = 8 cm

Area = 13 × 8 = 104 cm²

Linear pattern representing the area:

Let breadth = b cm

Area = 13b

4. Suppose the length of a rectangular box is 7 cm and breadth is 11 cm. Find the volume of the height is (i) 5 cm, (ii) 9 cm, ( iii) 13 cm. Find the linear pattern representing the volume of the rectangular box.

Solution :

 Length of a rectangular box = 7 cm, 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

Linear pattern representing the volume:

Let height = h cm

Volume = 7 × 11 × h

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.

Solution :

Pages read in 15 days = 20 × 15 = 300

Pages left after 15 days = 500 − 300 = 200 pages

Linear pattern:

Let d be the number of days.

Pages left = 500 − 20d

So, the linear expression is:

P = 500 − 20d

EXERCISE SET 2.4

1.Suppose a plant has height 1.75 feet and grows by 0.5 feet each month.
(i) Find the height after 7 months.
( ii ) Make a table of values for t varying from 0 to 10 months and show how the height, h, increases every month.
( iii ) Find an expression that relates h and t, and explain why it represents linear growth.

Solution : Initial height = 1.75 feet, each month = + 0.5 feet 

(i) Height after 7 months, h = 1.75 + 0.5 × 7 = 1.75 + 3.5 = 5.25 

Answer : 5.25 feet.

(ii) Table : [ 1 month = 1.75 + 0.5 = 2.25, 2 month = 1.75 + 2 × 0.5 = 2.75,........]

t ( months) : 0 ,          1,         2,         3,          4,          5,           6.        7,           8,        9         10

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) Expression : h = 1.75 + 0.5 t

It represents linear growth because the height increases by 0.5 feet every month.

2. A mobile phone bought for ₹10,000. Value decreases by ₹800 every year.
(i) Find the values of the phone after 3 years.
(ii) Make a table of values for t varying from 0 to 8 years and show how the value of the phone,V, depreciated with time.
(iii) Find an expression that relates V and t, and explain why it represents linear decay.

Solution : Initial cost = ₹10,000,  every year = - ₹800

(i)  Value after 3 years, V = 10,000 - 3 × 800 = 10,000 - 2,400 = 7,600

Answer : ₹7,600

(ii) Table : [ 0 year = 10,000, 1 year = 10,000 - 800 = 9,200....]

t ( years ) : 0,        1            2,       3            4,            5       6        7            8      

V ( ₹) :   10,000.   9,200   8,400   7,600   6,800  6,000  5,200   4,400   3,600

(iii) Expression : V = 10,000 - 800t

It represents linear decay because the value decreases by ₹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.
(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.

Solution : 

Initial population = 750

Every year = +50 people

(i) Population after 6 years, P = 750 + 6 × 50 = 750 + 300 = 1,050

Answer: 1050 people

(iii) Table :  [ 0 year = 750, 1 year = 750 + 50 = 800.....]

       t ( years  )  : 0,        1,    2     3      4,    5        6,        7        8          9,       10 

        P.                : 750.  800. 850 900 950 1,000 1,050 1,100 1,150 1,200. 1,250

(iii) Expression : P = 750 + 50 t

It represents linear growth because the population increases by 50 every year.

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.
(ii) After how many days will the balance run out ?
(iii) Make a table of values for x varying from 1 to 10 days and show how the balance b(x) ,reduces with time.

Solution : Recharge = ₹600

Balance reduces by ₹15 per day.

(i) Equation : b(x) = 600 - 15x

It is linear decay because the balance decreases by ₹15 every day.

(ii) After how many days will the balance become zero?

600 - 15x = 0

or, 600 = 15x

or,x = 600/15 = 40

Answer : 40 days.

(iii) Table : 

 x ( days)  : 1,      2,       3,          4,        5,           6,        7,           8,         9,             10 

b(x)(₹) :    585.    570.    555.  540.   525.       510.   495.       480.      465.         450

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.

Solution : Given,

y = ax + b

When x = 10 and y = 400,

10a + b = 400 ..........(1)

When x = 14 and y = 500,

14a + b = 500 ..........(2)

Subtract equation (1) from equation (2):

14a + b − (10a + b) = 500 − 400

4a = 100

a = 25

Substitute a = 25 into equation (1):

10 × 25 + b = 400

250 + b = 400

b = 150

Therefore,

a = 25

b = 150

Required equation:

y = 25x + 150
2. A gym charges a fixed monthly fee and an additional cost per hour for using the badminton court.
A student observed that:
When she used the 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 number of hours x, according to the relation y = ax + b
find the values of a and b .

Solution : Given,

y = ax + b

When x = 10 and y = 800,

10a + b = 800 ..........(1)

When x = 15 and y = 1100,

15a + b = 1100 ..........(2)

Subtract equation (1) from equation (2):

15a + b − (10a + b) = 1100 − 800

5a = 300

a = 60

Substitute a = 60 into equation (1):

10 × 60 + b = 800

600 + b = 800

b = 200

Therefore,

a = 60

b = 200

Required equation:

y = 60x + 200

3. Consider the relationship between temperature measured in degrees Celsius (°C) and degrees Fahrenheit (°F), given by ° C = a ° F + b.

Find a and b, given that:

Ice melts at 0°C and 32°F.

Water boils at 100°C and 212°F.

Solution : Given,

°C = a°F + b

When °C = 0 and °F = 32,

0 = 32a + b ..........(1)

When °C = 100 and °F = 212,

100 = 212a + b ..........(2)

Subtract equation (1) from equation (2):

100 = 180a

a = 100/180

a = 5/9

Substitute a = 5/9 into equation (1):

0 = 32 × 5/9 + b

b = -160/9

Therefore,

a = 5/9

b = -160/9

Required equation:

°C = (5/9)°F − 160/9

or

°C = (5/9)(°F − 32)

Final Answers:

a = 5/9, b = -160/9; Equation: °C = (5/9)(°F − 32)

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 

We observe that :

The graph of every equation of the form y = ax (where b = 0) always passes through the origin (0,0).

Among the lines y = 4x, y = 2x, and y = x, the line y = 4x is the steepest because it has the greatest value of the slope (a = 4).

Similarly, y = 2x( a = 2) is steeper than y = x ( a = 1) since 2 > 1.

Thus, for equations of the form y = ax, a larger value of a  ( slope) results in a steeper line.

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.

Answer: x³ – 7x² + 2x + 5

2. Find the values of the following polynomials.

(i) 5x² – 3x + 7, if x = 1

= 5(1)² – 3(1) + 7

= 5 – 3 + 7

= 9

Answer: 9

(ii) 4t³ – t² + 6, if t = a

= 4a³ – a² + 6

Answer: 4a³ – a² + 6

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.

5x/2 + 2/3 = –7/12

5x/2 = –7/12 – 2/3

5x/2 = –7/12 – 8/12

5x/2 = –15/12

5x/2 = –5/4

x = (–5/4) × (2/5)

x = –1/2

Answer: –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?

Solution :

Let the smaller number be x.

Then the larger number is 5x.

5x + 21 = 2(x + 21)

5x + 21 = 2x + 42

3x = 21

x = 7

Larger number = 5 × 7 = 35

Answer: 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.

Solution :

Linear pattern:

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

Answers: (i) ₹2300 (ii) ₹6800 Linear pattern: A = 800 + 250n

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 tens digit be x and the ones digit be y.

x – y = 3

Original number = 10x + y

Interchanged number = 10y + x

(10x + y) + (10y + x) = 143

11x + 11y = 143

x + y = 13

Solving:

x – y = 3

x + y = 13

2x = 16

x = 8

y = 5

Original number = 85

Interchanged number = 58

Answer: 85 and 58