Introduction to Linear Polynomials Class 9

20 Questions | 60 Marks

Raju bought 'x' red boxes with 4 pens each and 'y' blue boxes with 5 pencils each. If he got 3 extra pens, write an algebraic expression for the total number of pens and pencils. 3marks

A rectangular garden of length l metres and width w metres is to be fenced. Wire fencing along the length costs 100 per metre and wooden fencing along the width costs 80 per metre. Seeds cost `50 per square metre. Write an algebraic expression for the total cost. 3marks

Identify the degree of the following polynomials:(i)

7 x^{3} - 2 x^{2} + 5 x - 1

(ii)

15

(iii)

y^{2} + \frac{3}{4}

3marks

Write a polynomial of degree 2 where the coefficient of the

x^{2}

term is 5 and the coefficient of the x term is -3. 2marks

Find the perimeter of a square with side length 's' cm. If the side increases by 0.5 cm, how does the perimeter change? Represent this change using a linear polynomial. 3marks

A chess club charges a joining fee of 200 and 50 for every match played. If 'm' is the number of matches played, write a linear polynomial for the total cost. If a player paid `750, how many matches did they play? 3marks

Evaluate the linear polynomial

3 x - 5

for:(i)

x = 0

(ii)

x = - 2

2marks

The present age of a father is three times the present age of his son. After 5 years, their ages will add up to 70 years. Find their present ages. 5marks

A wire of length 20 cm is bent to form a rectangle of length x cm. Write an expression for the area of such a rectangle. What is the degree of this polynomial? 3marks

A student reads 20 pages of a book every day. Write a linear expression for the number of pages left after 'd' days, if the book has 500 pages. How many pages will be left after 15 days? 3marks

An auto-rickshaw fare is 25 for the first 2 km and 15 per km thereafter. Write a linear expression for the fare for a journey of 'n' km, where

n \geq 2

. What is the fare for travelling 8 km? 3marks

A mobile phone is bought for 10,000. Its value decreases by 800 every year.(i) Write an expression that relates the value 'v' and the time 't' in years.(ii) Find the value of the phone after 5 years. 3marks

A telecom company charges a fixed monthly fee and an additional cost per GB of data used. If 10 GB cost 350 and 20 GB cost 550, find the linear relationship

y = a x + b

representing the bill 'y' for 'x' GB of data. 5marks

Draw the graph of the linear equation

y = 3 x - 2

. Find the coordinates of the points where the line cuts the x-axis and the y-axis. 3marks

Plot the points (–1, –3), (0, 0), (1, 3), (3, 9) on a graph. Identify the linear relationship between x and y. 3marks

Observe the given pattern of square tiles:Stage 1: 1 tileStage 2: 3 tilesStage 3: 5 tilesFind a linear expression for the number of tiles at stage 'n'. How many tiles will be there in the 10th stage? 3marks

A plant has height 1.75 feet and grows by 0.5 feet each month.(i) Find an expression that relates the height 'h' and the time 't' in months.(ii) What is the height after 6 months? 3marks

Draw the graphs of the lines

y = 2 x + 1

and

y = 2 x + 4

on the same axes. What do you observe about these lines? 3marks

A gym charges a fixed monthly fee and an additional cost per hour for using a badminton court. If 10 hours cost 800 and 15 hours cost 1100, find the fixed monthly fee and the cost per hour. 5marks

Consider the linear equation

y = - x + 3

.(i) Draw its graph by selecting suitable points.(ii) Find the y-intercept of the line. 3marks

Answers

The algebraic expression is

4 x + 5 y + 3

The algebraic expression for the total cost is (200l + 160w + 50lw).

(i) Degree is 3.(ii) Degree is 0 (constant polynomial).(iii) Degree is 2.

A possible polynomial is

5 x^{2} - 3 x + c

, where 'c' is any constant. For example,

5 x^{2} - 3 x + 10

Perimeter of a square with side 's' is

4 s

. If the side increases by 0.5 cm, the new side is

(s + 0.5)

cm. The new perimeter is

4 (s + 0.5) = 4 s + 2

. The change in perimeter is

(4 s + 2) - 4 s = 2

cm. The linear polynomial representing the change is

P_{n e w} - P_{o l d} = (4 s + 2) - 4 s = 2

The linear polynomial for the total cost is

C (m) = 200 + 50 m

.If the player paid `750, then

200 + 50 m = 750

50 m = 750 - 200

50 m = 550

$m = \frac{550}{50} = 11$ . The player played 11 matches.

(i) For

x = 0

3 (0) - 5 = 0 - 5 = - 5

.(ii) For

x = - 2

3 (- 2) - 5 = - 6 - 5 = - 11

Let the son's present age be 'x' years. Then the father's present age is '3x' years.After 5 years, son's age will be

(x + 5)

and father's age will be

(3 x + 5)

.Their sum will be 70:

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

4 x + 10 = 70

4 x = 60

x = 15

Son's present age = 15 years.
Father's present age = $3 \times 15 = 45$ years.

The expression for the area of the rectangle is

x (10 - x) = 10 x - x^{2}

. The degree of this polynomial is 2.

The linear expression for the number of pages left after 'd' days is

500 - 20 d

.After 15 days, the number of pages left =

500 - 20 \times 15 = 500 - 300 = 200

pages.

For a journey of 'n' km (

n \geq 2

), the fare is the cost for the first 2 km plus the cost for the remaining

(n - 2)

km.Fare =

25 + 15 \times (n - 2) = 25 + 15 n - 30 = 15 n - 5

.For travelling 8 km (

n = 8

), the fare is

15 \times 8 - 5 = 120 - 5 = 115

. The fare is `115.

(i) The expression relating value 'v' and time 't' is

v (t) = 10000 - 800 t

.(ii) After 5 years,

v (5) = 10000 - 800 \times 5 = 10000 - 4000 = 6000

. The value is `6000.

We are given the linear relationship

y = a x + b

.When

x = 10

GB,

y = 350

. So,

350 = 10 a + b

(Equation 1).When

x = 20

GB,

y = 550

. So,

550 = 20 a + b

(Equation 2).Subtracting Equation 1 from Equation 2:

(550 - 350) = (20 a - 10 a) + (b - b)

200 = 10 a

$a = 20$ .
Substitute $a = 20$ into Equation 1:

350 = 10 (20) + b

350 = 200 + b

$b = 350 - 200 = 150$ .
The linear relationship is $y = 20 x + 150$ .

To draw the graph of

y = 3 x - 2

:For x-intercept, set

y = 0

0 = 3 x - 2 ⟹ 3 x = 2 ⟹ x = \frac{2}{3}

. Point:

(\frac{2}{3}, 0)

.For y-intercept, set

x = 0

y = 3 (0) - 2 ⟹ y = - 2

. Point:

(0, - 2)

.Plot these two points and draw a line through them.

Graph: (A sketch of the line passing through $(2 / 3, 0)$ and $(0, - 2)$ would be shown here.)
Slope = 3.
Y-intercept = -2.
The line cuts the y-axis at $(0, - 2)$ .

Plotting the points:(–1, –3)(0, 0)(1, 3)(3, 9)Observing the points, we can see that the y-coordinate is always 3 times the x-coordinate.The linear relationship is

y = 3 x

The sequence of the number of tiles is 1, 3, 5, ...The difference between consecutive terms is constant (2). This is an arithmetic progression.The linear expression for the number of tiles at stage 'n' is

T_{n} = a + (n - 1) d

, where

a = 1

(first term) and

d = 2

(common difference).

T_{n} = 1 + (n - 1) 2 = 1 + 2 n - 2 = 2 n - 1

.For the 10th stage (

n = 10

), the number of tiles will be

2 (10) - 1 = 20 - 1 = 19

(i) Let 'h' be the height in feet and 't' be the time in months.The initial height is 1.75 feet. The growth rate is 0.5 feet per month.The expression relating 'h' and 't' is

h (t) = 1.75 + 0.5 t

.(ii) After 6 months,

h (6) = 1.75 + 0.5 \times 6 = 1.75 + 3.00 = 4.75

feet.

To draw the graphs of

y = 2 x + 1

and

y = 2 x + 4

:For

y = 2 x + 1

:If

x = 0, y = 1

. Point (0, 1).If

x = 1, y = 3

. Point (1, 3).For

y = 2 x + 4

:If

x = 0, y = 4

. Point (0, 4).If

x = 1, y = 6

. Point (1, 6).

(A sketch of two parallel lines with slope 2, one cutting y-axis at 1 and the other at 4 would be shown here.)
Observation: The two lines are parallel because they have the same slope (a=2). They have different y-intercepts (b=1 and b=4).

Let the fixed monthly fee be b and the cost per hour be a. The relationship is

y = a x + b

.When

x = 10

hours,

y = ‘ 800. S o,

800 = 10a + b

(E q u a t i o n 1) . W h e n

x=15

h o u r s,

y=`1100. So,

1100 = 15 a + b

(Equation 2).Subtracting Equation 1 from Equation 2:

(1100 - 800) = (15 a - 10 a) + (b - b)

300 = 5 a

$a = \frac{300}{5} = 60$ .
Substitute $a = 60$ into Equation 1:

800 = 10 (60) + b

800 = 600 + b

$b = 800 - 600 = 200$ .
The fixed monthly fee is 200 and the cost per hour is 60.

For the linear equation

y = - x + 3

:(i) To draw the graph:If

x = 0, y = 3

. Point (0, 3).If

y = 0, 0 = - x + 3 ⟹ x = 3

. Point (3, 0).Plot points (0, 3) and (3, 0) and draw a line through them.(A sketch of the line passing through

(0, 3)

and

(3, 0)

would be shown here.)(ii) The y-intercept is the value of y when x=0, which is 3. The line cuts the y-axis at (0, 3).

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Introduction to Linear Polynomials — Important Questions

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