Kerala Syllabus 9th Standard Maths Text Book Questions & Answers chapter 3- Pairs of Equations

Pairs of Equations

Question 1.

In a rectangle of perimeter one metre, one side is 5 cm longer than the other. What are the lengths of the rectangle?

Solution: 22.5 cm & 27.5 cm are the lengths of the rectangle.

Question 2.

A class has 4 more girls than boys. On a day when only 8 boys were absent, the number of girls was twice that of the boys. How many girls and boys are there in the class?

Solution: 24 girls & 20 boys are the in the class.

Question 3.
A man invested 10000 rupees, split into two schemes, at annual rates of interest 8% and 9%. After one year he got 875 rupees as interest from both. How much did he invest in each?
Answer:
If one part is x then
the remaining part is 10000 – x
x×8100+(10000−x)×9100=100
8x + 90000 – 9x = 87500
90000 – 87500 = x
2500 = x
one part = 2500 and
remaining part = 7500

Question 4.
A three and a half metre long rod is to be cut into two pieces, one piece is to be bent into a square and the other into an equilateral triangle. The length of their sides must be the same. How should it be cut?
Answer:
Total length = 3½ m
Since the sides of a square and equilateral triangle are equal, all these 7 sides are equal.
∴ Length of one side
312÷7=72÷7=12m
Length of the rod for the square
=4×12 = 2m
Length of the rod for the equilateral triangle = 3×12 = 112m

Question 5.
The distance travelled in t seconds by an object starting with a speed of u metres/second and moving along a straight line with speed increasing at the rate of a metres/second every second is given by ut + ½ at² metres. An object moving in this manner travels 10 metres in 2 seconds and 28 metres in 4 seconds. With what speed did it start? At what rate does its speed change?
Answer:
If t = 2
ut + ½ at²= 10
2u + 2a= 10
u + a = 5 — (1)
If t = 4
4u + 8a = 28
u + 2a = 7 — (2)
from (1) and (2)


Textbook Page No. 40

Question 1.
Raju bought seven notebooks of two hundred pages and five of hundred pages, for 107 rupees. Joseph bought five notebooks of two hundred pages and seven of hundred pages, for 97 rupees. What is the price of each kind of notebook?
Answer:
Cost of 200 page note book = x
Cost of 100 page note book = y
7x + 5y= 107 …………(1)
5x + 7y = 97 …………(2)
(1) × 5 ⇒ 35x + 25y = 535 …………(3)
(2) × 7 ⇒ 35x + 49y = 679 …………(4)
(4) – (3) ⇒ 24y = 144
y = 14424 = 6
Substitute y = 6 in equation (1)
7x + 30 = 107; 7x = 77
x = 777 = 11
Price of the 200 pages notebook = Rs. 11
Price of the 100 pages notebook = Rs. 6

Question 2.
Four times a number and three times number added together make 43. Two times the second number, subtracted from three times the first give 11. What are the numbers?
Answer:
Let the first number = x and
the second number = y
4x + 3y = 43 …………(1)
3x – 2y = 11 …………(2)
(1) × 3 ⇒ 12x + 9y= 129 …………(3)
(2) × 4 ⇒ 12x – 8y = 44 …………(4)
(3) -(4) ⇒ 17y = 85; y = 8517 = 5
Substitute y = 5 in equation (1)
4x + 3y = 43
4x + 15 = 43
4x = 43 – 15 = 28
∴ x = 284 = 7, y = 5
First number = 7
Second number = 5

Question 3.
The sum of the digits of two – digit number is 11. The number got by interchanging the digits is 27 more than the original number. What is the number?
Answer:
If the numbers are x and y
x + y = 11 …………(1)
10x + y + 27 = 10y + x
9x – 9y = -27
X – y = -3 …………(2)
(1) + (2) 2x = 8; x = 4
x + y = 11
4 + y = 11
y = 7
∴ Required number is 47.

Question 4.
Four years ago, Rahim’s age was three times Ramu’s age. After two years, it would just be double. What are their ages now?
Answer:
Ramu’s present age = x
Rahim’s present age = y
4 years back,
Ramu’s age = x – 4
Rahim’s age = y – 4
3(x – 4) = y – 4
3x – 12 = y – 4
3x – y = 8 ……….(1)
After 2 years,
Ramu’s age = x + 2
Rahim’s age = y + 2
2(x + 2) = y + 2
2x + 4 = y + 2
2x – y = -2 ……….(2)
(1) – (2) ⇒ x = 10
3x – y = 8; 30 – y = 8; y = 22
x = 10, y = 22
Ramu’s present age = 10
Rahim’s present age = 22

Question 5.
If the length of a rectangle is in-creased by 5 metres and breadth decreased by 3 metres, the area would decrease by 5 square metres. If the length is increased by 3 metres and breadth increased by 2 metres, the area would increase by 50 square metres. What are the length and breadth?
Answer:
length = x; breadth = y
(x + 5)(y – 3) = xy – 5
xy – 3x + 5y – 15 = xy – 5
– 3x + 5y = + 10
3x – 5y = -10 ………..(1)
(x + 3)(y + 2) = xy + 50
xy + 2x + 3y + 6 = xy + 50
2x + 3y = 44 ………..(2)
(2) × 1 ⇒ 6x-10y = -20 ……….(3)
(3) × 2 ⇒ 6x + 9y = 132 …………(4)
(3)- (4) ⇒ -19y = -152
y = −152−19
a = 2
∴ u = 3

SCERT Class 9 physics Chapter - Equations of Motion - notes

 

State of rest : If the position of an object does not change with time, the object is said to be in a state of rest.
State of motion : If the position of an object changes with time,the object is said to be in a state of motion.

Distance  : The actual path traversed by an object in a certain interval of time is called distance traveled by that object.

Speed : It is  the distance travelled by an object in unit time. Speed is measured in meters/second.

Displacement  :  The change in  position when an object moves is calledthe displacement of that object. In other words It is the shortest distance between the initial and final positions of the object. 

Velocity : It is the displacement of an object in unit time. Velocity is measured in meters/second.

                          

Acceleration : It is the change in velocity in unit time. It is measured in meters/second².

                                                              

     
Uniform motion : If an object moves with uniform velocity, the motion is called uniform motion.As the velocity is same throughout the motion,the object travels equal distance in equal intervals of time.
                   
             Displacement  = velocity × time

Non-Uniform motion : If an object moves with non-uniform velocity, the motion is called non uniform motion. In non uniform motion,the object may have uniform acceleration or non uniform acceleration. 

Non-Uniformly accelerated motion :  If an object moves with non-uniform  acceleration, the motion is called non-uniformly accelerated motion. As the object is accelerated non uniformly throughout the motion,Graphical methods are used to find the velocity & displacement of the object.

Uniformly accelerated motion :  If an object moves with uniform  acceleration, the motion is called uniformly accelerated motion. As the object is accelerated uniformly throughout the motion,

    

Equations  of motion 

velocity - time relation   v = u + at 

position - time relation   S = ut + ½at²

If the velocity of an object is constant, the displacement = velocity × time
If it moves with uniform acceleration,the displacement is given by;

position - velocity  relation   v² - u² =  2aS 

Graphical Representation of Motion 

Motion of an object can be represented in 3 ways.

• Position - time graph or Distance - time graph 
• Velocity  - time graph 
• Acceleration - time graph 

Position - time graph or Distance - time graph

Position-time graph describes the motion of an object and helps to visualize the trajectory of the objects.

For an object,position is the dependent variable,which depends on the independent variable, time. 

In position time graph, the vertical Y axis represents the position of the object while the horizontal  X axis represents the time.

Slop of the position-time graph gives the velocity of the object. The steeper the slope is, the faster the motion is changing. 

Stationary object 

As the object is Stationary,position does not change with time.

The position time graph of a Stationary object is a straight line parallel to the time axis.

Slope of the line = velocity = zero 

Object moving with uniform velocity 

The position time graph of an object moving with uniform velocity is a straight line inclined to time axis.

The inclination of the line is the velocity of the object. 

Slope of the line = inclination of the line = velocity  

Object moving with uniform acceleration 

The position time graph of an object moving with uniform acceleration is a parabola.

Slope of the parabola = velocity of the object, changes every time.

Velocity - time graph

Velocity is the dependent variable which depends on the independent variable time.

In velocity time graph,Velocity of an object is plotted against time. 

Time is plotted on the horizontal axis (X axis ). Velocity is plotted on the vertical axis (Y axis).

Slop of the velocity time graph gives the acceleration of the object. The steeper the slope is, the faster the velocity is changing. 

Area under the velocity time graph gives the displacement of the object.

Stationary object 

As the object is Stationary, position does not change with time.

The velocity time graph of a Stationary object is a straight line on the time axis.

Velocity is zero on the time axis.

Slope of the line = acceleration = zero 

Area under the line = displacement = zero 

Object moving with uniform velocity 

The velocity time graph of an object moving with uniform velocity is a straight line parallel to the time axis, as the velocity is constant throughout the motion. 

Slope of the line = acceleration = zero  

Area under the line = displacement = velocity × time 

   

Object moving with uniform acceleration 

The velocity time graph of an object moving with uniform acceleration is a straight line inclined to time axis.

Slope of the line = inclination of the line = acceleration of the object = velocity/ time 

Area under the line = displacement = average velocity × time 

Object moving with unifincreasing acceleration 

The velocity time graph of an object moving with uniformly increasing acceleration is a parabola.

Slope of the curve =  acceleration of the object = velocity/ time 

Area under the curve = displacement = average velocity × time 

DRAW A TRIANGLE OF SIDES

DRAW A TRIANGLE OF SIDES

PAIRS OF EQUATION

 

PAIRS OF EQUATION

Maths Class-9