Category: Mathematics

Area and Perimeter is an important and basic topic in the Mensuration of 2-D or Planar Figures. The area is used to measure the space occupied by the planar figures. The perimeter is used to measure the boundaries of the closed figures. In Mathematics, these are two major formulas to solve the problems in the 2-dimensional shapes.

Each and every shape has two properties that are Area and Perimeter. Students can find the area and perimeter of different shapes like Circle, Rectangle, Square, Parallelogram, Rhombus, Trapezium, Quadrilateral, Pentagon, Hexagon, and Octagon. The properties of the figures will vary based on their structures, angles, and size. Scroll down this page to learn deeply about the area and perimeter of all the two-dimensional shapes.

Area and Perimeter Definition

Area: Area is defined as the measure of the space enclosed by the planar figure or shape. The Units to measure the area of the closed figure is square centimeters or meters.

Perimeter: Perimeter is defined as the measure of the length of the boundary of the two-dimensional planar figure. The units to measure the perimeter of the closed figures is centimeters or meters.

Formulas for Area and Perimeter of 2-D Shapes

1. Area and Perimeter of Rectangle:

• Area = l × b
• Perimeter = 2 (l + b)
• Diagnol = √l² + b²

Where, l = length
b = breadth

2. Area and Perimeter of Square:

• Area = s × s
• Perimeter = 4s

Where s = side of the square

3. Area and Perimeter of Parallelogram:

• Area = bh
• Perimeter = 2( b + h)

Where, b = base
h = height

4. Area and Perimeter of Trapezoid:

• Area = 1/2 × h (a + b)
• Perimeter = a + b + c + d

Where, a, b, c, d are the sides of the trapezoid
h is the height of the trapezoid

5. Area and Perimeter of Triangle:

• Area = 1/2 × b × h
• Perimeter = a + b + c

Where, b = base
h = height
a, b, c are the sides of the triangle

6. Area and Perimeter of Pentagon:

• Area = (5/2) s × a
• Perimeter = 5s

Where s is the side of the pentagon
a is the length

7. Area and Perimeter of Hexagon:

• Area = 1/2 × P × a
• Perimeter = s + s + s + s + s + s = 6s

Where s is the side of the hexagon.

8. Area and Perimeter of Rhombus:

• Area = 1/2 (d1 + d2)
• Perimeter = 4a

Where d1 and d2 are the diagonals of the rhombus
a is the side of the rhombus

9. Area and Perimeter of Circle:

• Area = Πr²
• Circumference of the circle = 2Πr

Where r is the radius of the circle
Π = 3.14 or 22/7

10. Area and Perimeter of Octagon:

• Area = 2(1 + √2) s²
• Perimeter = 8s

Where s is the side of the octagon.

Solved Examples on Area and Perimeter

Here are some of the examples of the area and perimeter of the geometric figures. Students can easily understand the concept of the area and perimeter with the help of these problems.

1. Find the area and perimeter of the rectangle whose length is 8m and breadth is 4m?

Solution:

Given,
l = 8m
b = 4m
Area of the rectangle = l × b
A = 8m × 4m
A = 32 sq. meters
The perimeter of the rectangle = 2(l + b)
P = 2(8m + 4m)
P = 2(12m)
P = 24 meters
Therefore the area and perimeter of the rectangle is 32 sq. m and 24 meters.

2. Calculate the area of the rhombus whose diagonals are 6 cm and 5 cm?

Solution:

Given,
d1 = 6cm
d2 = 5 cm
Area = 1/2 (d1 + d2)
A = 1/2 (6 cm + 5cm)
A = 1/2 × 11 cm
A = 5.5 sq. cm
Thus the area of the rhombus is 5.5 sq. cm

3. Find the area of the triangle whose base and height are 11 cm and 7 cm?

Solution:

Given,
Base = 11 cm
Height = 7 cm
We know that
Area of the triangle = 1/2 × b × h
A = 1/2 × 11 cm × 7 cm
A = 1/2 × 77 sq. cm
A = 38.5 sq. cm
Thus the area of the triangle is 38.5 sq. cm.

4. Find the area of the circle whose radius is 7 cm?

Solution:

Given,
Radius = 7 cm
We know that,
Area of the circle = Πr²
Π = 3.14
A = 3.14 × 7 cm × 7 cm
A = 3.14 × 49 sq. cm
A = 153.86 sq. cm
Therefore the area of the circle is 153.86 sq. cm.

5. Find the area of the trapezoid if the length, breadth, and height is 8 cm, 4 cm, and 5 cm?

Solution:

Given,
a = 8 cm
b = 4 cm
h = 5 cm
We know that,
Area of the trapezoid = 1/2 × h(a + b)
A = 1/2 × (8 + 4)5
A = 1/2 × 12 × 5
A = 6 cm× 5 cm
A = 30 sq. cm
Therefore the area of the trapezoid is 30 sq. cm.

6. Find the perimeter of the pentagon whose side is 5 meters?

Solution:

Given that,
Side = 5 m
The perimeter of the pentagon = 5s
P = 5 × 5 m
P = 25 meters
Therefore the perimeter of the pentagon is 25 meters.

FAQs on Area and Perimeter

1. How does Perimeter relate to Area?

The perimeter is the boundary of the closed figure whereas the area is the space occupied by the planar.

2. How to calculate the perimeter?

The perimeter can be calculated by adding the lengths of all the sides of the figure.

3. What is the formula for perimeter?

The formula for perimeter is the sum of all the sides.

In Maths Mensuration is nothing but a measurement of 2-D and 3-D Geometrical Figures. Mensuration is the study of the measurement of shapes and figures. We can measure the area, perimeter, and volume of geometrical shapes such as Cube, Cylinder, Cone, Cuboid, Sphere, and so on.

Keep reading this page to learn deeply about the mensuration. We can solve the problems easily, if and only we know the formulas of the particular shape or figure. This article helps to learn the mensuration formulae with examples. Learn the difference between the 2-D and 3-D shapes from here. Understand the concept of Mensuration by using various formulas.

Definition of Mensuration

Mensuration is the theory of measurement. It is the branch of mathematics that is used for the measurement of various figures like the cube, cuboid, square, rectangle, cylinder, etc. We can measure the 2 Dimensional and 3 Dimensional figures in the form of Area, Perimeter, Surface Area, Volume, etc.

What is a 2-D Shape?

The shape or figure with two dimensions like length and width is known as the 2-D shape. An example of a 2-D figure is a Square, Rectangle, Triangle, Parallelogram, Trapezium, Rhombus, etc. We can measure the 2-D shapes in the form of Area (A) and Perimeter (P).

What is 3-D Shape?

The shape with more or than two dimensions such as length, width, and height then it is known as 3-D figures. Examples of 3-Dimensional figures are Cube, Cuboid, Sphere, Cylinder, Cone, etc. The 3D figure is determined in the form of Total Surface Area (TSA), Lateral Surface Area (LSA), Curved Surface Area (CSA), and Volume (V).

Introduction to Mensuration

The important terminologies that are used in mensuration are Area, Perimeter, Volume, TSA, CSA, LSA.

• Area: The Area is an extent of two-dimensional figures that measure the space occupied by the closed figure. The units for Area is square units. The abbreviation for Area is A.
• Perimeter: The perimeter is used to measure the boundary of the closed planar figure. The units for Perimeter is cm or m. The abbreviation for Perimeter is P.
• Total Surface Area: The total surface area is the combination or sum of both lateral surface area and curved surface area. The units for the total surface area is square cm or m. The abbreviation for the total surface area is TSA.
• Lateral Surface Area: It is the measure of all sides of the object excluding top and base. The units for the lateral surface area is square cm or m. The abbreviation for the lateral surface area is LSA.
• Curved Surface Area: The area of a curved surface is called a Curved Surface Area. The units of the curved surface area are square cm or m. The abbreviation for the curved surface area is CSA.
• Volume: Volume is the measure of the three dimensional closed surfaces. The units for volume is cubic cm or m. The abbreviation for Volume is V.

Mensuration Formulas for 2-D Figures

Check out the formulas of 2-dimensional figures from here. By using these mensuration formulae students can easily solve the problems of 2D figures.

1. Rectangle:

• Area = length × width
• Perimeter = 2(l + w)

2. Square:

• Area = side × side
• Perimeter = 4 × side

3. Circle:

• Area = Πr²
• Circumference = 2Πr
• Diameter = 2r

4. Triangle:

• Area = 1/2 × base × height
• Perimeter = a + b + c

5. Isosceles Triangle:

• Area = 1/2 × base × height
• Perimeter = 2 × (a + b)

6. Scalene Triangle:

• Area = 1/2 × base × height
• Perimeter = a + b + c

7. Right Angled Triangle:

• Area = 1/2 × base × height
• Perimeter = b + h + hypotenuse
• Hypotenuse c = a²+b²

8. Parallelogram:

• Area = a × b
• Perimeter = 2(l + b)

9. Rhombus:

• Area = 1/2 × d1 × d2
• Perimeter = 4 × side

10. Trapezium:

• Area = 1/2 × h(a + b)
• Perimeter = a + b + c + d

11. Equilateral Triangle:

• Area = √3/4 × a²
• Perimeter = 3a

Mensuration Formulas of 3D Figures

The list of the mensuration formulae for 3-dimensional shapes is given below. Learn the relationship between the various parameters from here.

1. Cube:

• Lateral Surface Area = 4a²
• Total Surface Area = 6a²
• Volume = a³

2. Cuboid:

• Lateral Surface Area = 2h(l + b)
• Total Surface Area = 2(lb + bh + lh)
• Volume = length × breadth × height

3. Cylinder:

• Lateral Surface Area = 2Πrh
• Total Surface Area = 2Πrh + 2Πr²
• Volume = Πr²h

4. Cone:

• Lateral Surface Area = Πrl
• Total Surface Area = Πr(r + l)
• Volume = 1/3 Πr²h

5. Sphere:

• Lateral Surface Area = 4Πr²
• Total Surface Area = 4Πr²
• Volume = (4/3)Πr³

6. Hemisphere:

• Lateral Surface Area = 2Πr²
• Total Surface Area = 3Πr²
• Volume = (2/3)Πr³

Solved Problems on Mensuration

Here are some questions that help you to understand the concept of Mensuration. Use the Mensuration formulas to solve the problems.

1. Find the Length of the Rectangle whose Perimeter is 24 cm and Width is 3 cm?

Solution:

Given that,
Perimeter = 24 cm
Width = 3 cm
Perimeter of the rectangle = 2(l + w)
24 cm = 2(l + 3 cm)
2l + 6 = 24
2l + 6 = 24
2l = 24 – 6 = 18
2l = 18
l = 9 cm
Thus length of the rectangle = 9 cm

2. Calculate the volume of the Cuboid whole base area is 60 cm² and height is 5 cm.

Solution:

Given,
Base area = 60 cm²
Height = 5 cm
Volume of the Cuboid = base area × height
V = 60 cm² × 5 cm
V = 300 cm³
Thus the volume of the cuboid is 300 cm³.

3. Find the area of the Cube whose side is 10 centimeters.

Solution:

Given, side = 10 cm
Lateral Surface Area = 4a²
LSA = 4 × 10 × 10 = 400 cm²
Total Surface Area = 6a²
= 6 × 10 × 10 = 600 cm²
Volume of the cube = a³
V = 10 × 10 × 10 = 1000 cm³
Therefore the volume of the cube is 1000 cubic centimeters.

4. What is the lateral surface area of the sphere if the radius is 5 cm.

Solution:

Given,
The radius of the sphere = 5 cm
The formula for LSA of sphere = 4Πr²
Π = 3.14 or 22/7
LSA = 4 × 3.14 × 5 cm × 5 cm
LSA = 314 sq. cm
Thus the lateral surface area of the sphere is 314 sq. cm

5. What is the area of the parallelogram if the base is 15 cm and height is 10 cm.

Solution:

Given, Base = 15 cm
Height = 10 cm
We know that,
Area of parallelogram = bh
A = 15 cm × 10 cm
A = 150 sq. cm
Therefore the area of the parallelogram is 150 sq. cm.

FAQs on Mensuration

1. What is the use of Mensuration?

Mensuration is used to find the length, area, perimeter, and volume of the geometric figures.

2. What is the difference between 2D and 3D figures?

In 2D we can measure the area and perimeter. In 3D we can measure LSA, TSA, and Volume.

3. What is Mensuration in Math?

Mensuration is the branch of mathematics that studies the theory of measurement of 2D and 3D geometric figures or shapes.

In this article, we will learn the mathematical relation existing between Speed Distance and Time. Speed is the distance traveled by a moving object in unit time. Go through the entire article to learn about How is Speed Related to Time and Distance. Check out Solved Problems for finding Speed Time and Distance if few parameters are known. To help you understand the concept better we have provided Detailed Solutions.

If the distance is in km and the time taken to cover it is in hrs then the unit of Speed is given by km/hr. In the same way, if the distance is in m and the time taken to cover it is in sec then the unit of speed is given by m/sec. Speed can be either uniform or variable.

Example: If a Car travels 60 km in 1 hr then the Speed of it traveled is given by 60km/hr.

Types of Speed

There are two different types of Speeds and each one of their definition along with examples is explained in detail.

• Uniform Speed
• Variable Speed
• Average Speed

Uniform Speed: If an object travels the same distance in the same intervals of time then the object is said to be traveling with a Uniform Speed.

Example: If a car covers 60 km in 1st hour and 60 km in the 2nd hour, 60 km in the 3rd hour then the car is said to be moving with a uniform speed of 60kmph or 60 km/hr.

Variable Speed: If an object travels a different distance in the same intervals of time then the object is said to be moving with variable speed.

Example: If a car travels at 45 km for the 1st hour and 54 km for the 2nd hour and 65 km in the third hour then the car has a variable speed.

Average Speed: If a moving object travels different distances d₁, d₂, d₃, …. d_n with different speeds V₁, V₂, ……, V_n m/sec in time t₁, t₂, …., t_n seconds.

Average Speed of the Body = Total Distance Traveled/Total Time Taken

= d₁+d₂+d₃…..d_n/t₁+t₂+….t_n

Relation between Time Speed and Distance is given by the following

Speed = Distance/Time

Distance = Speed *Time

Time = Distance/Speed

Solved Problems on Relationship between Time Speed and Distance

1. A car travels a distance of 400 km in 5 hours. What is its speed in km/hr?

Solution:

We know the formula for Speed = Distance/Time

Speed = 400 km/5 hr

= 80 km/hr

The speed at which a car travels is 80 km/hr.

2. The distance between the two stations is 240 km. A train takes 2 hours to cover this distance. Calculate the speed of the train in km/hr and m/s?

Solution:

Distance between two stations = 240 km

Time = 2 hrs

Speed = Distance/Time

= 240 km/2 hr

= 120 km/hr

Speed of the train in m/sec is obtained by multiplying with 5/18

= 120*5/18

= 33.33 m/sec

3. Traveling at a speed of 45 kmph, how long is it going to take to travel 90 km?

Solution:

Speed = 45 kmph

Distance = 90 Km

we know the relation between speed distance and time is Speed = Distance/Time

45 kmph = 90 km/Time

Time = 90 km/45 kmph

= 2 hours

Time Speed and Distance is a major concept in Mathematics. Time and Distance are used extensively for questions relating to topics like circular motion, boats, and streams, motion in a straight line, clocks, races, etc. This article gives you a complete idea of the Relationship Between Time Speed and Distance, Units, Conversions, etc. Get Formula for Time and Distance, Solved Examples explaining the concept in detail.

Time Speed and Distance – Definition

Speed is a concept in motion that is all about how slow or fast an object travels. Speed is defined as the distance divided by Time. Speed, distance, and time are given to solve for one of the three variables when a piece of certain information is known.

Relationship between Time Speed and Distance

Speed = Distance/Time

Speed tells us how fast or slow an object travels and describes the distance traveled divided by the time taken to cover the distance.

From the above formula, Speed is directly proportional to Distance and inversely proportional to Time.

Units of Time Speed and Distance

Speed Distance and Time has different units for each of them and they are given as under

Time: seconds(s), minutes (min), hours (hr)

Distance: (meters (m), kilometers (km), miles, feet

Speed: m/s, km/hr

If Distance and Time Units are known Speed Units can be derived easily.

Time Speed and Distance Conversions

• To change between km/hour to m/sec, we multiply by 5/18. So, 1 km/hour = 5/18 m/sec
• To change between m/sec to km / hour, we multiply by 18/5. So, 1m /sec = 18/5 km/hour = 3.6 km/hour
• Similarly, 1 km/hr = 5/8 miles/hour
• 1 yard = 3 feet
• 1 mile= 1.609 kilometer
• 1 kilometer= 1000 meters = 0.6214 mile
• 1 mile = 1760 yards
• 1 mile = 5280 feet
• 1 hour= 60 minutes= 60*60 seconds= 3600 seconds
• 1 yard = 3 feet
• 1 mph = (1 x 5280) / (1 x 3600) = 22/15 ft/sec
• 1 mph = (1 x 1760) / (1 x 3600) = 22/45 yards/sec
• For a certain distance, if the ratio of speeds is a : b, then the ratio of times taken to cover the distance is given by b : a and vice versa.

Applications of Time Speed and Distance

Check out different models of questions asked on the concept of Time Speed and Distance. They are as under

Average Speed

Average Speed = Total Distance Travelled/Total Time Taken

Case 1: When Distance is Constant, Average Speed is given by = 2xy/x+y where x, y are two speeds at which the same distance is covered.

Case 2: When Time taken is Constant, Average Speed = (x+y)/2 where x, y are two speeds at which we traveled for the same time.

Examples

1. A person travels from one place to another at 40 km/hr and returns at 160 km/hr. If the total time taken is 5 hours, then find the Distance?

Solution:

Here the Distance is constant, so the Time taken will be inversely proportional to the Speed.

Ratio of Speed = 40:160

= 1:4

Ratio of Time = 4:1

Time taken = 5 hours

Therefore, time taken while going is 4 hours and while returning is 1 hour

Distance = Speed* Time

= 40*4

= 160 Km

Therefore, Distance Travelled is 160 Km.

2. Traveling at 3/2 nd of the original Speed a train is 20 minutes late. Find the usual Time taken by the train to complete the journey?

Solution:

Let the usual Speed be S1 and the usual Time is T₁. As the Distance to be covered in both cases is the same,

The ratio of usual Time to the Time taken when he is late will be the inverse of the usual Speed and the Speed when he is late

If the Speed is S₂ = S₁ then the Time taken T₂ = 3/2 T₁ Given T₂ – T₁ = 20 =>3/2 T1 – T1 = 20 => T1 = 40 minutes.

Inverse Proportionality of Speed & Time

Speed is inversely proportional to time when the distance is constant. If Speeds are in the ratio of m:n then the time taken will be n:m They are two methods of solving questions.

Using Inverse Proportionality
Using Constant Product Rule

Example

After traveling 60km, a train is meeting with an accident and travels at (2/3)rd of the usual Speed and reaches 30 min late. Had the accident happened 15km further on it would have reached 20 min late. Find the usual Speed?

Solution:

Using Inverse Proportionality

Here there are 2 cases

Case 1: accident happens at 60 km

Case 2: accident happens at 75 km

The difference between the two cases is only for the 15 km between 60 and 75. The time difference of 10 minutes is only due to these 15 km.

In case 1, 15 km between 60 and 75 is covered at (2/3)rd Speed.

In case 2, 15 km between 60 and 75 is covered at the usual Speed.

So the usual Time “t” taken to cover 15 km, can be found out as below. 3/2 t – t = 10 mins = > t = 20 mins, d = 15 km

so Usual Speed = 15km/20min = 15km/0.33hr= 45Km/hr

Using Constant Product Rule Method

Let the actual Time taken be T

There is a (1/3)rd decrease in Speed, this will result in a (1/2)nd increase in Time taken as Speed and Time are inversely proportional

The delay due to this decrease is 10 minutes

Thus 1/2 T= 10 and T=20 minutes

Also, Distance = 15 km

Thus Speed = 15/20 minutes = 15km/0.33hr = 45km/hr

Meeting Point

If two people travel from points A and B towards each other they meet at point P. Distance covered by them on the meeting is AB. Time taken by both to meet is the same. Since Time is constant Distance AP and BP will be in the ratio of their Speeds.

Consider the distance between A and B is d.

If two people walking towards each other from A and B. when they meet for the first time they cover a distance of “d ” and when they meet for the second time they cover a distance of “3d” and when they meet for the third time they cover a distance of “5d” …..

Example

1. Amar and Arun have to travel from Delhi to Jaipur in their respective cars. Arun is driving at 45 kmph while Amar is driving at 60 kmph. Find the Time taken by Amar to reach Jaipur if Arun takes 6 hrs?

Solution:

Since the Distance covered is constant in both cases, the Time taken will be inversely proportional to the Speed.

From the given data, the Speed of Amar and Arun are in ratio 45:60 or 3:4.

So the ratio of the Time taken by Arun to that taken by Amar will be in the ratio 4:3. So if Arun takes 6 hrs, Amar will take 4.5 hrs.

Solved Examples on Time and Distance

1. Seetha is driving a car with a speed of 60 km/hr for 1.5hr. How much distance does she travel?

Solution:

Speed = 60 Km/hr

Time = 1.5 hr

Distance = Speed *Time

= 60*1.5

= 90 Km

Therefore, Sheela Travels a distance of 90km.

2. While going to an office, Ram travels at a speed of 35 kmph, and on his way back, he travels at a speed of 40 kmph. What is his average speed for the whole journey?

Solution:

In this case, Distance is constant

Average Speed = 2xy/x+y where x, y is the speeds at which the distance is covered

Substitute the Speeds from the given data

Average Speed = (2*35*40)/35+40

= 37.33 km/hr

The Average Speed of the Whole Journey is 37.33Kmph

3. Ramu and Somesh are standing at two ends of a room with a width of 40 m. They start walking towards each other along the width of the room with a Speed of 4 m/s and 3 m/s respectively. Find the total distance traveled by Ramu when he meets Somesh for the third time?

Solution:

When Ramu meets Somesh for the third time he would have covered a distance of 5d i.e. 5*40m = 200m

The ratio of Speed of Ramu and Somesh is 4:3 so the Distance traveled by both of them will also be in the ratio of 4:3

Probability is a branch of mathematics that deals with the occurrence of random events. It is expressed from zero to one and predicts how likely events are to happen. In general, Probability is basically the extent to which something is likely to happen. You will learn about Probability Distribution where you will learn the possibility of outcomes for a random experiment.

Probability Definition

Probability is the measure of the likelihood of an event to occur.  In the case of events, we can’t predict with total certainty. We can only predict the cancer of an event to occur i.e. how likely it is to happen. Probability ranges between 0 to 1 in which 0 indicates the event to be an impossible one and 1 indicates a certain event.

Example: For instance, when we toss a coin there are only two possibilities either head or tail(H,T). If we toss two coins in the air there are three possible outcomes that are both the coins show heads, both the coins show tails, one is head and the other is tail i.e. (H, H), (T, T), (H, T).

Formula for Probability

Probability is defined as the possibility of an event to occur. The formula for Probability is given as the ratio of the number of favorable events to the total number of possible outcomes.

Probability of an event to happen = No. of Favourable Outcomes/ Total Number of Outcomes

This is the basic formula for Probability.

Probability Tree

Tree Diagram helps to organize and visualize different possible outcomes. Branches and ends are the two main positions of the tree. Each branch Probability is written on the branch and the ends contain the final outcome. Tree Diagram helps you to figure out when to multiply and add.

Types of Probability

There are three types of major probabilities. They are

• Theoretical Probability
• Experimental Probability
• Axiomatic Probability

Theoretical Probability: It depends on the possible chances of something to happen. Theoretical Probability mainly depends on the reasoning behind probability.

Experimental Probability: This kind of Probability depends on the observation of the experiment. Experimental Probability can be calculated on the number of possible outcomes to the total number of trials.

Axiomatic Probability: A Set of Rules or Axioms are Set applies to all types. Using the axiomatic approach to probability, the chances of occurrence or non-occurrence of the events can be quantified.

Conditional Probability is nothing but the likelihood of an event or outcome occurring based on the occurrence of a previous event or outcome.

Probability of an Event

Let us consider an Event E that occurs in r ways out of n possible ways. The probability of happening an event or its success is given by

P(E) = r/n

The probability of an event or its failure is given by

P(E’) = (n-r)/n = 1-(r/n)

E’ represents the event will not occur.

Therefore, we can say that

P(E)+P(E’) = 1

What are Equally Likely Events?

If the events have the same theoretical probability of happening then they are called Equally Likely Events. Results of Sample Space are said to be equally likely if all of them have the same probability of occurring. Below are some examples of Equally Likely Events.

• Getting 2 or 3 on throwing a die.
• Getting 1, 3, 4 on throwing a die

are all Equally Likely Events since the Probabilities of Each Event are Equal.

Complementary Events

In the Case of Such Events, there will only be two outcomes that state whether an event will occur or not. The complement of an event occurring is the exact opposite that the probability of an event is not occurring.

• It may or may not rain today
• Winning a lottery or not.
• You win the lottery or you don’t.

Probability Density Function

It is a probability function that represents the density of continuous random variables lying between a certain range of values. Standard Normal Distribution is used to create a database or statistics that are used in science to represent the real-valued variables whose values are unknown.

Additive Law of Probability

If E₁ and E₂ be any two events (not necessarily mutually exclusive events), then P(E₁ ∪ E₂) = P(E₁) + P(E₂) – P(E₁ ∩ E₂)

Probability Terms

Some of the Important Probability Terms are discussed here

Sample Space: Set of all possible outcomes that occur in any trail.

Example:

Tossing a Coin, Sample Space (S) = {H, T}

When you Roll a Die Sample Space (S) = {1, 2, 3, 4, 5, 6}

Sample Point: It is one of the Possible Outcome.

Example:

In a deck of cards, 3 of hearts is a sample point.

Experiment or Trial: Series of Actions where outcomes are always uncertain.

Example: Tossing a Coin, Choosing a Card from a Deck of Cards, Throwing a Dice.

Event: Single outcome of an experiment.

Example: Getting Tails while Tossing a Coin is an Event.

Outcome: Possible Result of an Experiment.

Head is a possible outcome when a coin is tossed.

Impossible Event: The Event can’t happen

While Tossing a Coin it is impossible to get head and tail at the same time.

Solved Examples on Probability

1. Find the Probability of getting 2 on rolling a die?

Solution:

Sample Space = {1, 2, 3, 4, 5, 6}

Number of Favourable Events = 1 i.e. {2}

Total Number of Outcomes = 6

Probability P = 1/6

Therefore, Probability of getting 2 on rolling a die is 1/6.

2. Two dice are rolled find the Probability that the Sum is

equal to 1

equal to 5

equal to 8

Solution:

In order to find the probability whose sum is equal to 5 we need to figure out the Sample Space

Practice the Questions based on the Uniform Rate of Growth and Depreciation from here. Learn about the Concept of Uniform Rate of Growth and Depreciation better by going through this entire article. You will find How to Apply Principal of Compound Interest on Combination of Uniform Rate of Growth and Depreciation. Check out Formula, Solved Examples on the concept of Uniform Rate of Increase or Decrease from this article. Get Step by Step Solutions for all the Problems provided and get a good hold on the concept.

How to find the Uniform Rate of Increase or Decrease?

Let us discuss how to find the Uniform Rate of Growth or Depreciation in detail in the below modules.

If a quantity P grows at the rate of r₁% in the first year and depreciates at r₂% in the second year and grows at r₃% in the third year then the quantity becomes Q after 3 years.

Take r/100 with a positive sign for each growth or appreciation of r % and r/100 with a negative sign for depreciation of r%.

Solved Examples on Uniform Rate of Growth or Depreciation

1. The current population of a town is 60,000. The population increases by 10 percent in the first year and decreases by 5% in the second year. Find the population after 2 years?

Solution:

Initial Population = 60,000

r₁ =10%, r₂ = 5%

Population after 2 Years Q = P(1+r₁/100)(1-r₂/100)

= 60,000(1+10/100)(1-5/100)

= 60,000(1+0.1)(1-0.05)

= 60,000(1.1)(0.95)

= 62,700

Therefore, the Population after 2 Years is 62,700.

2. The count of a certain breed of bacteria was found to increase at the rate of 4% per hour and then decrease by 2% per hour. Find the bacteria at the end of 2 hours if the count was initially 2,00,000.

Solution:

Since the Population of bacteria increases and decreases we use the formula

Q = P(1+r₁/100)(1-r₂/100)

= 2,00,000(1+4/100)(1-2/100)

= 2,00,000(1+0.04)(1-0.02)

= 2,00,000(1.04)(0.98)

= 2,03,840

Bacteria at the end of 2 hours is 2,03,840

3. The price of a car is $ 2,50,000. The value of the car depreciates by 10% at the end of the first year and after that, it depreciates by 15%. What will be the value of the car after 2 years?

Solution:

Initial Price of the Car = $ 2,50,000

r₁  = 10% r₂ = 15%

Since the Price of a Car depreciates we use the formula

Q = P(1-r₁/100)(1-r₂/100)

= 2,50,000(1-10/100)(1-15/100)

= 2,50,000(90/100)(85/100)

= 2,50,000(0.9)(0.85)

= $1,91,250

Value of Car after 2 Years is $1,91,250.

Get to know about the Uniform Rate of Depreciation along with its Formula and Solved Examples. In this article, we will discuss how to apply the Principal of Compound Interest on Problems of Uniform Rate of Depreciation. If the Rate of Decrease is Uniform then we call it a Uniform Decrease or Depreciation. Refer to the Solved Examples on How to find Rate of Depreciation or Decrease. We have listed the Step by Step Solutions for each and every problem making it easy for you to understand.

How to Calculate the Rate of Uniform Decrease or Depreciation?

Let us discuss in detail how to find the Uniform Rate of Decrease or Depreciation in the coming sections.

If the Present Value P of a Quantity Decreases at the rate of r % per unit of time then Value Q of the Quantity after n units of time is given by

Q = P(1-r/100)ⁿ

Depreciation in Value = P – Q

= P – P(1-r/100)ⁿ

= P{1-(1-r/100)ⁿ}

Efficiency of a Machine after regular use, Decrease in the Value of Furniture, Buildings, Decrease in the Number of Diseases, etc. come under Uniform Rate of Decrease or Depreciation.

Solved Examples on Uniform Rate of Depreciation

1. The value of a residential flat constructed at a cost of Rs.1,20,000 is depreciating at the rate of 5% per annum. What will be its value 4 years after construction?

Solution:

From the Given Data P = Rs. 1,20,000

r = 5%

n = 4 Years

We know the formula to find Q = P(1-r/100)ⁿ

Substitute the input values in the above formula and we have the equation as such

Q = 1,20,000(1-5/100)⁴

= 1,20,000(95/100)⁴

= 1,20,000(0.8145)

= Rs. 97,740

Therefore, the Value of a Residential Flat after 4 Years would be Rs. 97,740.

2. The price of a motor vehicle depreciates by 10% every year. By what percent will the price of the car reduce after 2 years?

Solution:

From the given data

Consider the Price be P

r = 10%

n = 2 Years

We know the formula to find Q = P(1-r/100)ⁿ

Substitute the input values in the above formula and we have the equation as such

Q = P(1-10/100)²

=P(90/100)²

= P(9/10)(9/10)

= 81P/100

Reduction in Price = P -81P/100

= 19P/100

Percent Reduction in Price = (19P/100)/P*100%

= 19%

3. The cost of a school bus depreciates by 8 % every year. If its present worth is $ 27,000. What will be its value after three years?

Solution:

From the given data

P = $27,000

r = 8%

n = 3 years

Formula to Calculate the Price of Depreciated Value Q = P(1-r/100)ⁿ

= 27,000(1-8/100)³

= 27,000(92/100)³

= 27,000(0.92)³

= 27,000(0.778)

= $ 21, 024

The Cost of a School Bus after the Depreciation is $ 21,024.

In order to simplify Rational Expressions involving the Sum or Difference go through the complete article. Have a glance at the step by step procedure for solving Rational Expressions involving the Sum or Difference. The Solved Examples on Rational Numbers will help you get a good grip on the concept. By the end of this article, you can solve problems of Rational Expressions including the Sum or Difference on your own.

How to Simplify Rational Expressions Involving the Sum or Difference?

Follow the below-mentioned guidelines to solve Rational Expressions involving the sum or difference. They are along the lines

Step 1: Firstly, find the LCM of Denominators of all the Numbers Involved.

Step 2: Write the Rational Number whose denominator is the LCM obtained in the earlier step. For obtaining the numerator simply divide the LCM obtained with all the denominators of the rational numbers. Multiply the numerators of the respective rational numbers with the quotients you got. Keep the Addition, Subtraction Signs as it as in the Expressions. Simplify the Expression to get an Integer as Numerator.

Step 3: Reduce the Rational Number to its lowest or simplest form if it is not present. The Rational Number Obtained is the required Rational Number.

Solved Examples Simplifying Rational Expressions involving the Sum or Difference

1. Simplify -3/2 + 9/6 – (-5)/4?

Solution:

Given Expression is -3/2 + 9/6 – (-5)/4

= -3/2+9/6+5/4

Find the LCM of Denominators i.e. 2, 6, 4

LCM(2, 6, 4) = 12

Express the Rational Numbers using the LCM obtained in terms of a common denominator.

-3/2 = -3*6/2*6 = -18/12

9/6 = 9*2/6*2 = 18/12

5/4 = 5*3/4*3 = 15/12

Placing the Rational Numbers in the Expression we get

= -18/12+18/12+15/12

= 15/12

Therefore, -3/2 + 9/6 – (-5)/4 on simplifying gives a Rational Number 15/12.

2. Simplify 7/10 – (-5)/14 + 9/-3?

Solution:

Given Rational Expression is 7/10 – (-5)/14 + 9/-3

As One of the Denominators is negative we rearrange it to get a positive denominator.

9/-3 = 9*(-1)/-3*(-1) = -9/3

The Rational Expression becomes 7/10+5/14-9/3

Find the LCM of Denominators 10, 14, 3

LCM(10, 14,3) = 210

Express the Rational Numbers in terms of a Common Denominator using the LCM obtained.

7/10 = 7*21/10*21 = 147/210

5/14 = 5*15/14*15 = 75/210

-9/3 = -9*70/3*70 = -630/210

Placing the Rational Numbers in the Expression we get

= 147/210+75/210-630/210

= (147+75-630)/210

= -408/210

= -68/35

Therefore, on simplifying 7/10 – (-5)/14 + 9/-3 we get -68/35.

Confused between Direct and Indirect Variation? Get clarity now!! We are providing detailed information and also the introduction to indirect proportion here. Direct Variation defines a linear relationship between 2 variables, inverse proportion defines another kind of relationship. Inversely proportion relationship will be described in longer form. Know every detail of Indirect Proportion or Variation here. Check the below sections to know all the information.

Inverse Variation

Mathematics is one of those subjects where you require problem solving and time management skills. With the given tips, you can easily make these possible to solve all the questions in given time. Ratios are the mathematical relationships which we use in the real world. These are explained based on fractions. If the fraction is represented as x:y, they are said to be in proportion which also states that 2 ratios are equal.

If we go with the example explanation, Suppose that we are constructing a bridge or a house, then the no. of days it takes to complete the building depends on the number of workers. Suppose we want to complete a house in less no. of days, then more number of workers are required. So, the no. of days is inversely proportional to no. of workers.

No. of days = 1/No. of Workers

Some of the examples in our day to day life also vary inversely. Some of these are in a stringed instrument, the frequency of vibration does vary inversely with the string length. The gravitational force between any two bodies would be inversely proprtional to power 2 of the distance.

If you are preparing for some of the competitive exams or bank exams, then you must definitely be perfect in all the topics and ratios and proportions. Most of the students will be confused between direct variation and indirect variation. So, you can check the example problems here to clear your confusion.

Definition of Inverse Variation

The mathematical expression or relationship between two variables that expresses by an equation in which the product of two quantities is equal to a constant value.

Sometimes, we notice that the variation in 1 value of one quantity differs or just opposite to the variation in another or second value. i.e. If the value of one quantity increases, the value of the another quantity decreases in the equal proportion and vice versa. In this case, two quantities are said to be inversely proportional.

Books for Inverse Variation

1. On the Law of Inverse Variation of Extension and Intension by Richard Milton Martin
2. Exam Prep for Thinking with Mathematical Models; Linear & Inverse Proportions by David Mason
3. Regularization of Inverse Problems by Heinz Werner Engl, ‎Martin Hanke, ‎A. Neubauer
4. Experimental Study Regarding Variation of Force in Inverse Proportions by Elsevier Limited
5. Thinking with Mathematical Models: Linear and Inverse Variation by Glenda Lappan
6. Fundamentals of Math Book 2: Algebra – Book 2 by Jerry Ortner
7. Inverse Problems, Image Analysis, and Medical Imaging by Jerry Zuhair Spinelli
8. New Mathematics Today by ANUBHUTI GANGAL

How to solve Inverse Variation Problems?

Step 1: Read the question once or twice carefully. The equation to solve Inverse Proportion is y=k/x. While solving word problems, use variables other than x, y. Use the variables which are relevant to the question or problem that is to be solved. Also, check the question carefully to know the inverse equations like squares, cubes, and square roots.

Step 2: Use the information given in the question to find the value of k. The constant value k is called constant of proportionality or constant of variation.

Step 3: Rewrite the formed equation from step 1 by substituting the values of k in the equation of step 2.

Step 4: Use the equation from step 3 and complete solving the remaining problem with the instructions given in the question.

Step 5: Do not forget to include the units at the end of the solution and also re-check the solution to avoid calculation mistakes.

Solved Example Questions

Question 1:

3 pipes take 60 minutes to water the field. How much time will it take to water the field with 6 pipes?

Solution:

Read the problem carefully

First, find out whether it is a directly proportional question or an inversely proportional question.

As more pipes are there, time reduces.

Hence pipes and minutes are inversely proprtional.

Here, we can write it as 3 pipes * 60 minutes = 6 pipes * n minutes

From the above equation, n = 30 minutes.

This is the mathematical representation of the given problem.

You can solve it even in a different way

Given for 3 pipes and asked to find for 6 pipes. It means the number of pipes is doubled. So, time will be reduced to half, that is 30 minutes.

Question 2:

4 friends consume 16kgs of rice in a month. The same amount of rice lasted for 20 days when a few more friends joined. How many additional members joined the group?

Solution:

It is given in the question that, 4 friends need 30 days to consume rice.

If rice is consumed within 20 days, how many additional friends are joined is the thing we need to find, let’s assume it to be x.

This a problem of inverse variation.

4 friends * 30 days = x friends * 20 days

From the above equation, x = 6.

So from the above solution, 6 friends need 20 days to consume 16kg of rice.

The number of additional members required to consume 16kgs of rice in 20 days is 2.

Question 3:

y varies inversely with the square of x. When x=2, y=10. Find x when y=20?

Solution:

In order to solve the above equation, we can translate the above sentence into a simple mathematical equation

y=k/(x^2). This implies y varies inversely with square of x by some constant.

Given x=2, y=3

Inverse Variation Equation is y=k/x²

10=k/2²

10=k/4

y=40/x² 

y=40x^-2

20=40/x²

20x²/20=40/20

x² = ₂

x=+-root(2)

We have provided all the details regarding inverse proportion here. Prepare all the topics and concepts from ratios and proportions. Stay tuned to our website for all the stunning updates. We wish you all the best for your future.

Students can find several questions on Compound Interest. Practice the Objective Questions of Compound Interest Over here and be prepared for the exams. Learn how to solve Compound Interest Problems by checking the Solved Examples. Use the Sample Problems over here covering various questions including the Compound Interest Formula.

Compound Interest Practice Test has questions when the Interest Rate is Compounded Annually, Half-Yearly, Quarterly, Various Rate of Interest, Amount Calculations, etc. Solve the Questions on CI and test your knowledge on the related areas and bridge the gap accordingly.

1. The compound interest on $ 20,000 at 5 % per annum for 3 years, compounded annually is?

Solution:

P = $20,000

R = 5%

n = 3 Years

A = P(1+R/100)ⁿ

= 20,000(1+5/100)³

=20,000(105/100)³

= 20,000(1.157)

= $23152

CI = A – P

= $23152 – $20,000

= $3152

2. The simple interest on a sum of money for 2 years at 3 % per annum is $ 6250. What will be the compound interest on the same sum at the same rate for the same period, compounded annually?

Solution:

From given data SI = $6250

T = 2 Years

R = 3%

SI = PTR/100

6250 = P*2*3/100

6250 = 6P/100

6250*100 = 6P

6P = 625000

P = $1,04,166

We know A = P(1+R/100)ⁿ

=1,04,166(1+3/100)²

=1,04,166(103/100)²

=1,10,509

Compound Interest = Amount – Principal

= $1,10,509 – $1,04,166

= $6343

3. If a sum of Rs. 10,000 lent for 10% per annum at compound interest then the sum of the amount will be Rs. 14,161 in

Solution:

We know A = P(1+R/100)ⁿ

Given P = Rs. 10,000

R = 10%

A = 14,161

n = ?

Substitute the input values in the formula of Amount we have

14161 = 10000(1+10/100)ⁿ

14161/10000 = (1+10/100)ⁿ

(11/10)⁴ = (11/10)ⁿ

n = 4 Years

4. The population of a city is 1,20,000. It increases by 5% in the first year and increases by 10% in the second year. What is the population of the town at the end of 2 yrs?

Solution:

The population of city = 1,20,000

The population of city after 2 years = P(1+R₁/100)(1+R₂/100)

= 1,20,000(1+5/100)(1+10/100)

= 1,20,000(1.05)(1.1)

= 1,38,600

Therefore, the Population of the city by the end of 2 years is 1,38,600.

5. The difference between simple interest and compound on Rs. 1500 for one year at 20% per annum reckoned half-yearly is

Solution:

Given Data is Principal = 1500

R = 20%

T = 1 year

SI = PTR/100

= (1500*1*20)/100

= Rs. 300

Amount A = P(1+R/100)ⁿ

=1500(1+(10/100))²

= 1500(1+1/10)²

=1500(1.1)²

= Rs. 1815

CI = A – P

= Rs. 1815 – Rs. 1500

= Rs. 315

Difference = CI – SI

= Rs. (315 – 300)

= Rs. 15

Direct Variation is an essential concept in Ratios and Proportions. We are providing the important formulas, explanations, and definitions here. Direct Proportion is one type of Ratios and Proportions. Go through the below sections to know the various details like Formulas Types, Definitions, Solved Questions, etc.

Importance of Direct Variation

Direct Proportion or Variation is the relationship between two different variables in which one variable is the constant of another variable. If the variable is directly proportional to another variable, then we define that one of the variables changes with the same ratio as the other increases. Also, if one variable decreases, then the ratio of the other variable decreases.

For example: If you save a huge amount of money every month, then you will increase your savings by a definite amount. This is called the constant of variation. Because there was a constant rate of increase. The constant rate of increase or decrease is called the “constant of variation”. Also, you will know more details regarding the direct variation in the upcoming sections. We also provide some tips, tricks, shortcuts, books, and solved questions.

Direct Variation Definition

Two quantities or equations are said to be variant if there is a consistent increase or decrease in quantity causes an increase or decrease in other quantity. In simple terms, Direct Variation is a relation between two numbers such that one number should be a constant multiple for another number. Mathematics generally deal with constant quantities or variable quantities.

If a value changes with different situations, it is called a variable and if a value does not change with different situations, it is called a constant. Consider an example. 22/7, 4, etc., are examples of constants. The population of a city/ town, speed of a car, etc., are examples of variables.  When a value of relative variable changes, there will be a change in the value of the variable, this is called variation.

The Direct Variation is like a simple relation between two variables. Consider an equation that says y varies directly with x if y=kx.

k is a constant called constant of proportionality or constant of variation.

It means that x is directly proportional to y, it implies if x increases, y increases, and if x decreases, y decreases. The ratio also will be the same.

So considering the above statements, the graph of the above direct variation equation is a straight line.

Books for Direct Variation

1. New Mathsahead: Book 7 (Rev. Edn.)
2. New Learning Composite Mathematics 8 by S.K. Gupta & Anubhuti Gangal
3. Maths Wiz Book by S.K. Gupta & Anubhuti Gangal
4. Direct Methods in the Calculus of Variations by Enrico Giusti
5. CALCULUS OF VARIATIONS WITH APPLICATIONS by A. S. GUPTA
6. Algebra: A Step-by-Step Guide by Jennifer Dagley
7. The Calculus of Variations by N.I. Akhiezer
8. CliffsNotes Algebra I Quick Review, 2nd Edition by Jerry Bobrow

How to find the Direct Variation?

Here are a few steps you need to follow in order to solve a direct variation problem

Step 1: Note down the formula for direct variation.

Step 2: In order to get variables, substitute the given values.

Step 3: Now, solve to get the constant of variation.

Step 4: Write the equation which satisfies x and y.

Solved Questions on Direct Variation

Question 1: A wooden box is made which is directly proportional to the no of wooden blocks. 120 wooden blocks are needed to make 30 boxes. How many wooden blocks are needed to prepare a box?

Solution:

In the above-given problem,

No of wooden blocks needed for 30 boxes = a= 120

Number of boxes = b = 30

No of wooden blocks needed for a box = y

The direct variation formula is

a=y*b

120=y*30

y=120/30

y=4

No of wooden blocks needed for a box = 4

Question 2: Given that a varies directly as b, with x constant of variation y=1/3, find a when b=12

Solution:

According to the given equation,

a=1/3b

Substitute the given b value,

a=1/3.12

a=4

Question 3: Suppose a varies directly as b and a=30 when b=6. What is the value of a when b=100?

Solution:

From the direct variation equation

b=ka

Substitute the given a and b values in the equation, and solve them for “k”

30=k*6

k=5

The equation is a=5b. Now substitute b=100 and find a

a=5.100

a=500

Question 4: Suppose that a car runs at a speed constantly and takes 3 hours to cover a distance of 180 km. How much time does the car take to cover a distance of 100km?

Solution:

Let T be the time taken to run the total distance.

Let S be the distance.

Suppose V is the speed of the car.

As per the Direct Variation equation S=kT where k is the constant

From the given question

S=180, T=3

Therefore, 180 = k*3 = 180/3 = 60

So, the constant speed of the car = 60km/hr

For 100km distance

S=kT

100=60*T

T=100/60=5/3hours=1 hour 40 mins

Therefore, the car takes 1 hour 40 mins to cover a distance of 100km.

Question 5: If X varies directly as Y and the value of X is 60 and Y is 40, find the equation that determines the direct variation of X and Y?

Solution:

As X varies directly with Y, the ratio of X  and Y is constant for any value of X and Y.

So, constant V=X/Y=60/40=3/2

Therefore, the equation that determines the direct proportion of X and Y is X=3/2Y.

Preparation Tips

• Attend as many as challenging questions you can.
• Solve most of the previously asked questions to become perfect.
• Know your strengths and weaknesses.
• Improve your problem-solving techniques.
• Know the mistakes you make beforehand and rectify them.
• Have good sleep and rest before the exam.
• Build self-confidence and ease of solving the problems.

The given Direct Variation will help you for your better preparation. Stay in touch with our website Learcbse.in to check the preparation tips, syllabus, various mathematical concepts, etc. Candidates have to note that we are providing various information for your reference. Click on the bookmark button to get the latest updates.