## Percentage Increase Calculator | Calculating Percentage Change

Percentage Increase is the amount of increase from the original number to the final number in terms of 100 parts of the original. Have a glance at the Percentage Increase Formula, How to Solve Percentage Increase Related Problems, etc. Check out the Solved Examples on Percentage Increase and learn how to approach them. Apply the Concept for Solving Some Real-Life Problems and get the Percentage Increase easily. Get to know the concept much better by going through the further modules.

## Percentage Increase Formula

The Formula to Calculate Percentage Increase is given by

Percentage Increase = ((New Value – Original Value)/Original Value)*100

To help you be clear with the concept of % increase we will provide you with some examples in the coming modules.

## How to Calculate the Percentage Increase?

Go through the following guidelines on how to find the Percentage Increase. They are as such

• Firstly, find the Percentage Increase i.e. (New Value – Original Value).
• Divide the Percentage Increase with the Original Value.
• Multiply the resultant fraction over 100 and place the percentage symbol after that.

### Example Questions on Increase Percentage

1. The Price of Paddy Increases from $10 to$13 Per Kg? Calculate the Percentage Increase in Price?

Solution:

Original Price of Paddy = $10 New Price of Paddy =$13

Increase in Price of Paddy = New Price – Original Price

= $13 –$10

= $3 Percentage Increase = (Increase in Price/Original Price)*100 = (3/10)*100 = 30% Therefore, Percentage Increase in the Price of Paddy is 30%. 2. Find the Increase in Value if 200 increases by 20%? Solution: Increase = 20%(200) = 20/100*200 = 40 Increase in Value = 200+40 = 240 Therefore, Increase in Value is 240. RSD Calculator is a small, quick & reliable and it works on all Android devices with Android. 3. Population of a Town increases from 15000 to 22000 in one year. Calculate the Percentage Increase in the Population? Solution: Population of the town initially = 15000 Population of town after one year = 22000 Increase in Population = 22000 – 15000 = 7000 Percentage Increase in Population = (Percentage Increase/Original Population)*100 = (7000/15000)*100 = 46.66% Therefore, the Increase in Population Percentage from 15000 to 22000 is 46.66% 4. Consider a$1,200 investment increased in value to $1,400 dollars in a year. What is the percent increase of the investment? Solution: Original Investment =$1200

Investment after one year = $1400 Increase in Investment =$1400 – $1200 =$200

Percentage Increase = (Increase in Investment/ Original Investment)*100

= (200/1200)*100

= 16.66%

5. If the Price of Petrol increases from Rs. 75/- to Rs. 80/- per litre. Find the Percentage Increase in Petrol?

Solution:

Original Price of Petrol = Rs. 75/-

New Price of Petrol = Rs. 80/-

Increase in Petrol Price = Rs. 80- Rs. 75

= Rs. 5

Percentage Increase = (Increase in Petrol Price/Original Price)*100

= (5/75)*100

= 100/15

= 6.66%

Therefore, the Percentage Increase in Petrol is 6.66%

## Area and Perimeter Definition, Formulas | How to find Area and Perimeter?

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

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,
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.

## Mensuration – Definition, Introduction, Formulas, Solved Problems

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.

## Relationship between Speed, Distance and Time | How is Speed Related to Time and Distance?

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, …. dn with different speeds V1, V2, ……, Vn m/sec in time t1, t2, …., tn seconds.

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

= d1+d2+d3…..dn/t1+t2+….tn

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

## Speed of Train | How to Calculate the Motion or Speed of a Train?

Learn What is meant by Speed and How to Calculate the Motion or Speed by going through the entire article. Check out Formula to find the Speed of a Train when Train Passes through a Stationary Object and Train Passing through a Stationary Object of Certain Length. We have mentioned Solved Examples explaining step by step for a better understanding of the concept. Learn about Time and Distance Concepts too from here and get a good hold of the concept by practicing the sample problems available on a regular basis.

## How to find the Speed of a Train?

There are two different scenarios to calculate the Speed of a Train. We explained each scenario in detail by taking a few examples. Understand the concept behind them easily. They are as under

• Train Passes through a Stationary Object
• Train Passes through a Stationary Object having Some Length

When a Train Passes through a Stationary Object

Consider the length of the train along with the engine be x. When the end of the train passes the object then the engine of the train travels the distance equal to the train length.

Time taken by train to pass the stationary object = length of the train/speed of the train

When a Train Passes through a Stationary Object having Some Length

When the end of the train passes the stationary object having a certain length, then the engine of the train moves a distance equal to the sum of both the length of the train and the stationary object.

Time taken by the train to pass stationary object = (length of the train+length of the stationary object)/speed of the train

### Solved Examples on Speed of a Train

1. Find the time taken by a train 125 m long, running at a speed of 108 km/hr in crossing the pole?

Solution:

Length of the Train = 125m

Speed of Train = 108 km/hr

= 108*5/18

= 30 m/sec

Time taken by train to cross the pole = 125m/30m/sec

= 4.1 sec

2. A train 320 m long is running at a speed of 60 km/hr. what time will it take to cross a 160 m long tunnel?

Solution:

Length of the train = 320m

Length of the tunnel = 160m

Length of Train+Length of Tunnel = (320+160) = 480m

Speed of train = 60 km/hr

= 60*5/18

= 16.66 m/sec

Time taken by train to cross the tunnel = 480m/16.66m/sec = 28.8 sec

3. A train is running at a speed of 75 km/hr. if it crosses a pole in just 15 second, what is the length of the train?

Solution:

Speed of the train = 75 km/hr

Speed of the train = 75 × 5/18 m/sec = 20.83 m/sec

Time taken by the train to cross the pole = 15 seconds

Therefore, length of the train = 20.83 m/sec × 15 sec = 312.5 m

4. A train 250 m long crosses the bridge 120 m in 20 seconds. Find the speed of the train in km/hr?

Solution:

Length of the train = 250m

Length of the Bridge = 120 m

Total Length = Length of Train + Lenth of Bridge

= 250+120

= 370 m

Time taken to cross the bridge = 20 seconds

Speed of the Train = Length/Time = 370m/20 sec = 18.5 m/sec

To convert m/sec to km/hr multiply with 18/5

Speed of the Train in km/hr = 18.5*18/5

= 66.6 km/hr

Therefore, the speed of the train is 66.6 km/hr