# Basic concepts of percentage

Basic concepts of percentage are essential for everyone. Application of concepts of percentage can be seen everywhere in arithmetic section of mathematics. With the intention of gaining knowledge of percentage, let us explore the topic.

Percentage can be defined as “a part out of each hundred”or in other words “for every hundred”.
For example, 20 out of 100 can be written as, \frac { 20 }{ 100 } or 20%.

Similarly, 1%=\frac { 1 }{ 100 }

50%=\frac { 50 }{ 100 }

99%=\frac { 99 }{ 100 }

We can express percentage in many ways:

• Express percentage as fraction:
“x%” can be expressed as fraction which= \frac { x }{ 100 }

to explain,

40%= \frac { 40 }{ 100 } = \frac { 2 }{ 5 }

• Express percentage as decimal:
We know, 2%= \frac { 2 }{ 100 }

to get it decimal form we have to divide 2 by 100,
i.e

⇒ 2%= \frac { 2 }{ 100 } =0.02

• Convert fraction into percentage:
Let \frac { a }{ b } be a fraction, to convert it into percentage we multiply it by 100 .
=( \frac { a }{ b }×100 )%
example,
percentage value of \frac { 1 }{ 4 } will be

⇒(\frac { 1 }{ 4 } ×100)%=25%

• Convert decimal into percentage:
In order to convert a decimal number into percentage, we multiply the decimal number with 100.
→1.5 can be in percentage as,

= 1.5× 100 = \frac { 15 }{ 10 } × 100  = 150%

Before moving further, let us solve some questions.
Question 1. Convert the following into fraction:
i) 75%      ii) 2%
Solution:
i) 75%=\frac { 75 }{ 100 } =\frac { 3 }{ 4 }

ii) 2%=\frac { 2 }{ 100 } =\frac { 1 }{ 50 }

Question 2: Convert the following into decimals:
i) 25%    ii) 20%
Solution:
i) 25%=\frac { 25 }{ 100 } =0.25

ii) 20% =\frac { 20 }{ 100 } =0.2

Question 3. What is 20% of  300 ?
Solution
: This can be written as,

= \frac { 20 }{ 100 } × 300

= 60

Question 4. Solve 2% of 50 + 5% of 10 .
Solution : This is,
=\frac { 2 }{ 100 } ×50 + \frac { 5 }{ 100 } × 10

=1+\frac { 1 }{ 2 }

= \frac { 3 }{ 2 } =1.5

Question 5. 3 is what percent of 50 ?
Solution: Required percentage = \frac { 3 }{ 50 } × 100

=6%

### Percentage and their fractional values

Following are some percentages and their fractional values that are often used in calculations.In fact, these values will be useful for solving questions swiftly.

We can find other fractions from these values. To enumerate,

⇒If  33\frac { 1 }{ 3 } %= \frac { 1 }{ 3 }
then, by multiplying both sides by 2 we will get,

{(33\frac { 1 }{ 3 }) × 2 }%=\frac { 1 }{ 3 } ×2

Thus 66\frac { 2 }{ 3 } % = \frac { 2 }{ 3 } .

See its quite simple. Similarly let us look another example.

⇒if 20% =\frac { 1 }{ 5 }

then 60% will be 3 times 20%

ie. 20% ×3=\frac { 1 }{ 5 } × 3

60%=\frac { 3 }{ 5 }

### Some important tricks for percentage:

1. If a is increased to a’, then percentage increase

#### percentage increase%= \frac { a-a' }{ a } ×100

2. When a is decreased to { a }_{ 0 } then,

#### percentage decrease%= \frac { a-{ a }_{ 0 } }{ a } ×  100

3. If amount is first increase by x% and then decreased by x%.The percentage change will be

#### Decrease of \frac { { a }^{ 2 } }{ 100 } %

4. If amount is first increase by x% and then decreased by y%.The percentage change will be ,

#### = (x-y-\frac { xy }{ 100 } )%

[ – sign for decrease and +sign for increase]

5. If amount is first decreased by x% and then increased by y%.The percentage change will be ,

#### = (-x+y-\frac { xy }{ 100 } )%

6. If amount is first increased by x% and then decreased by y%.Then  net percentage decrease will be ,
##### =(-x-y+\frac { xy }{ 100 } )%
7. If amount is first increased by x% and then increased by y%.Then  net percentage increase will be ,

### Comparision between two quantities:

1. If A is x% more than B, Then B is less than A by

#### (\frac { x }{ 100+x } ×100)%

2. If A is x% less than B, Then B is more than A by

### Price of commodity (for no change in expenditure)

• If the price of commodity is increase by x% then reduction in comsumption as not to inrease the expenditure is

#### (\frac { x }{ 100+x } ×100)%

• If the price of commodity is decrease by x% then increase in comsumption as not to decrease the expenditure is

### Population based questions

If the population of a town is P & it increases at a rate of R% per annum, Then

### DEPRECIATION

Let the present value of an object be P, if it depreciates at the rate of R% per annum , then

• The value of object after n years

#### =P({ 1-\frac { R }{ 100 } })^{ n }

• The value of object n years ago

### =\frac { P }{ ({ 1-\frac { R }{ 100 } ) }^{ n } }

These trick will be a boon in your performance and will increase your speed. Not only it will save your time but also will be beneficial in other sections of mathematics.

Do not try to memorize these formulas.The only way to learn them, is by using them in problems. Solve as many questions as you can.You can solve questions from previous years of your exam.

Concepts of percentage will be beneficial in sections like profit-loss,discount and Simple interest-compound interest. Make sure you practice a lot, because maths is nothing without practice.Once the basic concepts of percentage are clear there is no looking back.

At the same time if you want to learn other basic topics of mathematics,you can have a look.
BASIC TOPICS OF MATHEMATICS