## Percentage

A percentage is a fraction whose denominator is 100 and the numerator of the fraction is called the rate percent.
Percent is denoted by the sign '%'.
The term percent means for every hundred.

 If we have to find x % of a number, then x % of number = x times of that number 100
 For example: 20% of 700 = 20 × 700 = 20 × 7 = 140 100

### Some fast result :-

 10% of a number = Number 10
 11.11% of a number = Number 9
 12.50% of a number = Number 8
 14.28% of a number = Number 7
 6.66% of a number = Number 6
 20% of a number = Number 5
 25% of a number = Number 4
 33.33% of a number = Number 3
 50% of a number = Number 2
Example: Find 16.66% of 252.
 Solution: 16.66% of 252 = 1 × 252 = 42 6
Example: Find 14.28% of 504.
 Solution: 14.28% of 504 = 1 × 504 = 72 7
Example: Find 33.33% of 369.
 Solution: 33.33% of 369 = 1 × 369 = 123 3

### Conversion of Percent into Number :-

To convert any percent into number we should divided by 100.

Example : Express 20% in fraction.

 Solution: 20% = 20 = 1 100 5
Example : Express 75 % in fraction.
 Solution: 75% = 75 = 3 100 4
Example : - Express 64% in decimal.
 Solution: 64% = 64 = 0.64 100

### Conversion of Fraction into Percentage :-

To convert fraction into per cent we should multiply by 100 .

 Example : Convert 3 into percent. 5
 Solution: 3 × 100 = 60% 5
Example : Express 31/4 in percent.
 Solution: 31/4 × 100 = 13 × 100 = 13 × 25 = 325 4

#### Expressing one quantity as a percent with respect to other . The quantity to be expressed in percent × 100 % other quantity

Note : For this formula, both the quantity must be in same metric unit.
Example: 50 kg is what per cent of 250 kg?

 Solution: The quantity to be expressed in percent × 100 % other quantity
 = 50 × 100 % = 20% 250

Example: 20 g is what per cent of 2 kg?
Solution: First quantity = 20 g and second quantity = 2 kg = 2000 g

 ∴ The quantity to be expressed in percent × 100 % other quantity
 = 20 × 100 % = 1% 2000

### Increasing percent and Decreasing percent :-

 Percentage increase = Increase × 100 original value
Increase = New value - Original value

Example: Rent of the house is increased from ₹ 8000 to ₹ 10000. Express the increase in price as a percentage of the original rent.
Solution: Given, Initial rent = ₹ 8000 , New rent = ₹ 10000
Increase = New rent - Original rent
= 10000 - 8000 = ₹ 2000

 Percentage increase = Increase × 100 original value

 Percentage increase = 2000 × 100 = 25% 8000

 Percentage decrease = decrease × 100 original value
Decrease = Original value - New value

Example: The cost of a bike last year was ₹ 50000. Its cost this year is ₹ 45000. Find the percent decrease in its cost.
Solution: Given, last year cost = ₹ 50000
This year cost = ₹ 45000
Decrease = Original cost - New cost
Decrease = 50000 - 45000 = 5000

 Percentage decrease = decrease × 100 original value

 Percentage decrease = 5000 × 100 = 10% 50000

## Use of Percent :-

Trick: Successive change :-

 Net Percentage Change = A + B + A × B % 100

where, A = first change and B = second change

Example: If the salary of a person increased by 10% of income. And again increased by 10% in next year. Find net percent change in his total salary.
Solution: Let A = 10% and B = 10%

 Net Percentage Change = A + B + A × B % 100

 Net percent change in salary = 10 + 10 + ( 10 × 10 ) = 20 + 1 = 21% 100

Example: The price of an article is first increased by 20% and later on the price were decreased by 25% due to reduction in sales. Find the net percentage change in final price of article.
Solution:
Let A = 20% and B = -25%

 Net Percentage Change = A + B + A × B % 100

 Net percent change in final price = 20 - 25 + { 20 × ( - 25 ) } = -5 - 5 = -10% 100
10 % decrease in the price of an article.

Trick: To solve "Expenditure Constant" questions

 Reduction in consumption = x × 100 100 + x

 Increase in consumption = x × 100 100 - x
where, x = increase or decrease
Use '-'sign for decrease

Example: Price of sugar is increased by 10%. Find by how much % the consumer should the consumption of sugar to keep the expenditure constant.
Solution: Here, x = 10%
Using the above given formula , we have

 Reduction in consumption = 10 × 100 = 10 × 100 100 + 10 110

 Reduction in consumption = 100 = 9.09% 11

Example: Price of rice is decreased by 20 %. Find by how much per cent the consumption should be increased to keep the expenditure constant.
Solution: Here x = 20%
Using the above given formula , we have

 Increase in consumption = 20 × 100 = 20 × 100 = 25% 100 - 20 80

Trick: If the population of a town is P and it increases ( or decreases ) at the rate of r% per year, then

 Population after n year = P 1 + r n 100

 Population n year ago = P 1 - r n 100

Use '-' sign for decrement
Example: The population of a town is 472300. If it increases at the rate of 10 % per year, them what will be its population after 2 yr.
Solution:- given, P = 473200 , r = 10% , n = 2 yr

 Population after 2 year = P 1 + r n 100

 Population after 2 year = 472300 1 + 10 2 100

 Population after 2 year = 472300 1 + 1 2 10

 Population after 2 year = 472300 11 2 10

= 4723 × 11 × 11 = 571483

Trick: If the present population of a city is P and there is a increment or decrement of r1%, r2% and r3% in first, second and third year respectively, then

 Population of city after 3 yr = P 1 + r1  1 + r2  1 + r3 100 100 100

Use'-' sign for decrement
Example: Population of a city in 2010 was 200000. If in 2011 there is an increment of 10%, in 2012 there is a decrement of 20% and in 2013 there is increment of 30%, then find the population of city at the end of year 2013.
Solution:- given, P = 200000, r1 = 10%, r2 = 20%, r3 = 30%
Using above given formula ,

 Population of city at the end of year 2013 = P 1 + 10  1 - 20  1 + 30 100 100 100

 Population of city at the end of year 2013 = 200000 × 11 × 4 × 13 10 5 10

= 228800