Percentage

Percentage

Percentage is a number or ratio expressed as a fraction of 100.

Properties of Percentage
1. If we have to convert the percentage into a fraction than it is divided by 100.
2. If we have to convert the fraction into percentage we have to multiply with 100.
To express x% as a fraction: We have x% =(x/ 100)
To express a/b as a percent : We have ( a /b)=(a/b)*100%

Loss / Decrease condition
If there is an increase of X% and subsequently X% decrease then there is always loss/decrease in the condition.

Example sum
If Rohan salary is increased by 50% and subsequently decreased by 50%. How much percentage loss?

=(50*50)/100

=25% decrease

Percentage Increases

If the price of a commodity increases by R%, then the reduction in 

consumption so as not to increase the expenditure =

[(R/(100+R))*100]%

Example sum

If radha earning is 25% more than sita. Then sita earning is how 

many percentage less then by radha?

=[(25/(100+25))*100]%

= 20  %

Percentage Decreases
If the price of a commodity decreases by R%, then the increase in consumption so
as not to decrease the expenditure is :

=[(R/(100-R))*100]%

Example sum
If golu age is 20% less than gita than gita age is how many percentage more thangolu?
=(20/(100-20)*100)%
=25%

Results on Population

Let the population of a town be P now and suppose it increases at the rate of R% per annum,
1. Population after n years = P* (1 + ( R/100))^n
2. Population n years ago = P* (1 + (R/100 ))^n

Example sum
The population of a town is 176400. If it increases at the rate of 5% per annum, what will be its population 2 years hence, What was it 2 years ago?

Results on Depreciation
Let the present value of a machine be P.
Suppose it depreciates at the rate of R%per annum.
1. Value of the machine after n years = P *(1 - (R/ 100))^n
2. Value of the machine n years ago = P/ (1-(R/100))^n
If A is R % more than B, then B is less than A by {[R/(100+R)]*100}%
If A is R % less than B, then B is more than A by 

{[R/(100-R)]*100}%

Example sum
The value of a machine depreciates at the rate of 10% per annum. If its present value is Rs. 162000, what will be its worth after 2 years, What was the value of the machine 2 years ago ?
Value of the machine after 2 years = Rs. [162000 *(1-(10/100))^2]
= Rs. [162000* (100 - 10)/100)]
= Rs. 131220
Value of the machine after 2 years ago 

=Rs. [ 162000/(1- (10/100))^2]

= Rs. (162000*(10/9)*(10/9))

=Rs 200000

Easy tips to calculate percentage:

To find 5%, find 10% and divide it in two.

To find 15%, find 10%, then add 5%.

To find 20%, find 10% and double it.

To find 25%, find 50% and then halve it.

To find 60%, find 50% and add 10%.

To find 75%, find 50% and add 25%....and so on.

Example sum:
1.20% of 40 = 40 *(20/100) =8
Easy Method:
20% of 40 = 2*4 = 8 ( If zero exist in each term we multiply the 10's digit number )
2. 22%  of 30 = 30 * ( 22 / 100 ) =66/ 10
= 6.6

Easy Method:

22 % of 30 = 22 *3 = 66 ( In this case we consider the answer move 1 decimal place)
= 6.6

Extra Tricks:
If two articles are sold for the same selling price. On selling the first, the gain is x% and on selling the second the gain is y%, then net % profit

=[(2*( 100+x)(100+y))/((100+x)+(100+y))]- 100

Example sum:
1. A man sold two articles for Rs.600 each. On selling first, he gains 20% and on the other, he gains 30%. What is profit percent in the transaction?
Profit% = [(2* ( 100 + 20 ) * ( 100 +30 ))/((100+20)+(100+30))]-100

=[((2*120*130)/(120+130))]-100
=[31200/ 250]-100
= 124.8 - 100
= 24.8%