Percentage is a number or ratio expressed as a fraction of 100.
Properties of Percentage
1. If we have to convert percentage into fraction than it is divide by 100.
2. If we have to convert fraction into percentage we have to multiple with 100.
Loss / Decrease condition
If there is increase of X% and subsequently X% decrease then there is always loss / decrease in the condition.
Example sum
If rohan salary is increase by 50% and subsequently decrease by 50%. How much percentage loss?
= ( 50 * 50 ) / 100
= ( 5 * 5 ) %
= 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 than golu?
= ( 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 yaers ago ?
Population after 2 yaers = [ 176400 * ( 1 + ( 5 / 100 ) )2 ]
= 176400 * ( 105 / 100 )2
= 176400 * ( 21/ 20 )2
= ( 176400 * 21 * 21 ) / ( 20 * 20 )
= 194481
Population 2 yaers ago = 176400 / ( 1 * ( 5 / 100 ) )2
= ( 176400 * 20 * 20 ) / ( 21 * 21 )
= 160000
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
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 yeras, What was the value of the machine 2 yeras ago ?
Value of the machine after 2 years = Rs. [ 162000 * ( 1 - ( 10 / 100 ) )2 ]
= Rs. [ 162000 * ( 100 - 10 ) / 100 ) ]
= Rs. [ 16200 * ( 9 / 10 ) * ( 9 / 10 ) ]
= Rs. 131220
Value of the machine after 2 years ago = Rs. [ 162000 / ( 1 - ( 10 / 100 ) )2 ]
= Rs. [ 162000 / ( ( 100 - 10 ) / 100 )2]
= Rs. ( 162000 * ( 10 / 9 ) * ( 10 / 9 ) )
= Rs. 200000