Percentage: Formula And Examples Related To Population

In Case of population Increase
Eg. Let the population of a town be p now and suppose increase at the rate of R % per annum then,

Population after n years ( = P (1+{R over 100})^n )

Population n years ago  ( = {P over (1+{R over 100})^n})

Example Present population of a town 1.21 crore and it is increasing at 10 % per annum. Then find out the population after two years as well as before 2 years?

Sol. Given, Present population = 1.21 crore,
 

Two years after
 ( = P{(1+{R over 100})}^n )

 
( = 1.21(1+{10 over 100})^2 )
 

( = 1.21 times 1.21 = 1.4641 crore )

Two years before
 ( = {P over (1+{R over 100})^n} )
( = {1.21 over (1+{10 over 100})^2} )

( = { 1.21 over 1.21} =1 crore )

In Case of population decrease
Let the population of a town be p now and suppose decrease at the rate of R % per annum then

Population after n year

(= P(1-{R over 100})^n)

Population n years ago

( = {P over (1-{R over 100})^n })

Example: If the present population of a town is 72600 and it is decreased by 10 % per annum. What will be its population 2 years hence?

Sol.
As we Know Population after n year ( = P ({1+R over 100})^n )

Here given,
p = 72600,
R = 10%,
N = 2  years

Population after 2 years

( = 72600({1-10 over 100})^2 )

( = 72600 times{ 90 over 100} times {90 over 100} )
( = 58806 ( Answer ) )

Note : this formula is also used for the depreciation value of machine. 
Net % change (= {pm X pm Y pm {XY over 100}} )

‘ + ’ means increment

‘ – ’ means decrement

Example. if the length of a rectangle is increased by 30 % and breadth is decreased by 20 % find the net % change in the area of that rectangle.

Sol. We know that,  Net % change  (  =
x pm y pm {(x times y) over 100} )

( Let x = 30% ) ( ….( we put x = +30) )
( and y = 20% ) ( …( we put y =-20 ) )
Now net % change
 ( =  +30-20+{((+30) times (-20)) over 100})
 (= + 10 – 6 = 4 % )

(
Increase % = 4 % ( Answer) )