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
\( = +30-20+{((+30) \times (-20)) \over 100}\)
\(= + 10 – 6 = 4 % \)
\(
Increase \% = 4 \% \ ( Answer) \)