Question

The population of a place increased to 98,000 in 2019 at a rate of 7% per annum. Find the population in 2017.

Answer 1

$\huge\mathfrak\pink{Answer}$

New population =54,000

Time (n)=2 yrs

r=5%

Original population = ?

54000= p/(1− 5/100 ) ^(2)

54000= p/0.9025

P=48735

Population in 2001 was 48,735.

ii) A=54000(1+ 5/100) ^(2)

=54000(1.1025)=59,535

Population in 2005 will be 59,535.

Answer 2

Let the population of the place at 2017 be x
Therefore, the population in 2018= x+7% of x=107% of x
Therefore the population in 2019= 107% of (107% of x)=107*107*x/10000
According to the problem, (107^2)*x/10000=98000
=>x=98000*10000/107^2= 85597 (approx value)