Question

The population of a city increases at the rate 5% per year if the population in 2003 is 52920 what was the population in 2001

Answer 1

. Population in 2001 = 54000 / [1 + 5/100]^2 ; as it is 2 years ago

=> 54000 / 1.1025

Thus, Population in 2001 = 48,980

Answer 2

Answer:

The value of the population in 2001 is 48000 .

Step-by-step explanation:

The exponential increases function is given by

$y = a (1 + r)^{t}$

Where y is the final value , a is the initial value , r is the rate of interest in the decimal form and t is the time in years .

As given

The population of a city increases at the rate 5% per year if the population in 2003 is 52920 .

y = 52920

5% is written in the decimal form

$= \frac{5}{100}$

= 0.05

r = 0.05

As the years are taken from 2001 to 2003.

There is the difference of 2 years .

t = 2 years

Put all the values in the formula

$52920= a (1 + 0.05)^{2}$

$52920= a (1.05)^{2}$

$52920= 1.1025\ a$

$a = \frac{52920}{1.1025}$

a = 48000

Therefore the value of the population in 2001 is 48000 .