Question

the population of a city increases by 10% annually if the present population is 65000 what will be its population after 2 years

Answer 1

afterr 1st year the increase of population will be 65000 ×10/100
so, the population = 6500+65000=71500
on 2nd year , the current population is 71500
so increase =71500×10/100=7150
population=71500+7150=78650

Answer 2

Answer:

78650

Step-by-step explanation:

Since, the population is,

$A=P(1+r)^t$

Where,

P = initial population,

r = rate of increasing per period,

t = number of periods,

Here, P = 65000, r = 10% = 0.1, t = 2,

Hence, the population after 2 years would be,

$A=65000(1+0.1)^2$

$=65000(1.1)^2$

$=65000\times 1.21$

= 78650