Question

a district contains 64000 inhabitants if the population increases at the rate of 2 and the half per annum find the number of inhabitants at the end of 3 years

Answer 1

Initial Population (IP) = 64000
Rate (r) = 2.5% per annum
Time (n) = 3years

At the end of 3 years:
Population = IP * (1+r/100)^n
                  = 64000*(1+2.5/100)^3
                  =64000* 1.025^3
                  = 68921 people

Answer 2

Answer:  The number of inhabitants at the end of 3 years is 68921.

Step-by-step explanation:

Here, the initial population, P = 64000

Annual rate of interest, r = $2\frac{1}{2}\%=\frac{5}{2}\%$

And, time, t = 3 years

Hence, the population after 3 years is,

$A = P(1+\frac{r}{100})^t$

$=64000(1+\frac{5/2}{100})$

$=64000(1+\frac{5}{200})$

$=64000\times \frac{205}{200}$

$\frac{13120000}{200}=68921$

Hence, the number of inhabitants at the end of 3 years is 68921.