Question

11. A district contains 64000 inhabitants. If the
population increases at the rate of
5/2per
annum, find the number of inhabitants at the end
of 3 years.​

Answer 1

Answer:

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

$\huge\underline\mathtt{Solution}$

Given:

Principal (P)=64000

Rate of Interest (r)=5/2%

⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀=2.5 %

Time (n)=3 yrs

$\rule{300}{2}$

To Find:

The number of inhabitants at the end of 3 years.

Formula used:

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

Now,by putting the values in the formula, we get,

$\implies A = 64000(1 + \frac{2.5}{100} )^{3} \\ \implies A = 64000(1 + \frac{25}{1000}) ^{3} \\ \implies A = 64000( \frac{1025}{1000} )^{3} \\ \implies A = 64000 \times \frac{1025}{1000} \times \frac{1025}{1000} \times \frac{1025}{1000} \\ \implies A= \frac{68921000000}{1000000} \\ \implies A = 68921$

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

$\rule{300}{2}$

$\huge{\underline{\overline{\mathfrak{\pink{ThankYou}}}}}$