Question

The population of a town increases by 5% every year. If the population is 94,500 now, what was its population one year ago?​

Answer 1

Answer:

Answer is attached here.

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

The population of a town increases by 5% every year and the population is 94,500 now

So, it means, we have

Rate of increase, r = 5 % per annum

Present population, A = 94500

Time, n = 1 year

Let assume that the population one year ago be P.

We know,

Increase in population P at the rate of r % per annum for n years is given by

$\boxed{ \rm{ \:A \: = \: P \: {\bigg[1 + \dfrac{r}{100} \bigg]}^{n} }}$

where, A is the increased population.

So, on substituting the values, we get

$\rm \: 94500 = P {\bigg[1 + \dfrac{5}{100} \bigg]}^{1}$

$\rm \: 94500 = P {\bigg[1 + \dfrac{1}{20} \bigg]}$

$\rm \: 94500 = P {\bigg[\dfrac{20 + 1}{20} \bigg]}$

$\rm \: 94500 = P {\bigg[\dfrac{21}{20} \bigg]}$

$\rm \: P = 94500 \times \dfrac{20}{21}$

$\rm \: P = 4500 \times 20$

$\rm\implies \:P \: = \: 90000$

$\rule{190pt}{2pt}$

Additional information :-

1. Amount received on a certain sum of money of Rs P invested at the rate of r % per annum compounded annually for n years is given by

$\boxed{ \rm{ \:Amount \: = \: P \: {\bigg[1 + \dfrac{r}{100} \bigg]}^{n} \: \: }}$

2. Amount received on a certain sum of money of Rs P invested at the rate of r % per annum compounded semi - annually for n years is given by

$\boxed{ \rm{ \:Amount \: = \: P \: {\bigg[1 + \dfrac{r}{200} \bigg]}^{2n} \: \: }}$

3. Amount received on a certain sum of money of Rs P invested at the rate of r % per annum compounded quarterly for n years is given by

$\boxed{ \rm{ \:Amount \: = \: P \: {\bigg[1 + \dfrac{r}{400} \bigg]}^{4n} \: \: }}$

4. Amount received on a certain sum of money of Rs P invested at the rate of r % per annum compounded monthly for n years is given by

$\boxed{ \rm{ \:Amount \: = \: P \: {\bigg[1 + \dfrac{r}{1200} \bigg]}^{12n} \: \: }}$