Math, asked by hasini4697, 2 days ago

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

Answers

Answered by mathdude500
33

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

Given that,

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

We have to find

  • The population one year ago.

So, it means we have

  • Present population, A = 94500

  • Increase in population, r = 5 % per annum

  • Time, n = 1 year

We know,

If P is the population of the town increases every year at the rate of r % per annum for n years, then total population A is given by

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

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  \frac{20}{21}  \\

\rm\implies \:\boxed{ \rm{ \: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} \:  \: }} \\

Answered by StarFighter
30

Answer:

Given :-

  • The population of a town increases by 5% every year.
  • The present population is 94500.

\\

To Find :-

  • What is the population after one year ago.

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Formula Used :-

\clubsuit Amount Formula :

\bigstar \: \: \sf\boxed{\bold{\pink{A =\: P\bigg(1 + \dfrac{r}{100}\bigg)^n}}}\: \: \: \bigstar\\

where,

  • A = Present Population
  • P = Population after year
  • r = Rate of Increase
  • n = Time Period

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Solution :-

Given :

  • Present Population = 94500
  • Time Period = 1 year
  • Rate of Increase = 5% every year

According to the question by using the formula we get,

\implies \sf A =\: P\bigg(1 + \dfrac{r}{100}\bigg)^n\\

\implies \sf 94500 =\: P\bigg(1 + \dfrac{5}{100}\bigg)^1\\

\implies \sf 94500 =\: P\bigg(\dfrac{100 \times 1 + 5}{100}\bigg)^1\\

\implies \sf 94500 =\: P\bigg(\dfrac{100 + 5}{100}\bigg)^1\\

\implies \sf 94500 =\: P\bigg(\dfrac{105}{100}\bigg)^1\\

\implies \sf 94500 =\: P \times \dfrac{105}{100}\\

\implies \sf 94500 =\: \dfrac{P \times 105}{100}\\

\implies \sf 94500 =\: \dfrac{105P}{100}\\

By doing cross multiplication we get,

\implies \sf 105P =\: 94500(100)

\implies \sf 105P =\: 94500 \times 100

\implies \sf 105P =\: 9450000

\implies \sf P =\: \dfrac{9450000}{105}

\implies \sf P =\: \dfrac{\cancel{90000}}{\cancel{1}}

\implies \sf\bold{\red{P =\: 90000}}\\

\small \sf\bold{\purple{\underline{\therefore\: The\: population\: after\: one\: year\: is\: 90000\: .}}}\\

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