The population of a town increases by 5% every year. If the population is 94,500 now, what was its population one year ago?
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The population of a town increases by 5% every year. If the population is 94,500 now, what was its population one year ago?
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[tex]\large\underline{\sf{Solution-}}[/tex]
Given that,
We have to find
So, it means we have
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
[tex]\rm \: \boxed{ \rm{ \:A \: = \: P \: {\bigg[1 + \dfrac{r}{100} \bigg]}^{n} }} \\ [/tex]
So, on substituting the values, we get
[tex]\rm \: 94500 = P {\bigg[1 + \dfrac{5}{100} \bigg]}^{1} \\ [/tex]
[tex]\rm \: 94500 = P {\bigg[1 + \dfrac{1}{20} \bigg]} \\ [/tex]
[tex]\rm \: 94500 = P {\bigg[ \dfrac{20 + 1}{20} \bigg]} \\ [/tex]
[tex]\rm \: 94500 = P {\bigg[ \dfrac{21}{20} \bigg]} \\ [/tex]
[tex]\rm \: P = 94500 \times \frac{20}{21} \\ [/tex]
[tex]\rm\implies \:\boxed{ \rm{ \:P \: = \: 90000 \: }} \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
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
[tex]\boxed{ \rm{ \:Amount \: = \: P \: {\bigg[1 + \dfrac{r}{100} \bigg]}^{n} \: \: }} \\ [/tex]
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
[tex]\boxed{ \rm{ \:Amount \: = \: P \: {\bigg[1 + \dfrac{r}{200} \bigg]}^{2n} \: \: }} \\ [/tex]
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
[tex]\boxed{ \rm{ \:Amount \: = \: P \: {\bigg[1 + \dfrac{r}{400} \bigg]}^{4n} \: \: }} \\ [/tex]
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
[tex]\boxed{ \rm{ \:Amount \: = \: P \: {\bigg[1 + \dfrac{r}{1200} \bigg]}^{12n} \: \: }} \\ [/tex]
Answer:
Given :-
[tex]\\[/tex]
To Find :-
[tex]\\[/tex]
Formula Used :-
[tex]\clubsuit[/tex] Amount Formula :
[tex]\bigstar \: \: \sf\boxed{\bold{\pink{A =\: P\bigg(1 + \dfrac{r}{100}\bigg)^n}}}\: \: \: \bigstar\\[/tex]
where,
[tex]\\[/tex]
Solution :-
Given :
According to the question by using the formula we get,
[tex]\implies \sf A =\: P\bigg(1 + \dfrac{r}{100}\bigg)^n\\[/tex]
[tex]\implies \sf 94500 =\: P\bigg(1 + \dfrac{5}{100}\bigg)^1\\[/tex]
[tex]\implies \sf 94500 =\: P\bigg(\dfrac{100 \times 1 + 5}{100}\bigg)^1\\[/tex]
[tex]\implies \sf 94500 =\: P\bigg(\dfrac{100 + 5}{100}\bigg)^1\\[/tex]
[tex]\implies \sf 94500 =\: P\bigg(\dfrac{105}{100}\bigg)^1\\[/tex]
[tex]\implies \sf 94500 =\: P \times \dfrac{105}{100}\\[/tex]
[tex]\implies \sf 94500 =\: \dfrac{P \times 105}{100}\\[/tex]
[tex]\implies \sf 94500 =\: \dfrac{105P}{100}\\[/tex]
By doing cross multiplication we get,
[tex]\implies \sf 105P =\: 94500(100)[/tex]
[tex]\implies \sf 105P =\: 94500 \times 100[/tex]
[tex]\implies \sf 105P =\: 9450000[/tex]
[tex]\implies \sf P =\: \dfrac{9450000}{105}[/tex]
[tex]\implies \sf P =\: \dfrac{\cancel{90000}}{\cancel{1}}[/tex]
[tex]\implies \sf\bold{\red{P =\: 90000}}\\[/tex]
[tex]\small \sf\bold{\purple{\underline{\therefore\: The\: population\: after\: one\: year\: is\: 90000\: .}}}\\[/tex]
[tex]\\[/tex]
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