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(D) Let the population $3$ years ago be $P$.The population increases by $5 \%$ every year,which is similar to the compound interest formula.The formul

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$8000$

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(D) Let the population $3$ years ago be $P$.The population increases by $5 \%$ every year,which is similar to the compound interest formula.The formula is: $A = P(1 + \frac{r}{100})^n$Given: $A = 9261$,$r = 5 \%$,$n = 3$.$9261 = P(1 + \frac{5}{100})^3$$9261 = P(1 + \frac{1}{20})^3$$9261 = P(\frac{21}{20})^3$$9261 = P(\frac{9261}{8000})$$P = 9261 	imes \frac{8000}{9261}$$P = 8000$Therefore,the population $3$ years ago was $8000$.

The population of a town increases by $5 \%$ every year. If the present population is $9261,$ what was the population $3$ years ago?

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