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(A) Let the population at any time $t$ be $y$.It is given that the rate of increase of population is proportional to the number of inhabitants at any

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$31,250$

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(A) Let the population at any time $t$ be $y$.It is given that the rate of increase of population is proportional to the number of inhabitants at any time $t$.$\frac{dy}{dt} = ky$ (where $k$ is a constant)$\frac{dy}{y} = k dt$Integrating both sides,we get:$\ln y = kt + C$  ............$(1)$In the year $1999$,let $t = 0$ and $y = 20,000$.$\ln(20,000) = k(0) + C \Rightarrow C = \ln(20,000)$.In the year $2004$,$t = 5$ and $y = 25,000$.$\ln(25,000) = 5k + \ln(20,000)$$5k = \ln\left(\frac{25,000}{20,000}ight) = \ln\left(\frac{5}{4}ight)$$k = \frac{1}{5} \ln\left(\frac{5}{4}ight)$.In the year $2009$,$t = 10$.Substituting $t = 10, k = \frac{1}{5} \ln\left(\frac{5}{4}ight)$,and $C = \ln(20,000)$ in equation $(1)$:$\ln y = 10 	imes \frac{1}{5} \ln\left(\frac{5}{4}ight) + \ln(20,000)$$\ln y = 2 \ln\left(\frac{5}{4}ight) + \ln(20,000)$$\ln y = \ln\left(\left(\frac{5}{4}ight)^2 	imes 20,000ight)$$y = 20,000 	imes \frac{25}{16} = 1,250 	imes 25 = 31,250$.Thus,the population in $2009$ will be $31,250$.

The population of a village increases continuously at a rate proportional to the number of its inhabitants present at any time. If the population of the village was $20,000$ in $1999$ and $25,000$ in the year $2004,$ what will be the population of the village in $2009?$

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