Question

The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time. If the population of the village was $20000$ in $1999$ and $25000$ in the year $2004$ , the population of the village in $2009$ is

• A. 31250
• B. 26000
• C. 31000
• D. 31520

Answer 1

Answer: (A)

Let $P$ be the population at time $t$, then $\frac{dy}{dt}\propto y$
$\Rightarrow \frac{dy}{dt}=ky$, where $k$ is constant
$\Rightarrow \frac{dy}{y}=kdt$
On integrating both sides, we get $\log y=kt+C.....$(i)
In the year $1999,t=0,y=20000$
$\therefore$ From equation (i), $\log 20000=k(0)+C$
$\Rightarrow \log 20000=C....$(ii)
In the year $2004,t=5,y=25000$, so from Eq. (i)
$\Rightarrow \log 25000=k5+C$
$\Rightarrow \log 25000=5k+\log 20000$ [using Eq. (iii)]
$\Rightarrow 5k=\log \left(\frac{25000}{20000}\right)$
$=\log \left(\frac{5}{4}\right)$
$\Rightarrow k=\frac{1}{5}\log \frac{5}{4}$
For year $2009,t=10yrs$
Now, substituting the values of $t,k$ and $C$ in Eq. (i), we get
$\log y=10\times \frac{1}{5}\log \left(\frac{5}{4}\right)+\log (20000)$
$\Rightarrow \log y=\log \left[20000\times {\left(\frac{5}{4}\right)}^2\right](\because \log m+\log n=\log mn)$
$\Rightarrow y=20000\times \frac{5}{4}\times \frac{5}{4}$
$\Rightarrow y=31250$
Hence, the population of the village in 2009 will be 31250 .