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 $\frac{d y}{d t} \propto y$
$\Rightarrow \frac{d y}{d t}=k y$, where $k$ is constant
$\Rightarrow \frac{d y}{y}=k d t$
On integrating both sides, we get $\log y=k t+C .....$(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)
$\Rightarrow \log 25000 =k 5+C$
$\Rightarrow \log 25000 =5 k+\log 20000 $ [using Eq. (iii)]
$\Rightarrow 5 k =\log \left(\frac{25000}{20000}\right)$
$=\log \left(\frac{5}{4}\right)$
$\Rightarrow k=\frac{1}{5} \log \frac{5}{4}$
For year $2009, t=10 yrs$
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 m n) $
$\Rightarrow y=20000 \times \frac{5}{4} \times \frac{5}{4}$
$\Rightarrow y=31250$
Hence, the population of the village in 2009 will be 31250 .