Q15.    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 20, 000 in 1999 and 25000 in the year 2004, what will be the population of the village in 2009?

Answers (1)

Let n be the population of the village at any time t.

According to question,

\frac{dn}{dt} = kn\ \ (k\ is\ a\ constant)

\\ \implies \int \frac{dn}{n} = \int kdt \\ \implies \log n = kt + C

Now, at t=0, n = 20000 (Year 1999)

\\ \implies \log (20000) = k(0) + C \\ \implies C = \log2 + 4

Again, at t=5, n= 25000 (Year 2004)

\\ \implies \log (25000) = k(5) + \log2 + 4 \\ \implies \log 25 + 3 = 5k + \log2 +4 \\ \implies 5k = \log 25 - \log2 -1 =\log \frac{25}{20} \\ \implies k = \frac{1}{5}\log \frac{5}{4}

Using these values, at t =10 (Year 2009)

\\ \implies \log n = k(10)+ C \\ \implies \log n = \frac{1}{5}\log \frac{5}{4}(10) + \log2 + 4 \\ \implies \log n = \log(\frac{25.2.10000}{16}) = \log(31250) \\ \therefore n = 31250

Therefore, the population of the village in 2009 will be 31250.

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