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110. 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 25,000 in the year 2004, what will be the population of the village in 2009?

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Vishal Baghel | Contributor-Level 10

4 months ago

Let the population at any instant (t) be y.

It is given that the rate of increase of population is proportional to the number of inhabitants at any instant.

$∴ \frac{d y}{d t} α y$

$\Rightarrow \frac{d y}{d t} = k y$ (k is constant)

$\Rightarrow \frac{d y}{y} = k d t$

Integration both sides, we get:

$l o g y = k t + C . . . . . . . . . . (1)$

In the year $1999, t = 0 & y = 20000 .$

Therefore, we get:

$l o g 20000 = C . . . . . . . . . . (2)$

In the year $2004, t = 5 & y = 25000 .$

Therefore, we get:

$\begin{array}{l} \Rightarrow l o g 25000 = 5 k + l o g 20000 \\ \Rightarrow 5 k = l o g (\frac{25000}{20000}) = l o g (\frac{5}{4}) \\ \Rightarrow k = \frac{1}{5} l o g (\frac{5}{4}) . . . . . . . . . . (3) \end{array}$

In the year $2009, t = 10 y e a r s$

Now, on substituting the values of t, k, and C in equation (1), we get:

$\begin{array}{l} l o g y = 10 \times \frac{1}{5} l o g (\frac{5}{4}) + l o g (20000) \\ \Rightarrow l o g y = l o g [20000 \times {(\frac{5}{4})}^{2}] \\ \Rightarrow y = 20000 \times \frac{5}{4} \times \frac{5}{4} \\ \Rightarrow y = 31250 \end{array}$

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

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110. 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 25,000 in the year 2004, what will be the population of the village in 2009?

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