The population of a town increases at a rate proportional to the population at that time. If the population…

The population of a town increases at a rate proportional to the population at that time. If the population increases from 40 thousand to 80 thousand in 40 years, then the population in another 40 years will be
  1. 180000
  2. 128000
  3. 160000
  4. 256000

Solution

Let p be the population at time t years. Then $\frac{\mathrm{dp}}{\mathrm{dt}}=\mathrm{kp}$ $\Rightarrow \frac{\mathrm{dp}}{\mathrm{p}}=\mathrm{kdt}$
Integrating on both sides, we get $\log \mathrm{p}=\mathrm{kt}+\mathrm{c}$ when $\mathrm{t}=0, \mathrm{p}=40000$ $\begin{aligned} & \therefore \quad \log 40000=0+c \\ & \\ & \Rightarrow \mathrm{c}=\log 40000 \end{aligned}$ $\begin{aligned} \therefore \quad & \log \mathrm{p}=\mathrm{kt}+\log 40000 \\ & \Rightarrow \log \left(\frac{\mathrm{p}}{40000}\right)=\mathrm{kt} \end{aligned}$
When $\mathrm{t}=40$ years, $\mathrm{p}=80000$ $\begin{aligned} & \Rightarrow \log \left(\frac{80000}{40000}\right)=40 \mathrm{k} \\ & \Rightarrow \mathrm{k}=\frac{1}{40} \log 2 \end{aligned}$ $\therefore \quad \log \left(\frac{\mathrm{p}}{40000}\right)=\frac{1}{40} \log 2 \times \mathrm{t}$ $\therefore \quad$ Population after another 40 years, i.e., $\mathrm{t}=80$ years, we have $\begin{aligned} & \log \left(\frac{\mathrm{p}}{40000}\right)=\frac{1}{40} \log 2 \times 80 \\ & \Rightarrow \log \frac{\mathrm{p}}{40000}=2 \log 2 \\ & \Rightarrow \frac{\mathrm{p}}{40000}=4 \\ & \Rightarrow \mathrm{p}=16,0000 \end{aligned}$

Asked in: MHT CET 2024 (09 May Shift 1)

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