Q53 / 150 · MHT CET 2024 (09 May Shift 1) · Mathematics
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
- A180000
- B128000
- Correct answerC160000
- D256000
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 $\begin{aligned} & \log \left(\frac{\mathrm{p}}{40000}\right)=\frac{1}{40} \log 2 \times 80 \\ & \Rightarrow \log \frac{\mathrm{p}}{40000}=2 \log 2 \\ & \Rightarrow \frac{\mathrm{p}}{40000}=4 \\ & \Rightarrow \mathrm{p}=16,0000 \end{aligned}$
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 $\begin{aligned} & \log \left(\frac{\mathrm{p}}{40000}\right)=\frac{1}{40} \log 2 \times 80 \\ & \Rightarrow \log \frac{\mathrm{p}}{40000}=2 \log 2 \\ & \Rightarrow \frac{\mathrm{p}}{40000}=4 \\ & \Rightarrow \mathrm{p}=16,0000 \end{aligned}$
Hence, the correct answer is option: 'C'.