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HomeACSEE (Form 6)Advanced MathematicsDifferential Equations › A village grows at a rate proportional to its po…

A village grows at a rate proportional to its population. It was 20,000 in 1999 and 25,000 in 2004. Its population in 2009 is:

A30,000
B28,750
C32,500
D31,250
Answer & Solution
Correct answer: D. 31,250
1. The model is $\dfrac{dy}{dt}=ky$, whose solution is $\ln y=kt+c$. 2. At $t=0$, $y=20{,}000$, so $c=\ln 20{,}000$. 3. At $t=5$, $y=25{,}000$, giving $k=\dfrac15\ln\dfrac54$. 4. At $t=10$: $\ln y=\dfrac{10}{5}\ln\dfrac54+\ln 20{,}000$, so $y=20{,}000\times\left(\dfrac54\right)^{2}$. 5. That is $20{,}000\times 1.5625=31{,}250$. Answering $30{,}000$ assumes a constant increase of 5,000 per five years, which is linear rather than exponential growth. _Source: NECTA ACSEE 2023 Advanced Mathematics 142/2, Question 7: Differential Equations_
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