The population of a town was 3600 three years back. It is 4800 right now. What i

Question

The population of a town was 3600 three years back. It is 4800 right now. What is the rate of growth of population, if it has been constant over the years and has been compounding annually?

Answer: Option D
:
D
Final Value = Initial Value

{(1 + \frac{r}{100})}^{t}

Thus, Rate

= [{(\frac{F V}{I V})}^{\frac{1}{t}} - 1] \times 100

\Rightarrow \frac{F V}{I V} = \frac{48}{36} = 1.33

Going from answer options -
If 10 is the answer -

{1.1}^{3} = 1.331 \approx 1.33 = \frac{F V}{I V}

Therefore, rate is approximately 10%.

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