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?

15%

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12%

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10%

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Solution

The correct option is D 10%
Option (d)

Conventional approach :

$F i n a l V a l u e = I n i t i a l V a l u e {(1 + \frac{r}{100})}^{t}$

Thus, rate $= {(\frac{F V}{I V})}^{\frac{1}{t}} - 1$

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

Going from answer options

If 10 is the answer

${1.1}^{3} = 11^{2} \times 11 = 12 \times 11 \approx 1.33$

Therefore r is approximately $10 % .$

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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?

The correct option is D 10%Option (d) Conventional approach : Final Value=Initial Value(1+r100)t Thus, rate =(FVIV)1t−1 ⇒FVIV=4836=1.33 Going from answer options If 10 is the answer 1.13=112×11=12×11≈1.33 Therefore r is approximately 10%.