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%

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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Q. 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?

Q. The population of a town was 3600 three years back. It is 4800 right now. What will be the population three years down the line, if the rate of growth of population has been constant over the years?

Q. Mark (✓) against the correct answer:
The annual rate of growth in population of a town is 5%. If its present population is 4000, what will be its population after 2 years?
(a) 4441
(b) 4400
(c) 4410
(d) 4800

Q. The population of a town increases by

5 %

annually. If it is

15435

now, its population

2

years ago was:

Population of a town increases at a certain rate percent per annum. Present population of the town is 3600 and in 5 years it becomes 4800. How much will it be in 10 years?

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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% Final Value = Initial Value (1+r100)t Thus, Rate =[(FVIV)1t−1]×100 ⇒FVIV=4836=1.33 Going from answer options - If 10 is the answer - 1.13=1.331≈1.33=FVIV Therefore, rate is approximately 10%.