Question:
The population of a colony 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 and has been compounding annually?
The population of a colony 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 and has been compounding annually?
Updated On: Oct 10, 2024
- 6,000
- 6,400
- 7,200
- 9,600
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The Correct Option is B
Solution and Explanation
Let the rate of interest be R
According to question
4800 = 3600\((\)1 \(+\)\(\frac{R}{100}\)\()\)\(^3\)
\(\frac{4}{3}\)= \((\)1 \(+\)\(\frac{R}{100}\)\()\)\(^3\)........(i)
Now the population after 3 yr
= 48000\((\)1 \(+\)\(\frac{R}{100}\)\()\)\(^3\)
Now from equation (i)
= 4800 × \(\frac{4}{3}\) = 6400
So the correct option is (B)
According to question
4800 = 3600\((\)1 \(+\)\(\frac{R}{100}\)\()\)\(^3\)
\(\frac{4}{3}\)= \((\)1 \(+\)\(\frac{R}{100}\)\()\)\(^3\)........(i)
Now the population after 3 yr
= 48000\((\)1 \(+\)\(\frac{R}{100}\)\()\)\(^3\)
Now from equation (i)
= 4800 × \(\frac{4}{3}\) = 6400
So the correct option is (B)
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