The population of a town increases 5% annually. If its present population is 84,000, what will be its population in 2 years time?
92610
To determine the future population of the town after 2 years given an annual growth rate of 5%, we can use the formula for compound interest. The formula for the population after \(n\) years is given by:
\(P = P_0 (1 + r)^n\)
where:
- \(P_0\) is the current population (84,000).
- \(r\) is the growth rate (5% or 0.05).
- \(n\) is the number of years (2).
Plugging in the values:
\(P = 84000 \times (1 + 0.05)^2\)
\(P = 84000 \times (1.05)^2\)
First, calculate \((1.05)^2\):
\((1.05)^2 = 1.05 \times 1.05 = 1.1025\)
Now, calculate the final population:
\(P = 84000 \times 1.1025 = 92610\)
Therefore, the population of the town in 2 years will be 92,610, which matches the correct answer option "92610".
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Important Questions from Compound Interest
- A sum of money triples itself at a certain rate of compound interest in 11 years. In how many years will it amount to 27 times of itself?
A sum of ₹10,000 amounts to ₹12,321 in 2 years at a certain rate of interest per annum, compounded annually. The rate of interest per annum is:
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- A sum of money triples itself at a certain rate of compound interest in 17 years. In how many years will it amount to 81 times of itself?
If the interest earned during the $2^{nd}$ year on a certain sum is ₹3,342, and the rate of interest is 20% per annum compounded annually, then the sum is:
