Question

The population of a village increases at the rate of 5% every ten years. What was the population of the village 20 years ago if its present population is 8820?
(a) 7500
(b) 8000
(c) 8240
(d) 7250

Answer 1

Hint: Suppose the population of the village 20 years ago was x. So, the population of the village 10 years ago was = $x+x\times \dfrac{5}{100}=\dfrac{21x}{20}$. Similarly, present population = $\dfrac{21x}{20}+\dfrac{21x}{20}\times \dfrac{5}{100}=\dfrac{441x}{400}$. Equate this to 8820 and solve for x to get the final answer.
Complete step by step answer:
In this question, we are given that the population of a village increases at the rate of 5% every ten years.
We need to find the population of the village 20 years ago if its present population is 8820.
Suppose the population of the village 20 years ago was x.
Since the population of the village increases at the rate of 5% every ten years.
So, the population of the village 10 years ago was = $x+x\times \dfrac{5}{100}=\dfrac{21x}{20}$
And now after 10 more years, we will have the following:
The present population of village (after 20 years) is = $\dfrac{21x}{20}+\dfrac{21x}{20}\times \dfrac{5}{100}=\dfrac{441x}{400}$
As given present population of village = 8820
So, we have the following:
$\dfrac{441x}{400}=8820$
$x=\dfrac{8820\times 400}{441}=8000$
Therefore the population of the village 20 years ago was 8000.
Hence, option (b) is the correct answer.

Note: In this question, it is very important to use a variable to show the population of the village 20 years ago and work forwards from there. We can also solve the question by working backwards in the following way: Present population =$8820\div {{\left( 1+\dfrac{5}{100} \right)}^{2}}=\dfrac{8820\times 100\times 100}{105\times 105}=8000$.