Question

There is a continuous growth in population of a village at the rate of 5% per annum. If
its present population is 9261, what it was 3 years ago?​

Answer 1

Answer:

8000

Step-by-step explanation:

Let P be the population 3 years ago

Then present population = P x

(1+5/100)^3

9261=(P×21/20×21/20×21/20)

or P= (9261×20/21×20/21×20/21)

=8000

Hence, three years ago the population of the village was 8000

Answer 2

Given:

A village with

• Present population = 9261
• Growth rate = 5%
• Time = 3 years

What To Find:

We have to find the population of the village 3 years ago.

How To Find:

To find the population of the village 3 years ago, we will use the

• Formula of amount in compound interest i.e $\sf{A = P \left(1 + \dfrac{R}{100} \right)^T}$.
• Substitute the values and solve.

Solution:

Using the formula,

⇒ $\sf{A = P \left(1 + \dfrac{R}{100} \right)^T}$

Substitute the values,

⇒ $\sf{A = 9261 \left(1 + \dfrac{5}{100} \right)^3}$

Solve the brackets,

⇒ $\sf{A = 9261 \left( \dfrac{105}{100} \right)^3}$

Convert the fraction into decimal,

⇒ $\sf{A = 9261 ( 1.05 )^3}$

Remove the brackets,

⇒ A = 9261 × 1.05 × 1.05 × 1.05

Multiply the numbers,

⇒ A = 8000

∴ Therefore, the population of the village three years ago was 8000.