The population of a city increases at the rate of 5% pa. If the present population of the city is 185220, then what was its population 3yr ago?

Options

A. 181500
B. 183433
C. 160000
D. 127783

Answer

To find the population of the city 3 years ago given the current population and the annual growth rate, we can use the formula for compound interest. The formula to calculate the amount (A) after a certain time period with compound interest is:

\[ A = P \left(1 + \frac{r}{100}\right)^t \]

Where:
- \( A \) = the final amount (current population)
- \( P \) = the principal amount (population 3 years ago)
- \( r \) = rate of interest (growth rate)
- \( t \) = time in years

In this case, we need to rearrange the formula to solve for \( P \):

\[ P = \frac{A}{\left(1 + \frac{r}{100}\right)^t} \]

### Step-by-step Solution:

1. **Identify the given values:**
- Current population \( A = 185220 \)
- Growth rate \( r = 5\% \)
- Time period \( t = 3 \) years

2. **Convert the growth rate to a decimal:**
- \( r = 5\% = \frac{5}{100} = 0.05 \)

3. **Calculate \( 1 + \frac{r}{100} \):**
- \( 1 + \frac{5}{100} = 1 + 0.05 = 1.05 \)

4. **Calculate \( (1 + r)^t \):**
- \( (1.05)^3 \)
- \( (1.05)^3 = 1.157625 \) (using a calculator or by multiplying \( 1.05 \times 1.05 \times 1.05 \))

5. **Substitute the values into the formula for \( P \):**
- \( P = \frac{185220}{(1.05)^3} = \frac{185220}{1.157625} \)

6. **Calculate \( P \):**
- \( P \approx 160000 \) (using a calculator for the division)

### Final Answer:
The population of the city 3 years ago was approximately **160,000**.

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