The annual rate of growth in population of a town is 10%. If its present population is 33275, what was it 3 years ago?

Options

A. 25000
B. 27500
C. 30000
D. 26000

Answer

To find the population of the town 3 years ago given its present population and the annual growth rate, we will use the formula for compound interest. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Present Population (A) = 33,275
- Annual Growth Rate (R) = 10%
- Time (n) = 3 years

### Step 2: Write the formula for compound interest
The formula for the amount (A) in compound interest is:
\[ A = P \left(1 + \frac{R}{100}\right)^n \]
Where:
- A = Present amount (population now)
- P = Principal amount (population in the past)
- R = Rate of interest (growth rate)
- n = Time in years

### Step 3: Rearrange the formula to find the principal (P)
We need to solve for P (the population 3 years ago):
\[ P = \frac{A}{\left(1 + \frac{R}{100}\right)^n} \]

### Step 4: Substitute the known values into the formula
Substituting the values we have:
\[ P = \frac{33,275}{\left(1 + \frac{10}{100}\right)^3} \]
\[ P = \frac{33,275}{\left(1 + 0.10\right)^3} \]
\[ P = \frac{33,275}{(1.10)^3} \]

### Step 5: Calculate \( (1.10)^3 \)
Calculating \( (1.10)^3 \):
\[ (1.10)^3 = 1.10 \times 1.10 \times 1.10 = 1.331 \]

### Step 6: Substitute back to find P
Now substitute this back into the equation:
\[ P = \frac{33,275}{1.331} \]

### Step 7: Perform the division
Calculating the division:
\[ P \approx 25,000 \]

### Conclusion
The population of the town 3 years ago was approximately **25,000**.

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