The population of a town increas es by 5% in the first year, by 6% in the second year and by 8% in third year. The present popula tion of town is 120204. What was the population of the town 3 years ago?

Options

A. 102000
B. 100000
C. 98000
D. 101204

Answer

To find the population of the town three years ago, we need to work backwards from the current population, taking into account the percentage increases over the three years.

### Step-by-Step Solution:

1. **Identify the current population**:
The present population of the town is given as 120,204.

2. **Calculate the population before the third year**:
The population increases by 8% in the third year. To find the population before this increase, we can use the formula for reverse percentage increase:
\[ \text{Population before 3rd year} = \frac{\text{Current Population}}{1 + \text{Percentage Increase}} \]
Here, the percentage increase is 8%, which is 0.08 in decimal form.
\[ \text{Population before 3rd year} = \frac{120204}{1 + 0.08} = \frac{120204}{1.08} \approx 111,000 \]

3. **Calculate the population before the second year**:
The population increases by 6% in the second year. Again, we apply the reverse percentage increase:
\[ \text{Population before 2nd year} = \frac{\text{Population before 3rd year}}{1 + \text{Percentage Increase}} \]
Here, the percentage increase is 6%, or 0.06 in decimal form.
\[ \text{Population before 2nd year} = \frac{111000}{1 + 0.06} = \frac{111000}{1.06} \approx 104,720 \]

4. **Calculate the population before the first year**:
The population increases by 5% in the first year. We apply the reverse percentage increase once more:
\[ \text{Population before 1st year} = \frac{\text{Population before 2nd year}}{1 + \text{Percentage Increase}} \]
Here, the percentage increase is 5%, or 0.05 in decimal form.
\[ \text{Population before 1st year} = \frac{104720}{1 + 0.05} = \frac{104720}{1.05} \approx 99,000 \]

5. **Final Result**:
The population of the town three years ago was approximately **99,000**.