The population of a city increases by 20% at the end of every year. During which of the following years , the population would be doubled ?

Options

A. second
B. third
C. fourth
D. fifth

Explanation

To determine when the population of a city will double given that it increases by 20% at the end of every year, we can use the formula for compound interest. The formula is:

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

Where:
- \( A \) is the amount (final population),
- \( P \) is the principal (initial population),
- \( r \) is the rate of interest (percentage increase),
- \( t \) is the time in years.

### Step-by-Step Solution:

1. **Identify the Variables**:
- Let the initial population \( P \) be \( P \).
- The final population \( A \) when it doubles will be \( 2P \).
- The rate of increase \( r \) is 20%.

2. **Set Up the Equation**:
Substitute the known values into the compound interest formula:
\[ 2P = P \left(1 + \frac{20}{100}\right)^t \]

3. **Simplify the Equation**:
Since \( P \) is common on both sides, we can cancel it out:
\[ 2 = \left(1 + \frac{20}{100}\right)^t \]
This simplifies to:
\[ 2 = (1.2)^t \]

4. **Take Logarithm of Both Sides**:
To solve for \( t \), we can take the logarithm of both sides:
\[ \log(2) = t \cdot \log(1.2) \]

5. **Solve for \( t \)**:
Rearranging gives:
\[ t = \frac{\log(2)}{\log(1.2)} \]

6. **Calculate the Values**:
Using a calculator:
- \( \log(2) \approx 0.3010 \)
- \( \log(1.2) \approx 0.0792 \)

Now substitute these values:
\[ t \approx \frac{0.3010}{0.0792} \approx 3.8 \]

7. **Conclusion**:
Since \( t \) is approximately 3.8 years, the population will double sometime during the 4th year.

### Final Answer:
The population will double during the **4th year**.