Aman invests Rs 4000 at an annual compound interest rate. The investment's value after 3 years compared to its value after 5 years is in the ratio 25:36. Let the annual interest rate be $r$. The compound interest formula is:
\[ A = P \left(1 + \frac{r}{100}\right)^t \]
Here, $A$ is the final amount, $P$ is the principal, $r$ is the annual interest rate, and $t$ is the time in years.
We are given:
\[ \frac{A_3}{A_5} = \frac{25}{36} \]
Using the compound interest formula for 3 and 5 years:
\[ \frac{4000 \left(1 + \frac{r}{100}\right)^3}{4000 \left(1 + \frac{r}{100}\right)^5} = \frac{25}{36} \]
Simplifying the equation:
\[ \frac{\left(1 + \frac{r}{100}\right)^3}{\left(1 + \frac{r}{100}\right)^5} = \frac{25}{36} \]
Inverting both sides:
\[ \left(1 + \frac{r}{100}\right)^2 = \frac{36}{25} \]
Taking the square root of both sides:
\[ 1 + \frac{r}{100} = \frac{6}{5} \]
Solving for $r$:
\[ \frac{r}{100} = \frac{1}{5} \implies r = 20\% \]
The annual interest rate is 20%.
Next, we determine the minimum time for the investment to surpass Rs 20000, using the compound interest formula with $r=20\%$:
\[ 20000 = 4000 \left(1 + \frac{20}{100}\right)^t \]
\[ 5 = 1.2^t \]
Applying logarithms to both sides:
\[ \log(5) = t \log(1.2) \]
\[ t = \frac{\log(5)}{\log(1.2)} \approx \frac{0.69897}{0.07918} \approx 8.83 \]
Since the number of years must be an integer, the minimum number of years required is 9.
Conclusion: The investment will exceed Rs 20000 in a minimum of 9 years.