Let's denote:

• $P$ as the principal.
• $r$ as the rate of interest for simple interest (SI).
• $\frac{3}{2}r$ as the rate of interest for compound interest (CI) in the first case.

Case 1:

The rate of interest for CI is $\frac{3}{2}r$ and for SI is $r$. The difference between CI and SI for 2 years is Rs 4900.

Simple Interest (SI) formula for 2 years:

$\text{SI} = \frac{P \times r \times 2}{100}$

Compound Interest (CI) formula for 2 years:

$\text{CI} = P \times \left(1 + \frac{\frac{3}{2}r}{100}\right)^2 - P$

The difference between CI and SI is given as Rs 4900. Thus: $\text{CI} - \text{SI} = 4900$ Substitute the expressions for CI and SI into this equation.

Case 2:

The rate of interest for both CI and SI is $r$. The difference between CI and SI for 2 years is Rs 400.

Compound Interest (CI) formula when rate is $r$:

$\text{CI} = P \times \left(1 + \frac{r}{100}\right)^2 - P$

The difference between CI and SI is given as Rs 400. Thus: $\text{CI} - \text{SI} = 400$

Now, we need to solve these two equations to find the value of the principal $P$. Let's start by simplifying the compound interest formula for both cases.

Case 2 Simplification:

Expanding the CI formula: $\text{CI} = P \times \left[\left(1 + \frac{r}{100}\right)^2 - 1\right]$ $\text{CI} = P \times \left[\frac{r^2}{10000} + \frac{2r}{100}\right]$ The difference between CI and SI is: $P \times \left(\frac{r^2}{10000} + \frac{2r}{100}\right) - P \times \frac{2r}{100} = 400$ $P \times \frac{r^2}{10000} = 400 \quad \text{(1)}$

Case 1 Simplification:

Expanding the CI formula: $\text{CI} = P \times \left[\left(1 + \frac{3r}{200}\right)^2 - 1\right]$ $\text{CI} = P \times \left[\frac{9r^2}{40000} + \frac{6r}{200}\right]$ The difference between CI and SI is: $P \times \left(\frac{9r^2}{40000} + \frac{6r}{200}\right) - P \times \frac{2r}{100} = 4900$ $P \times \left(\frac{9r^2}{40000} + \frac{6r}{200} - \frac{2r}{100}\right) = 4900 \quad \text{(2)}$

Now, solving equations (1) and (2) will give us the value of $P$.

Let me calculate this for you.