when the rate of interest of Compound interest is 3/2 times of the rate of interst of Simple interest, the difference between Compound interest and Simple interest on a certain sum for 2 years is Rs 4900.When the rate of interest for Compound interest is equal to the rate of interest of Simple interest the difference between Compound interest and Simple interest on a certain sum for 2 years is Rs 400. Find the principal .

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.

When the rate of interest of Compound interest is 3/2 times of the rate of interest of Simple interest, the difference between Compound interest and Simple interest on a certain sum for 2 years is Rs 4900. When the rate of interest for Compound interest is equal to the rate of interest of Simple interest, the difference between Compound interest and Simple interest on a certain sum for 2 years is Rs 400. Find the principal.

Solve for the principal when the difference between compound interest and simple interest for 2 years is Rs 4900 at 3/2 times the simple interest rate, and Rs 400 when the rates are equal. This detailed solution covers compound and simple interest concepts, along with key formulas and algebraic equations. Suitable for students in Grades 10-12.

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