The compound interest and the simple interest for two years on a certain sum of money at a certain rate of interest are ₹ 2257.50, ₹ 2100, respectively. Find the principal and rate per cent:

Options

A. 12000 , 7%
B. `7500,8%`
C. `14000,10%`
D. `7000,15%`

Answer

To solve the problem, we need to find the principal amount (P) and the rate of interest (R) given the compound interest (CI) and simple interest (SI) for two years.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Compound Interest (CI) for 2 years = ₹2257.50
- Simple Interest (SI) for 2 years = ₹2100

2. **Understand the Relationship Between SI and CI:**
- The formula for Simple Interest (SI) for 2 years is:
\[ SI = \frac{P \times R \times T}{100} \]
where \( T = 2 \) years.
- Therefore, we can express SI as:
\[ SI = \frac{P \times R \times 2}{100} \]
- Given that SI = ₹2100, we can write:
\[ 2100 = \frac{P \times R \times 2}{100} \]
- Rearranging gives:
\[ P \times R = \frac{2100 \times 100}{2} = 105000 \quad (1) \]

3. **Calculate the Compound Interest:**
- The formula for Compound Interest (CI) for 2 years is:
\[ CI = P \left(1 + \frac{R}{100}\right)^2 - P \]
- This can be simplified to:
\[ CI = P \left[\left(1 + \frac{R}{100}\right)^2 - 1\right] \]
- Given that CI = ₹2257.50, we can write:
\[ 2257.50 = P \left[\left(1 + \frac{R}{100}\right)^2 - 1\right] \quad (2) \]

4. **Substituting \( P \) from Equation (1) into Equation (2):**
- From equation (1), we have \( P = \frac{105000}{R} \).
- Substitute \( P \) into equation (2):
\[ 2257.50 = \frac{105000}{R} \left[\left(1 + \frac{R}{100}\right)^2 - 1\right] \]

5. **Solving for R:**
- Rearranging gives:
\[ 2257.50R = 105000 \left[\left(1 + \frac{R}{100}\right)^2 - 1\right] \]
- Dividing both sides by 105000:
\[ \frac{2257.50R}{105000} = \left(1 + \frac{R}{100}\right)^2 - 1 \]
- Let’s simplify \( \frac{2257.50}{105000} \):
\[ \frac{2257.50}{105000} = 0.0215 \]
- Thus:
\[ 0.0215R = \left(1 + \frac{R}{100}\right)^2 - 1 \]
- Expanding the right-hand side:
\[ 0.0215R = \left(1 + \frac{2R}{100} + \frac{R^2}{10000}\right) - 1 \]
- This simplifies to:
\[ 0.0215R = \frac{2R}{100} + \frac{R^2}{10000} \]

6. **Solving the Quadratic Equation:**
- Rearranging gives:
\[ \frac{R^2}{10000} + \left(\frac{2}{100} - 0.0215\right)R = 0 \]
- Solving this quadratic equation will yield the value of \( R \).

7. **Finding the Principal:**
- Once \( R \) is found, substitute back into equation (1) to find \( P \).

### Final Results:
After solving the quadratic equation, we find:
- Rate of Interest (R) = 15%
- Principal (P) = ₹7000