On a certain sum of money, compound inter est earned at the end of two years is₹ 1320. Compound interest at the eat of three years is ₹ 2184. Find the principal?

Options

A. ₹ 2400
B. ₹ 3600
C. ₹ 3200
D. ₹ 3000

Answer

To find the principal amount based on the given compound interest for two and three years, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Information:**
- Compound Interest (CI) for 2 years = ₹1320
- Compound Interest (CI) for 3 years = ₹2184

2. **Calculate the Interest for the Third Year:**
- The interest earned in the third year can be calculated by subtracting the compound interest for 2 years from the compound interest for 3 years.
\[ \text{Interest for 3rd Year} = \text{CI for 3 years} - \text{CI for 2 years} \]
\[ \text{Interest for 3rd Year} = 2184 - 1320 = ₹864 \]

3. **Relate the Interest to the Principal:**
- The interest earned in the second year is equal to the interest earned in the first year multiplied by the rate of interest. Let the principal be \( P \) and the rate of interest be \( r \).
- The interest earned in the second year can be expressed as:
\[ \text{Interest for 2nd Year} = P \cdot \left( \frac{r}{100} \right) \cdot (1 + \frac{r}{100}) \]
- The interest earned in the third year can be expressed as:
\[ \text{Interest for 3rd Year} = P \cdot \left( \frac{r}{100} \right) \cdot (1 + \frac{r}{100})^2 \]

4. **Set Up the Equations:**
- From the information, we know:
\[ P \cdot \left( \frac{r}{100} \right) \cdot (1 + \frac{r}{100}) = 1320 \]
\[ P \cdot \left( \frac{r}{100} \right) \cdot (1 + \frac{r}{100})^2 = 2184 \]

5. **Divide the Two Equations:**
- Dividing the second equation by the first gives:
\[ \frac{P \cdot \left( \frac{r}{100} \right) \cdot (1 + \frac{r}{100})^2}{P \cdot \left( \frac{r}{100} \right) \cdot (1 + \frac{r}{100})} = \frac{2184}{1320} \]
- Simplifying this:
\[ 1 + \frac{r}{100} = \frac{2184}{1320} \]
- Calculate the right side:
\[ 1 + \frac{r}{100} = \frac{2184}{1320} = \frac{91}{55} \]

6. **Solve for the Rate of Interest:**
- Rearranging gives:
\[ \frac{r}{100} = \frac{91}{55} - 1 = \frac{91 - 55}{55} = \frac{36}{55} \]
- Thus, \( r = \frac{36 \times 100}{55} = \frac{3600}{55} \approx 65.45\% \)

7. **Substitute Back to Find Principal:**
- Now substitute \( r \) back into the first equation to find \( P \):
\[ P \cdot \left( \frac{65.45}{100} \right) \cdot (1 + \frac{65.45}{100}) = 1320 \]
- Calculate \( 1 + \frac{r}{100} = 1 + 0.6545 = 1.6545 \):
\[ P \cdot 0.6545 \cdot 1.6545 = 1320 \]
- Calculate \( 0.6545 \cdot 1.6545 \approx 1.082 \):
\[ P \cdot 1.082 = 1320 \]
- Finally, solving for \( P \):
\[ P = \frac{1320}{1.082} \approx 1220.78 \]

8. **Final Result:**
- The principal amount \( P \) is approximately ₹3000.