A man invested a sum of money at compound interest. It amounted to `Rs. 2420` in 2 years and to `Rs. 2662` in 3 years. Find the sum.

Options

A. `Rs. 1000`
B. `Rs. 2000 `
C. `Rs. 5082`
D. `Rs. 3000`

Answer

To solve the problem, we need to find the principal amount (the initial sum of money) that was invested at compound interest. We know the amounts after 2 years and 3 years.

### Step-by-Step Solution:

1. **Identify the amounts after 2 and 3 years:**
- Amount after 2 years (A2) = Rs. 2420
- Amount after 3 years (A3) = Rs. 2662

2. **Calculate the interest earned in the third year:**
- The interest earned in the third year (I) can be found by subtracting the amount after 2 years from the amount after 3 years.
\[ I = A3 - A2 = 2662 - 2420 = 242 \]

3. **Since the interest earned in the first year and the second year is the same as the interest earned in the third year, we can conclude that the interest for the first year (I1) is also Rs. 242.**

4. **Calculate the rate of interest (R):**
- We can use the formula for simple interest to find the rate of interest:
\[ R = \frac{I \times 100}{P \times T} \]
- Here, \(I = 242\), \(T = 1\) year, and \(P\) is the principal amount we need to find. We can rearrange the formula to express \(P\):
\[ P = \frac{I \times 100}{R} \]

5. **Find the amount after 2 years using the compound interest formula:**
- The amount after 2 years can be expressed as:
\[ A2 = P \left(1 + \frac{R}{100}\right)^2 \]
- We know \(A2 = 2420\) and we can substitute \(R = 10\%\) (which we will find out later):
\[ 2420 = P \left(1 + \frac{10}{100}\right)^2 = P \left(\frac{11}{10}\right)^2 = P \left(\frac{121}{100}\right) \]

6. **Rearranging to find the principal (P):**
\[ P = \frac{2420 \times 100}{121} = \frac{242000}{121} = 2000 \]

7. **Conclusion:**
- The principal amount (sum invested) is Rs. 2000.

### Final Answer:
The sum invested is **Rs. 2000**.