A sum of money is accumulating at compound interest at a certain rate of interest. If simple interest instead of compound were reckoned, the interest for the first two years would be diminished by 20 and that for the first three years, by 61. Find the sum.

Options

A. 7, 000
B. 8000
C. 7500
D. 6500

Answer

To solve the problem, we need to find the principal sum of money (P) given the differences in interest earned from compound interest (CI) and simple interest (SI) over two and three years.

### Step-by-Step Solution:

1. **Understanding the Problem:**
- Let the principal amount be \( P \).
- Let the rate of interest be \( r\% \).

2. **Formulas:**
- The formula for Simple Interest (SI) for \( n \) years is:
\[ SI = \frac{P \times r \times n}{100} \]
- The formula for Compound Interest (CI) for \( n \) years is:
\[ CI = P \left(1 + \frac{r}{100}\right)^n - P \]

3. **Interest for the First Two Years:**
- The difference between CI and SI for the first two years is given as 20:
\[ CI_2 - SI_2 = 20 \]
- Calculate \( SI_2 \):
\[ SI_2 = \frac{P \times r \times 2}{100} \]
- Calculate \( CI_2 \):
\[ CI_2 = P \left(1 + \frac{r}{100}\right)^2 - P \]
- Therefore, the equation becomes:
\[ P \left(1 + \frac{r}{100}\right)^2 - P - \frac{P \times r \times 2}{100} = 20 \]

4. **Interest for the First Three Years:**
- The difference between CI and SI for the first three years is given as 61:
\[ CI_3 - SI_3 = 61 \]
- Calculate \( SI_3 \):
\[ SI_3 = \frac{P \times r \times 3}{100} \]
- Calculate \( CI_3 \):
\[ CI_3 = P \left(1 + \frac{r}{100}\right)^3 - P \]
- Therefore, the equation becomes:
\[ P \left(1 + \frac{r}{100}\right)^3 - P - \frac{P \times r \times 3}{100} = 61 \]

5. **Setting Up the Equations:**
- From the first equation:
\[ P \left(1 + \frac{r}{100}\right)^2 - P - \frac{2Pr}{100} = 20 \]
- From the second equation:
\[ P \left(1 + \frac{r}{100}\right)^3 - P - \frac{3Pr}{100} = 61 \]

6. **Dividing the Two Equations:**
- To eliminate \( P \), we can divide the first equation by the second:
\[ \frac{20}{61} = \frac{\left(1 + \frac{r}{100}\right)^2 - 1 - \frac{2r}{100}}{\left(1 + \frac{r}{100}\right)^3 - 1 - \frac{3r}{100}} \]

7. **Simplifying the Equation:**
- Let \( x = 1 + \frac{r}{100} \).
- The equation simplifies to:
\[ \frac{20}{61} = \frac{x^2 - 1 - \frac{2r}{100}}{x^3 - 1 - \frac{3r}{100}} \]

8. **Finding the Rate \( r \):**
- After simplifying and solving the above equation, we find that \( r = 5\% \).

9. **Finding the Principal \( P \):**
- Using the first difference equation:
\[ P \left(1 + \frac{5}{100}\right)^2 - P - \frac{2P \times 5}{100} = 20 \]
- Substitute \( r = 5 \):
\[ P \left(1.05^2 - 1 - 0.1\right) = 20 \]
- Calculate \( 1.05^2 = 1.1025 \):
\[ P (1.1025 - 1 - 0.1) = 20 \]
\[ P (0.0025) = 20 \]
\[ P = \frac{20}{0.0025} = 8000 \]

### Final Answer:
The sum of money (P) is **8000**.