A sum of money deposited at compound interest amounts to Rs. 6690 after 3 years and Rs. 10,035 after 6 years. The sum is-

Options

A. Rs 4400
B. Rs. 4445
C. Rs. 4460
D. Rs. 4520

Answer

To find the sum of money deposited at compound interest, we can use the information given about the amounts after certain years.

Let the principal amount be \( P \) and the rate of interest be \( r \).

1. **Understanding the given amounts**:
- After 3 years, the amount is \( A_1 = 6690 \).
- After 6 years, the amount is \( A_2 = 10035 \).

2. **Using the formula for compound interest**:
The formula for the amount \( A \) after \( n \) years at compound interest is:
\[ A = P \left(1 + \frac{r}{100}\right)^n \]

3. **Setting up the equations**:
- For 3 years:
\[ 6690 = P \left(1 + \frac{r}{100}\right)^3 \quad \text{(1)} \]
- For 6 years:
\[ 10035 = P \left(1 + \frac{r}{100}\right)^6 \quad \text{(2)} \]

4. **Dividing the two equations**:
To eliminate \( P \), we can divide equation (2) by equation (1):
\[ \frac{10035}{6690} = \frac{P \left(1 + \frac{r}{100}\right)^6}{P \left(1 + \frac{r}{100}\right)^3} \]
This simplifies to:
\[ \frac{10035}{6690} = \left(1 + \frac{r}{100}\right)^3 \]

5. **Calculating the left-hand side**:
\[ \frac{10035}{6690} \approx 1.5 \]
Thus, we have:
\[ \left(1 + \frac{r}{100}\right)^3 = 1.5 \]

6. **Taking the cube root**:
Taking the cube root of both sides:
\[ 1 + \frac{r}{100} = \sqrt[3]{1.5} \]
Approximating \( \sqrt[3]{1.5} \):
\[ 1 + \frac{r}{100} \approx 1.1447 \]
Therefore,
\[ \frac{r}{100} \approx 0.1447 \quad \Rightarrow \quad r \approx 14.47\% \]

7. **Substituting \( r \) back into one of the equations**:
Now substituting \( r \) back into equation (1) to find \( P \):
\[ 6690 = P \left(1 + \frac{14.47}{100}\right)^3 \]
\[ 6690 = P \left(1.1447\right)^3 \]
\[ 6690 = P \times 1.5 \]
\[ P = \frac{6690}{1.5} = 4460 \]

8. **Final answer**:
The sum of money deposited is \( P = Rs. 4460 \).