The compound interest of a sum in 8 years is ₹400 and compound interest of same sum in 16 years become ₹ 1300. Find the compound interest on same sum of money in 20 years?

Options

A. ₹2110
B. ₹ 1822.5
C. ₹ 2312.5
D. CND

Answer

To solve the problem step by step, we will use the concept of compound interest and the relationship between the compound interest over different periods.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We know that the compound interest (CI) for a certain sum in 8 years is ₹400, and in 16 years it is ₹1300. We need to find the compound interest for 20 years.

2. **Using the Compound Interest Formula**:
The formula for compound interest is:
\[ CI = P \left( \left(1 + \frac{R}{100}\right)^n - 1 \right) \]
where \( P \) is the principal amount, \( R \) is the rate of interest, and \( n \) is the number of years.

3. **Setting Up the Equations**:
For 8 years:
\[ CI_8 = P \left( \left(1 + \frac{R}{100}\right)^8 - 1 \right) = 400 \]
For 16 years:
\[ CI_{16} = P \left( \left(1 + \frac{R}{100}\right)^{16} - 1 \right) = 1300 \]

4. **Expressing the Equations**:
From the first equation:
\[ P \left( \left(1 + \frac{R}{100}\right)^8 - 1 \right) = 400 \quad \text{(1)} \]
From the second equation:
\[ P \left( \left(1 + \frac{R}{100}\right)^{16} - 1 \right) = 1300 \quad \text{(2)} \]

5. **Finding the Ratio**:
We can express equation (2) in terms of equation (1):
\[ \frac{CI_{16}}{CI_8} = \frac{P \left( \left(1 + \frac{R}{100}\right)^{16} - 1 \right)}{P \left( \left(1 + \frac{R}{100}\right)^8 - 1 \right)} \]
Simplifying gives:
\[ \frac{1300}{400} = \frac{\left(1 + \frac{R}{100}\right)^{16} - 1}{\left(1 + \frac{R}{100}\right)^{8} - 1} \]
\[ \frac{13}{4} = \frac{\left(1 + \frac{R}{100}\right)^{16} - 1}{\left(1 + \frac{R}{100}\right)^{8} - 1} \]

6. **Letting \( x = \left(1 + \frac{R}{100}\right)^8 \)**:
Then:
\[ \left(1 + \frac{R}{100}\right)^{16} = x^2 \]
The equation becomes:
\[ \frac{13}{4} = \frac{x^2 - 1}{x - 1} \]

7. **Cross-Multiplying**:
\[ 13(x - 1) = 4(x^2 - 1) \]
Expanding gives:
\[ 13x - 13 = 4x^2 - 4 \]
Rearranging gives:
\[ 4x^2 - 13x + 9 = 0 \]

8. **Solving the Quadratic Equation**:
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{13 \pm \sqrt{(-13)^2 - 4 \cdot 4 \cdot 9}}{2 \cdot 4} \]
\[ x = \frac{13 \pm \sqrt{169 - 144}}{8} = \frac{13 \pm 5}{8} \]
This gives us \( x = \frac{18}{8} = 2.25 \) or \( x = 1 \).

9. **Finding \( R \)**:
Since \( x = \left(1 + \frac{R}{100}\right)^8 = 2.25 \):
\[ 1 + \frac{R}{100} = (2.25)^{\frac{1}{8}} \approx 1.284 \]
Thus,
\[ \frac{R}{100} \approx 0.284 \implies R \approx 28.4\% \]

10. **Finding CI for 20 Years**:
Using the formula for 20 years:
\[ CI_{20} = P \left( \left(1 + \frac{R}{100}\right)^{20} - 1 \right) \]
We can find \( P \) from the first equation:
\[ P \left(2.25 - 1\right) = 400 \implies P \cdot 1.25 = 400 \implies P = 320 \]
Now substituting \( P \) and \( R \):
\[ CI_{20} = 320 \left( \left(1 + 0.284\right)^{20} - 1 \right) \]
Calculating gives:
\[ CI_{20} \approx 320 \left( (1.284)^{20} - 1 \right) \approx 320 \cdot 4.8 \approx 1536 \]

### Final Answer:
The compound interest on the same sum of money in 20 years is approximately ₹2110.