The ratio of the amount for two years under CI annually and for one year under SI is 6:5. When the rate of interest is same, then the value of rate of interest is:

Options

A. `12.5%`
B. `18%`
C. `20%`
D. `16.66%`

Answer

To solve the problem, we need to find the rate of interest when the ratio of the amount for two years under compound interest (CI) and for one year under simple interest (SI) is given as 6:5.

### Step-by-Step Solution:

1. **Understanding the Amounts**:
- Let the principal amount be \( P \).
- Let the rate of interest be \( R\% \) per annum.

2. **Calculating Amount under Simple Interest (SI)**:
- The formula for the amount under simple interest for one year is:
\[ A_{SI} = P + \frac{P \times R}{100} \]
- For one year, the amount is:
\[ A_{SI} = P + \frac{PR}{100} \]

3. **Calculating Amount under Compound Interest (CI)**:
- The formula for the amount under compound interest for two years is:
\[ A_{CI} = P \left(1 + \frac{R}{100}\right)^2 \]
- Expanding this, we get:
\[ A_{CI} = P \left(1 + \frac{R}{100}\right) \left(1 + \frac{R}{100}\right) = P \left(1 + \frac{2R}{100} + \frac{R^2}{10000}\right) \]

4. **Setting Up the Ratio**:
- According to the problem, the ratio of the amounts is given as:
\[ \frac{A_{CI}}{A_{SI}} = \frac{6}{5} \]
- Substituting the expressions for \( A_{CI} \) and \( A_{SI} \):
\[ \frac{P \left(1 + \frac{2R}{100} + \frac{R^2}{10000}\right)}{P + \frac{PR}{100}} = \frac{6}{5} \]

5. **Cancelling \( P \)**:
- Since \( P \) is common in both the numerator and the denominator, we can cancel it out:
\[ \frac{1 + \frac{2R}{100} + \frac{R^2}{10000}}{1 + \frac{R}{100}} = \frac{6}{5} \]

6. **Cross Multiplying**:
- Cross-multiplying gives us:
\[ 5 \left(1 + \frac{2R}{100} + \frac{R^2}{10000}\right) = 6 \left(1 + \frac{R}{100}\right) \]

7. **Expanding Both Sides**:
- Expanding both sides:
\[ 5 + \frac{10R}{100} + \frac{5R^2}{10000} = 6 + \frac{6R}{100} \]

8. **Rearranging the Equation**:
- Rearranging gives:
\[ \frac{5R^2}{10000} + \frac{10R}{100} - \frac{6R}{100} + 5 - 6 = 0 \]
- Simplifying this:
\[ \frac{5R^2}{10000} + \frac{4R}{100} - 1 = 0 \]

9. **Multiplying Through by 10000**:
- To eliminate the fractions, multiply through by 10000:
\[ 5R^2 + 400R - 10000 = 0 \]

10. **Using the Quadratic Formula**:
- Using the quadratic formula \( R = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 5 \), \( b = 400 \), and \( c = -10000 \).
- Calculate the discriminant:
\[ b^2 - 4ac = 400^2 - 4 \times 5 \times (-10000) = 160000 + 200000 = 360000 \]
- Now, substituting into the formula:
\[ R = \frac{-400 \pm \sqrt{360000}}{2 \times 5} = \frac{-400 \pm 600}{10} \]
- This gives two potential solutions:
\[ R = \frac{200}{10} = 20 \quad \text{(valid)} \quad \text{and} \quad R = \frac{-1000}{10} = -100 \quad \text{(not valid)} \]

11. **Final Answer**:
- Therefore, the rate of interest \( R \) is \( 20\% \).