The ratio of CI for 3 years and SI for 1 year for a fixed amount at a rate of `r%` is 3.64. What is the value of r?

Options

A. `10%`
B. `15%`
C. `20%`
D. none of these

Answer

To solve the problem, we need to find the value of the rate \( r \% \) given that the ratio of Compound Interest (CI) for 3 years to Simple Interest (SI) for 1 year is 3.64.

### Step-by-Step Solution:

1. **Understanding Simple Interest (SI)**:
The formula for Simple Interest for 1 year is:
\[ SI = \frac{P \times r \times t}{100} \]
where \( P \) is the principal amount, \( r \) is the rate of interest, and \( t \) is the time in years. For 1 year, it simplifies to:
\[ SI = \frac{P \times r}{100} \]

2. **Understanding Compound Interest (CI)**:
The formula for Compound Interest for 3 years is:
\[ CI = P \left(1 + \frac{r}{100}\right)^3 - P \]
This can be simplified to:
\[ CI = P \left[\left(1 + \frac{r}{100}\right)^3 - 1\right] \]

3. **Setting up the ratio**:
According to the problem, the ratio of CI for 3 years to SI for 1 year is given as:
\[ \frac{CI}{SI} = 3.64 \]
Substituting the formulas we derived:
\[ \frac{P \left[\left(1 + \frac{r}{100}\right)^3 - 1\right]}{\frac{P \times r}{100}} = 3.64 \]
Simplifying this gives:
\[ \frac{\left(1 + \frac{r}{100}\right)^3 - 1}{\frac{r}{100}} = 3.64 \]

4. **Cross-multiplying**:
Cross-multiplying to eliminate the fraction:
\[ \left(1 + \frac{r}{100}\right)^3 - 1 = 3.64 \cdot \frac{r}{100} \]

5. **Expanding the left side**:
Using the binomial expansion:
\[ \left(1 + \frac{r}{100}\right)^3 = 1 + 3 \cdot \frac{r}{100} + 3 \cdot \left(\frac{r}{100}\right)^2 + \left(\frac{r}{100}\right)^3 \]
Therefore,
\[ \left(1 + \frac{r}{100}\right)^3 - 1 = 3 \cdot \frac{r}{100} + 3 \cdot \left(\frac{r}{100}\right)^2 + \left(\frac{r}{100}\right)^3 \]
Setting this equal to \( 3.64 \cdot \frac{r}{100} \):
\[ 3 \cdot \frac{r}{100} + 3 \cdot \left(\frac{r}{100}\right)^2 + \left(\frac{r}{100}\right)^3 = 3.64 \cdot \frac{r}{100} \]

6. **Rearranging the equation**:
Rearranging gives:
\[ 3 \cdot \left(\frac{r}{100}\right)^2 + \left(\frac{r}{100}\right)^3 = 0.64 \cdot \frac{r}{100} \]
Dividing through by \( \frac{r}{100} \) (assuming \( r \neq 0 \)):
\[ 3 \cdot \frac{r}{100} + \left(\frac{r}{100}\right)^2 = 0.64 \]

7. **Letting \( x = \frac{r}{100} \)**:
Substituting \( x \) gives:
\[ 3x + x^2 = 0.64 \]
Rearranging into standard quadratic form:
\[ x^2 + 3x - 0.64 = 0 \]

8. **Using the quadratic formula**:
The quadratic formula is:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = 3, c = -0.64 \):
\[ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-0.64)}}{2 \cdot 1} \]
\[ x = \frac{-3 \pm \sqrt{9 + 2.56}}{2} \]
\[ x = \frac{-3 \pm \sqrt{11.56}}{2} \]
\[ x = \frac{-3 \pm 3.4}{2} \]

9. **Calculating the values**:
This gives two potential solutions:
\[ x = \frac{0.4}{2} = 0.2 \quad \text{(valid solution)} \]
\[ x = \frac{-6.4}{2} = -3.2 \quad \text{(not valid)} \]

10. **Finding \( r \)**:
Since \( x = \frac{r}{100} \), we have:
\[ \frac{r}{100} = 0.2 \implies r = 20 \]

### Final Answer:
The value of \( r \) is \( 20\% \).