If the ratio of compound interest for 3 years and simple interest for 1 year on certain amount at certain rate of interest is 3.64:1 then find the rate of interest?

Options

A. `22%`
B. `21%`
C. `23%`
D. `20%`

Answer

To solve the problem, we need to find the rate of interest given that the ratio of compound interest (CI) for 3 years to simple interest (SI) for 1 year is 3.64:1.

### Step 1: Understand the formulas
- **Simple Interest (SI)** for 1 year is given by the formula:
\[ SI = \frac{P \times R \times T}{100} \]
where \( P \) is the principal amount, \( R \) is the rate of interest, and \( T \) is the time in years.

- **Compound Interest (CI)** for 3 years is given by the formula:
\[ CI = P \left(1 + \frac{R}{100}\right)^3 - P \]
This can be simplified to:
\[ CI = P \left[\left(1 + \frac{R}{100}\right)^3 - 1\right] \]

### Step 2: Set up the ratio
According to the problem, we know:
\[ \frac{CI}{SI} = \frac{3.64}{1} \]
Substituting the formulas for CI and SI, we get:
\[ \frac{P \left[\left(1 + \frac{R}{100}\right)^3 - 1\right]}{\frac{P \times R}{100}} = 3.64 \]
We can cancel \( P \) from both sides (assuming \( P \neq 0 \)):
\[ \frac{\left(1 + \frac{R}{100}\right)^3 - 1}{\frac{R}{100}} = 3.64 \]

### Step 3: Cross-multiply to eliminate the fraction
Cross-multiplying gives us:
\[ \left(1 + \frac{R}{100}\right)^3 - 1 = 3.64 \times \frac{R}{100} \]

### Step 4: Expand the left side
Using the binomial expansion:
\[ \left(1 + \frac{R}{100}\right)^3 = 1 + 3 \cdot \frac{R}{100} + 3 \cdot \left(\frac{R}{100}\right)^2 + \left(\frac{R}{100}\right)^3 \]
Thus,
\[ \left(1 + \frac{R}{100}\right)^3 - 1 = 3 \cdot \frac{R}{100} + 3 \cdot \left(\frac{R}{100}\right)^2 + \left(\frac{R}{100}\right)^3 \]

### Step 5: Substitute back into the equation
Now substituting this back into our equation:
\[ 3 \cdot \frac{R}{100} + 3 \cdot \left(\frac{R}{100}\right)^2 + \left(\frac{R}{100}\right)^3 = 3.64 \cdot \frac{R}{100} \]

### Step 6: Simplify the equation
We can multiply through by \( 100 \) to eliminate the fraction:
\[ 3R + 3R^2/100 + R^3/10000 = 3.64R \]
Rearranging gives:
\[ 3R + 3R^2/100 + R^3/10000 - 3.64R = 0 \]
\[ (3 - 3.64)R + 3R^2/100 + R^3/10000 = 0 \]
\[ -0.64R + 3R^2/100 + R^3/10000 = 0 \]

### Step 7: Multiply through by 10000 to clear denominators
\[ -6400R + 300R^2 + R^3 = 0 \]
This simplifies to:
\[ R^3 + 300R^2 - 6400R = 0 \]
Factoring out \( R \):
\[ R(R^2 + 300R - 6400) = 0 \]

### Step 8: Solve the quadratic equation
Now we can solve the quadratic equation:
\[ R^2 + 300R - 6400 = 0 \]
Using the quadratic formula:
\[ R = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-300 \pm \sqrt{300^2 + 4 \cdot 6400}}{2} \]
Calculating the discriminant:
\[ 300^2 + 4 \cdot 6400 = 90000 + 25600 = 115600 \]
Taking the square root:
\[ \sqrt{115600} = 340 \]
Thus:
\[ R = \frac{-300 \pm 340}{2} \]
Calculating the two possible values:
1. \( R = \frac{40}{2} = 20 \)
2. \( R = \frac{-640}{2} = -320 \) (not valid)

### Final Answer
The rate of interest \( R \) is \( 20\% \).

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