The C.I at a certain rate for Rs 'X' for 2 years is same as the S.I at the same rate for same Rs 'X' for four years. Find the rate. (in % p.a)

Options

A. 200
B. 300
C. 400
D. 500

Answer

To solve the problem, we need to equate the Compound Interest (C.I) for Rs. X at a certain rate for 2 years with the Simple Interest (S.I) at the same rate for Rs. X for 4 years. Let's go through the solution step by step.

### Step 1: Write the formulas for Simple Interest and Compound Interest
- The formula for Simple Interest (S.I) is:
\[ S.I = \frac{P \times R \times T}{100} \]
- The formula for Compound Interest (C.I) for 2 years is:
\[ C.I = P \left(1 + \frac{R}{100}\right)^2 - P \]

### Step 2: Set up the equation
Since the C.I for 2 years is equal to the S.I for 4 years, we can set up the equation:
\[ C.I = S.I \]
Substituting the formulas we have:
\[ P \left(1 + \frac{R}{100}\right)^2 - P = \frac{P \times R \times 4}{100} \]

### Step 3: Simplify the equation
We can factor out \(P\) from both sides:
\[ \left(1 + \frac{R}{100}\right)^2 - 1 = \frac{R \times 4}{100} \]
This simplifies to:
\[ \left(1 + \frac{R}{100}\right)^2 = 1 + \frac{4R}{100} \]

### Step 4: Expand the left side
Expanding the left side gives:
\[ 1 + 2 \cdot \frac{R}{100} + \left(\frac{R}{100}\right)^2 = 1 + \frac{4R}{100} \]

### Step 5: Cancel out the 1's and rearrange
Cancelling 1 from both sides:
\[ 2 \cdot \frac{R}{100} + \left(\frac{R}{100}\right)^2 = \frac{4R}{100} \]
Rearranging gives:
\[ \left(\frac{R}{100}\right)^2 - 2 \cdot \frac{R}{100} = 0 \]

### Step 6: Factor the equation
Factoring out \(\frac{R}{100}\):
\[ \frac{R}{100} \left(\frac{R}{100} - 2\right) = 0 \]

### Step 7: Solve for R
This gives us two solutions:
1. \(\frac{R}{100} = 0\) which implies \(R = 0\) (not a valid interest rate)
2. \(\frac{R}{100} - 2 = 0\) which implies \(R = 200\)

### Conclusion
Thus, the rate \(R\) is \(200\%\) per annum.