Reena took a loan of₹ 1200 with simple interest for as many years as the rate of interest. If she paid ₹432 as interest at the end of the loan period, what was the rate of interest ?

Options

A. `3.6`
B. 18
C. 6
D. None of these

Answer

To solve the problem step by step, we will use the formula for Simple Interest (SI):

**Step 1: Understand the formula for Simple Interest.**
The formula for Simple Interest is given by:
\[ \text{SI} = \frac{P \times R \times T}{100} \]
Where:
- \( P \) = Principal amount (the initial loan amount)
- \( R \) = Rate of interest (in percentage)
- \( T \) = Time (in years)

**Step 2: Identify the known values.**
From the question:
- Principal amount \( P = ₹1200 \)
- Total interest paid \( SI = ₹432 \)
- Time \( T \) is equal to the rate \( R \).

**Step 3: Substitute the known values into the formula.**
Since \( T = R \), we can replace \( T \) in the formula:
\[ 432 = \frac{1200 \times R \times R}{100} \]
This simplifies to:
\[ 432 = \frac{1200R^2}{100} \]

**Step 4: Simplify the equation.**
Multiply both sides by 100 to eliminate the fraction:
\[ 43200 = 1200R^2 \]

**Step 5: Divide both sides by 1200.**
\[ R^2 = \frac{43200}{1200} \]
Calculating the right side:
\[ R^2 = 36 \]

**Step 6: Take the square root of both sides.**
\[ R = \sqrt{36} \]
Thus,
\[ R = 6 \]

**Step 7: Conclusion.**
The rate of interest \( R \) is 6%.