A principal gets increased by 21% in 2 years at compound interest. What will be the simple interest on Rs. 1000 for 4 years at the same rate of interest ?

Options

A. Rs. 320
B. Rs. 400
C. Rs. 360
D. Rs. 420

Answer

To solve the problem step by step, we will first determine the rate of interest from the information given about the compound interest, and then we will calculate the simple interest.

### Step 1: Understand the compound interest increase
We know that the principal amount increases by 21% in 2 years due to compound interest. This means if we start with a principal amount of Rs. 100, after 2 years, it becomes Rs. 121.

### Step 2: Set up the equation for compound interest
Let the principal amount be \( P = 100X \) (where \( X \) is a common factor). After 2 years, the amount becomes:
\[ A = P \left(1 + \frac{r}{100}\right)^n \]
Substituting the values:
\[ 121X = 100X \left(1 + \frac{r}{100}\right)^2 \]

### Step 3: Simplify the equation
Dividing both sides by \( 100X \):
\[ \frac{121X}{100X} = \left(1 + \frac{r}{100}\right)^2 \]
This simplifies to:
\[ 1.21 = \left(1 + \frac{r}{100}\right)^2 \]

### Step 4: Take the square root
Taking the square root of both sides:
\[ \sqrt{1.21} = 1 + \frac{r}{100} \]
Calculating the square root:
\[ 1.1 = 1 + \frac{r}{100} \]

### Step 5: Solve for \( r \)
Subtracting 1 from both sides:
\[ 0.1 = \frac{r}{100} \]
Multiplying both sides by 100:
\[ r = 10\% \]

### Step 6: Calculate simple interest
Now, we will use the simple interest formula:
\[ SI = \frac{P \times r \times t}{100} \]
Where:
- \( P = 1000 \) (the principal)
- \( r = 10\% \) (the rate of interest)
- \( t = 4 \) (the time in years)

Substituting the values:
\[ SI = \frac{1000 \times 10 \times 4}{100} \]

### Step 7: Simplify the calculation
Calculating the simple interest:
\[ SI = \frac{40000}{100} = 400 \]

### Final Answer
The simple interest on Rs. 1000 for 4 years at the same rate of interest is **Rs. 400**.

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