The simple interest on a certain sum of money invested at a certain rate for 2 years amounts to Rs. 1200. The compound interest on the same sum of money invested at the same rate of interest for 2 years amounts to Rs. 1290. What was the principal ?
A. Rs. 12000
B. Rs. 16000
C. Rs. 6000
D. Rs. 4000

Options

A. C
B. A
C. D
D. B

Answer

To find the principal amount based on the given simple interest and compound interest, we can follow these steps:

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

- The formula for Compound Interest (CI) is:
\[ CI = P \left(1 + \frac{R}{100}\right)^T - P \]

### Step 2: Set up the equations
From the problem, we have:
1. Simple Interest for 2 years = Rs. 1200
\[ \frac{P \times R \times 2}{100} = 1200 \quad \text{(Equation 1)} \]

2. Compound Interest for 2 years = Rs. 1290
\[ P \left(1 + \frac{R}{100}\right)^2 - P = 1290 \quad \text{(Equation 2)} \]

### Step 3: Solve Equation 1 for \( R \)
From Equation 1:
\[ P \times R \times 2 = 1200 \times 100 \]
\[ P \times R = 60000 \quad \text{(Equation 3)} \]
Thus, we can express \( R \) as:
\[ R = \frac{60000}{P} \]

### Step 4: Substitute \( R \) in Equation 2
Substituting \( R \) from Equation 3 into Equation 2:
\[ P \left(1 + \frac{60000}{P \times 100}\right)^2 - P = 1290 \]
This simplifies to:
\[ P \left(1 + \frac{600}{P}\right)^2 - P = 1290 \]
\[ P \left(\frac{P + 600}{P}\right)^2 - P = 1290 \]
\[ P \left(\frac{(P + 600)^2}{P^2}\right) - P = 1290 \]
\[ \frac{(P + 600)^2}{P} - P = 1290 \]
\[ \frac{(P + 600)^2 - P^2}{P} = 1290 \]
\[ \frac{P^2 + 1200P + 360000 - P^2}{P} = 1290 \]
\[ \frac{1200P + 360000}{P} = 1290 \]
\[ 1200 + \frac{360000}{P} = 1290 \]
\[ \frac{360000}{P} = 90 \]
\[ P = \frac{360000}{90} \]
\[ P = 4000 \]

### Conclusion
The principal amount is Rs. 4000.