Two equal sums of money were invested, one at 4% and the other at 4.5%. At the end of 7 years, the simple interest received from the latter exceeded to that received from the former by 31.50. Each sum was

Options

A. 1200
B. 600
C. 750
D. 900

Answer

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

\[ \text{SI} = \frac{P \times R \times T}{100} \]

Where:
- \( P \) = Principal amount (the sum of money invested)
- \( R \) = Rate of interest per annum
- \( T \) = Time in years

Let’s denote the equal sums of money invested as \( P \).

### Step 1: Calculate the Simple Interest for both investments

1. **For the investment at 4%**:
\[ \text{SI}_1 = \frac{P \times 4 \times 7}{100} = \frac{28P}{100} = 0.28P \]

2. **For the investment at 4.5%**:
\[ \text{SI}_2 = \frac{P \times 4.5 \times 7}{100} = \frac{31.5P}{100} = 0.315P \]

### Step 2: Set up the equation based on the information given

According to the problem, the simple interest from the investment at 4.5% exceeds that from the investment at 4% by 31.50. Therefore, we can set up the equation:

\[ \text{SI}_2 - \text{SI}_1 = 31.50 \]

Substituting the values we calculated:

\[ 0.315P - 0.28P = 31.50 \]

### Step 3: Simplify the equation

Now, simplify the left side:

\[ 0.035P = 31.50 \]

### Step 4: Solve for \( P \)

To find \( P \), divide both sides by 0.035:

\[ P = \frac{31.50}{0.035} \]

Calculating the right side gives:

\[ P = 900 \]

### Conclusion

Each sum invested was \( \text{Rs. } 900 \).

---