Explanation:

- The sum grows from Rs1,250 to Rs1,458 in 3 years.

- Let's denote the original sum as \( P \).

- In the first year: the amount is Rs1,250.

- By the third year: the amount is Rs1,458.

- The amount after each year can be expressed using \( A = P(1 + r)^n \).

- In the first year: \( 1250 = P(1+r) \).

- In the third year: \( 1458 = P(1+r)^3 \).

Now let's solve:

- Dividing, \(\frac{1458}{1250} = \frac{P(1+r)^3}{P(1+r)} \)

- Simplifies to \( \frac{1458}{1250} = (1+r)^2 \)

- This allows us to find \( r \), then apply it to simple interest calculations:

Let's calculate the simple interest:

- Simple Interest formula: \( SI = \frac{P \times r \times 2}{100} \)

- Use \( r \) from the compound calculation for SI.

Correct Answer:

- Option 1: Rs600

- Option 2: Rs520

- Option 3: Rs500

- Option 4: Rs480

Your Conclusion:

- Correct Answer: Option 3: Rs500