Let the principal be \( P \) and the rate of interest be \( R \).
Using the formula for Simple Interest:
\[
\text{Amount} = P + \frac{P \times R \times T}{100}
\]
For 2 years, we have:
\[
720 = P + \frac{P \times R \times 2}{100} \quad \text{(Equation 1)}
\]
For 7 years, we have:
\[
1020 = P + \frac{P \times R \times 7}{100} \quad \text{(Equation 2)}
\]
Subtract Equation 1 from Equation 2:
\[
1020 - 720 = \left( \frac{P \times R \times 7}{100} \right) - \left( \frac{P \times R \times 2}{100} \right)
\]
\[
300 = \frac{P \times R \times 5}{100}
\]
\[
300 \times 100 = P \times R \times 5
\]
\[
30000 = P \times R \times 5
\]
\[
P \times R = 6000 \quad \text{(Equation 3)}
\]
Substitute this value into Equation 1:
\[
720 = P + \frac{6000 \times 2}{100}
\]
\[
720 = P + 120
\]
\[
P = 600
\]
Substitute \( P = 600 \) into Equation 3:
\[
600 \times R = 6000
\]
\[
R = 10% \text{ per annum}
\]