To solve for the time \( t \) using the formula \( I = P \times R \times t \), where:

- \( I = 7280 \) (Interest)
- \( P = 26000 \) (Principal)
- \( R = 0.07 \) (Rate of interest)

We can rearrange the formula to solve for \( t \):

\[
t = \frac{I}{P \times R}
\]

Substituting the values:

\[
t = \frac{7280}{26000 \times 0.07}
\]

Let's compute this.

\[
t = \frac{7280}{1820} = 4
\]

So, \( t = 4 \) years.

### Would you like more details or have any questions about this calculation?

Here are 5 related questions:
1. What would the time \( t \) be if the interest were different, say \( I = 10,000 \)?
2. How would the formula change if the interest rate were compounded annually?
3. What happens to the time \( t \) if the principal \( P \) is doubled?
4. Can we use the same formula for monthly interest rates?
5. How would you calculate \( t \) if the interest rate \( R \) were variable over time?

**Tip**: Always double-check the units for rate and time (e.g., annual rate with years). Misalignment of units can lead to incorrect results.

Learn how to calculate the time period needed to generate an interest of 7280 using the simple interest formula I = P × R × t, with principal 26000 and an interest rate of 0.07. This problem focuses on basic algebra and simple interest calculations, ideal for students in Grades 9-11.

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