Let's solve this step by step:

1. Total Number of Calculators in Both Sets: The school owns 2 sets of 30 calculators each. $\text{Total number of calculators} = 30 \times 2 = 60$

2. Unavailable Calculators: There are 2 calculators from one set and 6 calculators from the other set being repaired. $\text{Total unavailable calculators} = 2 + 6 = 8$

3. Available Calculators: Subtract the unavailable calculators from the total number of calculators. $\text{Available calculators} = 60 - 8 = 52$

4. Determine the Class Periods: The key question is to determine the number of students in each class period. If the number of students in any class period exceeds 52, then there won't be enough calculators for each student to use one without sharing.

Without knowing the exact number of students in each class period, the answer depends on whether any class has more than 52 students.

Conclusion:

• If any class period has more than 52 students, then there won't be enough calculators for that period.
• If no class period has more than 52 students, then there will be enough calculators for each student in every class period.

If you provide the number of students in each class period, I can give a more specific answer.

Would you like more details or have any questions?

Here are some related questions:

1. How could the school distribute the available calculators among students if the number exceeds 52?
2. What are alternative methods to ensure each student has a calculator if the number exceeds 52?
3. How does this scenario change if additional calculators are lost or repaired?
4. What strategies could be implemented to avoid such shortages in the future?
5. What other supplies might face similar issues, and how could they be managed?

Tip: Always check the exact requirements (like student count) to ensure you can meet the demand with available resources.