In a class of 100 students for a social activity, 50 liked to go for tree planting and 70 liked to go for cleanliness drive. 10 students did not like either. How many liked both?

Correct option is C

Given:

- Total students in the class: N = 100
- Students who liked tree planting: $$|T|$$​ = 50
- Students who liked cleanliness drive: |C| = 70
- Students who did not like either activity: 10 .

Formula Used:

Using the principle of inclusion and exclusion for sets:
​$$|T \cup C| = |T| + |C| - |T \cap C|$$​

- Where: $$|T \cup C|$$​ is the number of students who liked at least one activity.
- $$|T \cap C|$$​ is the number of students who liked both activities.

Solution:

1. Calculate the number of students who liked at least one activity:
​$$|T \cup C| = N - (\text{students who did not like either})$$​
​$$|T \cup C| = 100 - 10 = 90$$​

2. Substitute the known values into the formula:
​$$90 = 50 + 70 - |T \cap C|$$​

3. Solve for $$|T \cap C|$$​ :
​$$|T \cap C| = 50 + 70 - 90 = 30$$​

Final Answer:

Option C: 30