1. **Problem Statement:** A survey of 120 students found 60 liked cricket (C), 55 liked basketball (B), and 20 liked neither. 2. **Formula and Rules:** The union of two sets is given by: $$n(B \cup C) = n(B) + n(C) - n(B \cap C)$$ where $n(B \cap C)$ is the number of students who like both games. 3. **Calculate $n(B \cup C)$:** Since 20 students like neither, the number who like at least one game is: $$n(B \cup C) = 120 - 20 = 100$$ 4. **Find $n(B \cap C)$:** Using the union formula: $$100 = 55 + 60 - n(B \cap C)$$ $$n(B \cap C) = 55 + 60 - 100 = 15$$ 5. **Find number who like only cricket:** $$n(\text{only cricket}) = n(C) - n(B \cap C) = 60 - 15 = 45$$ 6. **Compare students who like both games and those who like neither:** Number who like both = 15 Number who like neither = 20 So, more students like neither than both. 7. **Venn Diagram Information:** - Only Cricket: 45 - Only Basketball: $n(B) - n(B \cap C) = 55 - 15 = 40$ - Both: 15 - Neither: 20 **Final answers:** - a) $n(B \cup C) = 100$ - b) Venn diagram as above - c) Number who like only cricket = 45 - d) Students liking both (15) are fewer than those liking neither (20)