In a group of 50 students, 25 play hockey, 30 play football and 8 play neither of these games. Using basic set theory and Venn diagram reasoning, what is the number of students who play both hockey and football?

Introduction / Context:
This question is a classic application of set theory and Venn diagrams in aptitude tests. You are given the total number of students in a group, the number who play hockey, the number who play football and the number who play neither game. The objective is to find how many students play both hockey and football. Understanding the inclusion exclusion principle is essential for solving such problems efficiently.

Given Data / Assumptions:
- Total number of students in the group is 50. - Students who play hockey: 25. - Students who play football: 30. - Students who play neither hockey nor football: 8. - Each student can play none, one or both of the two games. - We must find the number of students who play both hockey and football.

Concept / Approach:
Let H be the set of students who play hockey and F be the set of students who play football. The number of students who play at least one of the two games is the size of the union H ∪ F. From set theory, n(H ∪ F) = n(H) + n(F) - n(H ∩ F), where H ∩ F represents students who play both games. Additionally, students who play at least one game plus students who play neither must equal the total number of students. Using these relationships, we first compute n(H ∪ F) and then solve for n(H ∩ F).

Step-by-Step Solution:
Step 1: Let x be the number of students who play both hockey and football, that is x = n(H ∩ F). Step 2: Number of students who play at least one game is n(H ∪ F) = total students - students who play neither = 50 - 8 = 42. Step 3: Use the inclusion exclusion formula: n(H ∪ F) = n(H) + n(F) - n(H ∩ F). Step 4: Substitute known values: 42 = 25 + 30 - x. Step 5: Simplify the right side: 25 + 30 = 55, so 42 = 55 - x. Step 6: Rearrange to find x: x = 55 - 42 = 13.

Verification / Alternative check:
We can verify by reconstructing the Venn diagram numerically. In the hockey circle there are 25 students, of which 13 also play football. Therefore, only hockey players count is 25 - 13 = 12. In the football circle there are 30 students, of which 13 also play hockey. Therefore, only football players count is 30 - 13 = 17. Now total students who play at least one game should be 12 + 13 + 17 = 42, which matches 50 - 8. Adding 8 students who play neither gives 42 + 8 = 50, the total group size. All numbers are consistent, confirming that 13 is correct.

Why Other Options Are Wrong:
- If 10, 11 or 12 students played both games, the count of students who play at least one game would not equal 42 and would conflict with the given data. - If 9 students played both games, the union size would be 25 + 30 - 9 = 46, which is inconsistent with 50 total students and 8 non participants. Only 13 makes all counts align with the total and the given numbers for each game.

Common Pitfalls:
A major mistake is to add the two game counts directly, 25 + 30, without subtracting the intersection. This double counts students who play both games. Another common error is to confuse students who play neither with students who play exactly one game. Drawing a simple Venn diagram and labeling the regions with variables often makes the relationships much clearer and reduces the chance of algebra mistakes.

Final Answer:
The number of students who play both hockey and football is 13.