Question

In a college there are 150 students out of which 60% of the students play cricket, 30% of the students play football and 40% of the students play hockey. If it is know that no student plays all three sports and all the students play at least one sport, then what is the number of students that play exactly one sport?

• A. 45
• B. 65
• C. 105
• D. 85
• E. 75

Hint:

For Venn diagram problems, first list all the given information systematically. Use the total union formula to find the missing sum of intersections (like \(I_2\)). Then, use the formula for 'exactly one' or 'exactly two' regions. Drawing a Venn diagram can also help visualize the relationships.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem involves the application of set theory, specifically the Principle of Inclusion-Exclusion for three sets. We are given the total number of students and the number of students in each of three sets (sports), along with information about their intersections.
Step 2: Key Formula or Approach:
Let C, F, and H be the sets of students playing Cricket, Football, and Hockey.
1. Total students: \(n(C \cup F \cup H) = n(C) + n(F) + n(H) - (n(C \cap F) + n(F \cap H) + n(H \cap C)) + n(C \cap F \cap H)\)
2. Students playing exactly one sport: \(n(\text{exactly one}) = n(C \cup F \cup H) - n(\text{exactly two}) - n(\text{exactly three})\)
Alternatively, a more direct formula:
\(n(\text{exactly one}) = n(C)+n(F)+n(H) - 2(n(C \cap F) + n(F \cap H) + n(H \cap C)) + 3n(C \cap F \cap H)\)
Step 3: Detailed Explanation:
Total number of students = 150.
Since all students play at least one sport, \(n(C \cup F \cup H) = 150\).
Calculate the number of students playing each sport:
Number playing Cricket, \(n(C) = 60% \text{ of } 150 = 0.60 \times 150 = 90\).
Number playing Football, \(n(F) = 30% \text{ of } 150 = 0.30 \times 150 = 45\).
Number playing Hockey, \(n(H) = 40% \text{ of } 150 = 0.40 \times 150 = 60\).
Given information about intersections:
No student plays all three sports: \(n(C \cap F \cap H) = 0\).
Let \(I_2 = n(C \cap F) + n(F \cap H) + n(H \cap C)\) represent the sum of students playing in pairs of sports.
Using the Principle of Inclusion-Exclusion: \[ n(C \cup F \cup H) = n(C) + n(F) + n(H) - I_2 + n(C \cap F \cap H) \] Substitute the known values: \[ 150 = 90 + 45 + 60 - I_2 + 0 \] \[ 150 = 195 - I_2 \] \[ I_2 = 195 - 150 = 45 \] Now, we find the number of students playing exactly one sport. Let this be \(N_1\). \[ N_1 = n(C) + n(F) + n(H) - 2 \times I_2 + 3 \times n(C \cap F \cap H) \] Substitute the values we have: \[ N_1 = (90 + 45 + 60) - 2 \times (45) + 3 \times (0) \] \[ N_1 = 195 - 90 + 0 \] \[ N_1 = 105 \] Step 4: Final Answer:
The number of students that play exactly one sport is 105.