In a group of 110 students, 23 students did not participate in any of the two games: Badminton and Chess. 45 students participated in Badminton and 61 students participated in Chess. How many students participated in Badminton only?
26
Understanding the Student Games Participation
This question involves analyzing data about a group of students and their participation in two games: Badminton and Chess. We are given the total number of students, the number who did not participate in any game, the number who participated in Badminton, and the number who participated in Chess. We need to find out how many students participated only in Badminton.
Calculating Total Participants
First, let's find out how many students participated in at least one of the two games. We know the total number of students and the number who did not participate in any game.
Total students = 110
Students not participating in any game = 23
Students participating in at least one game = Total students - Students not participating in any game
Students participating in at least one game = 110 - 23 = 87
So, 87 students participated in either Badminton, Chess, or both.
Finding Students in Both Games
We are given the number of students who played Badminton and the number who played Chess. The group of 87 students is the union of those who played Badminton and those who played Chess. We can use the principle of inclusion-exclusion to find the number of students who participated in both games.
Let B be the set of students who participated in Badminton.
Let C be the set of students who participated in Chess.
We are given:
- |B| (Students in Badminton) = 45
- |C| (Students in Chess) = 61
- |B ∪ C| (Students in at least one game) = 87
The formula for the union of two sets is:
\(|B \cup C| = |B| + |C| - |B \cap C|\)
Where \(|B \cap C|\) is the number of students who participated in both Badminton and Chess.
We can rearrange the formula to find \(|B \cap C|\):
\(|B \cap C| = |B| + |C| - |B \cup C|\)
Now, substitute the values we have:
\(|B \cap C| = 45 + 61 - 87\)
\(|B \cap C| = 106 - 87\)
\(|B \cap C| = 19\)
So, 19 students participated in both Badminton and Chess.
Calculating Students in Badminton Only
We want to find the number of students who participated in Badminton only. This includes students who played Badminton but not Chess. This can be found by subtracting the number of students who played both games from the total number of students who played Badminton.
Students in Badminton only = (Total students in Badminton) - (Students in both Badminton and Chess)
Students in Badminton only = |B| - |B ∩ C|
Students in Badminton only = 45 - 19
Students in Badminton only = 26
Therefore, 26 students participated in Badminton only.
Summary of Steps
Let's summarize the steps taken:
- Calculated the total number of students participating in at least one game.
- Used the inclusion-exclusion principle to find the number of students participating in both games.
- Subtracted the number of students in both games from the total number in Badminton to find those in Badminton only.
Final Answer for Badminton Only Participants
Based on our calculations, the number of students who participated in Badminton only is 26.
| Category | Number of Students |
|---|---|
| Total Students | 110 |
| Did not participate in any game | 23 |
| Participated in at least one game (Badminton ∪ Chess) | 87 |
| Participated in Badminton (|B|) | 45 |
| Participated in Chess (|C|) | 61 |
| Participated in both Badminton and Chess (|B ∩ C|) | 19 |
| Participated in Badminton only (|B| - |B ∩ C|) | 26 |
Revision Table: Key Concepts for Set Problems
| Concept | Explanation | Formula (for two sets A and B) |
|---|---|---|
| Union (A ∪ B) | Elements in A or B or both. Represents participation in at least one activity. | \(|A \cup B| = |A| + |B| - |A \cap B|\) |
| Intersection (A ∩ B) | Elements common to both A and B. Represents participation in both activities. | \(|A \cap B| = |A| + |B| - |A \cup B|\) |
| Only A | Elements in A but not in B. Represents participation only in activity A. | \(|A \text{ only}| = |A| - |A \cap B|\) |
| Neither A nor B | Elements outside both A and B. Represents no participation in either activity. | Total - \(|A \cup B|\) |
Additional Information: Using Venn Diagrams
Problems like this can also be visualized using Venn diagrams. A Venn diagram for two sets (Badminton and Chess) would have two overlapping circles within a rectangle representing the total group.
- The overlapping region represents students in both games (|B ∩ C|). We found this is 19.
- The part of the Badminton circle that does not overlap represents students in Badminton only (|B| - |B ∩ C|). We found this is 45 - 19 = 26.
- The part of the Chess circle that does not overlap represents students in Chess only (|C| - |B ∩ C|). This would be 61 - 19 = 42.
- The region outside the circles within the rectangle represents students in neither game. We were given this is 23.
Checking the numbers: Students in Badminton only + Students in Chess only + Students in both + Students in neither = Total 26 + 42 + 19 + 23 = 110. This matches the total number of students, confirming our calculations.
Important Questions from Venn Diagram Problems
In a class of 100 students, every student has passed in one or more of the three subjects, i.e History, Economics and English. Among all the student, 24 students have passed in English only, 14 students have passed in History only 11 students have passed in both English and Economics only, and 12 students have passed in both English and History only. A total of 50 students have passed in History. If only 5 students have passed in all three subjects, then how many students have passed in Economies only?
60 students participated in one or more of the three competitions, i. e. Quiz, Extempore and Debate. A total of 22 students participated either in Quiz only or in Extempore only. 4 students participated in all three competitions. A total of 14 students participated in any of the two competitions only. How many students participated in Debated only?
In a class of 75 students, 40 students participate in Cricket, 28 students participate in Hockey, and 12 students participate in both Cricket and Hockey, whereas 19 students do not participate in any of the two sports. How many students participate only in Hockey?
How many students like french?
A. 30
B. 35
C. 40
D. 45
What is the ratio of students who like Spanish to those who like German?
A. 2/3
B. 1/2
C. 3/2
D. 4/9
