A university awarded medals in basketball, football, and volleyball. Only x students (x < 6) got medal in all the three sports and the medals went to a total of 15x students. It awarded 5x medals in basketball, (4x + 15) medals in football and (x + 25) medals in volleyball.
How many received medals in exactly one of three sports ?
21x − 40
Understanding the University Sports Medals Problem
This problem involves analyzing the distribution of medals awarded by a university in three sports: basketball, football, and volleyball. We are given information about the total number of students receiving medals, the number of medals awarded in each sport, and the number of students who received medals in all three sports. Our goal is to determine the number of students who received a medal in exactly one of these three sports.
Defining the Sets and Given Information
Let's define sets to represent the students who received medals in each sport:
- Let \(B\) be the set of students who received a medal in basketball.
- Let \(F\) be the set of students who received a medal in football.
- Let \(V\) be the set of students who received a medal in volleyball.
We are given the following information:
- Total number of students who received at least one medal (union of the sets): \(|B \cup F \cup V| = 15x\)
- Number of medals in basketball: \(|B| = 5x\)
- Number of medals in football: \(|F| = 4x + 15\)
- Number of medals in volleyball: \(|V| = x + 25\)
- Number of students who received medals in all three sports (intersection of all three sets): \(|B \cap F \cap V| = x\)
We are also told that \(x < 6\), but this information is not needed to find the expression for the number of students who received medals in exactly one sport.
Relating Medal Counts to Students by Category
Let's categorize the students based on the number of sports they received medals in:
- Let \(N_1\) be the number of students who received medals in exactly one sport.
- Let \(N_2\) be the number of students who received medals in exactly two sports.
- Let \(N_3\) be the number of students who received medals in exactly three sports.
We know from the problem that \(N_3 = |B \cap F \cap V| = x\).
The total number of students who received at least one medal is the sum of students in these categories:
\(|B \cup F \cup V| = N_1 + N_2 + N_3\)
Substituting the given values:
\(15x = N_1 + N_2 + x\)
This gives us a relationship between \(N_1\) and \(N_2\):
\(N_1 + N_2 = 15x - x = 14x\) (Equation 1)
Now consider the sum of the number of medals in each individual sport. When we sum \(|B|\), \(|F|\), and \(|V|\), students with exactly one medal (\(N_1\)) are counted once, students with exactly two medals (\(N_2\)) are counted twice, and students with exactly three medals (\(N_3\)) are counted three times.
So, the sum of individual medal counts is:
\(|B| + |F| + |V| = N_1 + 2N_2 + 3N_3\)
Substituting the given values:
\((5x) + (4x + 15) + (x + 25) = N_1 + 2N_2 + 3(x)\)
\(10x + 40 = N_1 + 2N_2 + 3x\)
Rearranging this equation gives us another relationship between \(N_1\) and \(N_2\):
\(N_1 + 2N_2 = 10x + 40 - 3x = 7x + 40\) (Equation 2)
Solving for Students with Exactly One Sport (\(N_1\))
We now have a system of two linear equations with two variables (\(N_1\) and \(N_2\)):
- \(N_1 + N_2 = 14x\)
- \(N_1 + 2N_2 = 7x + 40\)
To find \(N_1\), we can subtract Equation 1 from Equation 2:
\((N_1 + 2N_2) - (N_1 + N_2) = (7x + 40) - (14x)\)
\(N_1 + 2N_2 - N_1 - N_2 = 7x + 40 - 14x\)
\(N_2 = -7x + 40\)
Now substitute this value of \(N_2\) back into Equation 1:
\(N_1 + N_2 = 14x\)
\(N_1 + (-7x + 40) = 14x\)
\(N_1 - 7x + 40 = 14x\)
Add \(7x\) to both sides:
\(N_1 + 40 = 14x + 7x\)
\(N_1 + 40 = 21x\)
Subtract 40 from both sides to find \(N_1\):
\(N_1 = 21x - 40\)
Thus, the number of students who received medals in exactly one of the three sports is \(21x - 40\).
| Category of Students | Representation | Value based on x |
|---|---|---|
| Exactly one sport (\(N_1\)) | Students in B or F or V only | \(21x - 40\) |
| Exactly two sports (\(N_2\)) | Students in \(B \cap F\) or \(B \cap V\) or \(F \cap V\) only | \(40 - 7x\) |
| Exactly three sports (\(N_3\)) | Students in \(B \cap F \cap V\) | \(x\) |
| Total students with medals | \(N_1 + N_2 + N_3\) or \(|B \cup F \cup V|\) | \((21x - 40) + (40 - 7x) + x = 15x\) (Matches given) |
| Sum of individual medals | \(|B| + |F| + |V|\) or \(N_1 + 2N_2 + 3N_3\) | \((5x) + (4x+15) + (x+25) = 10x + 40\) or \((21x - 40) + 2(40 - 7x) + 3x = 21x - 40 + 80 - 14x + 3x = 10x + 40\) (Matches) |
The number of students who received medals in exactly one sport is \(21x - 40\).
Revision Table: University Sports Medals
| Concept | Description | Relevance to Problem |
|---|---|---|
| Set Theory | Mathematical theory dealing with collections of objects (sets). | Used to represent groups of students getting medals in specific sports. |
| Union of Sets (\(A \cup B \cup C\)) | The set of all elements in A, or B, or C, or any combination. | Represents total students with at least one medal (\(15x\)). |
| Intersection of Sets (\(A \cap B \cap C\)) | The set of elements common to A, B, and C. | Represents students with medals in all three sports (\(x\)). |
| Inclusion-Exclusion Principle | A counting technique to find the number of elements in a union of sets. | Used implicitly to relate total medal counts and category counts (\(N_1, N_2, N_3\)). |
| Exactly One Sport | Students in one sport's set, but not in the intersection with any other sport. | The value we were asked to calculate (\(N_1\)). |
Additional Information: Understanding Set Theory for Medal Problems
Problems like the university sports medals question are classic examples that can be solved using basic principles of set theory, often visualized with Venn diagrams. While we used algebraic manipulation of formulas derived from set principles, a Venn diagram helps understand the different regions representing students receiving medals in exactly one, exactly two, or exactly three sports.
- A Venn diagram for three sets (Basketball, Football, Volleyball) would have 8 distinct regions.
- The center region represents students in \(B \cap F \cap V\) (Exactly three sports, \(N_3\)).
- The three regions where circles overlap in pairs represent students in \(B \cap F\) only, \(B \cap V\) only, and \(F \cap V\) only. The total number of students in these three regions is \(N_2\).
- The three regions within each circle that do not overlap with any other circle represent students in B only, F only, and V only. The total number of students in these three regions is \(N_1\).
- The region outside all circles represents students who did not receive any medals in these three sports (not relevant to this specific problem as we deal with students who received at least one medal).
The sum of students in all 8 regions is the total number of students considered. The sum of students in the 7 regions within the circles is \(|B \cup F \cup V|\). The formulas we used, \(|B| + |F| + |V| = N_1 + 2N_2 + 3N_3\) and \(|B \cup F \cup V| = N_1 + N_2 + N_3\), are fundamental relationships derived from counting elements in these distinct regions.
Important Questions from Venn Diagrams
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