In a competitive examination, 250 students have registered. Out of these, 50 students have registered for Physics, 75 students for Mathematics and 35 students for both Mathematics and Physics. What is the number of students who have registered neither for Physics nor for Mathematics?
160
Understanding the Competitive Examination Problem
This question is about analyzing the registration data of students for different subjects in a competitive examination. We are given the total number of students registered, the number registered for Physics, the number registered for Mathematics, and the number registered for both subjects. Our goal is to find the number of students who did not register for either Physics or Mathematics.
Applying Set Theory to Competitive Exam Registrations
We can use the concepts of set theory to solve this problem. Let:
- \(U\) be the set of all students registered for the competitive examination.
- \(P\) be the set of students registered for Physics.
- \(M\) be the set of students registered for Mathematics.
We are given the following information:
| Description | Notation | Number of Students |
|---|---|---|
| Total students | \(|U|\) | 250 |
| Students registered for Physics | \(|P|\) | 50 |
| Students registered for Mathematics | \(|M|\) | 75 |
| Students registered for both Physics and Mathematics | \(|P \cap M|\) | 35 |
We want to find the number of students who registered for neither Physics nor Mathematics. In set theory terms, this is the number of students in the universal set \(U\) who are not in the union of sets \(P\) and \(M\). This can be written as \(|U| - |P \cup M|\).
Calculating Students Registered in At Least One Subject
First, we need to find the number of students who registered for Physics or Mathematics (or both). This is represented by the union of the sets, \(|P \cup M|\). We can use the Principle of Inclusion-Exclusion for two sets:
\(|P \cup M| = |P| + |M| - |P \cap M|\)
Now, let's substitute the given values into the formula:
\(|P \cup M| = 50 + 75 - 35\)
\(|P \cup M| = 125 - 35\)
\(|P \cup M| = 90\)
So, 90 students registered for at least one of the two subjects (Physics or Mathematics).
Finding Students Registered in Neither Physics nor Mathematics
The number of students who registered for neither subject is the total number of students minus the number of students who registered for at least one subject.
Number of students registered for neither = Total Students - \(|P \cup M|\)
Number of students registered for neither = \(250 - 90\)
Number of students registered for neither = \(160\)
Therefore, 160 students registered for neither Physics nor Mathematics in the competitive examination.
Revision Table: Competitive Exam Registrations
| Concept | Formula/Method | Value |
|---|---|---|
| Total Students | Given | 250 |
| Students in Physics (\(|P|\)) | Given | 50 |
| Students in Math (\(|M|\)) | Given | 75 |
| Students in Physics AND Math (\(|P \cap M|\)) | Given | 35 |
| Students in Physics OR Math (\(|P \cup M|\)) | \(|P| + |M| - |P \cap M|\) | \(50 + 75 - 35 = 90\) |
| Students in NEITHER Physics NOR Math | Total Students - \(|P \cup M|\) | \(250 - 90 = 160\) |
Additional Information on Set Theory and Competitive Exams
This problem demonstrates a common application of basic set theory in analyzing survey or registration data. Here are some related concepts:
- Only Physics: The number of students who registered for Physics but not Mathematics is \(|P| - |P \cap M|\). In this case, \(50 - 35 = 15\).
- Only Mathematics: The number of students who registered for Mathematics but not Physics is \(|M| - |P \cap M|\). In this case, \(75 - 35 = 40\).
- Venn Diagrams: These problems can also be visually represented using a Venn diagram, with overlapping circles for Physics and Mathematics within a rectangle representing the total students. The overlapping region is \(P \cap M\), the non-overlapping parts of the circles are "only P" and "only M", the union \(P \cup M\) is the area covered by both circles, and the area outside the circles within the rectangle is "neither P nor M".
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