Understanding the Student Exam Data

The problem provides information about the performance of 80 students in three subjects: Chemistry, Physics, and Mathematics. We are given the number of students who passed in individual subjects and various combinations of subjects. Our goal is to find the number of students who passed only in Mathematics.

Analyzing the Given Information

Let's break down the numbers provided:

• Total students: 80
• Students who did not appear: 8
• Students who appeared (passed at least one subject): \(80 - 8 = 72\)
• Passed in Chemistry (C): \(|C| = 24\)
• Passed in Physics (P): \(|P| = 35\)
• Passed in Mathematics (M): \(|M| = 29\)
• Passed in Chemistry and Physics and Mathematics (C & P & M): \(|C \cap P \cap M| = 2\)
• Passed in Chemistry and Physics but not Mathematics (C & P only): \(|C \cap P \setminus M| = 2\)
• Passed in Chemistry and Mathematics but not Physics (C & M only): \(|C \cap M \setminus P| = 2\)
• Passed in Physics and Mathematics but not Chemistry (P & M only): \(|P \cap M \setminus C| = 8\)

Finding Pairwise Intersections

We are given the number of students who passed in pairs of subjects, excluding those who passed all three. We can use this to find the total number of students who passed in each pair:

• Chemistry and Physics (|C & P|):
  \(|C \cap P| = |C \cap P \setminus M| + |C \cap P \cap M|\)
  \(|C \cap P| = 2 + 2 = 4\)
• Chemistry and Mathematics (|C & M|):
  \(|C \cap M| = |C \cap M \setminus P| + |C \cap P \cap M|\)
  \(|C \cap M| = 2 + 2 = 4\)
• Physics and Mathematics (|P & M|):
  \(|P \cap M| = |P \cap M \setminus C| + |C \cap P \cap M|\)
  \(|P \cap M| = 8 + 2 = 10\)

Calculating Students Passed in Mathematics Only

To find the number of students who passed only in Mathematics, we need to take the total number of students who passed Mathematics and subtract those who also passed in Chemistry or Physics or both along with Mathematics. This can be calculated as:

Students passed in Mathematics only = (Total passed in Mathematics) - (Passed in C & M only) - (Passed in P & M only) - (Passed in C & P & M)

Using the notation:

\(|M \text{ only}| = |M| - |C \cap M \setminus P| - |P \cap M \setminus C| - |C \cap P \cap M|\)

Plugging in the values we have:

\(|M \text{ only}| = 29 - 2 - 8 - 2\)

\(|M \text{ only}| = 29 - (2 + 8 + 2)\)

\(|M \text{ only}| = 29 - 12\)

\(|M \text{ only}| = 17\)

Alternatively, using the total pairwise intersections:

\(|M \text{ only}| = |M| - |C \cap M| - |P \cap M| + |C \cap P \cap M|\)
\(|M \text{ only}| = 29 - 4 - 10 + 2\)
\(|M \text{ only}| = 29 - 14 + 2\)
\(|M \text{ only}| = 15 + 2\)
\(|M \text{ only}| = 17\)

Both methods give the same result.

Summary of Student Groups

The calculation confirms that the number of students who passed in Mathematics only is 17.

Conclusion

Based on the given data and our step-by-step calculation using set theory principles, the number of students who passed in Mathematics only is 17.

Revision Table: Key Concepts

Additional Information: Venn Diagrams and Set Problems

Problems like this can often be visualized effectively using a Venn diagram. For three sets (Chemistry, Physics, Mathematics), a Venn diagram consists of three overlapping circles within a rectangle (representing the total universe of students). Each distinct region in the Venn diagram represents a specific combination of passing subjects (e.g., only Chemistry, Chemistry and Physics only, all three, etc.).

The steps to solve such problems using a Venn diagram typically involve:

1. Drawing the diagram with overlapping circles for each set.
2. Starting by filling in the innermost region: the intersection of all three sets.
3. Using the given information about pairwise intersections (excluding the triple intersection) to fill in the regions where only two subjects overlap.
4. Using the total for each individual subject to fill in the regions where only one subject is passed.
5. Finally, using the total number of students (or students who appeared) and summing up all regions to check consistency and potentially find the number of students outside all sets (those who didn't pass any subject among those who appeared).

In this problem, we started with the triple intersection (2), then the pairwise intersections excluding the triple intersection (2, 2, 8). We could then calculate the number who passed in only one subject by subtracting the relevant intersection numbers from the total for that subject.

For example, to find students who passed only in Mathematics:

Total in M = (Only M) + (M & C only) + (M & P only) + (M & C & P)
\(29 = (\text{Only M}) + 2 + 8 + 2\)
\(29 = (\text{Only M}) + 12\)
\((\text{Only M}) = 29 - 12 = 17\)

This reinforces the formula used in the main solution.