Analyzing Student Performance in Three Subjects

This problem asks us to determine the number of students who passed in Economics only, given information about student performance in three subjects: History, Economics, and English. We are given details about the total number of students and the number of students passing in various combinations of these subjects.

Understanding the Given Data for Student Passes

Let's list the information provided in the question:

• Total number of students: 100
• Every student passed in one or more of the three subjects.
• Number of students passed in English only: 24
• Number of students passed in History only: 14
• Number of students passed in both English and Economics only: 11
• Number of students passed in both English and History only: 12
• Total number of students passed in History: 50
• Number of students passed in all three subjects (History, Economics, and English): 5

We need to find the number of students who passed in Economics only.

Using Set Theory to Solve the Subject Pass Problem

We can represent the students who passed in each subject as sets. Let H be the set of students who passed in History, E be the set of students who passed in Economics, and N be the set of students who passed in English. The information can be translated into set notation:

• Total students = $|H \cup E \cup N| = 100$
• $|N \text{ only}| = 24$
• $|H \text{ only}| = 14$
• $|E \cap N \text{ only}| = 11$ (students in E and N, but not H)
• $|H \cap N \text{ only}| = 12$ (students in H and N, but not E)
• $|H| = 50$
• $|H \cap E \cap N| = 5$ (students in H, E, and N)

We need to find $|E \text{ only}|$.

Calculating Students Passing in History and Economics Only

We know the total number of students who passed in History ($|H| = 50$). This total includes students who passed History only, History and Economics only, History and English only, and all three subjects. In terms of disjoint regions in a Venn diagram:

$|H| = |H \text{ only}| + |H \cap E \text{ only}| + |H \cap N \text{ only}| + |H \cap E \cap N|$

We can plug in the known values:

$50 = 14 + |H \cap E \text{ only}| + 12 + 5$

$50 = 14 + 12 + 5 + |H \cap E \text{ only}|$

$50 = 31 + |H \cap E \text{ only}|$

Now, we can find the number of students who passed in both History and Economics only:

$|H \cap E \text{ only}| = 50 - 31$

$|H \cap E \text{ only}| = 19$

So, 19 students passed in both History and Economics only.

Finding Students Passing in Economics Only

Since every student passed in one or more subjects, the sum of the numbers of students in all the disjoint regions of the Venn diagram must equal the total number of students (100). The disjoint regions are:

• History only ($|H \text{ only}|$)
• Economics only ($|E \text{ only}|$)
• English only ($|N \text{ only}|$)
• History and Economics only ($|H \cap E \text{ only}|$)
• History and English only ($|H \cap N \text{ only}|$)
• Economics and English only ($|E \cap N \text{ only}|$)
• History, Economics, and English ($|H \cap E \cap N|$)

The sum is:

Total = $|H \text{ only}| + |E \text{ only}| + |N \text{ only}| + |H \cap E \text{ only}| + |H \cap N \text{ only}| + |E \cap N \text{ only}| + |H \cap E \cap N|$

We know all these values except $|E \text{ only}|$. Let's substitute them:

$100 = 14 + |E \text{ only}| + 24 + 19 + 12 + 11 + 5$

Now, let's sum the known values:

$14 + 24 + 19 + 12 + 11 + 5 = 85$

So the equation becomes:

$100 = 85 + |E \text{ only}|$$

To find $|E \text{ only}|$, subtract 85 from 100:

$|E \text{ only}| = 100 - 85$

$|E \text{ only}| = 15$

Therefore, 15 students passed in Economics only.

Checking the sum: $14 + 24 + 11 + 12 + 5 + 19 + 15 = 100$. The numbers add up correctly.

Conclusion: Students Passing in Economics Only

Based on our calculations using the provided data and set theory principles, the number of students who passed in Economics only is 15.

Revision Table: Key Data Points

Additional Information: Venn Diagrams and Set Operations

This problem is a classic application of set theory, often visualized using Venn diagrams. A Venn diagram for three sets (History, Economics, English) would have 8 distinct regions:

1. Outside all sets (students who passed none - 0 in this problem)
2. Only History
3. Only Economics
4. Only English
5. History and Economics only (not English)
6. History and English only (not Economics)
7. Economics and English only (not History)
8. History, Economics, and English (all three)

The problem provides the values for many of these regions directly, or indirectly allows us to calculate them. For example, "passed in both English and Economics only" means the intersection of E and N, excluding H. "Total passed in History" means the sum of the 'History only' region and the three intersection regions that include History ($H \cap E \text{ only}$, $H \cap N \text{ only}$, and $H \cap E \cap N$).

The key formula for the union of three sets is:

$|H \cup E \cup N| = |H| + |E| + |N| - |H \cap E| - |H \cap N| - |E \cap N| + |H \cap E \cap N|$

However, this formula uses the total intersection areas (e.g., $|H \cap E|$ includes those who passed all three), while the problem provides 'only' values for intersections. It's often simpler for 'only' data to sum the disjoint regions, as we did in the step-by-step solution.

Understanding the difference between $|A \cap B|$ and $|A \cap B \text{ only}|$ (or $|A \cap B \text{ without } C|$) is crucial for solving such problems. $|A \cap B| = |A \cap B \text{ only}| + |A \cap B \cap C|$.

In our case:

• $|H \cap E| = |H \cap E \text{ only}| + |H \cap E \cap N| = 19 + 5 = 24$
• $|H \cap N| = |H \cap N \text{ only}| + |H \cap E \cap N| = 12 + 5 = 17$
• $|E \cap N| = |E \cap N \text{ only}| + |H \cap E \cap N| = 11 + 5 = 16$

And the total number of students in each subject:

• $|H| = |H \text{ only}| + |H \cap E \text{ only}| + |H \cap N \text{ only}| + |H \cap E \cap N| = 14 + 19 + 12 + 5 = 50$ (Given)
• $|N| = |N \text{ only}| + |H \cap N \text{ only}| + |E \cap N \text{ only}| + |H \cap E \cap N| = 24 + 12 + 11 + 5 = 52$
• $|E| = |E \text{ only}| + |H \cap E \text{ only}| + |E \cap N \text{ only}| + |H \cap E \cap N| = 15 + 19 + 11 + 5 = 50$

Using the union formula to verify (though not strictly necessary for the solution):

$|H \cup E \cup N| = |H| + |E| + |N| - |H \cap E| - |H \cap N| - |E \cap N| + |H \cap E \cap N|$

$|H \cup E \cup N| = 50 + 50 + 52 - 24 - 17 - 16 + 5$

$|H \cup E \cup N| = 152 - 57 + 5$

$|H \cup E \cup N| = 95 + 5 = 100$

This confirms our calculated values for the individual regions are consistent with the total number of students.