Question

A group of 40 students appeared in an examination of 3 subjects – Mathematics, Physics & Chemistry. It was found that all students passed in at least one of the subjects, 20 students passed in Mathematics, 25 students passed in Physics, 16 students passed in Chemistry, at most 11 students passed in both Mathematics and Physics, at most 15 students passed in both Physics and Chemistry, at most 15 students passed in both Mathematics and Chemistry. The maximum number of students passed in all the three subjects is ________.

Answer 1

Correct Answer: 10

Using the principle of inclusion-exclusion for three sets \( M \), \( P \), and \( C \), we have:

\[ |M \cup P \cup C| = |M| + |P| + |C| - |M \cap P| - |P \cap C| - |M \cap C| + |M \cap P \cap C| \]

Given:

• \(|M| = 20\)
• \(|P| = 25\)
• \(|C| = 16\)
• \(|M \cap P| \leq 11\)
• \(|P \cap C| \leq 15\)
• \(|M \cap C| \leq 10\)

Since \(|M \cup P \cup C| = 40\), substitute the values and solve for \(|M \cap P \cap C|\):

\[ 40 = 20 + 25 + 16 - 11 - 15 - 10 + x \]

\[ x = 10 \]

Thus, the maximum number of students who passed in all three subjects is 10.

Answer 2

The problem asks for the maximum possible number of students who passed in all three subjects, given a set of conditions about the number of students who passed in various combinations of subjects.

Concept Used:

This problem is solved using the Principle of Inclusion-Exclusion for three sets. Let M, P, and C be the sets of students who passed in Mathematics, Physics, and Chemistry, respectively. The formula is:

\[ n(M \cup P \cup C) = n(M) + n(P) + n(C) - n(M \cap P) - n(P \cap C) - n(M \cap C) + n(M \cap P \cap C) \]

We also need to consider that the number of students in any specific region of a Venn diagram (e.g., passed in Mathematics and Physics only) must be non-negative.

Step-by-Step Solution:

Step 1: List the given information.

• Total number of students = 40. Since all students passed in at least one subject, \( n(M \cup P \cup C) = 40 \).
• Number of students who passed in Mathematics, \( n(M) = 20 \).
• Number of students who passed in Physics, \( n(P) = 25 \).
• Number of students who passed in Chemistry, \( n(C) = 16 \).
• Number of students who passed in Math and Physics, \( n(M \cap P) \le 11 \).
• Number of students who passed in Physics and Chemistry, \( n(P \cap C) \le 15 \).
• Number of students who passed in Math and Chemistry, \( n(M \cap C) \le 15 \).
• Let the number of students who passed in all three subjects be \( x \), so \( n(M \cap P \cap C) = x \). We need to maximize \(x\).

Step 2: Apply the Principle of Inclusion-Exclusion.

Substitute the known values into the formula:

\[ 40 = 20 + 25 + 16 - n(M \cap P) - n(P \cap C) - n(M \cap C) + x \] \[ 40 = 61 - (n(M \cap P) + n(P \cap C) + n(M \cap C)) + x \]

Rearranging the equation, we get a relationship between the sum of pairwise intersections and \(x\):

\[ n(M \cap P) + n(P \cap C) + n(M \cap C) = 21 + x \]

Step 3: Relate the intersections to the number of students passing in "exactly two" subjects.

Let's define:

• \(d\) = number of students who passed in Math and Physics only.
• \(e\) = number of students who passed in Physics and Chemistry only.
• \(f\) = number of students who passed in Math and Chemistry only.

From the Venn diagram, we know:

• \( n(M \cap P) = d + x \)
• \( n(P \cap C) = e + x \)
• \( n(M \cap C) = f + x \)

Step 4: Substitute these expressions back into the equation from Step 2.

\[ (d + x) + (e + x) + (f + x) = 21 + x \] \[ d + e + f + 3x = 21 + x \]

Now, solve for the sum of students passing in exactly two subjects:

\[ d + e + f = 21 - 2x \]

Step 5: Apply the non-negativity constraint to find the maximum value of \(x\).

The number of students in any category cannot be negative. Therefore, the sum of students passing in exactly two subjects must be greater than or equal to zero.

\[ d + e + f \ge 0 \]

Substituting the expression from Step 4:

\[ 21 - 2x \ge 0 \] \[ 21 \ge 2x \] \[ x \le \frac{21}{2} \] \[ x \le 10.5 \]

Since the number of students must be an integer, the maximum possible value for \(x\) is 10.

Step 6: Verify that \(x=10\) is consistent with all given conditions.

If \(x=10\):

• The number of students who passed in all three subjects is 10.
• From \(d+e+f = 21 - 2x\), we get \(d+e+f = 21 - 2(10) = 1\). This is possible (e.g., \(d=1, e=0, f=0\)).
• Check the intersection conditions:
  • \(n(M \cap P) = d + x = 1 + 10 = 11\), which satisfies \(n(M \cap P) \le 11\).
  • \(n(P \cap C) = e + x = 0 + 10 = 10\), which satisfies \(n(P \cap C) \le 15\).
  • \(n(M \cap C) = f + x = 0 + 10 = 10\), which satisfies \(n(M \cap C) \le 15\).

All conditions are satisfied for \(x=10\).

Final Computation & Result:

The inequality \(x \le 10.5\) shows that the maximum integer value for \(x\) is 10. We have also shown that a scenario where \(x=10\) is valid and consistent with all constraints.

The maximum number of students who passed in all the three subjects is 10.