From 50 students taking examinations in mathematics, physics and chemistry,37 passed mathematics, 24 physics and 43 chemistry. At most 19 passed mathematics and physics, at most 29 mathematics and chemistry and at most 20 physics and chemistry. Find the largest possible number that could have passed all three exams.

Options

A. 11
B. 12
C. 13
D. 14

Answer

To solve the problem step by step, let's define the variables and use the principle of inclusion-exclusion.

**Step 1: Define the variables.**
Let:
- \( N \) = Total number of students = 50
- \( N(M) \) = Number of students who passed Mathematics = 37
- \( N(P) \) = Number of students who passed Physics = 24
- \( N(C) \) = Number of students who passed Chemistry = 43
- \( N(M \cap P) \) = Number of students who passed both Mathematics and Physics
- \( N(M \cap C) \) = Number of students who passed both Mathematics and Chemistry
- \( N(P \cap C) \) = Number of students who passed both Physics and Chemistry
- \( N(M \cap P \cap C) \) = Number of students who passed all three subjects

**Step 2: Set up the inequalities based on the problem statement.**
From the problem, we know:
- \( N(M \cap P) \leq 19 \)
- \( N(M \cap C) \leq 29 \)
- \( N(P \cap C) \leq 20 \)

**Step 3: Use the principle of inclusion-exclusion.**
According to the principle of inclusion-exclusion, we have:
\[ N(M \cup P \cup C) = N(M) + N(P) + N(C) - N(M \cap P) - N(M \cap C) - N(P \cap C) + N(M \cap P \cap C) \]
Substituting the known values:
\[ 50 = 37 + 24 + 43 - N(M \cap P) - N(M \cap C) - N(P \cap C) + N(M \cap P \cap C) \]

**Step 4: Simplify the equation.**
This simplifies to:
\[ 50 = 104 - N(M \cap P) - N(M \cap C) - N(P \cap C) + N(M \cap P \cap C) \]
Rearranging gives:
\[ N(M \cap P) + N(M \cap C) + N(P \cap C) - N(M \cap P \cap C) = 54 \]

**Step 5: Substitute the maximum values.**
Let:
- \( x = N(M \cap P \cap C) \)

Then we can express:
\[ N(M \cap P) + N(M \cap C) + N(P \cap C) = 54 + x \]
Using the maximum values:
\[ 19 + 29 + 20 = 68 \]
Thus:
\[ 54 + x \leq 68 \]
This simplifies to:
\[ x \leq 14 \]

**Step 6: Conclusion.**
The largest possible number of students who passed all three exams is \( N(M \cap P \cap C) \leq 14 \).

Therefore, the answer is:
\[ \text{The largest possible number that could have passed all three exams is } 14. \]

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