In a school 45 students are participating in exams. Each student must take at least one of these three exams. 25 students take mathematics exam, 15 students take physics exam and 19 students take chemistry exam. There are 9 students who take at least two exams, then the number of students who participate in all the exams, is

We are given that 45 students are participating in exams, and each student must take at least one of the three exams: Mathematics, Physics, and Chemistry. We also have the following information:

- 25 students take the Mathematics exam.
- 15 students take the Physics exam.
- 19 students take the Chemistry exam.
- 9 students take at least two exams.

We need to find how many students take all three exams.

Let:
- \( n(M) = 25 \) be the number of students taking the Mathematics exam,
- \( n(P) = 15 \) be the number of students taking the Physics exam,
- \( n(C) = 19 \) be the number of students taking the Chemistry exam,
- \( n(M \cap P \cap C) = x \) be the number of students taking all three exams (this is what we need to find),
- \( n(M \cup P \cup C) = 45 \) be the total number of students (since every student takes at least one exam).

We are also told that 9 students take at least two exams. The number of students taking exactly two exams can be expressed as:

\[
n(M \cap P) + n(P \cap C) + n(C \cap M) - 3x = 9
\]

Using the inclusion-exclusion principle, the total number of students taking at least one exam is:

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

Substitute the values we know:

\[
45 = 25 + 15 + 19 - [n(M \cap P) + n(P \cap C) + n(C \cap M)] + x
\]

Simplify:

\[
45 = 59 - [n(M \cap P) + n(P \cap C) + n(C \cap M)] + x
\]

From this, we get:

\[
n(M \cap P) + n(P \cap C) + n(C \cap M) = 14 + x
\]

Now, substitute this into the equation for students taking at least two exams:

\[
14 + x - 3x = 9
\]

Simplify:

\[
14 - 2x = 9
\]

Solve for \( x \):

\[
14 - 9 = 2x
\]
\[
x = \frac{5}{2} = 2.5
\]
Step use

This problem involves 45 students participating in three exams: Mathematics, Physics, and Chemistry. Using set theory and the inclusion-exclusion principle, learn how to find the number of students who take all three exams when 9 students take at least two exams. Suitable for students in Grades 10-12.

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