In a class of 150 students, 75 students chose physics, 111 students chose mathematics and 40 students chose chemistry. All students chose at least one of the three subjects and at least one student chose all three subjects. The number of students who chose both physics and chemistry is equal to the number of students who chose both chemistry and mathematics, and this is half the number of students who chose both physics and mathematics. The maximum possible number of students who chose physics but not mathematics, is

The correct answer is Option 2 (35).

Let us solve this problem step-by-step using set theory principles and Venn diagram analysis.

Step 1: Define the Sets and Given Data
Let the three subjects be Physics (P), Mathematics (M), and Chemistry (C).
The total number of students in the class who chose at least one subject is given as:

$n(P\cup M\cup C)=150$

The number of students enrolled in each subject is:

$n\left(P\right)=75$

$n\left(M\right)=111$

$n\left(C\right)=40$

Let the region representing students who chose all three subjects be:

$n(P\cap M\cap C)=x$

We are given that $x\ge 1$ (at least one student chose all three subjects).

Step 2: Express the Intersections
Let the number of students who chose both Physics and Chemistry be $y$.
According to the problem statement:

1. The number of students who chose both Chemistry and Mathematics is also equal to $y$.

2. The number of students who chose both Physics and Chemistry ($y$) is half the number of students who chose both Physics and Mathematics.
Therefore, the number of students who chose both Physics and Mathematics is $2y$.

So we have:

$n(P\cap C)=y$

$n(C\cap M)=y$

$n(P\cap M)=2y$

Step 3: Apply the Principle of Inclusion-Exclusion
The standard formula for the union of three sets is:

$n(P\cup M\cup C)=n\left(P\right)+\mathrm{n}\left(M\right)+\mathrm{n}\left(C\right)-n(P\cap M)-\mathrm{n}(M\cap C)-\mathrm{n}(P\cap C)+n(P\cap M\cap C)$

Substitute the given values into the formula:

$150=75+111+40-2y-y-y+x$

$150=226-4y+x$

Rearranging the equation to find a relation between $x$ and $y$:

$4y-x=76$

Step 4: Express the Target Value
We need to find the maximum possible number of students who chose Physics but not Mathematics, denoted as $n(P\setminus M)$.

The number of students in Physics who are not in Mathematics is given by:

$n(P\setminus M)=n\left(P\right)-n(P\cap M)=75-2y$

To maximize $75-2y$, we must minimize $y$.

Step 5: Determine Constraints on $x$ and $y$
From $4y-\mathrm{x}=76$, we have:

$4y=76+x$

Since $x\ge 1$:

$4y\ge 76+1=77$

$y\ge \frac{77}{4}=19.25$

Since $y$ must be a whole number (representing a count of students), the minimum possible integer value for $y$ is:

$y_{\text{min}}=20$

When $y=20$, we find $x=4\left(20\right)-76=4$, which fully satisfies all set non-negativity conditions for all region counts in the Venn diagram.

Step 6: Calculate the Maximum Value
Substituting $y=20$ into our expression for $n(P\setminus M)$:

$\text{Maximum students}=75-2\left(20\right)=75-40=35$

Thus, the maximum possible number of students who chose physics but not mathematics is 35.