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 option is 35.

Let the sets of students who chose Physics, Mathematics, and Chemistry be denoted by $P$, $M$, and $C$ respectively. We are given the following values:
Total students: $|P\cup M\cup C|=150$
Students in Physics: $\left|P\right|=75$
Students in Mathematics: $\left|M\right|=111$
Students in Chemistry: $\left|C\right|=40$

Let us represent the regions of the Venn diagram for these three sets using the following variables:
• $p$: number of students who chose Physics only
• $m$: number of students who chose Mathematics only
• $h$: number of students who chose Chemistry only
• $a$: number of students who chose Physics and Mathematics but not Chemistry
• $b$: number of students who chose Physics and Chemistry but not Mathematics
• $c$: number of students who chose Mathematics and Chemistry but not Physics
• $x$: number of students who chose all three subjects (Physics, Mathematics, and Chemistry)

According to the problem, at least one student chose all three subjects:
$x\ge 1$

We are given that 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:
$|P\cap C|=|C\cap M|=\frac{1}{2}|P\cap M|$

Expressing these in terms of our variables:
$b+x=c+x\Rightarrow b=c$
$b+x=\frac{1}{2}(a+x)\Rightarrow a=2b+x$

Now, we write the equations for the cardinality of each set:
1. Chemistry: $h+b+c+x=40$
Substitute $c=b$:
$h+2b+x=40$   — (Equation 1)

2. Physics: $p+a+b+x=75$
Substitute $a=2b+x$:
$p+3b+2x=75$   — (Equation 2)

3. Mathematics: $m+a+c+x=111$
Substitute $a=2b+x$ and $c=b$:
$m+3b+2x=111$   — (Equation 3)

4. Total students choosing at least one subject:
$p+m+h+a+b+c+x=150$
Substitute the expressions for $a$ and $c$:
$p+m+h+4b+2x=150$   — (Equation 4)

Let us sum Equations 1, 2, and 3:
$(h+2b+x)+(p+3b+2x)+(m+3b+2x)=40+75+111$
$p+m+h+8b+5x=226$   — (Equation 5)

Subtract Equation 4 from Equation 5:
$(p+m+h+8b+5x)-(p+m+h+4b+2x)=226-150$
$4b+3x=76$

We want to find the maximum possible number of students who chose Physics but not Mathematics. This represents the region $P\backslash M=p+b$.

From Equation 2, we have:
$p=75-3b-2x$
Thus, the quantity to maximize is:
$p+b=75-2b-2x=75-2(b+x)$

To maximize $75-2(b+x)$, we must find the minimum possible value of $b+x$ subject to the constraints:
• $4b+3x=76$
• $x\ge 1$, where $x$ and $b$ are non-negative integers.
• All other region variables ( $p,m,h,a,b,c$) must be non-negative integers.

From $4b+3x=76$, we can write:
$3x=76-4b=4(19-b)$
Since 3 and 4 are coprime, $x$ must be a multiple of 4. Let $x=4k$ for an integer $k\ge 1$.
Substituting $x=4k$:
$4b+12k=76\Rightarrow b+3k=19\Rightarrow b=19-3k$

Since $b\ge 0$:
$19-3k\ge 0\Rightarrow k\le 6$

Thus, the possible values for $k$ are $1,2,3,4,5,6$. Let us check the values of $b+x$ and the resulting $p+b$ for each:
• For $k=1$: $x=4$, $b=16$ $\Rightarrow b+x=20$ $\Rightarrow p+b=75-2\left(20\right)=35$
• For $k=2$: $x=8$, $b=13$ $\Rightarrow b+x=21$ $\Rightarrow p+b=75-2\left(21\right)=33$
• For $k=3$: $x=12$, $b=10$ $\Rightarrow b+x=22$ $\Rightarrow p+b=75-2\left(22\right)=31$
• For $k=4$: $x=16$, $b=7$ $\Rightarrow b+x=23$ $\Rightarrow p+b=75-2\left(23\right)=29$
• For $k=5$: $x=20$, $b=4$ $\Rightarrow b+x=24$ $\Rightarrow p+b=75-2\left(24\right)=27$
• For $k=6$: $x=24$, $b=1$ $\Rightarrow b+x=25$ $\Rightarrow p+b=75-2\left(25\right)=25$

(Note: For all these cases, the remaining variables $p$, $m$, and $h$ are positive integers. For example, when $k=1$, we have $p=75-48-8=19$, $m=111-48-8=55$, and $h=40-32-4=4$.)

The maximum possible value for the number of students who chose Physics but not Mathematics ( $p+b$) is therefore 35 (attained when $k=1$).