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

✅ Correct Option: 2

Let $P \cap C = C \cap M = k$ and $P \cap M = 2k$ , and let $t$ be the number of students who chose all three subjects.

Using inclusion-exclusion:

$|P \cup M \cup C| = |P| + |M| + |C| - |P \cap M| - |P \cap C| - |M \cap C| + |P \cap M \cap C|$

$150 = 75 + 111 + 40 - 2k - k - k + t$

$150 = 226 - 4k + t$

$4k - t = 76 \quad \cdots (1)$

Physics but not Mathematics $= 75 - 2k$

To maximize this, $k$ must be minimized.

From $(1)$ : $t = 4k - 76$

Since at least one student chose all three subjects:

$t \geq 1$

$4k - 76 \geq 1$

$k \geq 19.25$

$k \geq 20$

Also, $t \leq k$ since students in all three cannot exceed those in $P \cap C$ :

$4k - 76 \leq k$

$3k \leq 76$

$k \leq 25$

So minimum $k = 20$ , giving $t = 4(20) - 76 = 4$ .

Physics but not Mathematics $= 75 - 2(20)$

$= 75 - 40$

$= 35$

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