Question

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

• A. \(30\)
• B. \(55\)
• C. \(35\)
• D. \(40\)

Hint:

For three-set Venn diagram problems:
Convert verbal relations about overlaps into equations using \(|A \cap B|\), \(|A \cap B \cap C|\), etc.
Use inclusion--exclusion to connect total strength with individual set sizes and intersections.
If asked for a maximum or minimum, express the required number in terms of a single variable and then optimize it under the given constraints.

Answer 1

The Correct Option is C

Approach: Set the pairwise overlaps in terms of one unknown, use inclusion-exclusion to link it to the triple-overlap, then minimise the relevant overlap — but only over values that keep every Venn region non-negative.

Step 1: Name the overlaps. Let \(n(P\cap C) = n(C\cap M) = x\) and \(n(P\cap M) = 2x\); let the triple be \(z = n(P\cap M\cap C) \ge 1\). (These are full pairwise intersections, triple included.)

Step 2: Inclusion-exclusion. With union \(= 150\) (everyone picks at least one):
\[ 150 = 75 + 111 + 40 - (2x + x + x) + z = 226 - 4x + z. \]
So \(4x = 76 + z\).

Step 3: What to maximise. "Physics but not Mathematics" \(= n(P) - n(P\cap M) = 75 - 2x\). To maximise this, make \(x\) as small as possible.

Step 4: Find the smallest valid \(x\). From \(4x = 76 + z\), we need \(76 + z\) divisible by 4, so \(z\) is a multiple of 4; with \(z \ge 1\), the smallest is \(z = 4\), giving \(x = 20\). (\(z=1,2,3\) give non-integer \(x\), so they are impossible.)

Intuition: Always confirm the candidate produces a legal Venn diagram — here \(x=20,\ z=4\) yields only-P \(=19\), only-M \(=55\), only-C \(=4\), all non-negative. ✓

Step 5: Answer. \[ 75 - 2x = 75 - 2(20) = \boxed{35}, \] matching option (c).

Answer 2

Let the sets be: \[ P = \text{Physics}, \quad M = \text{Mathematics}, \quad C = \text{Chemistry}. \] Given: \[ |P| = 75,\quad |M| = 111,\quad |C| = 40,\quad |P \cup M \cup C| = 150. \] Let \[ |P \cap C| = |M \cap C| = u,\quad |P \cap M| = 2u, \] and let \[ x = |P \cap M \cap C| \ge 1. \] Break the regions as: \[ \begin{aligned} a &= |P \cap M \text{ only}|, \\ b &= |P \cap C \text{ only}|, \\ c &= |M \cap C \text{ only}|, \\ p &= |P \text{ only}|, \\ m &= |M \text{ only}|, \\ q &= |C \text{ only}|. \end{aligned} \] Then: \[ |P \cap C| = b + x = u,\quad |M \cap C| = c + x = u,\quad |P \cap M| = a + x = 2u. \] So, \[ b + x = c + x \Rightarrow b = c, \] \[ a + x = 2u = 2(b + x) \Rightarrow a = 2b + x. \] 
Step 1: Use inclusion--exclusion on the whole class. Using: \[ |P \cup M \cup C| = |P| + |M| + |C| - |P \cap M| - |P \cap C| - |M \cap C| + |P \cap M \cap C|. \] Substitute: \[ 150 = 75 + 111 + 40 - (2u) - u - u + x \Rightarrow 150 = 226 - 4u + x \Rightarrow 4u - x = 76. \quad \cdots (1) \] Since \(u = b + x\), clearly \(u \ge x\) and all variables are non-negative integers. From (1), \[ x = 4u - 76. \] \[ x \ge 1 \Rightarrow 4u - 76 \ge 1 \Rightarrow u \ge 20. \] \[ x \le u \Rightarrow 4u - 76 \le u \Rightarrow 3u \le 76 \Rightarrow u \le \frac{76}{3} \Rightarrow u \le 25. \] So \(u\) is an integer with \[ 20 \le u \le 25. \]  
Step 2: Express the required quantity in terms of \(u\). We want the number of students who chose physics but not mathematics: \[ |P \cap M^c| = p + b. \] From set totals: \[ |P| = p + a + b + x = 75 \Rightarrow p = 75 - a - b - x. \] So: \[ p + b = 75 - a - b - x + b = 75 - a - x. \] But \(a = 2b + x\), and \(u = b + x\), so \(b = u - x\), hence: \[ a = 2(u - x) + x = 2u - x. \] Therefore: \[ p + b = 75 - (2u - x) - x = 75 - 2u. \] So, for any feasible \(u\), \[ |P \text{ but not } M| = 75 - 2u. \] To maximize this, we need to minimize \(u\), i.e., take the smallest feasible \(u\), which is \(u = 20\). Thus, \[ \max (p + b) = 75 - 2 \cdot 20 = 75 - 40 = 35. \] 
Step 3: Check feasibility for \(u = 20\). For \(u = 20\): \[ x = 4u - 76 = 80 - 76 = 4, \quad b = u - x = 16, \quad c = b = 16, \quad a = 2b + x = 36. \] Now: \[ \begin{aligned} |P| &: p + a + b + x = p + 36 + 16 + 4 = p + 56 = 75 \Rightarrow p = 19, \\ |M| &: m + a + c + x = m + 36 + 16 + 4 = m + 56 = 111 \Rightarrow m = 55, \\ |C| &: q + b + c + x = q + 16 + 16 + 4 = q + 36 = 40 \Rightarrow q = 4. \end{aligned} \] All are non-negative integers, so this configuration is valid. Thus the maximum possible number of students who chose physics but not mathematics is: \[ p + b = 19 + 16 = 35. \]