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

To solve this problem, we need to apply the principles of set theory and use the information given systematically to find the maximum number of students who chose physics but not mathematics.

Let's denote the following numbers:

• \(n(P)\): Number of students who chose Physics = 75
• \(n(M)\): Number of students who chose Mathematics = 111
• \(n(C)\): Number of students who chose Chemistry = 40

We also know:

• All students chose at least one subject, totaling 150 students.
• At least one student chose all three subjects.
• The number of students who chose both Physics and Chemistry is equal to those who chose both Chemistry and Mathematics, which is half of those who chose both Physics and Mathematics.

Let's determine the quantities for the students choosing multiple subjects using these equations:

Let:

• \(x\): Students choosing both Physics and Chemistry
• \(x\): Students choosing both Chemistry and Mathematics (same as above)
• \(2x\): Students choosing both Physics and Mathematics

To find the number of students who choose all three subjects, denote it as \(n(P \cap M \cap C) = z\). We know \(z \geq 1\).

Using the principle of inclusion-exclusion for three sets:

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

Substitute the known values:

\[150 = 75 + 111 + 40 - 2x - x - x + z\]

Simplify:

\[150 = 226 - 4x + z\]

Rearrange to find:

\[4x = 226 - 150 + z\]\[4x = 76 + z\]

Substitute the condition that all students chose at least one subject and find:

Maximize the students choosing Physics but not Mathematics, which is the set of students \(n(P) - n(P \cap M)\), or:

\[75 - 2x\]

To maximize \(75 - 2x\), minimize \(x\). From \(4x = 76 + z\) and knowing \(z \geq 1\), the smallest integer \(x\) can be is 20 (when \(z = 1\)).

Therefore:

\[75 - 2(20) = 75 - 40 = 35\]

However, considering the need to satisfy total count constraints, it restricts further fine-tuning. Thus, under the examined structure, the most feasible candidate realizing balance is:

Hence, maximum possible students choosing Physics but not Mathematics is 35.

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: \[ p + b = 19 + 16 = 35. \]