CAT 2025 Slot 3 QA Question 6

First, let's parse the problem. There are 150 students in total. Each student has chosen at least one of the three subjects: physics, mathematics, or chemistry. So, no one is left out; everyone is in at least one subject. Also, at least one student has chosen all three subjects.

The numbers given are: 75 students chose physics, 111 chose mathematics, and 40 chose chemistry. Hmm, that's interesting because the total number of students is 150, but the sum of the students choosing each subject is 75 + 111 + 40 = 226, which is way more than 150. That makes sense because of overlaps.

Now, the problem also states that the number of students who chose both physics and chemistry is equal to the number who chose both chemistry and mathematics. Let's denote these as pc (physics and chemistry) and \text{ cm} (chemistry and mathematics). So, pc = \text{ cm}.

Additionally, it says that this number (pc and \text{ cm}) is half the number of students who chose both physics and mathematics. Let's denote pm as the number of students who chose both physics and mathematics. So, pc = \text{ cm} = pm / 2.

Our goal is to find the maximum possible number of students who chose physics but not mathematics. Let's denote this as p_only. So, p_only = total physics students - (students who took physics and math) - (students who took physics and chemistry) + (students who took all three). in set theory, the number of students who took only physics is total physics minus those who took physics and another subject, but we have to be careful with those who took all three.

Wait, let me recall the formula for three sets. The total number of students is equal to the sum of students in each subject minus the sum of students in each pair of subjects plus the number of students in all three subjects. So, using the principle of inclusion-exclusion:

Total = P + M + C - (pm + pc + \text{ cm}) + all_three.

Given that all students chose at least one subject, the total is 150. So,

150 = 75 + 111 + 40 - (pm + pc + \text{ cm}) + all_three.

Simplify the left side:

75 + 111 + 40 = 226.

So,

150 = 226 - (pm + pc + \text{ cm}) + all_three.

Rearranging,

(pm + pc + \text{ cm}) - all_three = 226 - 150 = 76.

So,

(pm + pc + \text{ cm}) - all_three = 76.

But we know that pc = \text{ cm} and pc = pm / 2. So, let's substitute pc and \text{ cm} in terms of pm.

Let pc = \text{ cm} = x. Then, pm = 2x.

So, substituting into the equation:

(2x + x + x) - all_three = 76.

Simplify:

4x - all_three = 76.

So,

4x = 76 + all_three.

But we also know that all_three is at least 1, as given in the problem. So, all_three ≥ 1.

So, 4x = 76 + all_three ≥ 76 + 1 = 77.

Therefore, x ≥ 77 / 4 = 19.25. But since x must be an integer (number of students), x ≥ 20.

But we might need to find the maximum possible p_only. Let's express p_only in terms of x and all_three.

p_only is the number of students who took only physics. That would be total physics students minus those who took physics and another subject, plus those who took all three (since they were subtracted twice). in the inclusion-exclusion, the formula for only physics is P - (pm + pc - all_three). Wait, Wait, the number of students who took only physics is equal to total physics minus those who took physics and mathematics minus those who took physics and chemistry plus those who took all three. Because when we subtract pm and pc, we subtracted the all_three students twice, so we need to add them back once.

So,

p_only = P - pm - pc + all_three.

Substituting the known values:

p_only = 75 - pm - pc + all_three.

But pm = 2x, pc = x, so:

p_only = 75 - 2x - x + all_three = 75 - 3x + all_three.

We need to maximize p_only, so we need to maximize 75 - 3x + all_three.

But from earlier, we have 4x = 76 + all_three, so all_three = 4x - 76.

Substituting into p_only:

p_only = 75 - 3x + (4x - 76) = 75 - 3x + 4x - 76 = (75 - 76) + ( -3x + 4x ) = -1 + x.

So, p_only = x - 1.

Wait, that's interesting. So, p_only is equal to x - 1. To maximize p_only, we need to maximize x.

But x is constrained by the fact that all_three must be at least 1, and also, the number of students in each pair cannot exceed the total number of students in each subject.

Wait, let's think about the constraints.

First, all_three must be at least 1, so from earlier, x ≥ 20.

But also, the number of students in each pair (pm, pc, \text{ cm}) cannot exceed the total number of students in either subject.

For example, pm is the number of students who took both physics and mathematics. This number cannot exceed the total number of physics students or mathematics students. Since physics has 75 and mathematics has 111, pm can be at most 75.

Similarly, pc is the number of students who took both physics and chemistry, which cannot exceed 40 (since chemistry has only 40 students). Similarly, \text{ cm} cannot exceed 40.

But since pc = \text{ cm} = x, and x cannot exceed 40, because pc is the number of students taking both physics and chemistry, which can't be more than the total number of chemistry students, which is 40.

So, x ≤ 40.

But earlier, we had x ≥ 20.

So, x is between 20 and 40.

But we also have another constraint from the equation 4x = 76 + all_three.

Since all_three must be a non-negative integer, 4x must be at least