Students Solved All Four Questions

The problem asks for the smallest possible number of students who solved all four questions in an exam taken by 120 students. We are given the number of students who solved each individual question:

• Solved Question 1: 100 students
• Solved Question 2: 95 students
• Solved Question 3: 90 students
• Solved Question 4: 80 students

Let $N$ be the total number of students, so $N = 120$.

Let $Q_1, Q_2, Q_3, Q_4$ be the sets of students who solved Question 1, Question 2, Question 3, and Question 4, respectively. We are given:

• $|Q_1| = 100$
• $|Q_2| = 95$
• $|Q_3| = 90$
• $|Q_4| = 80$

We want to find the smallest possible value of $|Q_1 \cap Q_2 \cap Q_3 \cap Q_4|$.

Identifying the Constraint on the Number of Solvers

Any student who solved all four questions must necessarily be a student who solved Question 1, a student who solved Question 2, a student who solved Question 3, AND a student who solved Question 4.

This means the set of students who solved all four questions, $Q_1 \cap Q_2 \cap Q_3 \cap Q_4$, must be a subset of each individual set $Q_1, Q_2, Q_3, Q_4$.

Specifically, the number of students who solved all four questions cannot be more than the number of students who solved the question with the fewest successful attempts. In this case, the smallest group is the one that solved Question 4, with 80 students.

So, $|Q_1 \cap Q_2 \cap Q_3 \cap Q_4| \le |Q_4| = 80$.

Scenario Yielding 80 Solvers

Let's consider if it's possible for exactly 80 students to solve all four questions. This scenario occurs if the set of students who solved Question 4 is completely contained within the sets of students who solved Questions 1, 2, and 3.

If the set of students who solved Q4 ($Q_4$) is a subset of the set of students who solved Q1 ($Q_1$), a subset of Q2 ($Q_2$), and a subset of Q3 ($Q_3$), then any student who solved Q4 also solved Q1, Q2, and Q3. In this case, the set of students who solved all four questions is precisely the set of students who solved Q4.

This requires:

• $Q_4 \subseteq Q_1$
• $Q_4 \subseteq Q_2$
• $Q_4 \subseteq Q_3$

Let's check if this is consistent with the given numbers:

• $|Q_4| \le |Q_1|$? $80 \le 100$ (True)
• $|Q_4| \le |Q_2|$? $80 \le 95$ (True)
• $|Q_4| \le |Q_3|$? $80 \le 90$ (True)

Since these conditions are consistent with the given counts, a scenario where $Q_4 \subseteq Q_1$, $Q_4 \subseteq Q_2$, and $Q_4 \subseteq Q_3$ is possible. In such a scenario, the intersection of all four sets is simply $Q_4$ because $Q_1 \cap Q_2 \cap Q_3 \cap Q_4 = (Q_1 \cap Q_2 \cap Q_3) \cap Q_4$. If $Q_4$ is a subset of the others, the intersection is just $Q_4$.

In this possible scenario, the number of students who solved all four questions is $|Q_4| = 80$.

The remaining $120 - 80 = 40$ students did not solve all four. These 40 students must fill the remaining counts for Q1, Q2, and Q3 in a way consistent with the total counts.

• Number who solved Q1 but not all four: $100 - 80 = 20$.
• Number who solved Q2 but not all four: $95 - 80 = 15$.
• Number who solved Q3 but not all four: $90 - 80 = 10$.
• Number who solved Q4 but not all four: $80 - 80 = 0$.

These 20, 15, and 10 students (total 45) must be among the 40 students who did not solve all four. This requires careful distribution, involving overlaps among those who solved some but not all four questions. However, the key point is that the scenario where 80 students solved all four (because they were precisely the Q4 solvers who also solved Q1, Q2, and Q3) is possible and consistent with the total and individual question counts.

Smallest Possible Number

We have established that 80 is a possible number of students who solved all four questions. While set theory principles might suggest a lower theoretical minimum in general cases, in a scenario structured such that the smallest group of solvers is contained within the larger groups, the intersection is equal to the size of the smallest group.

Since the number of students who solved all four cannot exceed 80 (the size of the smallest group), and we have shown that 80 is a possible number, the smallest possible number in this context, considering the constraints imposed by the individual question counts within the total, is 80.