GATE CS 2022 Q4 — General Aptitude

Let $S$ represent the set of students and $I$ represent the set of inquisitive people. We are given that there is at least one student in the class, so $S \neq \emptyset$.

Let us express the four statements mathematically:

• Statement 1: All students are inquisitive. $\implies S \subseteq I$
• Statement 2: Some students are inquisitive. $\implies S \cap I \neq \emptyset$
• Statement 3: No student is inquisitive. $\implies S \cap I = \emptyset$
• Statement 4: Some students are not inquisitive. $\implies S \setminus I \neq \emptyset$

Now, let us analyze the compatibility of the statements:

• If Statement 1 is true ($S \subseteq I$), and since $S \neq \emptyset$, it implies that $S \cap I = S \neq \emptyset$. This directly contradicts Statement 3 ($S \cap I = \emptyset$). Thus, Statement 1 and Statement 3 cannot be true simultaneously.

Let's verify the other options to ensure they can be true simultaneously:

• Statement 1 and Statement 2: If all students are inquisitive ($S \subseteq I$) and $S \neq \emptyset$, then some students are inquisitive ($S \cap I \neq \emptyset$) is also true.
• Statement 2 and Statement 4: If some students are inquisitive and others are not, both statements are true simultaneously.
• Statement 3 and Statement 4: If no student is inquisitive ($S \cap I = \emptyset$), then since $S \neq \emptyset$, there are students who are not inquisitive, making Statement 4 true as well.

Hence, Statement 1 and Statement 3 cannot be true at the same time.