2023 CSAT — Q70

Decode the coded operators into plain inequalities first:

• $A + B$: $A$ neither smaller nor equal $\Rightarrow A > B$
• $A - B$: $A$ not greater $\Rightarrow A \le B$
• $A \times B$: $A$ not smaller $\Rightarrow A \ge B$
• $A \div B$: $A$ neither greater nor equal $\Rightarrow A < B$
• $A \pm B$: $A$ neither smaller nor greater $\Rightarrow A = B$

Now translate the Statement:

Chain them: $Q \le P \le T < R = S$, i.e.

Conclusion-1 ($Q \pm T$, i.e. $Q = T$): we only have $Q \le P \le T$, so $Q \le T$ but possibly $Q < T$. Equality is not forced. Does not follow.

Conclusion-2 ($S + Q$, i.e. $S > Q$): $S = R > T \ge P \ge Q$, so $S > Q$ strictly. Follows.

Answer: (b) Only Conclusion-2 follows from the Statement.

The discipline is decode every operator into an ordering symbol before reasoning — the words “neither smaller nor equal,” “not greater,” etc. each map to exactly one of $>, \le, \ge, <, =$. Then assemble the single chain $Q \le P \le T < R = S$. A “follows” conclusion must hold in every model: $S > Q$ always holds (strict, via $T < R$), but $Q = T$ can fail (when $Q < T$). Distinguishing a forced strict inequality from a merely-possible equality is the whole item.