2023 CSAT — Q77

Locate. On a free-standing critical-reasoning item that labels each statement true or false, the load-bearing move is to first convert every statement into what it actually asserts about the world — and a statement marked FALSE is information, not noise: read its negation as a fact. The false statement here is “At least one of R and S was selected,” so the fact is its negation: neither R nor S was selected.

Test (build the world, then check each conclusion against it). Lock the three facts: (i) exactly one of P, Q is in; (ii) R is out and S is out; (iii) at most two of R, S, T are in (an upper bound that, since R and S are already out, only caps T at “in or out”). The most that can be selected is one (P-or-Q) plus possibly T — two at the very most. Now test the conclusions against this ceiling. Conclusion 1 needs four selected; the world allows at most two, so it cannot be drawn. Conclusion 2 needs S selected; the negated false statement put S firmly out, so it cannot be drawn. When the facts forbid both conclusions, the answer is “neither.”

Eliminate by anatomy. (a) is the too-strong-for-the-facts trap — “at least four” ignores the ceiling the facts impose and reads the loose “at most two of R, S, T” as if it added selections rather than capping them. (b) is the reversed-truth-value trap — it reads the false “at least one of R and S” as though it were true, exactly inverting the truth-label the stem hands you. (c) compounds both. The transferable rule: on a true/false-labelled puzzle, a FALSE existential (“at least one…”) flips to a universal denial (“none…”), and a conclusion survives only if every admissible world forces it. Key: (d).

The middle statement is the labelled FALSE one — “At least one of R and S was selected” — and a false existential negates to its universal: NEITHER R NOR S was selected. The first (true) statement gives exactly ONE of {P, Q}. The third (true) statement is an upper bound only (“at most two of R, S, T”) and adds no selection. So the total selected is 1 (from P/Q) + 0 (R, S) + T’s status (0 or 1) ≤ 2. Conclusion 1 (“at least four”) is therefore impossible (forced FALSE). Conclusion 2 (“S was selected”) is the direct contradiction of the negated false statement (S is NOT selected), so it is forced FALSE. Neither conclusion can be drawn → (d).

Treat the three labels as hard facts.

• True: exactly one of P, Q selected.
• False: “At least one of R and S was selected” — its negation is the fact: neither R nor S was selected.
• True: at most two of R, S, T selected. With R and S already out, this only says T may or may not be in; it adds no one.

Maximum possible headcount = 1 (one of P, Q) + 0 (R, S) + at most 1 (T) = at most 2 selected.

Conclusion 1 — at least four selected. Impossible: the ceiling is two. Cannot be drawn.

Conclusion 2 — S was selected. Directly contradicts the negated false statement (S is out). Cannot be drawn.

Neither conclusion follows. Answer: (d) Neither 1 nor 2.

The whole item turns on one discipline the CSAT free-CR family rewards: a statement marked false is a fact in disguise — read its negation. “At least one of R and S was selected” is an existential; its denial is the universal “neither was selected,” which simultaneously kills Conclusion 2 and collapses the headcount that Conclusion 1 needs. The two “true” statements then only cap and partition; they never lift the count to four. A student who tracks truth-values mechanically — negate the false, treat bounds as bounds — reaches (d) without guessing.