2025 CSAT — Q73

Let $P$‘s and $Q$‘s goals at the $10$-minutes-left mark be $x$ and $x+3$ ($P$ trails by $3$), with $x \ge 0$ unknown. “Win” means $P$‘s final total strictly exceeds $Q$‘s.

Statement I alone ($P$ scores $4$ in the last $10$): $P_{\text{final}} = x+4$, but $Q$ may also score in the last $10$ — unknown. Insufficient.

Statement II alone ($Q_{\text{final}} = 4$): says nothing about $P$‘s late scoring. Insufficient.

Both together. $P_{\text{final}} = x + 4$ and $Q_{\text{final}} = 4$, so

$P$ wins $\iff x > 0$, and $x$ (P’s score with $10$ minutes left) is not determined:

• $x = 0$: final $4 - 4$ — a draw, $P$ does not win.
• $x = 1$: final $5 - 4$ — $P$ wins.

Both are consistent with all data (II needs $Q_{\text{final}}=4$, fine in both; $Q$‘s late goals $= 1 - x \ge 0$ holds for $x \in \{0,1\}$). Two opposite answers survive $\Rightarrow$ undecidable.

Answer: (d) The Question cannot be answered even using any of the Statements.

The decisive move is naming the hidden unknown: $P$‘s score when $10$ minutes remain ($x$) is never pinned, only the $3$-goal margin. Even with both statements, $P_{\text{final}} - Q_{\text{final}} = x$, so the result swings on $x$ — produce the counterexample pair $4{-}4$ (draw) vs $5{-}4$ (win) and sufficiency collapses. The trap (c) silently assumes $P$ trailed $0{-}3$; nothing states that. In DS, a yes/no question is answerable only if every admissible scenario gives the same verdict — one draw and one win means (d).