2024 CSAT — Q80

Treat the two equations as constraints that recover the unique assignment, then read off the answer.

Use $Q - S = 2S \Rightarrow Q = 3S$. Among $\{4,5,10,12,15\}$, which value tripled is also in the set?

• $S = 4 \Rightarrow Q = 12$ ✓ (both present)
• $S = 5 \Rightarrow Q = 15$ ✓ (both present)
• $S = 10 \Rightarrow Q = 30$ ✗

So two candidate cases. Now apply $T = R + S + 3$.

Case $S = 4,\ Q = 12$. Remaining values for $P, R, T$ are $\{5, 10, 15\}$, and $T = R + 4 + 3 = R + 7$. Testing $R \in \{5,10,15\}$ gives $T \in \{12,17,22\}$, none of which is an available value. Case fails.

Case $S = 5,\ Q = 15$. Remaining values for $P, R, T$ are $\{4, 10, 12\}$, and $T = R + 5 + 3 = R + 8$. Testing $R \in \{4,10,12\}$: $R = 4 \Rightarrow T = 12$ ✓ (available). Then $P = 10$ (the last remaining).

So $P = 10,\ Q = 15,\ R = 4,\ S = 5,\ T = 12$. Verify: $Q - S = 15 - 5 = 10 = 2S$ ✓ and $T = R + S + 3 = 4 + 5 + 3 = 12$ ✓.

Answer: (b) $2$.

The item is a constraint-satisfaction puzzle disguised as coding. The move that scores is using $Q = 3S$ to cut the five possibilities down to two cases, then letting the second equation act as the filter — $T = R + S + 3$ has no solution in the $S = 4$ branch, which is exactly the trap that pulls candidates toward “(c)” or “(d).” Always push both constraints through before declaring “insufficient data”: here the seemingly ambiguous setup is fully determined, and the unique assignment makes $P + R - T = 2$. The discipline — enumerate the few cases the first equation allows, then eliminate with the second — is the general key for letter-to-number coding under algebraic conditions.