The concept here is classic DS: one statement gives you a ratio, the other gives an absolute value. Neither alone is enough, but combined they pin down a unique answer.
Let d = distance from depot to Central Square. When the two buses meet, call the meeting point x km from the depot.
Both buses have been traveling the same time t. Bus S is 250 km from Central Square when they meet, so S traveled d - 250 km. That means x = d - 250. Bus R traveled all the way to Central Square (d km) and then turned back to the meeting point, so R traveled d + (d - x) = d + 250 km.
1. S is 250 km from Central Square. So S traveled d - 250 and R traveled d + 250. We have a relationship between the two travel distances, but d itself is free. d = 300 gives x = 50, d = 500 gives x = 250, both are consistent. Insufficient.
2. R traveled 3 times as fast as S. Since time is equal, R's distance = 3 times S's distance. So (d + 250) = 3(d - 250)... wait, that's combining with S1 already. With S2 alone, we just know the meeting point satisfies (2d - x) = 3x, giving d = 2x. Any d works here since x can be anything proportional. Insufficient.
Together: d = 2x (from S2) and x = d - 250 (from S1). Substituting: d = 2(d - 250) = 2d - 500. So d = 500 km.
The trap is S1. "250 km from the station" feels concrete enough to solve the problem, but without knowing the speed ratio, you can't anchor d. Think of it this way: if R were 4x faster instead of 3x, the meeting point would shift and d would be different even with the same 250 km gap.
Answer: C.
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