Train A leaves a station at 9:00 a.m. and travels at a... | SAT Linear equations in one variable Question, Answer Explained

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Math · AlgebraHard

Train A leaves a station at 9:00 a.m. and travels at a constant speed of $60$ miles per hour. Train B leaves the same station at 9:30 a.m. on a parallel track, traveling in the same direction at a constant speed of $80$ miles per hour. At what time will train B catch up to train A?

10:30 a.m.

11:00 a.m.

11:30 a.m.

12:00 p.m.

Show answer and explanation

Answer: B

Let $t$ be the number of hours after 9:00 a.m. Train A has gone $60t$ miles. Train B has traveled for $t - 0.5$ hours, so it has gone $80(t - 0.5)$ miles. Set these equal: $60t = 80t - 40$ , so $20t = 40$ and $t = 2$ . Train B catches up $2$ hours after 9:00 a.m., at 11:00 a.m.

Choice A adds train B's $1.5$ hours of travel to 9:00 a.m., but train B didn't leave until 9:30 a.m.

Why the other choices are wrong

A. This adds train B's 1.5 hours of travel to 9:00 a.m. Train B left at 9:30 a.m., so it catches up at 11:00 a.m.
C. This adds 2 hours to 9:30 a.m., but the 2 hours are measured from 9:00 a.m., when train A left.
D. This uses a 60 mile head start, but in half an hour train A travels only 30 miles. At a 20 mph difference, that takes 1.5 hours to close.

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